Mathematics Review - QUANTITATIVE ABILITY: TACTICS, STRATEGIES, PRACTICE, AND REVIEW

Barron's GRE, 18th Edition (2009)

Part 4. QUANTITATIVE ABILITY: TACTICS, STRATEGIES, PRACTICE, AND REVIEW

Chapter 14. Mathematics Review

The mathematics questions on the GRE General Test require a working knowledge of mathematical principles, including an understanding of the fundamentals of algebra, plane geometry, and arithmetic, as well as the ability to translate problems into formulas and to interpret graphs. The following review covers these areas thoroughly and will prove helpful.

This chapter is divided into 15 sections, labeled 14-A through 14-O. For each question on the Diagnostic Test and the five Model Tests, the Answer Key indicates which section of Chapter 14 you should consult if you need help on a particular topic.

How much time you initially devote to reviewing mathematics should depend on your math skills. If you have always been a good math student and you have taken some math in college and remember most of your high school math, you can skip the instructional parts of this chapter for now. If while doing the model tests in PART FIVE or on the accompanying CD-ROM, you find that you keep making mistakes on certain types of problems (averages, percents, circles, solid geometry, word problems, for example), or they take you too long, you should then study the appropriate sections here. Even if your math skills are excellent, and you don’t need the review, you should complete the sample questions in those sections; they are an excellent source of additional GRE questions. If you know that your math skills are not very good and you have not done much math since high school, then it is advisable to review all of this material, including working out the problems, before tackling the model tests.

No matter how good you are in math, you should carefully read and do the problems in Chapters 10, 11, 12, and 13. For many of these problems, two solutions are given: the most direct mathematical solution and a second solution using one or more of the special tactics taught in these chapters.

Arithmetic

To do well on the GRE, you need to feel comfortable with most topics of basic arithmetic. In the first five sections of Chapter 14, we will review the basic arithmetic operations, signed numbers, fractions, decimals, ratios, percents, and averages. Since the GRE uses these concepts to test your reasoning skills, not your ability to perform tedious calculations, we will concentrate on the concepts and not on arithmetic drill. The solutions to more than one-third of the mathematics questions on the GRE depend on your knowing the key facts in these sections. Be sure to review them all.

14-A. BASIC ARITHMETIC CONCEPTS

Let’s start by reviewing the most important sets of numbers and their properties. On the GRE the word number always means real number, a number that can be represented by a point on the number line.

Signed Numbers

The numbers to the right of 0 on the number line are called positive and those to the left of 0 are called negative. Negative numbers must be written with a negative sign (−2); positive numbers can be written with a plus sign (+2) but are usually written without a sign (2). All numbers can be called signed numbers.

KEY FACT A1

For any number a, exactly one of the following is true:

· a is negative

· a = 0

· a is positive

The absolute value of a number a, denoted |a|, is the distance between a and 0 on the number line. Since 3 is 3 units to the right of 0 on the number line and −3 is 3 units to the left of 0, both have an absolute value of 3:

· |3| = 3

· |−3| = 3

Two unequal numbers that have the same absolute value are called opposites. So, 3 is the opposite of −3 and −3 is the opposite of 3.

KEY FACT A2

The only number that is equal to its opposite is 0.

EXAMPLE 1

Column A		Column B
	a − b = −(a − b)
a		b

SOLUTION. Since −(a − b) is the opposite of a − b, a − b = 0, and so a = b. The answer is C.

In arithmetic we are basically concerned with the addition, subtraction, multiplication, and division of numbers. The third column of the following table gives the terms for the results of these operations.

Operation	Symbol	Result	Example
Addition	+	*Sum*	16 is the sum of 12 and 4 16 = 12 + 4
Subtraction	−	*Difference*	8 is the difference of 12 and 4 8 = 12 − 4
Multiplication*	×	*Product*	48 is the product of 12 and 4 48 = 12 × 4
Division	÷	*Quotient*	3 is the quotient of 12 and 4 3 = 12 ÷ 4

*Multiplication can be indicated also by a dot, parentheses, or the juxtaposition of symbols without any sign: 2² · 2⁴, 3(4), 3(x + 2), 3a, 4abc.

Given any two numbers a and b, we can always find their sum, difference, product, and quotient, except that we may never divide by zero.

· 0 ÷ 7 = 0

· 7 ÷ 0 is meaningless

EXAMPLE 2

What is the sum of the product and quotient of 8 and 8?

(A) 16 (B) 17 (C) 63 (D) 64 (E) 65

SOLUTION. Product: 8 × 8 = 64. Quotient: 8 ÷ 8 = 1. Sum: 64 + 1 = 65 (E).

KEY FACT A3

· The product of 0 and any number is 0. For any number a: a × 0 = 0.

· Conversely, if the product of two numbers is 0, at least one of them must be 0:

ab = 0 ⇒ a = 0 or b = 0.

EXAMPLE 3

Column A	Column B
The product of the integers from −7 to 2	The product of the integers from −2 to 7

SOLUTION. Do not multiply. Each column is the product of 10 numbers, one of which is 0. So, by KEY FACT A3, each product is 0. The columns are equal (C).

KEY FACT A4

The product and quotient of two positive numbers or two negative numbers are positive; the product and quotient of a positive number and a negative number are negative.

6 × 3 = 18 6 × (−3) = −18 (−6) × 3 = − 18 (−6) × (−3) = 18

6 ÷ 3 = 2 6 ÷ (−3) = −2 (−6) ÷ 3 = − 2 (−6) ÷ (−3) = 2

To determine whether a product of more than two numbers is positive or negative, count the number of negative factors.

KEY FACT A5

· The product of an even number of negative factors is positive.

· The product of an odd number of negative factors is negative.

EXAMPLE 4

Column A	Column B
(−1)(2)(−3)(4)(−5)	(1)(−2)(3)(−4)(5)

SOLUTION. Don’t waste time multiplying. The product in Column A is negative since it has 3 negative factors, whereas the product in Column B is positive since it has 2 negative factors. The answer is B.

KEY FACT A6

· The reciprocal of any nonzero number a is

· The product of any number and its reciprocal is 1:

KEY FACT A7

· The sum of two positive numbers is positive.

· The sum of two negative numbers is negative.

· To find the sum of a positive and a negative number, find the difference of their absolute values and use the sign of the number with the larger absolute value.

6 + 2 = 8 (−6) + (−2) = −8

To calculate either 6 + (−2) or (−6) + 2, take the difference, 6 − 2 = 4, and use the sign of the number whose absolute value is 6. So,

6 + (−2) = 4 (−6) + 2 = −4

KEY FACT A8

The sum of any number and its opposite is 0:

a + (−a) = 0.

Many of the properties of arithmetic depend on the relationship between subtraction and addition and between division and multiplication.

KEY FACT A9

· Subtracting a number is the same as adding its opposite.

· Dividing by a number is the same as multiplying by its reciprocal.

a − b = a + (−b) a ÷ b = a ×

Many problems involving subtraction and division can be simplified by changing them to addition and multiplication problems, respectively.

KEY FACT A10

To subtract signed numbers, change the problem to an addition problem, by changing the sign of what is being subtracted, and use KEY FACT A7.

In each case, the minus sign was changed to a plus sign, and either the 6 was changed to −6 or the −6 was changed to 6.

Integers

The *integers* are	{…, −4, −3, −2, −1, 0, 1, 2, 3, 4, …}.
The *positive integers* are	{1, 2, 3, 4, 5, …}.
The *negative integers* are	{…, −5, −4, −3, −2, −1}.

NOTE: The integer 0 is neither positive nor negative. Therefore, if a question on the GRE asks how many positive numbers have a certain property, and the only numbers with that property are −2, −1, 0, 1, and 2, the answer is 2.

Consecutive integers are two or more integers written in sequence in which each integer is 1 more than the preceding integer. For example:

22, 23 6, 7, 8, 9 −2, −1, 0, 1 n, n + 1, n + 2, n + 3

EXAMPLE 5

If the sum of three consecutive integers is less than 75, what is the greatest possible value of the smallest one?

(A) 23 (B) 24 (C) 25 (D) 26 (E) 27

SOLUTION. Let the numbers be n, n + 1, and n + 2. Then

n + (n + 1) + (n + 2) = 3n + 3 ⇒ 3n + 3 < 75 ⇒ 3n < 72 ⇒ n < 24.

So, the most n can be is 23 (A).

CAUTION

Never assume that number means integer: 3 is not the only number between 2 and 4; there are infinitely many, including 2.5, 3.99, , π and .

EXAMPLE 6

If 2 < x < 4 and 3 < y < 7, what is the largest integer value of x + y?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11

SOLUTION. If x and y are integers, the largest value is 3 + 6 = 9. However, although x + y is to be an integer, neither x nor y must be. If x = 3.8 and y = 6.2, then x + y = 10 (D).

The sum, difference, and product of two integers are always integers; the quotient of two integers may be an integer, but it is not necessarily one. The quotient 23 ÷ 10 can be expressed as or or 2.3. If the quotient is to be an integer, we can also say that the quotient is 2 and there is a remainder of 3. It depends upon our point of view. For example, if 23 dollars is to be divided among 10 people, each one will get $2.30 (2.3 dollars); but if 23 books are to be divided among 10 people, each one will get 2 books and there will be 3 left over (the remainder).

EXAMPLE 7

How many positive integers less than 100 have a remainder of 3 when divided by 7?

SOLUTION. To leave a remainder of 3 when divided by 7, an integer must be 3 more than a multiple of 7. For example, when 73 is divided by 7, the quotient is 10 and the remainder is 3: 73 = 10 × 7 + 3. So, just take the multiples of 7 and add 3. (Don’t forget that 0 is a multiple of 7.)

0 × 7 + 3 = 3; 1 × 7 + 3 = 10; 2 × 7 + 3 = 17;

…; 13 × 7 + 3 = 94

A total of 14 numbers.

If a and b are integers, the following four terms are synonymous:

a is a *divisor* of b	a is a *factor* of b
b is a *divisible* by a	b is a *multiple* of a

They all mean that when b is divided by a there is no remainder (or, more precisely, the remainder is 0). For example:

3 is a divisor of 12	3 is a factor of 12
12 is divisible by 3	12 is multiple of 3

KEY FACT A11

Every integer has a finite set of factors (or divisors) and an infinite set of multiples.

The factors of 12:	−12, −6, −4, −3, −2, −1, 1, 2, 3, 4, 6, 12
The multiples of 12:	…, −48, −36, −24, −12, 0, 12, 24, 36, 48, …

The only positive divisor of 1 is 1. All other positive integers have at least 2 positive divisors: 1 and itself, and possibly many more. For example, 6 is divisible by 1 and 6, as well as 2 and 3, whereas 7 is divisible only by 1 and 7. Positive integers, such as 7, that have exactly 2 positive divisors are called prime numbers or primes. The first few primes are

2, 3, 5, 7, 11, 13, 17, 19, 23.

Memorize this list — it will come in handy. Note that 1 is not a prime.

KEY FACT A12

Every integer greater than 1 that is not a prime can be written as a product of primes.

To find the prime factorization of any integer, find any two factors; if they’re both primes, you are done; if not, factor them. Continue until each factor has been written in terms of primes. A useful method is to make a factor tree.

For example, here are the prime factorizations of 108 and 240:

EXAMPLE 8

For any positive integer a, let a denote the smallest prime factor of a.

Which of the following is equal to 35?

(A) 10 (B) 15 (C) 45 (D) 55 (E)75

SOLUTION. Check the first few primes; 35 is not divisible by 2 or 3, but is divisible by 5, so 5 is the smallest prime factor of 35: 35 = 5. Now check the five choices: 10 = 2, and 15, 45, and 75 are all equal to 3. Only 55 = 5. The answer is D.

The least common multiple (LCM) of two or more integers is the smallest positive integer that is a multiple of each of them. For example, the LCM of 6 and 10 is 30. Infinitely many positive integers are multiples of both 6 and 10, including 60, 90, 180, 600, 6000, and 66,000,000, but 30 is the smallest one. The greatest common factor (GCF) or greatest common divisor (GCD) of two or more integers is the largest integer that is a factor of each of them. For example, the only positive integers that are factors of both 6 and 10 are 1 and 2, so the GCF of 6 and 10 is 2. For small numbers, you can often find their GCF and LCM by inspection. For larger numbers, KEY FACTS A13 and A14 are useful.

HELPFUL HINT

It is usually easier to find the GCF than the LCM. For example, you might see immediately that the GCF of 36 and 48 is 12. You could then use KEY FACT A13 to find the LCM: since GCF × LCM = 36 × 48, then

KEY FACT A13

The product of the GCF and LCM of two numbers is equal to the product of the two numbers.

KEY FACT A14

To find the GCF or LCM of two or more integers, first get their prime factorizations.

· The GCF is the product of all the primes that appear in each factorization, using each prime the smallest number of times it appears in any of the factorizations.

· The LCM is the product of all the primes that appear in any of the factorizations, using each prime the largest number of times it appears in any of the factorizations.

For example, let’s find the GCF and LCM of 108 and 240. As we saw:

108 = 2 × 2 × 3 × 3 × 3 and 240 = 2 × 2 × 2 × 2 × 3 × 5.

· GCF. The primes that appear in both factorizations are 2 and 3: 2 appears twice in the factorization of 108 and 4 times in the factorization of 240, so we take it twice; 3 appears 3 times in the factorization of 108, but only once in the factorization of 240, so we take it just once. The GCF = 2 × 2 × 3 = 12.

· LCM. Take one of the factorizations and add to it any primes from the other that are not yet listed. So, start with 2 × 2 × 3 × 3 × 3 (108) and look at the primes from 240: there are four 2s; we already wrote two 2s, so we need two more; there is a 3 but we already have that; there is a 5, which we need. So, the LCM = (2 × 2 × 3 × 3 × 3) × (2 × 2 × 5) = 108 × 20 = 2160.

EXAMPLE 9

What is the smallest number that is divisible by both 34 and 35?

SOLUTION. We are being asked for the LCM of 34 and 35. By KEY FACT A13, . But the GCF is 1 since no number greater than 1 divides evenly into both 34 and 35. So, the LCM is 34 × 35 = 1190.

The even numbers are all the multiples of 2:

{…, −4, −2, 0, 2, 4, 6, …}

The odd numbers are the integers not divisible by 2:

{…, −5, −3, −1, 1, 3, 5, …}

NOTE:

· The terms odd and even apply only to integers.

· Every integer (positive, negative, or 0) is either odd or even.

· 0 is an even integer; it is a multiple of 2. (0 = 0 × 2)

· 0 is a multiple of every integer. (0 = 0 × n)

· 2 is the only even prime number.

KEY FACT A15

The tables below summarize three important facts:

1. If two integers are both even or both odd, their sum and difference are even.

2. If one integer is even and the other odd, their sum and difference are odd.

3. The product of two integers is even unless both of them are odd.

Exponents and Roots

Repeated addition of the same number is indicated by multiplication:

17 + 17 + 17 + 17 + 17 + 17 + 17 = 7 × 17

Repeated multiplication of the same number is indicated by an exponent:

17 × 17 × 17 × 17 × 17 × 17 × 17 = 17⁷

In the expression 17⁷, 17 is called the base and 7 is the exponent.

At some time, you may have seen expressions such as 2⁻⁴, , or even . On the GRE, although the base, b, can be any number, the exponents you will see will almost always be positive integers.

KEY FACT A16

For any number b: b¹ = b, and bⁿ = b × b × … × b, where b is used as a factor n times.

i. 2⁵ × 2³ = (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2) = 2⁸ = 2⁵⁺³

ii.

iii. (2²)³ = (2 × 2)³ = (2 × 2) × (2 × 2) × (2 × 2) = 2⁶ = 2^2×3

iv. 2³ × 7³ = (2 × 2 × 2) × (7 × 7 × 7) = (2 × 7)(2 × 7) (2 × 7) = (2 × 7)³

These four examples illustrate the following important laws of exponents given in KEY FACT A17.

KEY FACT A17

For any numbers b and c and positive integers m and n:

CAUTION

In (i) and (ii) the bases are the same and in (iv) the exponents are the same. None of these rules applies to expressions such as 7⁵ × 5⁷, in which both the bases and the exponents are different.

EXAMPLE 10

If 2^x = 32, what is x²?

(A) 5 (B) 10 (C) 25 (D) 100 (E) 1024

SOLUTION. To solve 2^x = 32, just count (and keep track of) how many 2s you need to multiply to get 32: 2 × 2 × 2 × 2 × 2 = 32, so x = 5 and x² = 25 (C).

EXAMPLE 11

If 3^a × 3^b = 3¹⁰⁰, what is the average (arithmetic mean) of a and b?

SOLUTION. Since 3^a × 3^b = 3^a+b, we see that a + b = 100 ⇒ = 50.

The next KEY FACT is an immediate consequence of KEY FACTS A4 and A5.

KEY FACT A18

For any positive integer n:

· 0ⁿ = 0

· if a is positive, then aⁿ is positive

· if a is negative and n is even, then aⁿ is positive

· if a is negative and n is odd, then aⁿ is negative.

EXAMPLE 12

Column A	Column B
(−13)¹⁰	(−13)²⁵

SOLUTION. Column A is positive and Column B is negative. So Column A is greater.

Squares and Square Roots

The exponent that appears most often on the GRE is 2. It is used to form the square of a number, as in πr² (the area of a circle), a² + b² = c² (the Pythagorean theorem), or x² − y² (the difference of two squares). Therefore, it is helpful to recognize the perfect squares, numbers that are the squares of integers. The squares of the integers from 0 to 15 are as follows:

There are two numbers that satisfy the equation x² = 9: x = 3 and x = −3. The positive one, 3, is called the (principal) square root of 9 and is denoted by the symbol. Clearly, each perfect square has a square root: and = 12. But, it is an important fact that every positive number has a square root.

KEY FACT A19

For any positive number a, there is a positive number b that satisfies the equation b² = a. That number is called the square root of a and we write b = .

So, for any positive number a: = × = a.

The only difference between and is that the first square root is an integer, while the second one isn’t. Since 10 is a little more than 9, we should expect that is a little more than = 3. In fact, (3.1)² = 9.61, which is close to 10, and (3.16)² = 9.9856, which is very close to 10. So, ≈ 3.16. On the GRE you will never have to evaluate such a square root; if the solution to a problem involves a square root, that square root will be among the answer choices.

EXAMPLE 13

What is the circumference of a circle whose area is 10π?

SOLUTION. Since the area of a circle is given by the formula A = πr², we have

πr² = 10π ⇒ r² = 10 ⇒ r = .

The circumference is given by the formula C = 2πr, so C = 2π (D). (See Section 14-L on circles.)

KEY FACT A20

For any positive numbers a and b:

CAUTION

For example:

CAUTION

Although it is always true that ()² = a, = a is true only if a is positive:

= 5, not −5.

EXAMPLE 14

SOLUTION. Column A: Since x¹⁰x¹⁰ = x²⁰, = x¹⁰. Column B: (x⁵)² = x¹⁰. The columns are equal (C).

PEMDAS

When a calculation requires performing more than one operation, it is important to carry them out in the correct order. For decades students have memorized the sentence “Please Excuse My Dear Aunt Sally,” or just the first letters, PEMDAS, to remember the proper order of operations. The letters stand for:

· Parentheses: first do whatever appears in parentheses, following PEMDAS within the parentheses if necessary.

· Exponents: next evaluate all terms with exponents.

· Multiplication and Division: then do all multiplications and divisions in order from left to right — do not multiply first and then divide.

· Addition and Subtraction: finally, do all additions and subtractions in order from left to right — do not add first and then subtract.

Here are some worked-out examples.

1.	12 + 3 × 2 = 12 + 6 = 18	[Multiply before you add.]
	(12 + 3) × 2 = 15 × 2 = 30	[First add in the parentheses.]
2.	12 ÷ 3 × 2 = 4 × 2 = 8	[Just go from left to right.]
	12 ÷ (3 × 2) = 12 ÷ 6 = 2	[First multiply inside the parentheses.]
3.	5 × 2³ = 5 × 8 = 40	[Do exponents first.]
	(5 × 2)³ = 10³ = 1000	[First multiply inside the parentheses.]
4.	4 + 4 ÷ (2 + 6) = 4 + 4 ÷ 8 = 4 + .5 = 4.5
	[First add in the parentheses, then divide, and finally add.]
5.	100 − 2²(3 + 4 × 5) = 100 − 2²(23) = 100 − 4(23) = 100 − 92 = 8
	[First evaluate what’s inside the parentheses (using PEMDAS); then take the exponent; then multiply; and finally subtract.]

There is an important situation when you shouldn’t start with what’s in the parentheses. Consider the following two examples.

i. What is the value of 7(100 − 1) ?

Using PEMDAS, you would write 7(100 − 1) = 7(99), and then multiply: 7 × 99 = 693. But you can do this even quicker in your head if you think of it this way: 7(100 − 1) = 700 − 7 = 693.

ii. What is the value of (77 + 49) ÷ 7?

If you followed the rules of PEMDAS, you would first add, 77 + 49 = 126, and then divide, 126 ÷ 7 = 18. This is definitely more difficult and time-consuming than mentally doing = 11 + 7 = 18.

Both of these examples illustrate the very important distributive law.

KEY FACT A21

The distributive law

For any real numbers a, b, and c:

· a(b + c) = ab + ac

· a(b − c) = ab − ac

and if a ≠ 0

HELPFUL HINT

Many students who use the distributive law with multiplication forget about it with division. Don’t you do that.

EXAMPLE 15

Column A	Column B
5(a − 7)	5a − 7

EXAMPLE 16

SOLUTION 15. By the distributive law, Column A = 5a − 35. The result of subtracting 35 from a number is always less than the result of subtracting 7 from that number. Column B is greater.

SOLUTION 16.	Column A	Column B
By the distributive law:		10 + x
Subtract 10 from each column:		x

The columns are equal if x = 0, but not if x = 1.

The answer is D.

EXAMPLE 17

If a = 9 × 8321 and b = 9 × 7321, what is the value of a − b?

SOLUTION. Remember, you will never have to do tedious multiplications on the GRE, so there must be an easier way to solve this. If you think, you can do it in your head. Remember the distributive law—in far less time than it takes to write the equation on your scrap paper, you should realize that

a − b = 9(8321) − 9(7321) = 9(8321 − 7321) = 9(1000) = 9000.

Inequalities

The number a is greater than the number b, denoted a > b, if a is to the right of b on the number line. Similarly, a is less than b, denoted a < b, if a is to the left of b on the number line. Therefore, if a is positive, a > 0, and if a is negative, a < 0. Clearly, if a > b, then b < a.

The following KEY FACT gives an important alternate way to describe greater than and less than.

KEY FACT A22

· For any numbers a and b:

a > b means that a − b is positive.

· For any numbers a and b:

a < b means that a − b is negative.

KEY FACT A23

· For any numbers a and b, exactly one of the following is true:

a > b or a = b or a < b.

The symbol ≥ means greater than or equal to and the symbol ≤ means less than or equal to. The statement “x ≥ 5” means that x can be 5 or any number greater than 5; the statement “x ≤ 5” means that x can be 5 or any number less than 5. The statement “2 < x < 5” is an abbreviation for the statement “2 < x and x < 5.” It means that x is a number between 2 and 5 (greater than 2 and less than 5).

Inequalities are very important on the GRE, especially on the quantitative comparison questions where you have to determine which of two quantities is the greater one. KEY FACTS A24 and A25 give some important facts about inequalities.

If the result of performing an arithmetic operation on an inequality is a new inequality in the same direction, we say that the inequality has been preserved. If the result of performing an arithmetic operation on an inequality is a new inequality in the opposite direction, we say that the inequality has been reversed.

KEY FACT A24

· Adding a number to an inequality or subtracting a number from an inequality preserves it.

If a < b, then a + c < b + c and a − c < b − c.

3 < 7 ⇒ 3 + 100 < 7 + 100 (103 < 107)
3 < 7 ⇒ 3 − 100 < 7 − 100 (−97 < − 93)

· Adding inequalities in the same direction preserves them.

If a < b and c < d, then a + c < b + d.

3 < 7 and 5 < 10 ⇒ 3 + 5 < 7 + 10 (8 < 17)

· Multiplying or dividing an inequality by a positive number preserves it.

If a < b, and c is positive, then ac < bc and

3 < 7 ⇒ 3 × 100 < 7 × 100 (300 < 700)
3 < 7 ⇒ 3 ÷ 100 < 7 ÷

· Multiplying or dividing an inequality by a negative number reverses it.

If a < b, and c is negative, then ac > bc and .

3 < 7 ⇒ 3 × (−100) > 7 × (−100) (−300 > −700)
3 < 7 ⇒ 3 × (−100) > 7 ÷ (−100)

· Taking negatives reverses an inequality.

If a < b, then −a > −b and if a > b, then −a < −b.

3 < 7 ⇒ −3 > −7 and 7 > 3 ⇒ −7 < −3

· If two numbers are each positive or negative, then taking reciprocals reverses an inequality.

If a and b are both positive or both negative and a < b, then .

HELPFUL HINT

Be sure you understand KEY FACT A24; it is very useful. Also, review the important properties listed in KEY FACTS A25 and A26. These properties come up often on the GRE.

KEY FACT A25

Important inequalities for numbers between 0 and 1.

· If 0 < x < 1, and a is positive, then xa < a. For example: .85 × 19 < 19.

· If 0 < x < 1, and m and n are positive integers with m > n, then

x^m < xⁿ < x. For example, .

· If 0 < x < 1, then > x. For example, .

· If 0 < x < 1, then > x. In fact, > 1. For example, > 1 > 0.2.

KEY FACT A26

Properties of Zero

· 0 is the only number that is neither positive nor negative.

· 0 is smaller than every positive number and greater than every negative number.

· 0 is an even integer.

· 0 is a multiple of every integer.

· For every number a: a + 0 = a and a − 0 = a.

· For every number a: a × 0 = 0.

· For every positive integer n: 0ⁿ = 0.

· For every number a (including 0): a ÷ 0 and are meaningless symbols. (They are undefined.)

· For every number a other than 0: 0 ÷ a = = 0.

· 0 is the only number that is equal to its opposite: 0 = −0.

· If the product of two or more numbers is 0, at least one of them is 0.

KEY FACT A27

Properties of 1

· For any number a: 1 × a = a and = a.

· For any integer n: 1ⁿ = 1.

· 1 is a divisor of every integer.

· 1 is the smallest positive integer.

· 1 is an odd integer.

· 1 is not a prime.

Practice Exercises—Basic Arithmetic

Multiple-Choice Questions

1. For how many positive integers, a, is it true that a² ≤ 2a?
(A) None
(B) 1
(C) 2
(D) 4
(E) More than 4

2. If 0 < a < b < 1, which of the following are true?
  I. a − b is negative
II.  is positive
III.  is positive
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

3. If the product of 4 consecutive integers is equal to one of them, what is the largest possible value of one of the integers?
(A) 0 (B) 3 (C) 4 (D) 6 (E) 24

4. At 3:00 A.M. the temperature was 13° below zero. By noon it had risen to 32°. What was the average hourly increase in temperature?
(A)
(B)
(C) 5°
(D) 7.5°
(E) 45°

5. If a and b are negative, and c is positive, which of the following statements are true?
I. a − b < a − c
II. If a < b, then
III.
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

6. If −7 ≤ x ≤ 7 and 0 ≤ y ≤ 12, what is the greatest possible value of y − x?
(A) −19 (B) 5 (C) 7 (D) 17 (E) 19

7. If (7^a)(7^b) = , what is d in terms of a, b, c?
(A)
(B) c − a − b
(C) a + b − c
(D) c − ab
(E)

8. If each of and can be replaced by +, −, or ×, how many different values are there for the expression 2 2 2?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 9

9. A number is “terrific” if it is a multiple of 2 or 3. How many terrific numbers are there between −11 and 11?
(A) 6 (B) 7 (C) 11 (D) 15 (E) 17

10. If x y represents the number of integers greater than x and less than y, what is the value of −π ?
(A) 2 (B) 3 (C) 4 (D) 5 (E) 6

Questions 11 and 12 refer to the following definition. For any positive integer n, (n) represents the number of positive divisors of n.

11. Which of the following are true?
I. (5) = (7)
II. (5) · (7) = (35)
III. (5) + (7) = (12)
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

12. What is the value of (((12)))?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 6

13. If p and q are primes greater than 2, which of the following must be true?
I. p + q is even
II. pq is odd
III. p² − q² is even
(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

14. If 0 < x < 1, which of the following lists the numbers in increasing order?

15. Which of the following is equal to (7⁸ × 7⁹)¹⁰?
(A) 7²⁷
(B) 7⁸²
(C) 7¹⁷⁰
(D) 49¹⁷⁰
(E) 49⁷²⁰

Quantitative Comparison Questions

	Column A	Column B
16.	The product of the odd integers between −8 and 8	The product of the even integers between −9 and 9

	a and b are nonzero integers
17.	a + b	ab

18.

The remainder when a positive integer is divided by 7

19.

24 ÷ 6 × 4

20.

x − 17

	n is an integer greater than 1 that leaves a remainder of 1 when it is divided by 2, 3, 4, 5, and 6
21.	n	60

22.

The number of primes that are divisible by 2

The number of primes that are divisible by 3

n is a positive integer
23.	The number of different prime factors of n	The number of different prime factors of n²

24.

The number of even positive factors of 30

The number of odd positive factors of 30

n is a positive integer
25.	(−10)ⁿ	(−10)ⁿ^{+ 1}

ANSWER KEY

1. C	6. E	11. C	16. A	21. A
2. D	7. B	12. C	17. D	22. C
3. B	8. A	13. E	18. B	23. C
4. C	9. D	14. B	19. A	24. C
5. D	10. D	15. C	20. A	25. D

Answer Explanations

1. (C) Since a is positive, we can divide both sides of the given inequality by a:

a² ≤ 2a ⇒ a ≤ 2 ⇒ a = 1 or 2.

2. (D) Since a < b, a − b is negative (I is true). Since a and b are positive, so is their product, ab; and the reciprocal of a positive number is positive (II is true).

and have just seen that the numerator is negative and the denominator positive; so the value of the fraction is negative (III is false).

3. (B) If all four integers were negative, their product would be positive, and so could not equal one of them. If all of the integers were positive, their product would be much greater than any of them (even 1 × 2 × 3 × 4 = 24). So, the integers must include 0, in which case their product is 0. The largest set of four consecutive integers that includes 0 is 0, 1, 2, 3.

4. (C) In the 9 hours from 3:00 to 12:00, the temperature rose 32 − (−13) = 32 + 13 = 45 degrees. So, the average hourly increase was 45° ÷ 9 = 5°.

5. (D) Since b is negative and c is positive, b < c ⇒ −b > −c ⇒ a − b > a − c

(I is false). Since c is positive, dividing by c preserves the inequality. (II is true.) Since b is negative, is negative, and so is less than which is positive (III is true).

6. (E) To make y − x as large as possible, let y be as big as possible (12), and subtract the smallest amount possible (x = −7): 12 − (−7) = 19.

7. (B) (7^a)(7^b) = 7^a^+b, and Therefore, a + b = c − d ⇒ a + b + d = c ⇒ d = c − a − b

8. (A) Just list the 9 possible outcomes of replacing and by +, −, and ×, and see that there are 4 different values: −2, 2, 6, 8.

2 + 2 + 2 = 6 2 − 2 − 2 = −2 2 × 2 × 2 = 8

2 + 2 − 2 = 2 2 − 2 × 2 = −2 2 × 2 + 2 = 6

2 + 2 × 2 = 6 2 − 2 + 2 = 2 2 × 2 − 2 = 2

9. (D) There are 15 “terrific” numbers: 2, 3, 4, 6, 8, 9, 10, their opposites, and 0.

10. (D) There are 5 integers (1, 0, −1, −2, −3) that are greater than −3.14 (−π) and less than 1.41

11. (C) Since 5 and 7 have two positive factors each, (5) = (7). (I is true.)

Since 35 has 4 divisors (1, 5, 7, and 35) and (5) · (7) = 2 × 2 = 4. (II is true.)

Since the positive divisors of 12 are 1, 2, 3, 4, 6, and 12, (12) is 6, which is not equal to 2 + 2. (III is false.)

12. (C) (((12))) = ((6)) = (4) = 3

13. (E) All primes greater than 2 are odd, so p and q are odd, and p + q is even (I is true). The product of two odd numbers is odd (II is true). Since p and q are odd, so are their squares, and so the difference of the squares is even (III is true).

14. (B) For any number, x, between 0 and 1:

x² < x and x <

15. (C) First, multiply inside the parentheses:

7⁸ × 7⁹ = 7¹⁷; then, raise to the 10th power: (7¹⁷)¹⁰ = 7¹⁷⁰.

16. (A) Since the product in Column A has 4 negative factors (−7, −5, −3, −1), it is positive. The product in Column B also has 4 negative factors, but be careful—it also has the factor 0, and so Column B is 0.

17. (D) If a and b are each 1, then a + b = 2, and ab = 1; so, Column A is greater. But, if a and b are each 3, a + b = 6, and ab = 9, and Column B is greater.

18. (B) The remainder is always less than the divisor.

19. (A) According to PEMDAS, you divide and multiply from left to right (do not do the multiplication first): 24 ÷ 6 × 4 = 4 × 4 = 16.

20. (A) By the distributive law,

which is greater than x − 17 (the larger the number you subtract, the smaller the difference.)

21. (A) The LCM of 2, 3, 4, 5, 6 is 60; and all multiples of 60 are divisible by each of them. So, n could be 61 or 1 more than any multiple of 60.

22. (C) The only prime divisible by 2 is 2, and the only prime divisible by 3 is 3. The number in each column is 1.

23. (C) If you make a factor tree for n², the first branches would be n and n. Now, when you factor each n, you get exactly the same prime factors. (See the example below.)

24. (C) Just list the factors of 30: 1, 2, 3, 5, 6, 10, 15, 30. Four of them are odd and four are even.

25. (D) If n is even, then n + 1 is odd, and consequently (−10)ⁿ is positive, whereas (−10)ⁿ⁺¹ is negative. If n is odd, exactly the opposite is true.

14-B. FRACTIONS AND DECIMALS

Several questions on the GRE involve fractions or decimals. The 23 KEY FACTS in this section cover all of the important facts you need to know for the GRE.

When a whole is divided into n equal parts, each part is called one nth of the whole, written For example, if a pizza is cut (divided) into 8 equal slices, each slice is one eighth of the pizza; a day is divided into 24 equal hours, so an hour is one twenty-fourth of a day; and an inch is one twelfth of a foot.

· If Donna slept for 5 hours, she slept for five twenty-fourths of a day.

· If Taryn bought 8 slices of pizza, she bought eight eighths of a pie.

· If Aviva’s shelf is 30 inches long, it measures thirty twelfths of a foot.

Numbers such as and in which one integer is written over a second integer, are called fractions. The center line is called the fraction bar. The number above the bar is called the numerator, and the number below the bar is called the denominator.

CAUTION

The denominator of a fraction can never be 0.

· A fraction, such as in which the numerator is less than the denominator, is called a proper fraction. Its value is less than 1.

· A fraction, such as in which the numerator is more than the denominator, is called an improper fraction. Its value is greater than 1.

· A fraction, such as in which the numerator and denominator are the same, is also improper, but it is equal to 1.

It is useful to think of the fraction bar as a symbol for division. If three pizzas are divided equally among eight people, each person gets of a pizza. If you actually divide 3 by 8, you get that

KEY FACT B1

Every fraction, proper or improper, can be expressed in decimal form (or as a whole number) by dividing the numerator by the denominator.

Note that any number beginning with a decimal point can be written with a 0 to the left of the decimal point. In fact, some calculators will express 3 ÷ 8 as .375, whereas others will print 0.375.

Unlike the examples above, when most fractions are converted to decimals, the division does not terminate after 2 or 3 or 4 decimal places; rather it goes on forever with some set of digits repeating itself.

However, on the GRE, you do not need to be concerned with this. On both multiple-choice and quantitative comparison questions, all numbers written as decimals terminate.

Comparing Fractions and Decimals

KEY FACT B2

To compare two decimals, follow these rules.

· Whichever number has the greater number to the left of the decimal point is greater: since 11 > 9, 11.001 > 9.896 and since 1 > 0, 1.234 > 0.8. (Recall that if a decimal is written without a number to the left of the decimal point, you may assume that a 0 is there. So, 1.234 > .8.)

· If the numbers to the left of the decimal point are equal (or if there are no numbers to the left of the decimal point), proceed as follows:

1. If the numbers do not have the same number of digits to the right of the decimal point, add zeros to the end of the shorter one.

2. Now, compare the numbers ignoring the decimal point.

For example, to compare 1.83 and 1.823, add a 0 to the end of 1.83, forming 1.830. Now compare them, thinking of them as whole numbers: since, 1830 > 1823, then 1.830 > 1.823.

EXAMPLE 1

Column A	Column B
.2139	.239

SOLUTION. Do not think that Column A is greater because 2139 > 239. Be sure to add a 0 to the end of .239 (forming .2390) before comparing. Now, since 2390 > 2139, Column B is greater.

KEY FACT B3

There are two methods of comparing fractions:

1. Convert them to decimals (by dividing), and use KEY FACT B2.

2. Cross-multiply.

For example, to compare and we have two choices.

1. Write and . Since .375 > .333, then

2. Cross-multiply: Since 3 × 3 > 8 × 1, then

KEY FACT B4

When comparing fractions, there are three situations in which it is easier just to look at the fractions, and not use either method in KEY FACT B3.

1. If the fractions have the same denominator, the fraction with the larger numerator is greater. Just as $9 is more than $7, and 9 books are more than 7 books, 9 tenths are more than 7 tenths:

2. If the fractions have the same numerator, the fraction with the smaller denominator is greater. If you divide a cake into 5 equal pieces, each piece is larger than the pieces you would get if you had divided the cake into 10 equal pieces: and similarly

3. Sometimes the fractions are so familiar or easy to work with, you just know the answer. For example, and

KEY FACT B5

KEY FACTS B2, B3, and B4 apply to positive decimals and fractions.

· Clearly, any positive number is greater than any negative number:

· For negative decimals and fractions, use KEY FACT A24, which states that if a > b, then −a < −b:

and

EXAMPLE 2

Which of the following lists the fractions and in order from least to greatest?

SOLUTION. Quickly, convert each to adecimal, writing down the first few decimal places: and It is now easy to order the decimals: .625 < .650 < .666. The answer is C.

ALTERNATIVE SOLUTION. Cross-multiply.

· Since 8 × 2 > 3 × 5, then

· Since 8 × 13 > 20 × 5, then

· Since 20 × 2 > 3 × 13, then

EXAMPLE 3

Column A	Column B
0 < x < y
	0

SOLUTION. By KEY FACT B4, and so by KEY FACT A22, is positive. Column A is greater.

Equivalent Fractions

If Bill and Al shared a pizza, and Bill ate the pizza and Al ate of it, they had exactly the same amount.

We express this idea by saying that and are equivalent fractions: they have the exact same value.

NOTE: If you multiply both the numerator and denominator of by 4 you get ; and if you divide both the numerator and denominator of by 4 you get .

This illustrates the next KEY FACT.

KEY FACT B6

Two fractions are equivalent if multiplying or dividing both the numerator and denominator of the first one by the same number gives the second one.

Consider the following two cases.

1. When the numerator and denominator of are each multiplied by 15, the products are 3 × 15 = 45 and 8 × 15 = 120. Therefore, and are equivalent fractions.

2. Since 2 must be multiplied by 14 to get 28, but 3 must be multiplied by 15 to get 45, then and are not equivalent fractions.

KEY FACT B7

To determine if two fractions are equivalent, cross-multiply. The fractions are equivalent if and only if the two products are equal.

For example, since 120 × 3 = 8 × 45, then and are equivalent.

Since 45 × 2 ≠ 3 × 28, then and are not equivalent fractions.

A fraction is in lowest terms if no positive integer greater than 1 is a factor of both the numerator and denominator. For example, is in lowest terms, since no integer greater than 1 is a factor of both 9 and 20; but is not in lowest terms, since 3 is a factor of both 9 and 24.

KEY FACT B8

Every fraction can be reduced to lowest terms by dividing the numerator and the denominator by their greatest common factor (GCF). If the GCF is 1, the fraction is already in lowest terms.

EXAMPLE 4

For any positive integer n: n! means the product of all the integers from 1 to n.

What is the value of

SOLUTION. Clearly, you do not want to calculate 6! (1·2·3·4·5·6 = 720) and 8! (1·2·3·4·5·6·7·8 = 40,320) and then have to reduce Here’s the easy solution:

Arithmetic Operations with Decimals

KEY FACT B9

To add or subtract decimal numbers, make sure that the decimal points are lined up, and then add or subtract normally, ignoring the decimal points. Finally, place a decimal point in the answer immediately below the other decimal points.

For example, 3.2 + 7 + 1.125 = 11.325 and 3.456 − 1.28 = 2.176

KEY FACT B10

To multiply decimal numbers, multiply normally, ignoring the decimal points. Then count the total number of digits to the right of the decimal points in both factors, and place a decimal point in the product that many places from the right.

KEY FACT B11

To divide decimal numbers, count the number of digits to the right of the decimal point in the divisor, and move the decimal point in both the divisor and the dividend that many places to the right (adding zeros if necessary). Now, divide normally and if there is a decimal point in the dividend, place a decimal point in the quotient directly above the one in the dividend.

For example, 35 ÷ 1.25 = 28 and .035 ÷ 1.25 = .028

Multiplying and dividing by powers of 10 is particularly easy and can be accomplished just by moving the decimal point.

KEY FACT B12

To multiply any decimal or whole number by a power of 10, move the decimal point as many places to the right as there are 0s in the power of 10, filling in with 0s, if necessary.

KEY FACT B13

To divide any decimal or whole number by a power of 10, move the decimal point as many places to the left as there are 0s in the power of 10, filling in with 0s, if necessary.

EXAMPLE 5

Column A	Column B
3.75 × 10⁴	37,500,000 ÷ 10³

SOLUTION. In Column A, move the decimal point 4 places to the right: 37,500. In Column B, move the decimal point 3 places to the left: 37,500. The answer is C.

Arithmetic Operations with Fractions

KEY FACT B14

To multiply two fractions, multiply their numerators and multiply their denominators:

KEY FACT B15

To multiply a fraction by any other number, write that number as a fraction whose denominator is 1:

Before multiplying fractions, reduce. You may reduce by dividing any numerator and any denominator by a common factor.

EXAMPLE 6

Express the product, in lowest terms.

SOLUTION. If you multiply the numerators and denominators you get which is a nuisance to reduce. It is better to use TACTIC B1 and reduce first:

When a problem requires you to find a fraction of a number, multiply.

EXAMPLE 7

If of the 350 sophomores at Monroe High School are girls, and of them play on a team, how many sophomore girls do not play on a team?

SOLUTION. There are sophomore girls.

Of these, play on a team. So, 200 − 175 = 25 do not play on a team.

The reciprocal of any nonzero number x is that number y such that xy = 1. Since then is the reciprocal of x. Similarly, the reciprocal of the fraction is the fraction since

KEY FACT B16

To divide any number by a fraction, multiply that number by the reciprocal of the fraction.

EXAMPLE 8

In the meat department of a supermarket, 100 pounds of chopped meat was divided into packages, each of which weighed of a pound.

How many packages were there?

SOLUTION.

KEY FACT B17

· To add or subtract fractions with the same denominator, add or subtract the numerators and keep the denominator:

· To add or subtract fractions with different denominators, first rewrite the fractions as equivalent fractions with the same denominators:

NOTE: The easiest common denominator to find is the product of the denominators (6 × 4 = 24, in this example), but the best denominator to use is the least common denominator, which is the least common multiple (LCM) of the denominators (12, in this case). Using the least common denominator minimizes the amount of reducing that is necessary to express the answer in lowest terms.

KEY FACT B18

If is the fraction of a whole that satisfies some property, then is the fraction of that whole that does not satisfy it.

EXAMPLE 9

In a jar, of the marbles are red, are white, and are blue.

What fraction of the marbles are neither red, white, nor blue?

SOLUTION. The red, white, and blue marbles constitute

of the total, so of the marbles are neither red, white, nor blue.

EXAMPLE 10

Lindsay ate of a cake and Emily ate of it. What fraction of the cake was still uneaten?

EXAMPLE 11

Lindsay ate of a cake and Emily ate of what was left. What fraction of the cake was still uneaten?

CAUTION: Be sure to read questions carefully. In Example 10, Emily ate of the cake. In Example 11, however, she only ate of the that was left after Lindsay had her piece: she ate of the cake.

SOLUTION 10: of the cake was eaten, and was uneaten.

SOLUTION 11: of the cake was eaten, and the other was uneaten.

Arithmetic Operations with Mixed Numbers

A mixed number is a number such as which consists of an integer followed by a fraction. It is an abbreviation for the sum of the number and the fraction; so, is an abbreviation for Every mixed number can be written as an improper fraction, and every improper fraction can be written as a mixed number:

KEY FACT B19

1. To write a mixed number as an improper fraction, multiply the whole number (3) by the denominator (2), add the numerator (1), and write the sum over the denominator (2):

2. To write an improper fraction as a mixed number, divide the numerator by the denominator; the quotient (3) is the whole number and the remainder (1) is placed over the denominator to form the fractional part

KEY FACT B20

To add mixed numbers, add the integers and add the fractions:

KEY FACT B21

To subtract mixed numbers, subtract the integers and the fractions. However, if the fraction in the second number is greater than the fraction in the first number, you first have to borrow 1 from the integer part. For example, since we can’t subtract until we borrow 1 from the 5:

Now, you have

KEY FACT B22

To multiply or divide mixed numbers, change them to improper fractions:

CAUTION

Be aware that is not rather:

Complex Fractions

A complex fraction is a fraction, such as which has one or more fractions in its numerator or denominator or both.

KEY FACT B23

There are two ways to simplify a complex fraction:

· Multiply every term in the numerator and denominator by the least common multiple of all the denominators that appear in the fraction.

· Simplify the numerator and the denominator, and then divide.

To simplify multiply each term by 12, the LCM of 6 and 4:

Practice Exercises—Fractions and Decimals

Multiple-Choice Questions

1. A biology class has 12 boys and 18 girls. What fraction of the class are boys?

2. For how many integers, a, between 30 and 40 is it true that and are all in lowest terms?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

3. What fractional part of a week is 98 hours?

4. What is the value of the product

5. If of a number is 22, what is of that number?

6. Jason won some goldfish at the state fair.

During the first week, of them died, and during the second week, of those still alive at the end of the first week died. What fraction of the original goldfish were still alive after two weeks?

7. of 24 is equal to of what number?

8. If 7a = 3 and 3b = 7, what is the value of

9. What is the value of

10. Which of the following are greater than x when

(A) I only

(B) I and II only

(D) II and III only

(E) I, II, and III

11. One day at Lincoln High School, of the students were absent, and of those present went on a field trip. If the number of students staying in school that day was 704, how many students are enrolled at Lincoln High?

(A) 840

(B) 960

(D) 1600

(E) 3520

12. If a = 0.87, which of the following are less than a?

(A) None

(B) I only

(D) III only

(E) II and III only

13. For what value of x is

(A) .001 (B) .01 (C) .1 (D) 10 (E) 100

14. If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all the fractions whose numerators are in A and whose denominators are in B, what is the product of all of the numbers in C?

15. For the final step in a calculation, Ezra accidentally divided by 1000 instead of multiplying by 1000. What should he do to his incorrect answer to correct it?

(A) Multiply it by 1000.

(B) Multiply it by 100,000.

(D) Square it.

(E) Double it.

Quantitative Comparison Questions

	Column A	Column B
16.


17.	xy

18.

	Judy needed 8 pounds of chicken. At the supermarket, the only packages available weighed of a pound each.
19.	The number of packages Judy needed to buy.	11

20.


21.

22.

23.


24.

25.

ANSWER KEY

1. A	6. C	11. B	16. C	21. C
2. C	7. A	12. C	17. A	22. A
3. E	8. A	13. D	18. A	23. A
4. A	9. A	14. A	19. C	24. A
5. E	10. B	15. C	20. B	25. B

Answer Explanations

1. (A) The class has 30 students, of whom 12 are boys. So, the boys make up of class.

2. (C) If a is even, then is not in lowest terms, since both a and 8 are divisible by 2. Therefore, the only possibilities are 31, 33, 35, 37, and 39; but and , so only 3 integers—31, 33, and 37—satisfy the given condition.

3. (E) There are 24 hours in a day and 7 days in a week, so there are 24 × 7 = 168 hours in a week: .

4. (A) Reduce each fraction and multiply:

5. (E) Don’t bother writing an equation for this one; just think. We know that of the number is 22, and of a number is twice as much as of it: 2 × 22 = 44.

6. (C) The algebra way is to let x = the number of goldfish Jason won. During the first week x died, so x were still alive. During week two, of them died and of them survived:

On the GRE, the best way is to assume that the original number of goldfish was 40, the LCM of the denominators (see TACTIC 3, Chapter 12).

Then, 8 died the first week ( of 40), and 12 of the 32 survivors ( of 32) died the second week. In all, 8 + 12 = 20 died; the other 20( the original number) were still alive.

7. (A) If x is the number, then , which means (dividing by 15) that x = 1, and so x = 7.

8. (A) 7a = 3 and 3b = 7 ⇒ a = and b = ⇒

9. (A) Don’t start by doing the arithmetic. This is just .

Now, replacing a with gives .

10. (B) The reciprocal of a positive number less than 1 is greater than 1 (I is true).

, which is greater than 1 (II is true). Since is positive and is negative, when x = < 0 and, therefore, less than x (III is false).

11. (B) If s is the number of students enrolled, is the number who were absent, and is the number who were present. Since of them went on a field trip, of them stayed in school. Therefore,

12. (C) Since a < 1, > a (I is false). Since a < 1, a² < a (II is true). The reciprocal of a positive number less than 1 is greater than 1 (III is false).

13. (D) There are two easy ways to do this. The first is to see that (34.56)(7.89) has 4 decimal places, whereas (.3456)(78.9) has 5, so the numerator has to be divided by 10. The second is to round off and calculate mentally: since 30 × 8 = 240, and .3 × 80 = 24, we must divide by 10.

14. (A) Nine fractions are formed:

Note that although some of these fractions are equivalent, we do have nine distinct fractions. When you multiply, the three 2s and the three 3s in the numerators cancel with the three 2s and three 3s in the denominators. So, the numerator is 1 and the denominator is 4 × 4 × 4 = 64.

15. (C) Multiplying Ezra’s incorrect answer by 1000 would undo the final division he made. At that point he should have multiplied by 1000. So, to correct his error, he should multiply again by 1000. In all, Ezra should multiply his incorrect answer by 1000 × 1000 = 1,000,000.

16. (C) Each column equals .

17. (A) Column A:

Column B: .

Finally, .

18. (A) Column A:

19. (C) . Since 10 packages wouldn’t be enough, she had to buy 11. (10 packages would weigh only pounds.)

20. (B) You don’t need to multiply on this one: since is less than , which is already less than .

21. (C) Column B is the sum of 2 complex fractions:

Simplifying each complex fraction, by multiplying numerator and denominator by 6, or treating these as the quotient of 2 fractions, we get , which is exactly the value of Column A.

22. (A) When two fractions have the same numerator, the one with the smaller denominator is bigger, and 2¹⁰⁰ < 3¹⁰⁰.

23. (A) Since Column A is the product of 4 negative numbers, it is positive, and so is greater than Column B, which, being the product of 3 negative numbers, is negative.

24. (A) Column A: . Since Column B is the reciprocal of Column A, Column B = .

25. (B) If 0 < x < 1, then x² < x < . In this question, x = .

14-C. PERCENTS

The word percent means hundredth. We use the symbol “%” to express the word “percent.” For example, “17 percent” means “17 hundredths,” and can be written with a % symbol, as a fraction, or as a decimal:

KEY FACT C1

· To convert a percent to a decimal, drop the % symbol and move the decimal point two places to the left, adding 0s if necessary. (Remember that we assume that there is a decimal point to the right of any whole number.)

· To convert a percent to a fraction, drop the % symbol, write the number over 100, and reduce.

KEY FACT C2

· To convert a decimal to a percent, move the decimal point two places to the right, adding 0s if necessary, and add the % symbol.

· To convert a fraction to a percent, first convert the fraction to a decimal, then convert the decimal to a percent, as indicated above.

You should be familiar with the following basic conversions:

Knowing these conversions can help solve many problems more quickly. For example, the fastest way to find 25% of 32 is not to multiply 32 by 0.25; rather, it is to know that 25% = , and that of 32 is 8.

Many questions involving percents can actually be answered more quickly in your head than by using paper and pencil. Since 10% = , to take 10% of a number, just divide by 10 by moving the decimal point one place to the left: 10% of 60 is 6. Also, since 5% is half of 10%, then 5% of 60 is 3 (half of 6); and since 30% is 3 times 10%, then 30% of 60 is 18 (3 × 6).

Practice doing this, because improving your ability to do mental math will add valuable points to your score on the GRE.

Solving Percent Problems

Consider the following three questions:

(i) What is 45% of 200?

(ii) 90 is 45% of what number?

(iii) 90 is what percent of 200?

The arithmetic needed to answer each of these questions is very easy, but unless you set a question up properly, you won’t know whether you should multiply or divide. In each case, there is one unknown, which we will call x. Now just translate each sentence, replacing “is” by “=” and the unknown by x.

EXAMPLE 1

Charlie gave 20% of his baseball cards to Kenne and 15% to Paulie. If he still had 520 cards, how many did he have originally?

(A) 555 (B) 700 (C) 800 (D) 888 (E) 1000

SOLUTION. Originally, Charlie had 100% of the cards (all of them). Since he gave away 35% of them, he has 100% − 35% = 65% of them left. So, 520 is 65% of what number?

520 = .65x ⇒ x = 520 ÷ .65 = 800 (C).

EXAMPLE 2

After Ruth gave 110 baseball cards to Alison and 75 to Susanna, she still had 315 left. What percent of her cards did Ruth give away?

SOLUTION. Ruth gave away a total of 185 cards and had 315 left. Therefore, she started with 185 + 315 = 500 cards. So, 185 is what percent of 500?

Ruth gave away 37% of her cards, (C).

Since percent means hundredth, the easiest number to use in any percent problem is 100:

KEY FACT C3

For any positive number a: a% of 100 is a.

For example: 91.2% of 100 is 91.2; 300% of 100 is 300; and % of 100 is .

In any problem involving percents, use the number 100. (It doesn’t matter whether or not 100 is a realistic number—a country can have a population of 100; an apple can cost $100; a man can run 100 miles per hour.)

EXAMPLE 3

In 1985 the populations of town A and town B were the same. From 1985 to 1995 the population of town A increased by 60% while the population of town B decreased by 60%. In 1995, the population of town B was what percent of the population of town A?

(A) 25% (B) 36% (C) 40% (D) 60% (E) 120%

SOLUTION. On the GRE, do not waste time with a nice algebraic solution. Simply, assume that in 1985 the population of each town was 100. Then, since 60% of 100 is 60, in 1995, the populations were 100 + 60 = 160 and 100 − 60 = 40. So, in 1995, town B’s population was = 25% of town A’s (A).

Since a% of b is , and b% of a is , we have the result shown in KEY FACT C4.

KEY FACT C4

For any positive numbers a and b: a% of b = b% of a.

KEY FACT C4 often comes up on the GRE in quantitative comparison questions: Which is greater, 13% of 87 or 87% of 13? Don’t multiply — they’re equal.

Percent Increase and Decrease

KEY FACT C5

· The percent increase of a quantity is

· The percent decrease of a quantity is

For example:

· If the price of a lamp goes from $80 to $100, the actual increase is $20, and the percent increase is .

· If a $100 lamp is on sale for $80, the actual decrease in price is $20, and the percent decrease is .

Notice that the percent increase in going from 80 to 100 is not the same as the percent decrease in going from 100 to 80.

KEY FACT C6

If a < b, the percent increase in going from a to b is always greater than the percent decrease in going from b to a.

KEY FACT C7

· To increase a number by k%, multiply it by (1 + k%).

· To decrease a number by k%, multiply it by (1 − k%).

For example:

· The value of a $1600 investment after a 25% increase is $1600(1 + 25%) = $1600(1.25) = $2000.

· If the investment then loses 25% of its value, it is worth $2000(1 − 25%) = $2000(.75) = $1500.

Note that, after a 25% increase followed by a 25% decrease, the value is $1500, $100 less than the original amount.

KEY FACT C8

An increase of k% followed by a decrease of k% is equal to a decrease of k% followed by an increase of k%, and is always less than the original value. The original value is never regained.

EXAMPLE 4

Column A	Column B
Store B always sells CDs at 60% off the list price. Store A sells its CDs at 40% off the list price, but often runs a special sale during which it reduces its prices by 20%.
The price of a CD when it is on sale at store A	The price of the same CD at store B

SOLUTION. Assume the list price of the CD is $100. Store B always sells the CD for $40 ($60 off the list price). Store A normally sells the CD for $60 ($40 off the list price), but on sale reduces its price by 20%. Since 20% of 60 is 12, the sale price is $48 ($60 − $12). The price is greater at Store A.

Notice that a decrease of 40% followed by a decrease of 20% is not the same as a single decrease of 60%; it is less. In fact, a decrease of 40% followed by a decrease of 30% wouldn’t even be as much as a single decrease of 60%.

KEY FACT C9

· A decrease of a% followed by a decrease of b% always results in a smaller decrease than a single decrease of (a + b)%.

· An increase of a% followed by an increase of b% always results in a larger increase than a single increase of (a + b)%.

· An increase (or decrease) of a% followed by another increase (or decrease) of a% is never the same as a single increase (or decrease) of 2a%.

EXAMPLE 5

Column A	Column B
Sally and Heidi were both hired in January at the same salary. Sally got two 40% raises, one in July and another in November. Heidi got one 90% raise in October.
Sally’s salary at the end of the year	Heidi’s salary at the end of the year

SOLUTION. Since this is a percent problem, assume their salaries were $100. Column A: Sally’s salary rose to 100(1.40) = 140, and then to 140(1.40) = $196. Column B: Heidi’s salary rose to 100(1.90) = $190. Column A is greater.

EXAMPLE 6

In January, the value of a stock increased by 25%, and in February, it decreased by 20%. How did the value of the stock at the end of February compare with its value at the beginning of January?

(A) It was less. (B) It was the same. (C) It was 5% greater. (D) It was more than 5% greater. (E) It cannot be determined from the information given.

SOLUTION. Assume that at the beginning of January the stock was worth $100. Then at the end of January it was worth $125. Since 20% of 125 is 25, during February its value decreased from $125 to $100. The answer is B.

KEY FACT C10

· If a number is the result of increasing another number by k%, to find the original number, divide by (1 + k%).

· If a number is the result of decreasing another number by k%, to find the original number, divide it by (1 − k%).

For example, if the population of a town in 1990 was 3000, and this represents an increase of 20% since 1980, to find the population in 1980, divide 3000 by (1 + 20%): 3000 ÷ 1.20 = 2500.

EXAMPLE 7

From 1989 to 1990, the number of applicants to a college increased 15% to 5060. How many applicants were there in 1989?

(A) 759 (B) 4301 (C) 4400 (D) 5819 (E) 5953

SOLUTION. The number of applicants in 1989 was 5060 ÷ 1.15 = 4400 (C).

CAUTION

Percents over 100%, which come up most often on questions involving percent increases, are often confusing for students. First of all, be sure you understand that 100% of a number is that number, 200% of a number is 2 times the number, and 1000% of a number is 10 times the number. If the value of an investment goes from $1000 to $5000, it is now worth 5 times, or 500%, as much as it was originally; but there has only been a 400% increase in value:

EXAMPLE 8

The population of a country doubled every 10 years from 1960 to 1990. What was the percent increase in population during this time?

(A) 200% (B) 300% (C) 700% (D) 800% (E) 1000%

SOLUTION. The population doubled three times (once from 1960 to 1970, again from 1970 to 1980, and a third time from 1980 to 1990). Assume that the population was originally 100. Then it increased from 100 to 200 to 400 to 800. So the population in 1990 was 8 times the population in 1960, but this was an increase of 700 people, or 700% (C).

Practice Exercises — Percents

Multiple-Choice Questions

1. If 25 students took an exam and 4 of them failed, what percent of them passed?
(A) 4%
(B) 21%
(C) 42%
(D) 84%
(E) 96%

2. Amanda bought a $60 sweater on sale at 5% off. How much did she pay, including 5% sales tax?
(A) $54.15
(B) $57.00
(C) $57.75
(D) $59.85
(E) $60.00

3. What is 10% of 20% of 30%?
(A) 0.006%
(B) 0.6%
(C) 6%
(D) 60%
(E) 6000%

4. If c is a positive number, 500% of c is what percent of 500c?
(A) 0.01 (B) 0.1 (C) 1 (D) 10 (E) 100

5. What percent of 50 is b?

6. 8 is % of what number?
(A) .24
(B) 2.4
(C) 24
(D) 240
(E) 2400

7. During his second week on the job, Mario earned $110. This represented a 25% increase over his earnings of the previous week. How much did he earn during his first week of work?
(A) $82.50
(B) $85.00
(C) $88.00
(D) $137.50
(E) $146.67

8. At Bernie’s Bargain Basement everything is sold for 20% less than the price marked. If Bernie buys radios for $80, what price should he mark them if he wants to make a 20% profit on his cost?
(A) $96
(B) $100
(C) $112
(D) $120
(E) $125

9. Mrs. Fisher was planning on depositing a certain amount of money each month into a vacation fund. She then decided not to make any contributions during June and July. To make the same annual contribution that she had originally planned, by what percent should she increase her monthly deposits?
(A) 16%
(B) 20%
(C) 25%
(D) 33%
(E) It cannot be determined from the information given.

10. The price of a loaf of bread was increased by 20%. How many loaves can be purchased for the amount of money that used to buy 300 loaves?
(A) 240 (B) 250 (C) 280 (D) 320 (E) 360

11. If 1 micron = 10,000 angstroms, then 100 angstroms is what percent of 10 microns?
(A) 0.0001%
(B) 0.001%
(C) 0.01%
(D) 0.1%
(E) 1%

12. There are twice as many girls as boys in an English class. If 30% of the girls and 45% of the boys have already handed in their book reports, what percent of the students have not yet handed in their reports?
(A) 25%
(B) 35%
(C) 50%
(D) 65%
(E) 75%

13. An art dealer bought a Ming vase for $1000 and later sold it for $10,000. By what percent did the value of the vase increase?
(A) 10%
(B) 90%
(C) 100%
(D) 900%
(E) 1000%

14. During a sale a clerk was putting a new price tag on each item. On one jacket, he accidentally raised the price by 15% instead of lowering the price by 15%. As a result the price on the tag was $45 too high. What was the original price of the jacket?
(A) $60
(B) $75
(C) $90
(D) $105
(E) $150

15. On a test consisting of 80 questions, Eve answered 75% of the first 60 questions correctly. What percent of the other 20 questions does she need to answer correctly for her grade on the entire exam to be 80%?
(A) 85%
(B) 87.5%
(C) 90%
(D) 95%
(E) 100%

Quantitative Comparison Questions

Column A

Column B

16.

n% of 25 is 50

17.

18.

The price of cellular phone 1 is 20% more than the price of cellular phone 2.

19.

20.

21.

Bank A pays 5% interest on its savings accounts.
Bank B pays 4% interest on its savings accounts.

22.

A solution that is 20% sugar is made sweeter by doubling the amount of sugar.

23.

b is an integer greater than 1, and b equals n% of b²

24.

After Ali gave Lior 50% of her money, she had 20% as much as he did.

25.

ANSWER KEY

1. D	6. E	11. D	16. C	21. D
2. D	7. C	12. D	17. A	22. A
3. B	8. D	13. D	18. D	23. B
4. C	9. B	14. E	19. B	24. D
5. E	10. B	15. D	20. C	25. C

Answer Explanations

1. (D) If 4 students failed, then the other 25 − 4 = 21 students passed, and .

2. (D) Since 5% of 60 is 3, Amanda saved $3, and thus paid $57 for the sweater. She then had to pay 5% sales tax on the $57: .05 × 57 = 2.85, so the total cost was $57 + $2.85 = $59.85.

3. (B) 10% of 20% of 30% = .10 × .20 × .30 = .006 = .6%.

4. (C) 500% of c = 5c, which is 1% of 500c.

5. (E) .

6. (E) .

7. (C) To find Mario’s earnings during his first week, divide his earnings from the second week by 1.25: 110 ÷ 1.25 = 88.

8. (D) Since 20% of 80 is 16, Bernie wants to get $96 for each radio he sells. What price should the radios be marked so that after a 20% discount, the customer will pay $96? If x represents the marked price, then .80x = 96⇒ x = 96 ÷ .80 = 120.

9. (B) Assume that Mrs. Fisher was going to contribute $100 each month, for an annual total of $1200. Having decided not to contribute for 2 months, the $1200 will have to be paid in 10 monthly deposits of $120 each. This is an increase of $20, and a percent increase of

10. (B) Assume that a loaf of bread used to cost $1 and that now it costs $1.20 (20% more). Then 300 loaves of bread used to cost $300. How many loaves costing $1.20 each can be bought for $300? 300 ÷ 1.20 = 250.

11. (D) 1 micron = 10,000 angstroms ⇒ 10 microns = 100,000 angstroms; dividing both sides by 1000, we get 100 angstroms = (10 microns); and = .001 = 0.1%.

12. (D) Assume that there are 100 boys and 200 girls in the class. Then, 45 boys and 60 girls have handed in their reports. So 105 students have handed them in, and 300 − 105 = 195 have not handed them in. What percent of 300 is 195?

13. (D) The increase in the value of the vase was $9,000. So the percent increase is

14. (E) If p represents the original price, the jacket was priced at 1.15p instead of .85p. Since this was a $45 difference, 45 = 1.15p − .85p = .30p ⇒ p = 45 ÷ .30 = $150.

15. (D) To earn a grade of 80% on the entire exam, Eve needs to correctly answer 64 questions (80% of 80). So far, she has answered 45 questions correctly (75% of 60). Therefore, on the last 20 questions she needs 64 − 45 = 19 correct answers; and = 95%.

16. (C) Column A: 400% of 3 = 4 × 3 = 12.
Column B: 300% of 4 = 3 × 4 = 12.

17. (A) Since n% of 25 is 50, then 25% of n is also 50, and 50% of n is twice as much: 100. If you don’t see that, just solve for n:
.

18. (D) A 25% discount on a $10 television is much less than $25, whereas a 25% discount on a $1000 television is much more than $25. (They would be equal only if the regular price of the television were $100.)

19. (B) Assume that the price of cellular phone 2 is $100; then the price of cellular phone 1 is $120, and on sale at 20% off it costs $24 less: $96.

20. (C) For any numbers a and b: a% of b is equal to b% of a.

21. (D)

Multiply by 100:

The columns are equal if a and b are equal, and unequal otherwise.

22. (A) Bank B would have to increase its rate from 4% to 5%, an actual increase of 1%. This represents a percent increase of

23. (B) Assume a vat contains 100 ounces of a solution, of which 20% or 20 ounces is sugar (the remaining 80 ounces being water). If the amount of sugar is doubled, there would be 40 ounces of sugar and 80 ounces of water. The sugar will then comprise of the solution.

24. (D) If b = 2, then b² = 4, and 2 = 50% of 4; in this case, the columns are equal. If b = 4, b² = 16, and 4 is not 50% of 16; in this case, the columns are not equal.

25. (C) Avoid the algebra and just assume Ali started with $100. After giving Lior $50, she had $50 left, which was 20% or one-fifth of what he had. So, Lior had 5 × $50 = $250, which means that originally he had $200.
Column A: 75% of $200 = $150.
Column B: 150% of $100 = $150.
The columns are equal.

14-D. RATIOS AND PROPORTIONS

A ratio is a fraction that compares two quantities that are measured in the same units. The first quantity is the numerator and the second quantity is the denominator.

For example, if there are 4 boys and 16 girls on the debate team, we say that the ratio of the number of boys to the number of girls on the team is 4 to 16, or .

This is often written 4:16. Since a ratio is just a fraction, it can be reduced or converted to a decimal or a percent. The following are all different ways to express the same ratio:

CAUTION

Saying that the ratio of boys to girls on the team is 1:4 does not mean that of the team members are boys. It means that for each boy on the team there are 4 girls; so for every 5 members of the team, there are 4 girls and 1 boy. Boys, therefore, make up of the team, and girls .

KEY FACT D1

If a set of objects is divided into two groups in the ratio of a:b, then the first group contains of the objects and the second group contains of the objects.

EXAMPLE 1

Last year, the ratio of the number of tennis matches that Central College’s women’s team won to the number of matches they lost was 7:3. What percent of their matches did the team win?

SOLUTION. The team won = 70% of their matches.

EXAMPLE 2

If 45% of the students at a college are male, what is the ratio of male students to female students?

Reminder: In problems involving percents the best number to use is 100.

SOLUTION. Assume that there are 100 students. Then 45 of them are male, and 100 − 45 = 55 of them are female. So, the ratio of males to females is .

If we know how many boys and girls there are in a club, then, clearly, we know not only the ratio of boys to girls, but several other ratios too. For example, if the club has 7 boys and 3 girls: the ratio of boys to girls is , the ratio of girls to boys is , the ratio of boys to members is , the ratio of members to girls is , and so on.

However, if we know a ratio, we cannot determine how many objects there are. For example, if a jar contains only red and blue marbles, and if the ratio of red marbles to blue marbles is 3:5, there may be 3 red marbles and 5 blue marbles, but not necessarily. There may be 300 red marbles and 500 blue ones, since the ratio 300:500 reduces to 3:5. In the same way, all of the following are possibilities for the distribution of marbles.

The important thing to observe is that the number of red marbles can be any multiple of 3, as long as the number of blue marbles is the same multiple of 5.

KEY FACT D2

If two numbers are in the ratio of a:b, then for some number x, the first number is ax and the second number is bx. If the ratio is in lowest terms, and if the quantities must be integers, then x is also an integer.

In any ratio problem, write the letter x after each number and use some given information to solve for x.

EXAMPLE 3

If the ratio of men to women in a particular dormitory is 5:3, which of the following could not be the number of residents in the dormitory?

(A) 24 (B) 40 (C) 96 (D) 150 (E) 224

SOLUTION. If 5x and 3x are the number of men and women in the dormitory, respectively, then the number of residents in the dormitory is 5x + 3x = 8x. So, the number of students must be a multiple of 8. Of the five choices, only 150 (D) is not divisible by 8.

NOTE:Assume that the ratio of the number of pounds of cole slaw to the number of pounds of potato salad consumed in the dormitory’s cafeteria was 5:3. Then, it is possible that a total of exactly 150 pounds was eaten: 93.75 pounds of cole slaw and 56.25 pounds of potato salad. In Example 3, 150 wasn’t possible because there had to be a whole number of men and women.

EXAMPLE 4

The measures of the two acute angles in a right triangle are in the ratio of 5:13. What is the measure of the larger angle?

(A) 25° (B) 45° (C) 60° (D) 65° (E) 75°

SOLUTION. Let the measure of the smaller angle be 5x and the measure of the larger angle be 13x. Since the sum of the measures of the two acute angles of a right triangle is 90° (KEY FACT J1), 5x + 13x = 90 ⇒ 18x = 90⇒ x = 5.

Therefore, the measure of the larger angle is 13 × 5 = 65° (D).

Ratios can be extended to three or four or more terms. For example, we can say that the ratio of freshmen to sophomores to juniors to seniors in a college marching band is 6:8:5:8, which means that for every 6 freshmen in the band there are 8 sophomores, 5 juniors, and 8 seniors.

NOTE: TACTIC D1 applies to extended ratios, as well.

EXAMPLE 5

The concession stand at Cinema City sells popcorn in three sizes: large, super, and jumbo. One day, Cinema City sold 240 bags of popcorn, and the ratio of large to super to jumbo was 8:17:15. How many super bags of popcorn were sold that day?

(A) 48 (B) 90 (C) 102 (D) 108 (E) 120

SOLUTION. Let 8x, 17x, and 15x be the number of large, super, and jumbo bags of popcorn sold, respectively. Then 8x + 17x + 15x = 240 ⇒ 40x = 240 ⇒ x = 6.

The number of super bags sold was 17 × 6 = 102 (C).

KEY FACT D3

KEY FACT D1 applies to extended ratios, as well. If a set of objects is divided into 3 groups in the ratio a:b:c, then the first group contains of the objects, the second , and the third .

EXAMPLE 6

If the ratio of large to super to jumbo bags of popcorn sold at Cinema City was 8:17:15, what percent of the bags sold were super?

(A) 20% (B) 25% (C) 33% (D) 37.5% (E) 42.5%

SOLUTION. Super bags made up = 42.5% of the total (E).

A jar contains a number of red (R), white (W), and blue (B) marbles. Suppose that R:W = 2:3 and W:B = 3:5. Then, for every 2 red marbles, there are 3 white ones, and for those 3 white ones, there are 5 blue ones. So, R:B = 2:5, and we can form the extended ratio R:W:B = 2:3:5.

If the ratios were R:W = 2:3 and W:B = 4:5, however, we wouldn’t be able to combine them as easily. From the diagram below, you see that for every 8 reds there are 15 blues, so R:B = 8:15.

To see this without drawing a picture, we write the ratios as fractions: and . Then, we multiply the fractions:

Not only does this give us R:B = 8:15, but also, if we multiply both W numbers, 3 × 4 =12, we can write the extended ratio: R:W:B = 8:12:15.

EXAMPLE 7

Column A	Column B
Jar A and jar B each have 70 marbles, all of which are red, white, or blue. In jar A, R:W = 2:3 and W:B = 3:5. In jar B, R:W = 2:3 and W:B = 4:5.
The number of white marbles in jar A	The number of white marbles in jar B

SOLUTION. From the discussion immediately preceding this example, in jar A the extended ratio R:W:B is 2:3:5, which implies that the white marbles constitute of the total: × 70 = 21.

In jar B the extended ratio R:W:B is 8:12:15, so the white marbles are of the total: × 70 = 24. The answer is B.

A proportion is an equation that states that two ratios are equivalent. Since ratios are just fractions, any equation such as in which each side is a single fraction is a proportion. Usually the proportions you encounter on the GRE involve one or more variables.

Solve proportions by cross-multiplying: if , then ad = bc.

Setting up a proportion is a common way of solving a problem on the GRE.

EXAMPLE 8

If , what is the value of x?

(A) 12 (B) 24 (C) 36 (D) 42 (E) 48

SOLUTION. Cross-multiply: 3(84) = 7x ⇒ 252 = 7x ⇒ x = 36 (C).

EXAMPLE 9

If , what is the value of ?

SOLUTION. Cross-multiply: 16(x + 2) = 17x ⇒ 16x + 32 = 17x ⇒ x = 32.

So, = 2 (D).

EXAMPLE 10

A state law requires that on any field trip the ratio of the number of chaperones to the number of students must be at least 1:12. If 100 students are going on a field trip, what is the minimum number of chaperones required?

(A) 6 (B) 8 (C) 8 (D) 9 (E) 12

SOLUTION. Let x represent the number of chaperones required, and set up a pro-portion: . Cross-multiply: 100 = 12x ⇒ x = 8. This, of course, is not the answer since, clearly, the number of chaperones must be a whole number. Since x is greater than 8, 8 chaperones would not be enough. The answer is 9 (D).

A rate is a fraction that compares two quantities measured in different units. The word “per” often appears in rate problems: miles per hour, dollars per week, cents per ounce, students per classroom, and so on.

Set up rate problems just like ratio problems. Solve the proportions by cross- multiplying.

EXAMPLE 11

Brigitte solved 24 math problems in 15 minutes. At this rate, how many problems can she solve in 40 minutes?

(A) 25 (B) 40 (C) 48 (D) 60 (E) 64

SOLUTION. Handle this rate problem exactly like a ratio problem. Set up a proportion and cross-multiply:

When the denominator in the given rate is 1 unit (1 minute, 1 mile, 1 dollar), the problem can be solved by a single division or multiplication. Consider Examples 12 and 13.

EXAMPLE 12

If Stefano types at the rate of 35 words per minute, how long will it take him to type 987 words?

EXAMPLE 13

If Mario types at the rate of 35 words per minute, how many words can he type in 85 minutes?

SOLUTION 12. Set up a proportion and cross-multiply:

SOLUTION 13. Set up a proportion and cross-multiply:

Notice that in Example 12, all we did was divide 987 by 35, and in Example 13, we multiplied 35 by 85. If you realize that, you don’t have to introduce x and set up a proportion. You must know, however, whether to multiply or divide. If you’re not absolutely positive which is correct, write the proportion; then you can’t go wrong.

CAUTION: In rate problems it is essential that the units in both fractions be the same.

EXAMPLE 14

If 3 apples cost 50¢, how many apples can you buy for $20?

(A) 20 (B) 60 (C) 120 (D) 600 (E) 2000

SOLUTION. We have to set up a proportion, but it is not . In the first fraction, the denominator represents cents, whereas in the second fraction, the denominator represents dollars. The units must be the same. We can change 50 cents to 0.5 dollar or we can change 20 dollars to 2000 cents:

On the GRE, some rate problems involve only variables. They are handled in exactly the same way.

EXAMPLE 15

If a apples cost c cents, how many apples can be bought for d dollars?

SOLUTION. First change d dollars to 100d cents, and set up a proportion:. Now cross-multiply: 100ad = cx ⇒ x = (E).

Most students find problems such as Example 15 very difficult. If you get stuck on such a problem, use TACTIC 2, Chapter 11, which gives another strategy for handling these problems.

Notice that in rate problems, as one quantity increases or decreases, so does the other. If you are driving at 45 miles per hour, the more hours you drive, the further you go; if you drive fewer miles, it takes less time. If chopped meat cost $3.00 per pound, the less you spend, the fewer pounds you get; the more meat you buy, the more it costs.

In some problems, however, as one quantity increases, the other decreases. These cannot be solved by setting up a proportion. Consider the following two examples, which look similar but must be handled differently.

EXAMPLE 16

A hospital needs 150 pills to treat 6 patients for a week. How many pills does it need to treat 10 patients for a week?

EXAMPLE 17

A hospital has enough pills on hand to treat 10 patients for 14 days. How long will the pills last if there are 35 patients?

SOLUTION 16. Example 16 is a standard rate problem. The more patients there are, the more pills are needed.

The ratio or quotient remains constant: .

SOLUTION 17. In Example 17, the situation is different. With more patients, the supply of pills will last for a shorter period of time; if there were fewer patients, the supply would last longer. It is not the ratio that remains constant, it is the product.

There are enough pills to last for 10 × 14 = 140 patient-days:

There are many mathematical situations in which one quantity increases as another decreases, but their product is not constant. Those types of problems, however, do not appear on the GRE.

If one quantity increases as a second quantity decreases, multiply them; their product will be a constant.

EXAMPLE 18

If 15 workers can pave a certain number of driveways in 24 days, how many days will 40 workers take, working at the same rate, to do the same job?

(A) 6 (B) 9 (C) 15 (D) 24 (E) 40

SOLUTION. Clearly, the more workers there are, the less time it will take, so use TACTIC D4: multiply. The job takes 15 × 24 = 360 worker-days:

Note that it doesn’t matter how many driveways have to be paved, as long as the 15 workers and the 40 workers are doing the same job. Even if the question had said, “15 workers can pave 18 driveways in 24 days,” the number 18 would not have entered into the solution. This number would be important only if the second group of workers was going to pave a different number of driveways.

EXAMPLE 19

If 15 workers can pave 18 driveways in 24 days, how many days would it take 40 workers to pave 22 driveways?

(A) 6 (B) 9 (C) 11 (D) 15 (E) 18

SOLUTION. This question is similar to Example 18, except that now the jobs that the two groups of workers are doing are different. The solution, however, starts out exactly the same way. Just as in Example 18, 40 workers can do in 9 days the same job that 15 workers can do in 24 days. Since that job is to pave 18 driveways, 40 workers can pave 18 ÷ 9 = 2 driveways every day. So, it will take 11 days for them to pave 22 driveways (C).

Practice Exercises—Ratios and Proportions

Multiple-Choice Questions

1. If of the employees in a supermarket are not college graduates, what is the ratio of the number of college graduates to those who are not college graduates?
(A) 1:3 (B) 3:7 (C) 3:4 (D) 4:3 (E) 3:1

2. If , what is the value of a²?

3. If 80% of the applicants to a program were rejected, what is the ratio of the number accepted to the number rejected?

4. Scott can read 50 pages per hour. At this rate, how many pages can he read in 50 minutes?
(A) 25
(B)
(C)
(D) 48
(E) 60

5. If all the members of a team are juniors or seniors, and if the ratio of juniors to seniors on the team is 3:5, what percent of the team members are seniors?
(A) 37.5%
(B) 40%
(C) 60%
(D) 62.5%
(E) It cannot be determined from the information given.

6. The measures of the three angles in a triangle are in the ratio of 1:1:2. Which of the following must be true?
I. The triangle is isosceles.
II. The triangle is a right triangle.
III. The triangle is equilateral.
(A) None
(B) I only
(C) II only
(D) I and II only
(E) I and III only

7. What is the ratio of the circumference of a circle to its radius?

8. The ratio of the number of freshmen to sophomores to juniors to seniors on a college basketball team is 4:7:6:8. What percent of the team are sophomores?
(A) 16%
(B) 24%
(C) 25%
(D) 28%
(E) 32%

9. At Central State College the ratio of the number of students taking Spanish to the number taking French is 7:2. If 140 students are taking French, how many are taking Spanish?
(A) 40
(B) 140
(C) 360
(D) 490
(E) 630

10. If a:b = 3:5 and a:c = 5:7, what is the value of b:c?
(A) 3:7
(B) 21:35
(C) 21:25
(D) 25:21
(E) 7:3

11. If x is a positive number and , then x =
(A) 3 (B) 4 (C) 6 (D) 12 (E) 36

12. In the diagram below, b:a = 7:2. What is b − a?

(A) 20 (B) 70 (C) 100 (D) 110 (E) 160

13. A snail can move i inches in m minutes. At this rate, how many feet can it move in h hours?

14. Gilda can grade t tests in hours. At this rate, how many tests can she grade in x hours?

15. A club had 3 boys and 5 girls. During a membership drive the same number of boys and girls joined the club. How many members does the club have now if the ratio of boys to girls is 3:4?
(A) 12 (B) 14 (C) 16 (D) 21 (E) 28

16. If , what is the value of x?
(A) (B) 3 (C) 7 (D) 17 (E) 136

17. If 4 boys can shovel a driveway in 2 hours, how many minutes will it take 5 boys to do the job?
(A) 60 (B) 72 (C) 96 (D) 120 (E) 150

18. If 500 pounds of mush will feed 20 pigs for a week, for how many days will 200 pounds of mush feed 14 pigs?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

Quantitative Comparison Questions

	Column A	Column B
The ratio of red to blue marbles in a jar was 3:5. The same number of red and blue marbles were added to the jar.
19.	The ratio of red to blue marbles now	3:5

Three associates agreed to split the $3000 profit of an investment in the ratio of 2:5:8.
20.	The difference between the largest and the smallest share	$1200

The ratio of the number of boys to girls in the chess club is 5:2. The ratio of the number of boys to girls in the glee club is 11:4.
21.	The number of boys in the chess club	The number of boys in the glee club

Sally invited the same number of boys and girls to her party. Everyone who was invited came, but 5 additional boys showed up. This caused the ratio of girls to boys at the party to be 4:5.
22.	The number of people she invited to her party	40

A large jar is full of marbles. When a single marble is drawn at random from the jar, the probability that it is red is .
23.	The ratio of the number of red marbles to non-red marbles in the jar

3a = 2b and 3b = 5c
24.	The ratio of a to c	1

The radius of circle II is 3 times the radius of circle I
25.		3π

ANSWER KEY

1. A	6. D	11.C	16.D	21.D
2. D	7. E	12.C	17.C	22.C
3. B	8. D	13.A	18.A	23.A
4. B	9. D	14.B	19.A	24.A
5. D	10.D	15.B	20.C	25.B

Answer Explanations

1. (A) Of every 4 employees, 3 are not college graduates, and 1 is a college graduate. So the ratio of graduates to nongraduates is 1:3.

2. (D) Cross-multiplying, we get:

3. (B) If 80% were rejected, 20% were accepted, and the ratio of accepted to rejected is 20:80 = 1:4.

4. (B) Set up a proportion:

5. (D) Out of every 8 team members, 3 are juniors and 5 are seniors. Seniors, therefore, make up of the team.

6. (D) It is worth remembering that if the ratio of the measures of the angles of a triangle is 1:1:2, the angles are 45-45-90 (see Section 14-J). Otherwise, the first step is to write x + x + 2x = 180 ⇒ 4x = 180 ⇒ x = 45. Since two of the angles have the same measure, the triangle is isosceles, and since one of the angles measures 90°, it is a right triangle. I and II are true, and, of course, III is false.

7. (E) By definition, π is the ratio of the circumference to the diameter of a circle (see Section 14-L). Therefore, .

8. (D) The fraction of the team that is sophomores is .

9. (D) Let the number of students taking Spanish be 7x, and the number taking French be 2x. Then, 2x = 140 ⇒ x = 70 ⇒ 7x = 490.

10. (D) Since . So,

Alternatively, we could write equivalent ratios with the same value for a:
a:b = 3:5 = 15:25 and a:c = 5:7 = 15:21.
So, when a = 15, b = 25, and c = 21.

11. (C) To solve a proportion, cross-multiply:

12. (C) Let b = 7x and a = 2x. Then,
7x + 2x = 180 ⇒ 9x = 180 ⇒ x = 20 ⇒ b = 140 and a = 40 ⇒ b − a = 140 − 40 = 100.

13. (A) Set up the proportion, keeping track of units: .

14. (B) Gilda grades at the rate of
.
Since she can grade tx tests each hour, in x hours she can grade x(tx) = tx² tests.

15. (B) Suppose that x boys and x girls joined the club. Then, the new ratio of boys to girls would be (3 + x):(5 + x), which we are told is 3:4. So, .

Therefore, 3 boys and 3 girls joined the other 3 boys and 5 girls: a total of 14.

16. (D) Cross-multiplying, we get:

17. (C) Since 4 boys can shovel the driveway in 2 hours, or 2 × 60 = 120 minutes, the job takes 4 × 120 = 480 boy-minutes; and so 5 boys

18. (A) Since 500 pounds will last for 20 pig-weeks = 140 pig-days, 200 pounds will last for × 140 pig-days = 56 pig-days, and .

19. (A) Assume that to start there were 3x red marbles and 5x blue ones and that y of each color were added.

20. (C) The shares are 2x, 5x, and 8x, and their sum is 3000:

21. (D) Ratios alone can’t answer the question, “How many?” There could be 5 boys in the chess club or 500. We can’t tell.

22. (C) Assume that Sally invited x boys and x girls. When she wound up with x girls and x + 5 boys, the girl:boy ratio was 4:5. So,

Sally invited 40 people (20 boys and 20 girls).

23. (A) If the probability of drawing a red marble is , 3 out of every 7 marbles are red, and 4 out of every 7 are non-red. So the ratio of red:non-red = 3:4, which is greater than .

24. (A) Multiplying the first equation by 3 and the second by 2 to get the same coefficient of b, we have: 9a = 6b and .

25. (B) Assume the radius of circle I is 1 and the radius of circle II is 3. Then the areas are π and 9π, respectively. So, the area of circle II is 9 times the area of circle I, and 3π > 9.

14-E. AVERAGES

The average of a set of n numbers is the sum of those numbers divided by n.

If the weights of three children are 80, 90, and 76 pounds, respectively, to calculate the average weight of the children, you would add the three weights and divide by 3:

The technical name for this type of average is “arithmetic mean,” and on the GRE those words always appear in parentheses—for example, “What is the average (arithmetic mean) of 80, 90, and 76?”

Usually, on the GRE, you are not asked to find an average; rather, you are given the average of a set of numbers and asked for some other information. The key to solving all of these problems is to first find the sum of the numbers. Since , multiplying both sides by n yields the equation: sum = nA.

If you know the average, A, of a set of n numbers, multiply A by n to get their sum.

EXAMPLE 1

One day a supermarket received a delivery of 25 frozen turkeys. If the average (arithmetic mean) weight of a turkey was 14.2 pounds, what was the total weight, in pounds, of all the turkeys?

SOLUTION. Use TACTIC E1: 25 × 14.2 = 355.

NOTE: We do not know how much any individual turkey weighed nor how many turkeys weighed more or less than 14.2 pounds. All we know is their total weight.

EXAMPLE 2

Sheila took five chemistry tests during the semester and the average (arithmetic mean) of her test scores was 85. If her average after the first three tests was 83, what was the average of her fourth and fifth tests?

(A) 83 (B) 85 (C) 87 (D) 88 (E) 90

SOLUTION.

· Use TACTIC E1: On her five tests, Sheila earned 5 × 85 = 425 points.

· Use TACTIC E1 again: On her first three tests she earned 3 × 83 = 249 points.

· Subtract: On her last two tests Sheila earned 425 − 249 = 176 points.

· Calculate her average on her last two tests: .

NOTE: We cannot determine Sheila’s grade on even one of the tests.

KEY FACT E1

• If all the numbers in a set are the same, then that number is the average.

• If the numbers in a set are not all the same, then the average must be greater than the smallest number and less than the largest number. Equivalently, at least one of the numbers is less than the average and at least one is greater.

If Jessica’s test grades are 85, 85, 85, and 85, her average is 85. If Gary’s test grades are 76, 83, 88, and 88, his average must be greater than 76 and less than 88. What can we conclude if, after taking five tests, Kristen’s average is 90? We know that she earned exactly 5 × 90 = 450 points, and that either she got a 90 on every test or at least one grade was less than 90 and at least one was over 90. Here are a few of the thousands of possibilities for Kristen’s grades:

(a) 90, 90, 90, 90, 90 (b) 80, 90, 90, 90, 100 (c) 83, 84, 87, 97, 99 (d) 77, 88, 93, 95, 97 (e) 50, 100, 100, 100, 100

In (b), 80, the one grade below 90, is 10 points below, and 100, the one grade above 90, is 10 points above. In (c), 83 is 7 points below 90, 84 is 6 points below 90, and 87 is 3 points below 90, for a total of 7 + 6 + 3 = 16 points below 90; 97 is 7 points above 90, and 99 is 9 points above 90, for a total of 7 + 9 = 16 points above 90.

These differences from the average are called deviations, and the situation in these examples is not a coincidence.

KEY FACT E2

The total deviation below the average is equal to the total deviation above the average.

EXAMPLE 3

If the average (arithmetic mean) of 25, 31, and x is 37, what is the value of x? (A) 31 (B) 37 (C) 43 (D) 55 (E) 56

SOLUTION 1. Use KEY FACT E2. Since 25 is 12 less than 37 and 31 is 6 less than 37, the total deviation below the average is 12 + 6 = 18. Therefore, the total deviation above must also be 18. So, x = 37 + 18 = 55 (D).

SOLUTION 2. Use TACTIC E1. Since the average of the three numbers is 37, the sum of the 3 numbers is 3 × 37 = 111. Then,

KEY FACT E3

Assume that the average of a set of numbers is A. If a number x is added to the set and a new average is calculated, then the new average will be less than, equal to, or greater than A, depending on whether x is less than, equal to, or greater than A, respectively.

EXAMPLE 4

Column A	Column B
The average (arithmetic mean) of the integers from 0 to 12	The average (arithmetic mean) of the integers from 1 to 12

HELPFUL HINT

Remember TACTIC 5 from Chapter 12. We don’t have to calculate the averages, we just have to compare them.

SOLUTION 1. Column B is the average of the integers from 1 to 12, which is surely greater than 1. In Column A we are taking the average of those same 12 numbers and 0. Since the extra number, 0, is less than the Column B average, the Column A average must be lower [KEY FACT E3]. The answer is B.

SOLUTION 2. Clearly the sum of the 13 integers from 0 to 12 is the same as the sum of the 12 integers from 1 to 12. Since that sum is positive, dividing by 13 yields a smaller quotient than dividing by 12 [KEY FACT B4].

Although in solving Example 4 we didn’t calculate the averages, we could have:

0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78 and .

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78 and .

Notice that the average of the 13 consecutive integers 0, 1,…,12 is the middle integer, 6, and the average of the 12 consecutive integers 1, 2,…,12 is the average of the two middle integers, 6 and 7. This is a special case of KEY FACT E4.

KEY FACT E4

Whenever n numbers form an arithmetic sequence (one in which the difference between any two consecutive terms is the same): (i) if n is odd, the average of the numbers is the middle term in the sequence and (ii) if nis even, the average of the numbers is the average of the two middle terms.

For example, in the arithmetic sequence 6, 9, 12, 15, 18, the average is the middle number, 12; in the sequence 10, 20, 30, 40, 50, 60, the average is 35, the average of the two middle numbers — 30 and 40.

EXAMPLE 5

On Thursday, 20 of the 25 students in a chemistry class took a test and their average was 80. On Friday, the other 5 students took the test, and their average was 90. What was the average (arithmetic mean) for the entire class?
(A) 80 (B) 82 (C) 84 (D) 85 (E) 88

SOLUTION. The class average is calculated by dividing the sum of all 25 test grades by 25.

• The first 20 students earned a total of:	20 × 80 = 1600 points
• The other 5 students earned a total of:	5 × 90 = 450 points
• Add: altogether the class earned:	1600 + 450 = 2050 points
• Calculate the class average:	= 82 (B).

Notice that the answer to Example 5 is not 85, which is the average of 80 and 90. This is because the averages of 80 and 90 were earned by different numbers of students, and so the two averages had to be given different weights in the calculation. For this reason, this is called a weighted average.

KEY FACT E5

To calculate the weighted average of a set of numbers, multiply each number in the set by the number of times it appears, add all the products, and divide by the total number of numbers in the set.

So, the solution to Example 5 should look like this:

Problems involving average speed will be discussed in Section 14-H, but we mention them briefly here because they are closely related to problems on weighted averages.

HELPFUL HINT

Without doing any calculations, you should immediately realize that since the grade of 80 is being given more weight than the grade of 90, the average will be closer to 80 than to 90 — certainly less than 85.

EXAMPLE 6

For the first 3 hours of his trip, Justin drove at 50 miles per hour. Then, due to construction delays, he drove at only 40 miles per hour for the next 2 hours. What was his average speed, in miles per hour, for the entire trip?
(A) 40 (B) 43 (C) 46 (D) 48 (E) 50

SOLUTION. This is just a weighted average:

Note that in the fractions above, the numerator is the total distance traveled and the denominator the total time the trip took. This is always the way to find an average speed. Consider the following slight variation on Example 6.

EXAMPLE 6A

For the first 100 miles of his trip, Justin drove at 50 miles per hour, and then due to construction delays, he drove at only 40 miles per hour for the next 120 miles. What was his average speed, in miles per hour, for the entire trip?

SOLUTION. This is not a weighted average. Here we immediately know the total distance traveled, 220 miles. To get the total time the trip took, we find the time for each portion and add: the first 100 miles took 100 ÷ 50 = 2 hours, and the next 120 miles took 120 ÷ 40 = 3 hours. So the average speed was miles per hour.

Notice that in Example 6, since Justin spent more time traveling at 50 miles per hour than at 40 miles per hour, his average speed was closer to 50; in Example 6a, he spent more time driving at 40 miles per hour than at 50 miles per hour, so his average speed was closer to 40.

Two other terms that are associated with averages are median and mode. In a set of n numbers that are arranged in increasing order, the median is the middle number (if n is odd), or the average of the two middle numbers (if n is even). The mode is the number in the set that occurs most often.

EXAMPLE 7

During a 10-day period, Jorge received the following number of phone calls each day: 2, 3, 9, 3, 5, 7, 7, 10, 7, 6. What is the average (arithmetic mean) of the median and mode of this set of data?
(A) 6 (B) 6.25 (C) 6.5 (D) 6.75 (E) 7

SOLUTION. The first step is to write the data in increasing order:

2, 3, 3, 5, 6, 7, 7, 7, 9, 10.

· The median is 6.5, the average of the middle two numbers.

· The mode is 7, the number that appears more times than any other.

· The average of the median and the mode is .

Practice Exercises—Averages

Multiple-Choice Questions

1. Michael’s average (arithmetic mean) on 4 tests is 80. What does he need on his fifth test to raise his average to 84?
(A) 82 (B) 84 (C) 92 (D) 96 (E) 100

2. Maryline’s average (arithmetic mean) on 4 tests is 80. Assuming she can earn no more than 100 on any test, what is the least she can earn on her fifth test and still have a chance for an 85 average after seven tests?
(A) 60 (B) 70 (C) 75 (D) 80 (E) 85

3. Sandrine’s average (arithmetic mean) on 4 tests is 80. Which of the following cannot be the number of tests on which she earned exactly 80 points?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

4. What is the average (arithmetic mean) of the positive integers from 1 to 100, inclusive?
(A) 49 (B) 49.5 (C) 50 (D) 50.5 (E) 51

5. If 10a + 10b = 35, what is the average (arithmetic mean) of a and b?
(A) 1.75 (B) 3.5 (C) 7 (D) 10 (E) 17.5

6. If x + y = 6, y + z = 7, and z + x = 9, what is the average (arithmetic mean) of x, y, and z?
(A) (B) (C) (D) 11 (E) 22

7. If the average (arithmetic mean) of 5, 6, 7, and w is 8, what is the value of w?
(A) 8 (B) 12 (C) 14 (D) 16 (E) 24

8. What is the average (arithmetic mean) of the measures of the five angles in a pentagon?
(A) 36°
(B) 72°
(C) 90°
(D) 108°
(E) 144°

9. If a + b = 3(c + d), which of the following is the average (arithmetic mean) of a, b, c, and d?
(A)
(B)
(C)
(D)
(E) c + d

10. In the diagram below, lines and m are not parallel.

If A represents the average (arithmetic mean) of the degree measures of all eight angles, which of the following is true?
(A) A = 45
(B) 45 < A < 90
(C) A = 90
(D) 90 < A < 180
(E) A = 180

11. What is the average (arithmetic mean) of 2¹⁰ and 2²⁰?
(A) 2¹⁵
(B) 2⁵ + 2¹⁰
(C) 2⁹ + 2¹⁹
(D) 2²⁹
(E) 30

12. Let M be the median and m the mode of the following set of numbers: 10, 70, 20, 40, 70, 90. What is the average (arithmetic mean) of M and m?
(A) 50 (B) 55 (C) 60 (D) 62.5 (E) 65

Quantitative Comparison Questions

	Column A	Column B
13.	The average (arithmetic mean) of the measures of the three angles of an equilateral triangle	The average (arithmetic mean) of the measures of the three angles of a right triangle

10 students took a test and the average grade was 80.
No one scored exactly 80.

14.

The number of grades over 80

15.

The average (arithmetic mean) of 2x and 2y

180

There are the same number of boys and girls in a club.

The average weight of the boys is 150 pounds.

The average weight of the girls is 110 pounds.

16.

The number of boys weighing over 150

The number of girls weighing over 110

The average (arithmetic mean) of
22, 38, x, and y is 15.
x > 0

17.

18.

The average (arithmetic mean) of the even numbers between 1 and 11

The average (arithmetic mean) of the odd numbers between 2 and 12

19.

The average (arithmetic mean) of 17, 217, 417

The average (arithmetic mean) of 0, 17, 217, 417

y > 0

20.

The average (arithmetic mean) of x and y

The average (arithmetic mean) of x, y, and 2y

ANSWER KEY

1. E	5. A	9. E	13. C	17. B
2. C	6. A	10. C	14. D	18. B
3. D	7. C	11. C	15. C	19. A
4. D	8. D	12. D	16. D	20. D

Answer Explanations

1. (E) Use TACTIC E1. For Michael’s average on five tests to be an 84, he needs a total of 5 × 84 = 420 points. So far, he has earned 4 × 80 = 320 points. Therefore, he needs 100 points more.
Alternative solution. Use KEY FACT E2.
Assume Michael’s first 4 tests were all 80s. His total deviation below 84 is 4 × 4 = 16. So, his total deviation above 84 must also be 16. He needs 84 + 16 = 100.

2. (C) Use TACTIC E1. So far, Maryline has earned 320 points. She can survive a low grade on test five if she gets the maximum possible on both the sixth and seventh tests. So, assume she gets two 100s. Then her total for tests 1, 2, 3, 4, 6, and 7 would be 520. For her seven-test average to be 85, she needs a total of 7 × 85 = 595 points. Therefore, she needs at least 595 − 520 = 75 points.
Alternative solution. Use KEY FACT E2.
Assume Maryline’s first four tests were all 80s. Then her total deviation below 85 would be 4 × 5 = 20. Her maximum possible deviation above 85 (assuming 100s on tests 6 and 7) is 15 + 15 = 30. So, on test 5 she can deviate at most 10 more points below 85: 85 − 10 = 75.

3. (D) Since Sandrine’s 4-test average is 80, she earned a total of 4 × 80 = 320 points. Could Sandrine have earned a total of 320 points with:

0 grades of 80?	Easily; for example, 20, 100, 100, 100 or 60, 70, 90, 100.
1 grade of 80?	Lots of ways; 80, 40, 100, 100, for instance.
2 grades of 80?	Yes; 80, 80, 60, 100.
4 grades of 80?	Sure: 80, 80, 80, 80.
3 grades of 80?	NO! 80 + 80 + 80 + x = 320 ⇒ x = 80, as well.

4. (D) Clearly, the sequence of integers from 1 to 100 has 100 terms, and so by KEY FACT E4, we know that the average of all the numbers is the average of the two middle ones: 50 and 51. The average, therefore, is 50.5.

5. (A) Since 10a + 10b = 35, dividing both sides of the equation by 10, we get that a + b = 3.5. Therefore, the average of a and b is 3.5 ÷ 2 = 1.75.

6. (A) Whenever a question involves three equations, add them:
Divide by 2: x + y + z = 11
The average of x, y, and z is .

7. (C) Use TACTIC E1: the sum of the 4 numbers is 4 times their average:
5 + 6 + 7 + w = 4 × 8 = 32 ⇒ 18 + w = 32 ⇒ w = 14.
Alternative solution. Use KEY FACT E2: 5 is 3 below 8, 6 is 2 below 8, and 7 is 1 below 8, for a total deviation of 3 + 2 + 1 = 6 below the average of 8. To compensate, w must be 6 more than 8: 6 + 8 = 14.

8. (D) The average of the measures of the five angles is the sum of their measures divided by 5. The sum is (5 − 2) × 180 = 3 × 180 = 540 (see Section 14-K). So, the average is 540 ÷ 5 = 108.

9. (E) Calculate the average:

10. (C) Since a + b + c + d = 360, and e + f + g + h = 360 (see Section 14-I), the sum of the measures of all 8 angles is 360 + 360 = 720, and their average is 720 ÷ 8 = 90.

11. (C) The average of 2¹⁰ and 2²⁰ is

12. (D) Arrange the numbers in increasing order: 10, 20, 40, 70, 70, 90. M, the median, is the average of the middle two numbers: ; the mode, m, is 70, the number that appears most frequently. The average of M and m, therefore, is the average of 55 and 70, which is 62.5.

13. (C) In any triangle, the sum of the measures of the three angles is 180°, and the average of their measures is 180 ÷ 3 = 60.

14. (D) From KEY FACT E1, we know only that at least one grade was above 80. In fact, there may have been only one (9 grades of 79 and 1 grade of 89, for example). But there could have been five or even nine (for example, 9 grades of 85 and 1 grade of 35).
Alternative solution. The ten students scored exactly 800 points. Ask, “Could they be equal?” Could there be exactly five grades above 80? Sure, five grades of 100 for 500 points and five grades of 60 for 300 points. Must they be equal? No, eight grades of 100 and two grades of 0 also total 800.

15. (C) The average of 2x and 2y is , which equals 180.

16. (D) It is possible that no boy weighs over 150 (if every single boy weighs exactly 150); on the other hand, it is possible that almost every boy weighs over 150. The same is true for the girls.

17. (B) Use TACTIC E1:
22 + 38 + x + y = 4(15) = 60 ⇒ 60 + x + y = 60 ⇒ x + y = 0.
Since it is given that x is positive, y must be negative.

18. (B) Don’t calculate the averages. Each number in Column A (2, 4, 6, 8, 10) is less than the corresponding number in Column B (3, 5, 7, 9, 11), and so the Column A average must be less than the Column B average.
Alternative solution. Observe that the numbers in each column form an arithmetic sequence, so by KEY FACT E4 the averages are just the middle numbers (6 and 7).

19. (A) You don’t have to calculate the averages. The average of the set of numbers in Column A is clearly positive, and by KEY FACT E3, adding 0 to that set must lower the average.

20. (D) Use KEY FACT E3: If x < y, then the average of x and y is less than y, and surely less than 2y. So, 2y has to raise the average. On the other hand, if x is much larger than y, then 2y would lower the average.

Algebra

For the GRE you need to know only a small portion of the algebra normally taught in a high school elementary algebra course and none of the material taught in an intermediate or advanced algebra course. Sections 14-F, 14-G, and 14-H review only those topics that you absolutely need for the GRE.

14-F. POLYNOMIALS

Even though the terms monomial, binomial, trinomial, and polynomial are not used on the GRE, you must be able to work with simple polynomials, and the use of these terms will make it easier for us to discuss the important concepts.

A monomial is any number or variable or product of numbers and variables. Each of the following is a monomial:

3 −4 x y 3x −4xyz 5x³ 1.5xy² a³b⁴

The number that appears in front of the variables in a monomial is called the coefficient. The coefficient of 5x³ is 5. If there is no number, the coefficient is 1 or −1, because x means 1x and −ab² means −1ab².

On the GRE, you could be asked to evaluate a monomial for specific values of the variables.

EXAMPLE 1

What is the value of −3a²b when a = −4 and b = 0.5?

(A) −72 (B) −24 (C) 24 (D) 48 (E) 72

SOLUTION. Rewrite the expression, replacing the letters a and b with the numbers −4 and 0.5, respectively. Make sure to write each number in parentheses. Then evaluate: −3(−4)²(0.5) = −3(16)(0.5) = −24 (B).

CAUTION

Be sure you follow PEMDAS (see Section 14-A): handle exponents before the other operations. In Example 1, you cannot multiply −4 by −3, get 12, and then square the 12; you must first square −4.

A polynomial is a monomial or the sum of two or more monomials. Each monomial that makes up the polynomial is called a term of the polynomial. Each of the following is a polynomial:

2x² 2x² + 3 3x² − 7 x² + 5x − 1 a²b + b²a x² − y² w² − 2w + 1

The first polynomial in the above list is a monomial; the second, third, fifth, and sixth polynomials are called binomials, because each has two terms; the fourth and seventh polynomials are called trinomials, because each has three terms. Two terms are called like terms if they have exactly the same variables and exponents; they can differ only in their coefficients: 5a²b and −3a²b are like terms, whereas a²b and b²a are not.

The polynomial 3x² + 4x + 5x + 2x² + x − 7 has 6 terms, but some of them are like terms and can be combined:

3x² + 2x² = 5x² and 4x + 5x + x = 10x.

So, the original polynomial is equivalent to the trinomial 5x² + 10x − 7.

KEY FACT F1

The only terms of a polynomial that can be combined are like terms.

KEY FACT F2

To add two polynomials, put a plus sign between them, erase the parentheses, and combine like terms.

HELPFUL HINT

To add, subtract, multiply, and divide polynomials, use the usual laws of arithmetic. To avoid careless errors, before performing any arithmetic operations, write each polynomial in parentheses.

EXAMPLE 2

What is the sum of 5x² + 10x − 7 and 3x² − 4x + 2?

SOLUTION. (5x² + 10x − 7) + (3x² − 4x + 2)
= 5x² + 10x − 7 + 3x² − 4x + 2
= (5x² + 3x²) + (10x − 4x) + (−7 + 2)
= 8x² + 6x − 5.

KEY FACT F3

To subtract two polynomials, change the minus sign between them to a plus sign and change the sign of every term in the second parentheses. Then just use KEY FACT F2 to add them: erase the parentheses and then combine like terms.

CAUTION

Make sure you get the order right in a subtraction problem.

EXAMPLE 3

Subtract 3x² − 4x + 2 from 5x² + 10x − 7.

SOLUTION. Be careful. Start with the second polynomial and subtract the first:

(5x² + 10x − 7) − (3x² − 4x + 2) = (5x² + 10x − 7) + (−3x² + 4x − 2) = 2x² + 14x − 9.

EXAMPLE 4

What is the average (arithmetic mean) of 5x² + 10x − 7, 3x² − 4x + 2, and 4x² + 2?

SOLUTION. As in any average problem, add and divide:

(5x² + 10x − 7) + (3x² − 4x + 2) + (4x² + 2) = 12x² + 6x − 3, and by the distributive law (KEY FACT A21), .

KEY FACT F4

To multiply monomials, first multiply their coefficients, and then multiply their variables (letter by letter), by adding the exponents (see Section 12A).

EXAMPLE 5

What is the product of 3xy²z³ and −2x²y²?

SOLUTION. (3xy²z³)(−2x²y²) = 3(−2)(x)(x²)(y²)(y²)(z³) = −6x³y⁴z³.

All other polynomials are multiplied by using the distributive law.

KEY FACT F5

To multiply a monomial by a polynomial, just multiply each term of the polynomial by the monomial.

EXAMPLE 6

What is the product of 2a and 3a² − 6ab + b²?

SOLUTION. .

On the GRE, the only other polynomials that you could be asked to multiply are two binomials.

KEY FACT F6

To multiply two binomials, use the so-called FOIL method, which is really nothing more than the distributive law: Multiply each term in the first parentheses by each term in the second parentheses and simplify by combining terms, if possible.

EXAMPLE 7

What is the value of (x − 2)(x + 3) − (x − 4)(x + 5)?

SOLUTION. First, multiply both pairs of binomials:

(x − 2)(x + 3) = x² + 3x − 2x − 6 = x² + x − 6
(x − 4)(x + 5) = x² + 5x − 4x − 20 = x² + x − 20

Now, subtract:

(x² + x − 6) − (x² + x − 20) = x² + x − 6 − x² − x + 20 = 14.

KEY FACT F7

The three most important binomial products on the GRE are these:

· (x − y)(x + y) = x² + xy − yx − y² = x² − y²

· (x − y)² = (x − y)(x − y) = x² − xy − yx + y² = x² − 2xy + y²

· (x + y)² = (x + y)(x + y) = x² + xy + yx + y² = x² + 2xy + y²

HELPFUL HINT

If you memorize these, you won’t have to multiply them out each time you need them.

EXAMPLE 8

If a − b = 7 and a + b = 13, what is the value of a² − b²?

(A) −120 (B) 20 (C) 91 (D) 120 (E) 218

SOLUTION. In Section 14-G, we will review how to solve such a pair of equations; but even if you know how, you should not do it here. You do not need to know the values of a and b to answer this question. The moment you see a² − b², you should think (a − b)(a + b). Then:

a² − b² = (a − b)(a + b) = (7)(13) = 91 (C).

EXAMPLE 9

If x² + y² = 36 and (x + y)² = 64, what is the value of xy?

(A) 14 (B) 28 (C) 100 (D) 128 (E) 2304

SOLUTION.

64 = (x + y)² = x² + 2xy + y² = x² + y² + 2xy = 36 + 2xy.

Therefore, 2xy = 64 − 36 = 28 ⇒ xy = 14 (A).

On the GRE, the only division of polynomials you might have to do is to divide a polynomial by a monomial. You will not have to do long division of polynomials.

KEY FACT F8

To divide a polynomial by a monomial, use the distributive law. Then simplify each term by reducing the fraction formed by the coefficients to lowest terms and applying the laws of exponents.

EXAMPLE 10

What is the quotient when 32a²b + 12ab³c is divided by 8ab?

SOLUTION. By the distributive law, .

Now reduce each fraction: .

On the GRE, the most important way to use the three formulas in KEY FACT F7 is to recognize them in reverse. In other words, whenever you see x² − y², you should realize that it can be rewritten as (x − y)(x + y). This process, which is the reverse of multiplication, is called factoring.

EXAMPLE 11

Column A		Column B
The value of x² + 4x + 4 when x = 95.9		The value of x2 − 4x + 4 when x = 99.5

SOLUTION. Obviously, you don’t want to plug in 95.9 and 99.5 (remember that the GRE never requires you to do tedious arithmetic). Recognize that x² + 4x + 4 is equal to (x + 2)² and that x² − 4x + 4 is equal to (x − 2)². So, Column A is just (95.9 + 2)² = 97.9², whereas Column B is (99.5 − 2)² = 97.5². Column A is greater.

EXAMPLE 12

What is the value of (1,000,001)² − (999,999)²?

SOLUTION. Do not even consider squaring 999,999. You know that there has to be an easier way to do this. In fact, if you stop to think, you can get the right answer in a few seconds. This is just a² − b² where a = 1,000,001 and b= 999,999, so change it to (a − b)(a + b):

(1,000,001)² − (999,999)² = (1,000,001 − 999,999)(1,000,001 + 999,999) = (2)(2,000,000) = 4,000,000.

Although the coefficients of any of the terms in a polynomial can be fractions, as in , the variable itself cannot be in the denominator. An expression such as , which does have a variable in the denominator, is called an algebraic fraction. Fortunately, you should have no trouble with algebraic fractions, since they are handled just like regular fractions. The rules that you reviewed in Section 14-B for adding, subtracting, multiplying, and dividing fractions apply to algebraic fractions, as well.

EXAMPLE 13

What is the sum of the reciprocals of x² and y²?

SOLUTION. To add , you need a common denominator, which is x²y².

Multiply the numerator and denominator of by y² and the numerator and denominator of by x², and then add:

Often, the way to simplify algebraic fractions is to factor the numerator or the denominator or both. Consider the following example, which is harder than anything you will see on the GRE, but still quite manageable.

EXAMPLE 14

What is the value of when x = 9999?

SOLUTION. Don’t use FOIL to multiply the denominator. That’s going the wrong way. We want to simplify this fraction by factoring everything we can. First factor an x out of the numerator and notice that what’s left is the difference of two squares, which can be factored. Then factor out the 3 in the second factor in the denominator:

So, instead of plugging 9999 into the original expression, plug it into 9999 ÷ 3 = 3333.

Practice Exercises—Polynomials

Multiple-Choice Questions

1. What is the value of when a = 117 and b = 118?
(A) −1
(B) 1
(C) 117.5
(D) 175
(E) 235

2. If a² − b² = 21 and a² + b² = 29, which of the following could be the value of ab?
I. −10
II. 5
III. 10
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) II and III only

3. What is the average (arithmetic mean) of x² + 2x − 3, 3x² − 2x − 3, and 30 − 4x²?

4. What is the value of x² + 12x + 36 when x = 994?
(A) 11,928
(B) 98,836
(C) 100,000
(D) 988,036
(E) 1,000,000

5. If c² + d² = 4 and (c − d)² = 2, what is the value of cd?
(A) 1 (B) (C) 2 (D) 3 (E) 4

6. What is the value of
(2x + 3)( x + 6) − (2x − 5)(x + 10)?
(A) 32
(B) 16
(C) 68
(D) 4x² + 30x + 68
(E) 4x² + 30x − 32

7. If and ab = c, what is the average of a and b?

8. If x² − y² = 28 and x − y = 8, what is the average of x and y?
(A) 1.75 (B) 3.5 (C) 7 (D) 8 (E) 10

9. Which of the following is equal to

10. If = 100, what is the value of + a²?
(A) 10 (B) 64 (C) 98 (D) 100 (E) 102

Quantitative Comparison Questions

	Column A	Column B
	n < 0
11.	−2n²	(−2n)²

	d < c
12.	(c − d)(c + d)	(c − d)(c − d)

	x = −3 and y = 2
13.	−x²y³	0

14.

(r + s)(r − s)

r (s + r) − s (r + s)

15.

4x + 8

ANSWER KEY

1. E	4. E	7. B	10. C	13. B
2. D	5. A	8. A	11. B	14. C
3. E	6. C	9. B	12. D	15. D

Answer Explanations

1. (E) = a + b = 117 + 118 = 235.

2. (D) Adding the two equations, we get that 2a² = 50 ⇒ a² = 25 ⇒ b² = 4. So, a = 5 or −5 and b = 2 or −2. The only possibilities for their product are 10 and −10. (Only I and III are true.)

3. (E) To find the average, take the sum of the three polynomials and then divide by 3. Their sum is (x² + 2x − 3) + (3x² − 2x − 3) + (30 − 4x²) = 24, and 24 ÷ 3 = 8.

4. (E) You can avoid messy, time-consuming arithmetic if you recognize that x² + 12x + 36 = (x + 6)². The value is (994 + 6)² = 1000² = 1,000,000.

5. (A) Start by squaring c − d: 2 = (c − d)² = c² − 2cd + d² = c² + d² − 2cd = 4 − 2cd. So, 2 = 4 − 2cd ⇒ 2cd = 2 ⇒ cd = 1.

6. (C) First multiply out both pairs of binomials: (2x + 3)(x + 6) = 2x² + 15x + 18 and (2x − 5)(x + 10) = 2x² + 15x − 50.
Now subtract:
(2x² + 15x + 18) − (2x² + 15x − 50) = 18 − (−50) = 68.

7. (B) ⇒ 1 = a + b ⇒ .

8. (A) x² − y² = (x − y)(x + y) ⇒ 28 = 8(x + y) ⇒ x + y = 28 ÷ 8 = 3.5. Finally, the average of x and y is = 1.75.

9. (B) Expand each square:
Similarly,
Subtract:

10. (C) 100 = + 2 + a² ⇒ + a² = 98.

11. (B) Since n is negative, n² is positive, and so −2n² is negative. Therefore, Column A is negative, whereas Column B is positive.

Column A Column B

12. (D) c > d ⇒ c − d is
positive, so divide
If d = 0 the columns are equal;
if d = 1, they aren’t.

13. (B) Column A: −(−3)²2³ = − (9)(8) = −72.

14. (C) Column B: r(s + r) − s(r + s) = rs + r² − sr − s² = r² − s²
Column A: (r + s)(r − s) = r² − s².

15. (D) Column A: = 5(x + 2).
Column B: 4x + 8 = 4(x + 2). If x = −2, both columns are 0; for any other value of x the columns are unequal.

14-G. SOLVING EQUATIONS AND INEQUALITIES

The basic principle that you must adhere to in solving any equation is that you can manipulate it in any way, as long as you do the same thing to both sides. For example, you may always add the same number to each side; subtract the same number from each side; multiply or divide each side by the same number (except 0); square each side; take the square root of each side (if the quantities are positive); or take the reciprocal of each side. These comments apply to inequalities, as well, except you must be very careful, because some procedures, such as multiplying or dividing by a negative number and taking reciprocals, reverse inequalities (see Section 14-A).

Most of the equations and inequalities that you will have to solve on the GRE have only one variable and no exponents. The following simple six-step method can be used on all of them.

EXAMPLE 1

If + 3(x − 2) = 2(x + 1) + 1, what is the value of x?

SOLUTION. Follow the steps outlined in the following table.

Step	What to Do	Example 1
1	Get rid of fractions and decimals by multiplying both sides by the Lowest Common Denominator (LCD).	Multiply each term by 2: x + 6(x − 2) = 4(x + 1) + 2.
2	Get rid of all parentheses by using the distributive law.	x + 6x − 12 = 4x + 4 + 2.
3	Combine like terms on each side.	7x − 12 = 4x + 6.
4	By adding or subtracting, get all the variables on one side.	Subtract 4x from each side: 3x − 12 = 6.
5	By adding or subtracting, get all the plain numbers on the other side.	Add 12 to each side: 3x = 18.
6	Divide both sides by the coefficient of the variable.*	Divide both sides by 3: x = 6.

*Note: If you start with an inequality and in Step 6 you divide by a negative number, remember to reverse the inequality (see KEY FACT A24).

Example 1 is actually harder than any equation on the GRE, because it required all six steps. On the GRE that never happens. Think of the six steps as a list of questions that must be answered. Ask if each step is necessary. If it isn’t, move on to the next one; if it is, do it.

Let’s look at Example 2, which does not require all six steps.

EXAMPLE 2

For what real number n is it true that 3(n − 20) = n?
(A) −10 (B) 0 (C) 10 (D) 20 (E) 30

SOLUTION. Do whichever of the six steps are necessary.

Step	Question	Yes/No	What to Do
1	Are there any fractions or decimals?	No
2	Are there any parentheses?	Yes	Get rid of them: 3n − 60 = n.
3	Are there any like terms to combine?	No
4	Are there variables on both sides?	Yes	Subtract n from each side: 2n − 60 = 0.
5	Is there a plain number on the same side as the variable?	Yes	Add 60 to each side: 2n = 60.
6	Does the variable have a coefficient?	Yes	Divide both sides by 2: n = 30.

Memorize the six steps in order and use this method whenever you have to solve this type of equation or inequality.

EXAMPLE 3

Three brothers divided a prize as follows. The oldest received of it, the middle brother received of it, and the youngest received the remaining $120.

What was the value of the prize?

SOLUTION. If x represents the value of the prize, then

Solve this equation using the six-step method.

Step	Question	Yes/No	What to Do
1	Are there any fractions or decimals?	Yes	To get rid of them, multiply by 15. 15(120) = 15(x) 6x + 5x + 1800 = 15x
2	Are there any parentheses?	No
3	Are there any like terms to combine?	Yes	Combine them: 11x + 1800 = 15x.
4	Are there variables on both sides?	Yes	Subtract 11x from each side: 1800 = 4x.
5	Is there a plain number on the same side as the variable?	No
6	Does the variable have a coefficient?	Yes	Divide both sides by 4: x = 450.

Sometimes on the GRE, you are given an equation with several variables and asked to solve for one of them in terms of the others.

HELPFUL HINT

In applying the six-step method, you shouldn’t actually write out the table, as we did in Examples 1–4, since it would be too time consuming. Instead, use the method as a guideline and mentally go through each step, doing whichever ones are required.

When you have to solve for one variable in terms of the others, treat all of the others as if they were numbers, and apply the six-step method.

EXAMPLE 4

If a = 3b − c, what is the value of b in terms of a and c?

SOLUTION. To solve for b, treat a and c as numbers and use the six-step method with b as the variable.

Step	Question	Yes/No	What to Do
1	Are there any fractions or decimals?	No
2	Are there any parentheses?	No
3	Are there any like terms to combine?	No
4	Are there variables on both sides?	No	Remember: the only variable is b.
5	Is there a plain number on the same side as the variable?	Yes	Remember: we’re considering c as a number, and it is on the same side as b, the variable. Add c to both sides: a + c = 3b.
6	Does the variable have a coefficient?	Yes	Divide both sides by 3:

Sometimes when solving equations, you may see a shortcut. For example, to solve 7(w − 3) = 42, it saves time to start by dividing both sides by 7, getting w − 3 = 6, rather than by using the distributive law to eliminate the parentheses. Similarly, if you have to solve a proportion such as it is easier to cross-multiply, getting 5x = 21, than to multiply both sides by 35 to get rid of the fractions (although that’s exactly what cross-multiplying accomplishes). Other shortcuts will be illustrated in the problems at the end of the section. If you spot such a shortcut, use it; but if you don’t, be assured that the six-step method always works.

EXAMPLE 5

If x − 4 = 11, what is the value of x − 8?
(A) −15 (B) −7 (C) −1 (D) 7 (E) 15

SOLUTION. Going immediately to Step 5, add 4 to each side: x = 15. But this is not the answer. You need the value not of x, but of x − 8: 15 − 8 = 7 (D).

As in Example 5, on the GRE you are often asked to solve for something other than the simple variable. In Example 5, you could have been asked for the value of x² or x + 4 or (x − 4)², and so on.

As you read each question on the GRE, on your scrap paper write down whatever you are looking for, and circle it. This way you will always be sure that you are answering the question that is asked.

HELPFUL HINT

Very often, solving the equation is not the quickest way to answer the question. Consider Example 6.

EXAMPLE 6

If 2x − 5 = 98, what is the value of 2x + 5?

SOLUTION. The first thing you should do is write 2x + 5 on your paper and circle it. The fact that you are asked for the value of something other than x should alert you to look at the question carefully to see if there is a shortcut.

· The best approach here is to observe that 2x + 5 is 10 more than 2x − 5, so the answer is 108 (10 more than 98).

· Next best would be to do only one step of the six-step method, add 5 to both sides: 2x = 103. Now, add 5 to both sides: 2x + 5 = 103 + 5 = 108.

· The worst method would be to divide 2x = 103 by 2, get x = 51.5, and then use that to calculate 2x + 5.

EXAMPLE 7

If w is an integer, and the average (arithmetic mean) of 3, 4, and w is less than 10, what is the greatest possible value of w?
(A) 9 (B) 10 (C) 17 (D) 22 (E) 23

SOLUTION. Set up the inequality: < 10. Do Step 1 (get rid of fractions by multiplying by 3): 3 + 4 + w < 30. Do Step 3 (combine like terms): 7 + w < 30. Finally, do Step 5 (subtract 7 from each side): w < 23. Since w is an integer, the most it can be is 22.

The six-step method also works when there are variables in denominators.

EXAMPLE 8

For what value of x is
(A) 5 (B) 10 (C) 20 (D) 30 (E) 50

SOLUTION. Multiply each side by the LCD, 5x:

Now solve normally: 20 + 3x = 50 ⇒ 3x = 30 ⇒ x = 10 (B).

EXAMPLE 9

If x is positive, and y = 5x² + 3, which of the following is an expression for x in terms of y?

SOLUTION. The six-step method works when there are no exponents. However, we can treat x² as a single variable, and use the method as far as possible:

Now take the square root of each side; since x is positive, the only solution is (B).

CAUTION

Doing the same thing to each side of an equation does not mean doing the same thing to each term of the equation.
Study Examples 10 and 11 carefully.

EXAMPLE 10

If , what is a in terms of b and c?

NOTE: You cannot just take the reciprocal of each term; the answer is not a = b + c. Here are two solutions.

SOLUTION 1. First add the fractions on the right hand side:

Now, take the reciprocal of each side: .

SOLUTION 2. Use the six-step method. Multiply each term by abc, the LCD:

EXAMPLE 11

If a > 0 and a² + b² = c², what is a in terms of b and c?

SOLUTION. a² + b² = c² ⇒ a² = c² − b². Be careful: you cannot now take the square root of each term and write, a = c − b. Rather, you must take the square root of each side: = .

There are a few other types of equations that you could have to solve on the GRE. Fortunately, they are quite easy. You probably will not have to solve a quadratic equation. However, if you do, you will not need the quadratic formula, and you will not have to factor a trinomial. Here are two examples.

EXAMPLE 12

If x is a positive number and x² + 64 = 100, what is the value of x?
(A) 6 (B) 12 (C) 13 (D) 14 (E) 36

SOLUTION. When there is an x²-term, but no x-term, we just have to take the square root:

x² + 64 = 100 ⇒ x² = 36 ⇒ x = = 6 (A).

EXAMPLE 13

What is the largest value of x that satisfies the equation 2x² − 3x = 0?
(A) 0 (B) 1.5 (C) 2 (D) 2.5 (E) 3

SOLUTION. When an equation has an x²-term and an x-term but no constant term, the way to solve it is to factor out the x and to use the fact that if the product of two numbers is 0, one of them must be 0 (KEY FACT A3):

The largest value is 1.5 (B).

In another type of equation that occasionally appears on the GRE, the variable is in the exponent. These equations are particularly easy and are basically solved by inspection.

EXAMPLE 14

If 2^x^{+ 3} = 32, what is the value of 3^x^{+ 2}?
(A) 5 (B) 9 (C) 27 (D) 81 (E) 125

SOLUTION. How many 2s do you have to multiply together to get 32? If you don’t know that it’s 5, just multiply and keep track. Count the 2s on your fingers as you say to yourself, “2 times 2 is 4, times 2 is 8, times 2 is 16, times 2 is 32.” Then

2^x^{+ 3} = 32 = 2⁵ ⇒ x + 3 = 5 ⇒ x = 2.

Therefore, x + 2 = 4, and 3^x^{+ 2} = 3⁴ = 3 × 3 × 3 × 3 = 81 (D).

Occasionally, both sides of an equation have variables in the exponents. In that case, it is necessary to write both exponentials with the same base.

EXAMPLE 15

If 4^w^{+ 3} = 8^w^{− 1}, what is the value of w?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 9

SOLUTION. Since it is necessary to have the same base on each side of the equation, write 4 = 2² and 8 = 2³. Then

4^w^{+ 3} = (2²)^w^{+ 3} = 2^{2(w + 3)} = 2^{2w + 6} and 8^w^{− 1} = (2³)^w^{− 1} = 2^{3(w − 1)} = 2^{3w − 3}.

So, 2^{2w + 6} = 2^{3w − 3} ⇒ 2w + 6 = 3w − 3 ⇒ w = 9 (E).

Systems of Linear Equations

The equations x + y = 10 and x − y = 2 each have lots of solutions (infinitely many, in fact). Some of them are given in the tables below.

However, only one pair of numbers, x = 6 and y = 4, satisfy both equations simultaneously: 6 + 4 = 10 and 6 − 4 = 2. This then is the only solution of the system of equations: .

A system of equations is a set of two or more equations involving two or more variables. To solve such a system, you must find values for each of the variables that will make each equation true. In an algebra course you learn several ways to solve systems of equations. On the GRE, the most useful way to solve them is to add or subtract (usually add) the equations. After demonstrating this method, we will show in Example 19 one other way to handle some systems of equations.

To solve a system of equations, add or subtract them. If there are more than two equations, add them.

EXAMPLE 16

Column A		Column B
	x + y = 10
	x − y = 2
x		y

SOLUTION. Add the two equations:

Replacing x with 6 in x + y = 10 yields y = 4. So, Column A is greater.

HELPFUL HINT

On the GRE, most problems involving systems of equations do not require you to solve the system. They usually ask for something other than the values of each variable. Read the questions very carefully, circle what you need, and do no more than is required.

EXAMPLE 17

If 3a + 5b = 10 and 5a + 3b = 30, what is the average (arithmetic mean) of a and b?

· (A) 2.5

· (B) 4

· (C) 5

· (D) 20

· (E) It cannot be determined from the information given.

SOLUTION. Add the two equations:

Divide both sides by 8: a + b = 5

The average of a and b is:

NOTE: It is not only unnecessary to first solve for a and b (a = 7.5 and b = −2.5.), but, because that procedure is so much more time-consuming, it would be foolish to do so.

EXAMPLE 18

Column A		Column B
	7a − 3b = 200
	7a + 3b = 100
a		b

SOLUTION. Don’t actually solve the system. Add the equations:

14a = 300 ⇒ 7a = 150.

So, replacing 7a with 150 in the second equation, we get 150 + 3b = 100; so 3b, and hence b, must be negative, whereas a is positive. Therefore, a > b, and Column A is greater.

Occasionally on the GRE, it is as easy, or easier, to solve the system by substitution.

HELPFUL HINT

Remember TAC-TIC 5,Chapter 12. On quantitative comparison questions, you don’t need to know the value of the quantity in each column; you only need to know which one is greater.

If one of the equations in a system of equations consists of a single variable equal to some expression, substitute that expression for the variable in the other equation.

EXAMPLE 19

Column A		Column B
	x + y = 10
	y = x − 2
x		y

SOLUTION. Since the second equation states that a single variable (y), is equal to some expression (x − 2), substitute that expression for y in the first equation: x + y = 10 becomes x + (x − 2) = 10. Then, 2x − 2 = 10, 2x = 12, and x= 6. As always, to find the value of the other variable (y), plug the value of x into one of the two original equations: y = 6 − 2 = 4. Column A is greater.

Practice Exercises — Equations/Inequalities

Multiple-Choice Questions

1. If 4x + 12 = 36, what is the value of x + 3?
(A) 3 (B) 6 (C) 9 (D) 12 (E) 18

2. If 7x + 10 = 44, what is the value of 7x − 10?

3. If 4x + 13 = 7 − 2x, what is the value of x?

4. If x − 4 = 9, what is the value of x² − 4?
(A) 21 (B) 77 (C) 81 (D) 165 (E) 169

5. If ax − b = c − dx, what is the value of x in terms of a, b, c, and d?

6. If , what is the value of x?
(A) 3 (B) 18 (C) 27 (D) 54 (E) 72

7. If 3x − 4 = 11, what is the value of (3x − 4)²?
(A) 22 (B) 36 (C) 116 (D) 121 (E) 256

8. If 64¹² = 2^a^{− 3}, what is the value of a ?
(A) 9 (B) 15 (C) 69 (D) 72 (E) 75

9. If the average (arithmetic mean) of 3a and 4b is less than 50, and a is twice b, what is the largest possible integer value of a ?
(A) 9 (B) 10 (C) 11 (D) 19 (E) 20

10.If , then a =

11.If x = 3a + 7 and y = 9a², what is y in terms of x ?

12.If 4y − 3x = 5, what is the smallest integer value of x for which y > 100?
(A) 130
(B) 131
(C) 132
(D) 395
(E) 396

Quantitative Comparison Questions

	Column A	Column B
	a + b = 13
	a − b = 13
13.	b	13


14.	a	b

	4x² = 3x
15.	x	1

	a + b = 1
	b + c = 2
	c + a = 3
16.	The average (arithmetic mean) of a, b, and c	1

	3x − 4y = 5
	y = 2x
17.	x	y


18.	x	12

	3r − 5s = 17
	2r − 6s = 17
19.	The average (arithmetic mean) of r and s	10


	c and d are positive
20.	c	d

ANSWER KEY

1. C	5. A	9. D	13. B	17. A
2. D	6. D	10. C	14. A	18. A
3. C	7. D	11. A	15. B	19. B
4. D	8. E	12. C	16. C	20. B

Answer Explanations

1. (C) The easiest method is to recognize that x + 3 is of 4x + 12 and, therefore, equals of 36, which is 9. If you don’t see that, solve normally: 4x + 12 = 36 ⇒ 4x = 24 ⇒ x = 6 ⇒x + 3 = 9.

2. (D) Subtracting 20 from each side of 7x + 10 = 44 gives 7x − 10 = 24. If you don’t see that, subtract 10 from each side, getting 7x = 34. Then subtract 10 to get 7x − 10 = 24. The worst alternative is to divide both sides of 7x = 34 by 7 to get x = ; then you have to multiply by 7 to get back to 34, and then subtract 10.

3. (C) Add 2x to each side: 6x + 13 = 7. Subtract 13 from each side: 6x = −6. Divide by 6: x = −1.

4. (D) x − 4 = 9 ⇒x = 13 ⇒x² = 169 ⇒ x² − 4 = 165.

5. (A) Treat a, b, c, and d as constants, and use the six-step method to solve for x:
ax − b = c − dx ⇒ ax − b + dx = c ⇒
ax + dx = c + b ⇒ x(a + d) = b + c ⇒
.

6. (D) Multiply both sides by 18, the LCD:

6x + 3x + 2x = 594 ⇒ 11x = 594 ⇒x = 54.

It’s actually easier not to multiply out
18 × 33; leave it in that form, and then divide
by .

7. (D) Be alert. Since you are given the value of 3x − 4, and want the value of (3x − 4)², just square both sides: 11² = 121. If you don’t see that, you’ll waste time solving 3x − 4 = 11 (x = 5), only to use that value to calculate that 3x − 4 is equal to 11, which you aready knew.

8. (E) 2^a^{− 3} = 64¹² = (2⁶)¹² = 2⁷² ⇒
a − 3 = 72 ⇒a = 75.

9. (D) Since a = 2b, 2a = 4b. Therefore, the average of 3a and 4b is the average of 3a and 2a, which is 2.5a. Therefore, 2.5a < 50 ⇒
a < 20. So the largest integer value of a is 19.

10.(C) Taking the reciprocal of each side, we get
a − b = . So a = b + .

11.(A) x = 3a + 7 ⇒x − 7 = 3a⇒a = .
Therefore,
.

12.(C) Solving for y yields y = .
Then, since y > 100:
> 100 ⇒ 5 + 3x > 400 ⇒
3x > 395 ⇒ x > 131.666.
The smallest integer value of x is 132.

13.(B) Adding the two equations, we get that 2a = 26. Therefore, a = 13 and b = 0.

14.(A) Express each side of = 8 as a power of 2:
8 = 2³ and = 2^{(a−1) − (b+1)} = 2^a^−b−2.
Therefore, a − b − 2 = 3 ⇒a = b + 5, and so a is greater.

15. (B) 4x² = 3x⇒ 4x² − 3x = 0 ⇒
x(4x − 3) = 0 ⇒

There are two possible values of x, both of which are less than 1.

16.(C) When we add all three equations, we get
2a + 2b + 2c = 6 ⇒a + b + c = 3 ⇒
.

17. (A) Use substitution. Replace y in the first equation with 2x:

18.(A) Multiply both sides by 6, the LCD:

−12 > −x⇒x > 12.

19.(B) The first thing to try is to add the equations. That yields 5r − 11s = 24, which does not appear to be useful. So now try to subtract the equations. That yields r + s = 10. So the average of r and s is .

20.(B) Multiply both sides by cd, the LCD of the fractions:

14-H. WORD PROBLEMS

On a typical GRE you will see several word problems, covering almost every math topic for which you are responsible. In this chapter you have already seen word problems on consecutive integers in Section A; fractions and percents in Sections B and C; ratios and rates in Section D; and averages in Section E. Later in this chapter you will see word problems involving probability, circles, triangles, and other geometric figures. A few of these problems can be solved with just arithmetic, but most of them require basic algebra.

To solve word problems algebraically, you must treat algebra as a foreign language and learn to translate “word for word” from English into algebra, just as you would from English into French or Spanish or any other language. When translating into algebra, we use some letter (often x) to represent the unknown quantity we are trying to determine. It is this translation process that causes difficulty for some students. Once translated, solving is easy using the techniques we have already reviewed. Consider the following pairs of typical GRE questions. The first ones in each pair (1a and 2a) would be considered easy, whereas the second ones (1b and 2b) would be considered harder.

EXAMPLE 1A

What is 4% of 4% of 40,000?

EXAMPLE 1B

In a lottery, 4% of the tickets printed can be redeemed for prizes, and 4% of those tickets have values in excess of $100. If the state prints 40,000 tickets, how many of them can be redeemed for more than $100?

EXAMPLE 2A

If x + 7 = 2(x − 8), what is the value of x?

EXAMPLE 2B

In 7 years Erin will be twice as old as she was 8 years ago. How old is Erin now?

Once you translate the words into arithmetic expressions or algebraic equations, Examples 1a and 1b and 2a and 2b are identical. The problem that many students have is doing the translation. It really isn’t very difficult, and we’ll show you how. First, though, look over the following English to algebra “dictionary.”

English Words	Mathematical Meaning	Symbol
Is, was, will be, had, has, will have, is equal to, is the same as	Equals	=
Plus, more than, sum, increased by, added to, exceeds, received, got, older than, farther than, greater than	Addition	+
Minus, fewer, less than, difference, decreased by, subtracted from, younger than, gave, lost	Subtraction	−
Times, of, product, multiplied by	Multiplication	×
Divided by, quotient, per, for	Division	÷,
More than, greater than	Inequality	>
At least	Inequality	≥
Fewer than, less than	Inequality	<
At most	Inequality	≤
What, how many, etc.	Unknown quantity	x (or some some other variable)

Let’s use our dictionary to translate some phrases and sentences.

1. The sum of 5 and some number is 13.	5 + x = 13
2. John was 2 years younger than Sam.	J = S − 2
3. Bill has at most $100.	B ≤ 100
4. The product of 2 and a number exceeds that number by 5 (is 5 more than).	2N = N + 5

In translating statements, you first must decide what quantity the variable will represent. Often it’s obvious. Other times there is more than one possibility.

Let’s translate and solve the two questions from the beginning of this section, and then we’ll look at a few new ones.

EXAMPLE 1B

SOLUTION. Let x = the number of tickets worth more than $100. Then

x = 4% of 4% of 40,000 = .04 × .04 × 40,000 = 64,

which is also the solution to Example 1a.

EXAMPLE 2B

In 7 years Erin will be twice as old as she was 8 years ago. How old is Erin now?

HELPFUL HINT

In all word problems on the GRE, remember to write down and circle what you are looking for. Don’t answer the wrong question!

SOLUTION. Let x = Erin’s age now. Then 8 years ago she was x − 8, and 7 years from now she will be x + 7. So,

x + 7 = 2(x − 8) ⇒ x + 7 = 2x − 16 ⇒ 7 = x − 16 ⇒ x = 23,

which is also the solution to Example 2a.

Most algebraic word problems on the GRE are not too difficult, and if you can do the algebra, that’s usually the best way. But if, after studying this section, you still get stuck on a question during the test, don’t despair. Use the tactics that you learned in Chapter 11, especially TACTIC 1—backsolving.

HELPFUL HINT

In problems involving ages, remember that “years ago” means you need to subtract, and “years from now” means you need to add.

Age Problems

EXAMPLE 3

In 1980, Judy was 3 times as old as Adam, but in 1984 she was only twice as old as he was. How old was Adam in 1990?
(A) 4 (B) 8 (C) 12 (D) 14 (E) 16

SOLUTION. Let x be Adam’s age in 1980 and fill in the table below.

Year	Judy	Adam
1980	3x	x
1984	3x + 4	x + 4

Now translate: Judy’s age in 1984 was twice Adam’s age in 1984: 3x + 4 = 2(x + 4)

3x + 4 = 2x + 8 ⇒ x + 4 = 8 ⇒ x = 4

HELPFUL HINT

It is often very useful to organize the data from a word problem in a table.

So, Adam was 4 in 1980. However, 4 is not the answer to this question. Did you remember to circle what you’re looking for? The question could have asked for Adam’s age in 1980 (Choice A) or 1984 (Choice B) or Judy’s age in any year whatsoever (Choice C is 1980 and Choice E is 1984); but it didn’t. It asked for Adam’s age in 1990. Since he was 4 in 1980, then 10 years later, in 1990, he was 14 (D).

Distance Problems

Distance problems all depend on three variations of the same formula:

These are usually abbreviated, .

EXAMPLE 4

How much longer, in seconds, is required to drive 1 mile at 40 miles per hour than at 60 miles per hour?

SOLUTION. The time to drive 1 mile at 40 miles per hour is given by

The time to drive 1 mile at 60 miles per hour is given by t = hour = 1 minute.

The difference is minute = 30 seconds.

Note that this solution used the time formula given, but required only arithmetic, not algebra. Example 5 requires an algebraic solution.

EXAMPLE 5

Avi drove from his home to college at 60 miles per hour. Returning over the same route, there was a lot of traffic, and he was only able to drive at 40 miles per hour. If the return trip took 1 hour longer, how many miles did he drive each way?
(A) 2 (B) 3 (C) 5 (D) 120 (E) 240

SOLUTION. Let x = the number of hours Avi took going to college and make a table.

	rate	time	distance
Going	60	x	60x
Returning	40	x + 1	40(x + 1)

Since he drove the same distance going and returning,

60x = 40(x + 1) ⇒ 60x = 40x + 40 ⇒ 20x = 40 ⇒ x = 2.

Now be sure to answer the correct question. When x = 2, Choices A, B, and C are the time in hours that it took going, returning, and round-trip; Choices D and E are the distances each way and round-trip. You could have been asked for any of the five. If you circled what you’re looking for, you won’t make a careless mistake. Avi drove 120 miles each way, and so the correct answer is D.

The d in d = rt stands for “distance,” but it could really be any type of work that is performed at a certain rate, r, for a certain amount of time, t. Example 5 need not be about distance. Instead of driving 120 miles at 60 miles per hour for 2 hours, Avi could have read 120 pages at a rate of 60 pages per hour for 2 hours; or planted 120 flowers at the rate of 60 flowers per hour for 2 hours; or typed 120 words at a rate of 60 words per minute for 2 minutes.

Examples 6 and 7 illustrate two additional word problems of the type that you might find on the GRE.

EXAMPLE 6

Lindsay is trying to collect all the cards in a special commemorative set of baseball cards. She currently has exactly of the cards in that set. When she gets 10 more cards, she will then have of the cards. How many cards are in the set?
(A) 30 (B) 60 (C) 120 (D) 180 (E) 240

SOLUTION. Let x be the number cards in the set. First, translate this problem from English into algebra: . Now, use the six-step method of Section 14-G to solve the equation. Multiply by 12 to get, 3x + 120 = 4x, and then subtract 3x from each side: x = 120 (D).

EXAMPLE 7

Jen, Ken, and Len have a total of $390. Jen has 5 times as much as Len, and Ken has as much as Jen. How much money does Ken have?
(A) $40 (B) $78 (C) $150 (D) $195 (E) $200

HELPFUL HINT

You often have a choice as to what to let the variable represent. Don’t necessarily let it represent what you’re looking for; rather, choose what will make the problem easiest to solve.

Suppose, for example, that in this problem you let x represent the amount of money that Ken has. Then since Ken has as much as Jen, Jen has as much as Ken: x; and Jen would have of that: . It is much easier here to let x represent the amount of money Len has.

SOLUTION. Let x represent the amount of money Len has. Then 5x is the amount that Jen has, and (5x) is the amount that Ken has. Since the total amount of money is $390, x + 5x + x = 390.

Multiply by 4 to get rid of the fraction: 4x + 20x + 15x = 1560.

Combine like terms and then divide: 39x = 1560 ⇒ x = 40.

So Len has $40, Jen has 5 × 40 = $200, and Ken has (200) = $150 (C).

Practice Exercises — Word Problems

Multiple-Choice Questions

1. Howard has three times as much money as Ronald. If Howard gives Ronald $50, Ronald will then have three times as much money as Howard. How much money do the two of them have together?
(A) $75
(B) $100
(C) $125
(D) $150
(E) $200

2. In the afternoon, Beth read 100 pages at the rate of 60 pages per hour; in the evening, when she was tired, she read another 100 pages at the rate of 40 pages per hour. What was her average rate of reading for the day?
(A) 45 (B) 48 (C) 50 (D) 52 (E) 55

3. If the sum of five consecutive integers is S, what is the largest of those integers in terms of S?

4. As a fund-raiser, the school band was selling two types of candy: lollipops for 40 cents each and chocolate bars for 75 cents each. On Monday, they sold 150 candies and raised 74 dollars. How many lollipops did they sell?
(A) 75 (B) 90 (C) 96 (D) 110 (E) 120

5. A jar contains only red, white, and blue marbles. The number of red marbles is the number of white ones, and the number of white ones is the number of blue ones. If there are 470 marbles in all, how many of them are blue?
(A) 120
(B) 135
(C) 150
(D) 184
(E) 200

6. The number of shells in Judy’s collection is 80% of the number in Justin’s collection. If Justin has 80 more shells than Judy, how many shells do they have altogether?
(A) 180
(B) 320
(C) 400
(D) 720
(E) 800

7. What is the greater of two numbers whose product is 900, if the sum of the two numbers exceeds their difference by 30?
(A) 15 (B) 60 (C) 75 (D) 90 (E) 100

8. On a certain project the only grades awarded were 80 and 100. If 10 students completed the project and the average of their grades was 94, how many earned 100?
(A) 2 (B) 3 (C) 5 (D) 7 (E) 8

9. If x years ago Adam was 12, and x years from now he will be 2x years old, how old will he be 3x years from now?
(A) 18
(B) 24
(C) 30
(D) 54
(E) It cannot be determined from the information given.

10. Since 1950, when Barry was discharged from the army, he has gained 2 pounds every year. In 1980 he was 40% heavier than in 1950. What percent of his 1995 weight was his 1980 weight?
(A) 80 (B) 85 (C) 87.5 (D) 90 (E) 95

Quantitative Comparison Questions

	Column A	Column B
	Lindsay is twice as old as she was 10 years ago.
	Kimberly is half as old as she will be in 10 years.
11.	Lindsay’s age now	Kimberly’s age now

	Boris spent of his take-home pay on
	Saturday and of what was left on Sunday.
	The rest he put in his savings account.
12.	The amount of his take-home pay that he spent	The amount of his take-home pay that he saved

	In 8 years, Tiffany will be 3 times as old as she is now.
13.	The number of years until Tiffany will be 6 times as old as she is now	16

	Rachel put exactly 50 cents worth of postage on an envelope using only 4-cent stamps and 7-cent stamps.
14.	The number of 4-cent stamps she used	The number of 7-cent stamps she used

	Car A and Car B leave from the same spot at the same time. Car A travels due north at 40 mph. Car B travels due east at 30 mph.
15.	Distance from Car A to Car B 9 hours after they left	450 miles

ANSWER KEY

1. B	4. D	7. B	10. C	13. A
2. B	5. E	8. D	11. A	14. D
3. E	6. D	9. D	12. C	15. C

Answer Explanations

After the gift, Ronald will have 3 times as much money as Howard: x + 50 = 3(3x − 50) ⇒ x + 50 = 9x − 150 ⇒ 8x = 200 ⇒ x = 25. So Ronald has $25 and Howard has $75, for a total of $100.

2. (B) Beth’s average rate of reading is determined by dividing the total number of pages she read (200) by the total amount of time she spent reading. In the afternoon she read for hours, and in the evening for hours, for a total time of hours. So, her average rate was pages per hour.

3. (E) Let the 5 consecutive integers be n, n + 1, n + 2, n + 3, n + 4. Then, S = n + n + 1 + n + 2 + n + 3 + n + 4 = 5n + 10 ⇒ 5n = S − 10 ⇒ n = . Choice A, therefore, is the smallest of the integers; the largest is

4. (D) Let x represent the number of chocolate bars sold; then 150 − x is the number of lollipops sold. We must use the same units, so we could write 75 cents as .75 dollars or 74 dollars as 7400 cents. Let’s avoid the decimals: xchocolates sold for 75x cents and (150 − x) lollipops sold for 40(150 − x) cents. So, 7400 = 75x + 40(150 − x) = 75x + 6000 − 40x = 6000 + 35x ⇒ 1400 = 35x ⇒ x = 40 and 150 − 40 = 110.

5. (E) If b is the number of blue marbles, then there are white ones, and red ones. Therefore,

6. (D) If x is the number of shells in Justin’s collection, then Judy has .80x. Since Justin has 80 more shells than Judy, x = .80x + 80 ⇒ .20x = 80 ⇒x = 80 ÷ .20 = 400.
So Justin has 400 and Judy has 320: a total of 720.

7. (B) If x represents the greater and y the smaller of the two numbers, then (x + y) = 30 + (x − y) ⇒ y = 30 − y ⇒ 2y = 30 ⇒ y = 15; and since xy = 900, x = 900 ÷ 15 = 60.

8. (D) If x represents the number of students earning 100, then 10 − x is the number of students earning 80. So

9. (D) Since x years ago, Adam was 12, he is now years from now, he will be . But, we are told that at that time he will be 2x years old. So, 12 + x = 2x ⇒ x = 12. Thus, he is now 12 + 6 = 18, and 3x or 36 years from now he will be 18 + 36 = 54.

10. (C) Let x be Barry’s weight in 1950. By 1980, he had gained 60 pounds (2 pounds per year for 30 years) and was 40% heavier: 60 = .40x ⇒ x = 60 ÷ .4 = 150. So in 1980, he weighed 210. Fifteen years later, in 1995, he weighed 240: = 87.5%.

11. (A) You can do the simple algebra, but you should realize that Lindsay is as old now as Kimberly will be in 10 years. If x represents Lindsay’s age now, x = 2(x − 10) ⇒ x = 2x − 20 ⇒ x = 20. Similarly, Kimberly is now 10 and will be 20 in 10 years.

12. (C) Let x represent the amount of Boris’s take-home pay. On Saturday, he spent x and still had x; but on Sunday, he spent of that: = x. So he spent of his take-home pay each day. He spent and saved .

13. (A) If x represents Tiffany’s age now, then in 8 years she will be x + 8, and so

x + 8 = 3x ⇒ 8 = 2x ⇒ x = 4.

Tiffany will be 6 times as old 20 years from now, when she will be 24.

14. (D) If x and y represent the number of 4-cent stamps and 7-cent stamps that Rachel used, respectively, then 4x + 7y = 50. This equation has infinitely many solutions but only 2 in which x and y are both positive integers:
y = 2 and x = 9 or y = 6 and x = 2.

15. (C) Draw a diagram. In 9 hours Car A drove 360 miles north and Car B drove 270 miles east. These are the legs of a right triangle, whose hypotenuse is the distance between them. You can use the Pythagorean theorem if you don’t recognize that this is just a 3-4-5 right triangle: the legs are 90 × 3 and 90 × 4, and the hypotenuse is 90 × 5 = 450.

Geometry

Although about 30% of the math questions on the GRE have to do with geometry, there are only a relatively small number of facts you need to know — far less than you would learn in a geometry course — and, of course, there are no proofs. In the next six sections we will review all of the geometry that you need to know to do well on the GRE. We will present the material exactly as it appears on the GRE, using the same vocabulary and notation, which might be slightly different from the terminology you learned in your high school math classes. In particular, the word “congruent” and the symbol “≅” are not used — angles or line segments that have the same measure are considered “equal.” The numerous multiple-choice and quantitative comparison examples will show you exactly how these topics are treated on the GRE.

14-I. LINES AND ANGLES

An angle is formed by the intersection of two line segments, rays, or lines. The point of intersection is called the vertex. On the GRE, angles are always measured in degrees.

KEY FACT I1

Angles are classified according to their degree measures.

· An acute angle measures less than 90°.

· A right angle measures 90°.

· An obtuse angle measures more than 90° but less than 180°.

· A straight angle measures 180°.

x < 90
acute angle

x = 90
right angle

90 < x < 180
obtuse angle

x = 180
straight angle

NOTE: The small square in the second angle in the figure above is always used to mean that the angle is a right angle. On the GRE, if an angle has a square in it, it must measure exactly 90°, whether or not you think that the figure has been drawn to scale.

KEY FACT I2

If two or more angles form a straight angle, the sum of their measures is 180°.

a + b = 180

w + x + y + z = 180

EXAMPLE 1

In the figure below, R, S, and T are all on line l,. What is the average of a, b, c, d, and e?

(A) 18 (B) 36 (C) 45 (D) 90 (E) 180

SOLUTION. Since ∠RST is a straight angle, by KEY FACT I2, the sum of a, b, c, d, and e is 180, and so their average is = 36 (B).

In the figure below, since a + b + c + d = 180 and e + f + g = 180, a + b + c + d + e + f + g = 180 + 180 = 360.

It is also true that u + v + w + x + y + z = 360, even though none of the angles forms a straight angle.

KEY FACT I3

The sum of all the measures of all the angles around a point is 360°.

Note: This fact is particularly important when the point is the center of a circle, as we shall see in Section 14-L.

a + b + c + d = 360

When two lines intersect, four angles are formed. The two angles in each pair of opposite angles are called vertical angles.

KEY FACT I4

Vertical angles have equal measures.

EXAMPLE 2

In the figure at the right, what is the value of x?

(A) 6 (B) 8 (C) 10 (D) 20 (E) 40

SOLUTION. Since the measures of vertical angles are equal, 3x + 10 = 5(x – 2) ⇒ 3x + 10 = 5x – 10 ⇒ 3x + 20 = 5x ⇒ 20 = 2x ⇒ x = 10 (C).

KEY FACT I5

If one of the angles formed by the intersection of two lines (or line segments) is a right angle, then all four angles are right angles.

a = b = c = 90

Two lines that intersect to form right angles are called perpendicular.

In the figures below, line divides ∠ABC into two equal parts, and line k divides line segment DE into two equal parts. Line is said to bisect the angle, and line kbisects the line segment. Point M is called the midpoint of segment DE.

EXAMPLE 3

In the figure at the right, lines k, , and m intersect at O. If line m bisects ∠AOB, what is the value of x?

(A) 25 (B) 35 (C) 45 (D) 50 (E) 60

SOLUTION. m∠AOB + 130 = 180 ⇒ m∠AOB = 50; and since m bisects ∠AOB, x = 25 (A).

Two lines that never intersect are said to be parallel. Consequently, parallel lines form no angles. However, if a third line, called a transversal, intersects a pair of parallel lines, eight angles are formed, and the relationships among these angles are very important.

KEY FACT I6

If a pair of parallel lines is cut by a transversal that is perpendicular to the parallel lines, all eight angles are right angles.

KEY FACT I7

If a pair of parallel lines is cut by a transversal that is not perpendicular to the parallel lines,

· Four of the angles are acute and four are obtuse;

· The four acute angles are equal: a = c =e = g;

· The four obtuse angles are equal: b = d = f = h;

· The sum of any acute angle and any obtuse angle is 180°: for example, d + e = 180, c +f = 180, b + g = 180, ….

KEY FACT I8

If a pair of lines that are not parallel is cut by a transversal, none of the properties listed in KEY FACT I7 is true.

You must know KEY FACT I7 — virtually every GRE has at least one question based on it. However, you do not need to know the special terms you learned in high school for these pairs of angles; those terms are not used on the GRE.

EXAMPLE 4

In the figure below, AB is parallel to CD. What is the value of x?

(A) 37 (B) 45 (C) 53 (D) 63 (E) 143

SOLUTION. Let y be the measure of ∠BED. Then by KEY FACT I2:
37 + 90 + y = 180 ⇒ 127 + y = 180 ⇒ y = 53.

Since AB is parallel to CD, by KEY FACT I7, x = y⇒x = 53 (C).

EXAMPLE 5

In the figure below, lines and k are parallel. What is the value of a + b?

(A) 45 (B) 60 (C) 75 (D) 90 (E) 135

SOLUTION. It is impossible to determine the value of either a or b. We can, however, find the value of a + b. We draw a line through the vertex of the angle parallel to and k. Then, looking at the top two lines, we see that a = x, and looking at the bottom two lines, we see that b = y. So, a + b = x + y = 45 (A).

Alternative solution. Draw a different line and use a Key Fact from Section 14-J on triangles. Extend one of the line segments to form a triangle. Since and k are parallel, the measure of the third angle in the triangle equals a. Now, use the fact that the sum of the measures of the three angles in a triangle is 180° or, even easier, that the given 45° angle is an external angle of the triangle, and so is equal to the sum of a and b.

Practice Exercises — Lines and Angles

Multiple-Choice Questions

1. In the figure below, what is the average (arithmetic mean) of the measures of the five angles?

(A) 36 (B) 45 (C) 60 (D) 72 (E) 90

2. In the figure below, what is the value of ?

(A) 1 (B) 10 (C) 11 (D) 30 (E) 36

3. In the figure below, what is the value of b?

(A) 9 (B) 18 (C) 27 (D) 36 (E) 45

4. In the figure below, what is the value of x if y:x = 3:2?

(A) 18 (B) 27 (C) 36 (D) 45 (E) 54

5. What is the measure of the angle formed by the minute and hour hands of a clock at 1:50?

(A) 90°

(B) 95°

(D)115°

(E)120°

6. Concerning the figure below, if a = b, which of the following statements must be true?

I. c = d

II. and k are parallel

III. m and are perpendicular

(A) none

(B) I only

(D) I and III only

(E) I, II, and III

7. In the figure below, a:b = 3:5 and c:b = 2:1. What is the measure of the largest angle?

(A) 30 (B) 45 (C) 50 (D) 90 (E) 100

8. A, B, and C are points on a line with B between A and C. Let M and N be the midpoints of AB and BC, respectively. If AB:BC = 3:1, what is MN:BC?

(A) 1:2

(B) 2:3

(D) 3:2

(E) 2:1

9. In the figure below, lines k and are parallel. What is the value of y – x?

(A) 15 (B) 30 (C) 45 (D) 60 (E) 75

10. In the figure below, line m bisects ∠AOC and line bisects ∠AOB. What is the measure of ∠DOE?

(A) 75 (B) 90 (C) 100 (D) 105 (E) 120

Quantitative Comparison Questions

ANSWER KEY

1. D	4. C	7. E	10. B	13. D
2. C	5. D	8. E	11. D	14. C
3. D	6. B	9. C	12. B	15. C

Answer Explanations

1. (D) The markings in the five angles are irrelevant. The sum of the measures of the five angles is 360°, and 360 ÷ 5 = 72. If you calculated the measure of each angle you should have gotten 36, 54, 72, 90, and 108; but you would have wasted time.

2. (C) From the diagram, we see that 6a = 180, which implies that a = 30, and that 5b = 180, which implies that b = 36. So, = 11.

3. (D) Since vertical angles are equal, the two unmarked angles are 2b and 4a. Since the sum of all six angles is 360°,
360 = 4a +2b + 2a + 4a + 2b + b = 10a + 5b.
However, since vertical angles are equal,
b = 2a⇒ 5b = 10a. Hence,
360 = 10a + 5b = 10a + 10a = 20a ⇒
a = 18 ⇒b = 36.

4. (C) Since x + y + 90 = 180, x + y = 90. Also, since y:x = 3:2, y = 3t and x = 2t. Therefore,
3t + 2t = 90 ⇒ 5t = 90 ⇒
t = 18 ⇒x = 2(18) = 36.

5. (D) For problems such as this, always draw a diagram. The measure of each of the 12 central angles from one number to the next on the clock is 30°. At 1:50 the minute hand is pointing at 10, and the hour hand has gone of the way from 1 to 2. So from 10 to 1 on the clock is 90°, and from 1 to the hour hand is (30°) = 25°, for a total of 90° + 25° = 115°.

6. (B) No conclusions can be made about the lines; they could form any angles whatsoever. (II and III are both false.) I is true:

c = 180 – a = 180 – b = d.

7. (E) Since a:b = 3:5, then a = 3x and b = 5x. c:b = c:5x = 2:1 ⇒c = 10x. Then,

3x + 5x + 10x = 180 ⇒ 18x = 180 ⇒
x = 10 ⇒c = 10x = 100.

8. (E) If a diagram is not provided on a geometry question, draw one on your scrap paper. From the figure below, you can see that MN:BC = 2:1.

9. (C) Since the lines are parallel, the angle marked y and the sum of the angles marked x and 45 are equal: y = x + 45 ⇒y – x = 45.

10. (B) Let x = m∠AOC, and y = m∠AOB.

Then, x + y = m∠AOC + m∠AOB =

(180) = 90.

11. (D) No conclusion can be made: x could equal 50 or be more or less.

12. (B) Since m∠A + 32 + 75 = 180, m∠A = 73; and since AB is parallel to CD, a = 73, whereas, because vertical angles are equal, b = 75.

13. (D) Column A Column B
a + b + c + d 2a + 2b

Subtract a and b
from each column:
Since b = d,
subtract them:

c + d

c

a + b

a

There is no way to determine whether a is less than, greater than, or equal to c.

14. (C) Whether the lines are parallel or not,
a + b = c + d = e + f = g + h = 180.
Each column is equal to 360.

15. (C) Extend line segment AB to form a transversal. Since w + z = 180 and w + (x + y) = 180, it follows that z = x + y.

14-J. TRIANGLES

More geometry questions on the GRE pertain to triangles than to any other topic. To answer them, there are several important facts that you need to know about the angles and sides of triangles. The KEY FACTS in this section are extremely useful. Read them carefully, a few times if necessary, and make sure you learn them all.

KEY FACT J1

In any triangle, the sum of the measures of the three angles is 180°:

FIGURE 1

Figure 1 (a–e) illustrates KEY FACT J1 for five different triangles, which will be discussed below.

EXAMPLE 1

In the figure below, what is the value of x?

(A) 25 (B) 35 (C) 45 (D) 55 (E) 65

SOLUTION. Use KEY FACT J1 twice: first, for ΔCDE and then for ΔABC.

· m∠DCE + 120 + 35 = 180 ⇒ m∠DCE + 155 = 180 ⇒ m∠DCE = 25.

· Since vertical angles are equal, m∠ACB = 25 (see KEY FACT I6).

· x + 90 + 25 = 180 ⇒x + 115 = 180 ⇒x = 65 (E).

EXAMPLE 2

In the figure at the right, what is the value of a?

(A) 45 (B) 60 (C) 75 (D) 120 (E) 135

SOLUTION. First find the value of b: 180 = 45 + 75 + b = 120 + b ⇒ b = 60.

Then, a + b = 180 ⇒a = 180 – b = 180 – 60 = 120 (D).

In Example 2, ∠BCD, which is formed by one side of ΔABC and the extension of another side, is called an exterior angle. Note that to find a we did not have to first find b; we could have just added the other two angles: a = 75 + 45 = 120. This is a useful fact to remember.

KEY FACT J2

The measure of an exterior angle of a triangle is equal to the sum of the measures of the two opposite interior angles.

KEY FACT J3

In any triangle:

· the longest side is opposite the largest angle;

· the shortest side is opposite the smallest angle;

· sides with the same length are opposite angles with the same measure.

CAUTION

In KEY FACT J3 the condition “in any triangle” is crucial. If the angles are not in the same triangle, none of the conclusions hold. For example, in the figures below, AB and DE are not equal even though they are each opposite a 90° angle, and QS is not the longest side in the figure, even though it is opposite the largest angle in the figure.

Consider triangles ABC, JKL, and RST in Figure 1.

· In ΔABC: BC is the longest side since it is opposite angle A, the largest angle (71°). Similarly, AB is the shortest side since it is opposite angle C, the smallest angle (44°). So AB < AC < BC.

· In ΔJKL: angles J and L have the same measure (45°), so JK = KL.

· In ΔRST: since all three angles have the same measure (60°), all three sides have the same length: RS = ST = TR.

EXAMPLE 3

Which of the following statements concerning the length of side YZ is true?
(A) YZ < 9
(B) YZ = 9
(C) 9 < YZ < 10
(D) YZ = 10
(E) YZ > 10

SOLUTION.

· By KEY FACT J1, m∠X + 70 + 58 = 180 ⇒ m∠X = 52.

· So, X is the smallest angle.

· Therefore, by KEY FACT J3, YZ is the shortest side. SoYZ < 9 (A).

Classification of Triangles

Acute triangles are triangles such as ABC and RST, in which all three angles are acute. An acute triangle could be scalene, isosceles, or equilateral.

Obtuse triangles are triangles such as DEF, in which one angle is obtuse and two are acute. An obtuse triangle could be scalene or isosceles.

Right triangles are triangles such as GHI and JKL, which have one right angle and two acute ones. A right triangle could be scalene or isosceles. The side opposite the 90° angle is called the hypotenuse, and by KEY FACT J3, it is the longest side. The other two sides are called the legs.

If x and y are the measures of the acute angles of a right triangle, then by KEY FACT J1, 90 + x + y = 180 ⇒ x + y = 90.

KEY FACT J4

In any right triangle, the sum of the measures of the two acute angles is 90°.

EXAMPLE 4

SOLUTION. Since the diagram indicates that ΔABC is a right triangle, then, by KEY FACT J1, x + y = 90. So the average of x and y = = 45.

The columns are equal (C).

The most important facts concerning right triangles are the Pythagorean theorem and its converse, which are given in KEY FACT J5 and repeated as the first line of KEY FACT J6.

KEY FACT J5

Let a, b , and c be the sides of ΔABC, with a ≤ b ≤ c. If ΔABC is a right triangle, a² + b² = c²; and if a² + b² = c², then ΔABC is a right triangle.

KEY FACT J6

Let a, b, and c be the sides of ΔABC, with a ≤ b ≤ c.

· a² + b² = c² if and only if angle C is a right angle. (ΔABC is a right triangle.)

· a² + b² < c² if and only if angle C is obtuse. (ΔABC is an obtuse triangle.)

· a² + b² > c² if and only if angle C is acute.(ΔABC is an acute triangle.)

EXAMPLE 5

Which of the following are not the sides of a right triangle?

(A) 3, 4, 5 (B) 1, 1, (C) 1, , 2 (D) , , (E) 30, 40, 50

SOLUTION. Just check each choice.

	A: 3² + 4² = 9 + 16 = 25 = 5²	These are the sides of a right triangle.
	B: 1² + 1² = 1 + 1 = 2 = ()²	These are the sides of a right triangle.
	C: 1² + ()² = 1 + 3 = 4 = 2²	These are the sides of a right triangle.
	D: ()² + ()² = 3 + 4 = 7 ≠ ()²	These arenot the sides of a right triangle.
	E: 30² + 40² = 900 + 1600 = 2500 = 50²	These are the sides of a right triangle.

The answer is D.

Below are the right triangles that appear most often on the GRE. You should recognize them immediately whenever they come up in questions. Carefully study each one, and memorize KEY FACTS J7–J11.

On the GRE, the most common right triangles whose sides are integers are the 3-4-5 triangle (A) and its multiples (B).

KEY FACT J7

For any positive number x, there is a right triangle whose sides are 3x, 4x, 5x.

For example:

x = 1	3, 4, 5	x = 5	15, 20, 25
x = 2	6, 8, 10	x = 10	30, 40, 50
x = 3	9, 12, 15	x = 50	150, 200, 250
x = 4	12, 16, 20	x = 100	300, 400, 500

NOTE: KEY FACT J7 applies even if x is not an integer. For example:

x = .5

1.5, 2, 2.5

x = π

3π, 4π, 5π

The only other right triangle with integer sides that you should recognize immediately is the one whose sides are 5, 12, 13, (C).

Let x = length of each leg, and h = length of the hypotenuse, of an isosceles right triangle (D). By the Pythagorean theorem (KEY FACT J5), x² + x² = h².

So, 2x² = h², and h = .

KEY FACT J8

In a 45-45-90 right triangle, the sides are x, x, and x. So,

· by multiplying the length of a leg by , you get the hypotenuse.

· by dividing the hypotenuse by , you get the length of each leg.

KEY FACT J9

The diagonal of a square divides the square into two isosceles right triangles.

The last important right triangle is the one whose angles measure 30°, 60°, and 90°. (E)

KEY FACT J10

An altitude divides an equilateral triangle into two 30-60-90 right triangles.

Let 2x be the length of each side of equilateral ΔABC in which altitude AD is drawn. Then ΔABD is a 30-60-90 right triangle, and its sides are x, 2x, and h.

By the Pythagorean theorem,

x² + h² = (2x)² = 4x².

So h² = 3x², and h = .

KEY FACT J11

In a 30-60-90 right triangle the sides are x, x, and 2x.

If you know the length of the shorter leg (x),

· multiply it by to get the longer leg, and

· multiply it by 2 to get the hypotenuse.

If you know the length of the longer leg (a),

· divide it by to get the shorter leg, and

· multiply the shorter leg by 2 to get the hypotenuse.

If you know the length of the hypotenuse (h),

· divide it by 2 to get the shorter leg, and

· multiply the shorter leg by to get the longer leg.

EXAMPLE 6

What is the area of a square whose diagonal is 10?

(A) 20 (B) 40 (C) 50 (D) 100 (E) 200

SOLUTION. Draw a diagonal in a square of side s, creating a 45-45-90 right triangle. By KEY FACT J8:

s = and A = s² = = 50.

The answer is C.

EXAMPLE 7

In the diagram at the right, if BC = , what is the value of CD?

SOLUTION. Since ΔABC and ΔDAC are 30-60-90 and 45-45-90 right triangles, respectively, use KEY FACTS J11 and J8.

· Divide the longer leg, BC, by to get the shorter leg, AB: = .

· Multiply AB by 2 to get the hypotenuse: AC = 2.

· Since AC is also a leg of isosceles right ΔDAC, to get hypotenuse CD, multiply

AC by : CD = 2 × = 2 × 2 = 4 (E).

KEY FACT J12

Triangle Inequality

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

The best way to remember this is to see that x + y, the length of the path from A to C through B, is greater than z, the length of the direct path from A to C.

NOTE: If you subtract x from each side of x + y > z, you see that z – x < y.

KEY FACT J13

The difference of the lengths of any two sides of a triangle is less than the length of the third side.

EXAMPLE 8

If the lengths of two of the sides of a triangle are 6 and 7, which of the following could be the length of the third side?
I. 1
II. 5
III. 15
(A) None (B) I only (C) II only (D) I and II only (E) I, II, and III

SOLUTION. Use KEY FACTS J12 and J13.

· The third side must be less than 6 + 7 = 13. (III is false.)

· The third side must be greater than 7 – 6 = 1. (I is false.)

· Any number between 1 and 13 could be the length of the third side. (II is true.)

The answer is C.

The following diagram illustrates several triangles, two of whose sides have lengths of 6 and 7.

On the GRE, two other terms that appear regularly in triangle problems are perimeter and area (see Section 14-K).

EXAMPLE 9

In the figure at the right, what is the perimeter of ΔABC?
(A) 20 + 10
(B) 20 + 10
(C) 25
(D) 30
(E) 40

SOLUTION. First, use KEY FACTS J3 and J1 to find the measures of the angles.

· Since AB = AC, m∠B = m∠C. Represent each of them by x.

· Then, x + x + 60 = 180 ⇒ 2x = 120 ⇒x = 60.

· Since the measure of each angle of ΔABC is 60, the triangle is equilateral.

· So BC = 10, and the perimeter is 10 + 10 + 10 = 30 (D).

KEY FACT J14

The area of a triangle is given by A = bh, where b is the base and h is the height.

NOTE:

1. Any side of the triangle can be taken as the base.

2. The height or altitude is a line segment drawn to the base or, if necessary, to an extension of the base from the opposite vertex.

3. In a right triangle, either leg can be the base and the other the height.

4. The height may be outside the triangle. [See the figure below.]

In the figure below:

· If AC is the base, BD is the height.

· If AB is the base, CE is the height.

· If BC is the base, AF is the height.

EXAMPLE 10

What is the area of an equilateral triangle whose sides are 10?

(A) 30 (B) 25 (C) 50 (D) 50 (E) 100

SOLUTION. Draw an equilateral triangle and one of its altitudes.

· By KEY FACT J10, ΔABD is a 30-60-90 right triangle.

· By KEY FACT J11, BD = 5 and AD = 5.

· The area of ΔABC = (10)(5) = 25 (B).

Replacing 10 by s in Example 10 yields a very useful result.

KEY FACT J15

If A represents the area of an equilateral triangle with side s, then A = .

Practice Exercises — Triangles

Multiple-Choice Questions

1. In the triangle above, what is the value of x?

(A) 20 (B) 30 (C) 40 (D) 50 (E) 60

2. If the difference between the measures of the two smaller angles of a right triangle is 8°, what is the measure, in degrees, of the smallest angle?

(A) 37 (B) 41 (C) 42 (D) 49 (E) 53

3. What is the area of an equilateral triangle whose altitude is 6?

(A)18

(B) 12

(D) 36

(E) 24

4. Two sides of a right triangle are 12 and 13. Which of the following could be the length of the third side?

I. 5

II. 11

III.

(A) I only

(B) II only

(D) I and III

(E) I, II, and III

5. What is the value of PS in the triangle above?

(A)5

(B)10

(D)13

(E)12

6. If the measures of the angles of a triangle are in the ratio of 1:2:3, and if the length of the smallest side of the triangle is 10, what is the length of the longest side?

(A)10

(B)10

(D)20

(E)30

7. What is the value of x in the figure above?

(A) 80 (B) 100 (C) 115 (D) 120 (E) 130

8. In the figure above, what is the value of w?

(A) 100 (B) 110 (C) 120 (D) 130 (E) 140

Questions 9–10 refer to the following figure.

9. What is the area of ΔBED?

(A) 12 (B) 24 (C) 36 (D) 48 (E) 60

10. What is the perimeter of ΔBED?

(A) 19 + 5 (B)28 (C)17 + (D)32 (E)36

Questions 11–12 refer to the following figure.

11. What is the area of ΔDFH?

(A) 3 (B) 4.5 (C) 6 (D) 7.5 (E) 10

12. What is the perimeter of ΔDFH?

(A)8 + (B)8 + (C)16 (D)17 (E)18

13. Which of the following expresses a true relationship between x and y in the figure above?

(A) y = 60 – x

(B) y = x

(D) y = 180 – 3x

(E) x = 90 – 3y

Questions 14–15 refer to the following figure.

14. What is the perimeter of ΔABC?

(A) 48 (B) 48 + 12 (C) 48 + 12 (D)60 (E)60 + 6

15. What is the area of ΔABC?

(A) 108 (B) 54 + 72 (C) 54 + 72 (D)198 (E)216

Quantitative Comparison Questions

Column A

Column B

The lengths of two sides of a triangle are 7 and 11.

Column A

Column B

Questions 19–20 refer to the following figure.

90 < x

Questions 22–23 refer to the following figure in which the horizontal and vertical lines divide square ABCD into 16 smaller squares.

ANSWER KEY

1. D	6. D	11. B	16. A	21. B
2. B	7. C	12. B	17. C	22. A
3. B	8. B	13. A	18. D	21. A
4. D	9. B	14. C	19. A	24. C
5. D	10. D	15. C	20. B	25. B

Answer Explanations

1. (D) x + 2x + 30 = 180 ⇒ 3x + 30 = 180 ⇒ 3x = 150 ⇒x = 50.

2. (B) Draw a diagram and label it.

Then write the equations: x + y = 90 and x – y = 8.

Add the equations:

x + y = 90
+ x – y = 8
2x = 98

So x = 49 and y = 90 – 49 = 41.

3. (B) Draw altitude AD in equilateral ΔABC.

By KEY FACT J11, BD = = 2, and BD is one half the base. So, the area is 2 × 6 = 12.

4. (D) If the triangle were not required to be a right triangle, by KEY FACTS J11 and J12 any number greater than 1 and less than 25 could be the length of the third side. But for a right triangle, there are only two possibilities:

• If 13 is the hypotenuse, then the legs are 12 and 5. (I is true.) (If you didn’t recognize the 5-12-13 triangle, use the Pythagorean theorem: 12² + x² = 13², and solve.)

• If 12 and 13 are the two legs, then use the Pythagorean theorem to find the hypotenuse: 12² + 13² = c² ⇒ c² = 144 + 169 = 313 ⇒c = .
(III is true.)
An 11-12-13 triangle is not a right triangle. So II is false.

5. (D) Use the Pythagorean theorem twice, unless you recognize the common right triangles in this figure (which you should). Since PR = 20 and QR = 16, ΔPQR is a 3x-4x-5x right triangle with x = 4. So PQ = 12, and ΔPQS is a right triangle whose legs are 5 and 12. The hypotenuse, PS, therefore, is 13.

6. (D) If the measures of the angles are in the ratio of 1:2:3,
x + 2x + 3x = 180 ⇒ 6x = 180 ⇒x = 30. So the triangle is a 30-60-90 right triangle, and the sides are a, 2a, and a. Since a = 10, then 2a, the length of the longest side, is 20.

7. (C) Label the other angles in the triangle.

50 + a + b = 180 ⇒a + b = 130, and since the triangle is isosceles, a = b. Therefore, a and b are each 65, and x = 180 – 65 = 115.

8. (B) Here, 50 + 90 + a = 180 ⇒ a = 40, and since vertical angles are equal, b = 40. Then, 40 + 30 + w = 180 ⇒w = 110.

9. (B) You could calculate the area of the rectangle and subtract the area of the two white right triangles, but you shouldn’t. It is easier to solve this problem if you realize that the shaded area is a triangle whose base is 4 and whose height is 12. The area is (4)(12) = 24.

10. (D) Since both BD and ED are the hypotenuses of right triangles, their lengths can be calculated by the Pythagorean theorem, but these are triangles you should recognize: the sides of ΔDCE are 5-12-13, and those of ΔBAD are 9-12-15 (3x-4x-5x, with x = 3). So the perimeter of ΔBED is 4 + 13 + 15 = 32.

11. (B) Since ΔDGH is a right triangle whose hypotenuse is 5 and one of whose legs is 3, the other leg, GH, is 4. Since GF = DE is 7, HF is 3. Now, ΔDFH has a base of 3 (HF) and a height of 3 (DG), and its area is (3)(3) = 4.5.

12. (B) In ΔDFH, we already have that DH = 5 and HF = 3; we need only find DF, which is the hypotenuse of ΔDEF. By the Pythagorean theorem, (DF)² = 3² + 7² = 9 + 49 = 58 ⇒DF = .

So the perimeter is 3 + 5 + = 8 + .

13. (A) x + 2x + 3y = 180 ⇒ 3x + 3y = 180 ⇒ x + y = 60 ⇒y = 60 – x.

14. (C) ΔABD is a right triangle whose hypotenuse is 15 and one of whose legs is 9, so this is a 3x-4x-5x triangle with x = 3; so AD = 12. Now ΔADC is a 30-60-90 triangle, whose shorter leg is 12. Hypotenuse AC is 24, and leg CDis 12. So the perimeter is 24 + 15 + 9 + 12 = 48 + 12.

15. (C) From the solution to 14, we have the base (9 + 12) and the height (12) of ΔABC. Then, the area is (12)(9 + 12) = 54 + 72.

16. (A) Any side of a triangle must be greater than the difference of the other two sides (KEY FACT J13), so the third side is greater than 11 – 7 = 4.

17. (C) Draw a diagram. A diagonal of a square is the hypotenuse of each of the two 45-45-90 right triangles formed. The ratio of the length of the hypotenuse to the length of the leg in such a triangle is :1, so the columns are equal.

18. (D) BC can be any positive number less than 20 (by KEY FACTS J12 and J13, BC > 10 – 10 = 0 and BC < 10 + 10 = 20). So the perimeter can be any number greater than 20 and less than 40.

19. (A) Since OA and OB are radii, they are each equal to 5. With no restrictions on x, AB could be any positive number less than 10, and the bigger x is, the bigger AB is. If x were 90, AB would be 5, but we are told that x > 90, so AB > 5 > 7.

20. (B) Since AB must be less than 10, the perimeter is less than 20.

21. (B) Don’t calculate either area. The length of a side of an equilateral triangle is greater than the length of an altitude. So the triangle in Column B is larger (its sides are greater than 10).

22. (A) Column A: The perimeter of the shaded region consists of 12 line segments, each of which is the hypotenuse of a 45-45-90 right triangle whose legs are 1. So each line segment is , and the perimeter is 12. Column B: The perimeter of the square is 16. To compare 12 and 16, square them: (12)² = 144 × 2 = 288; 16² = 256.

23. (A) The white region consists of 12 right triangles, each of which has an area of , for a total area of 6. Since the area of the large square is 16, the area of the shaded region is 16 – 6 = 10.

24. (C) Since a = 180 – 145 = 35 and b = 180 – 125 = 55, a + b = 35 + 55 = 90. Therefore, 180 = a + b + c = 90 + c⇒c = 90.

25. (B) Since 65 + 45 = 110, m∠P = 70. Since ∠P is the largest angle, QR, the side opposite it, is the largest side.

14-K. QUADRILATERALS AND OTHER POLYGONS

A polygon is a closed geometric figure made up of line segments. The line segments are called sides and the endpoints of the line segments are called vertices (each one is called a vertex). Line segments inside the polygon drawn from one vertex to another are called diagonals. The simplest polygons, which have three sides, are the triangles, which you just studied in Section J. A polygon with four sides is called a quadrilateral. The only other terms you should be familiar with are pentagon, hexagon, octagon, and decagon, which are the names for polygons with five, six, eight, and ten sides, respectively.

In this section we will present a few facts about polygons and quadrilaterals in general, but the emphasis will be on reviewing the key facts you need to know about four special quadrilaterals.

Every quadrilateral has two diagonals. If you draw in either one, you will divide the quadrilateral into two triangles. Since the sum of the measures of the three angles in each of the triangles is 180°, the sum of the measures of the angles in the quadrilateral is 360°.

KEY FACT K1

In any quadrilateral, the sum of the measures of the four angles is 360°.

In exactly the same way, any polygon can be divided into triangles by drawing in all of the diagonals emanating from one vertex.

Notice that the pentagon is divided into three triangles, and the hexagon is divided into four triangles. In general, an n-sided polygon is divided into (n − 2) triangles, which leads to KEY FACT K2.

KEY FACT K2

The sum of the measures of the n angles in a polygon with n sides is (n – 2) × 180°.

EXAMPLE 1

In the figure at the right, what is the value of x?
(A) 60 (B) 90 (C) 100 (D) 120 (E) 150

SOLUTION. Since ΔDEF is equilateral, all of its angles measure 60°; also, since the two angles at vertex D are vertical angles, their measures are equal. Therefore, the measure of ∠D in quadrilateral ABCD is 60°. Finally, since the sum of the measures of all four angles of ABCD is 360°,

60 + 90 + 90 + x = 360 ⇒ 240 + x = 360 ⇒ x = 120 (D).

In the polygons in the figure that follows, one exterior angle has been drawn at each vertex. Surprisingly, if you add the measures of all of the exterior angles in any of the polygons, the sums are equal.

100 + 120 + 140 = 360

65 + 110 + 130 + 350 = 360

60 + 60 + 60 + 60 = 360

KEY FACT K3

In any polygon, the sum of the exterior angles, taking one at each vertex, is 360°.

A regular polygon is a polygon in which all of the sides are the same length and each angle has the same measure. KEY FACT K4 follows immediately from this definition and from KEY FACTS K2 and K3.

KEY FACT K4

In any regular polygon the measure of each interior angle is and the measure of each exterior angle is .

EXAMPLE 2

What is the measure of each interior angle in a regular decagon?
(A) 36 (B) 72 (C) 108 (D) 144 (E) 180

SOLUTION 1. The measure of each of the 10 interior angles is

SOLUTION 2. The measure of each of the 10 exterior angles is 36 (360 ÷ 10). Therefore, the measure of each interior angle is 180 – 36 = 144.

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

KEY FACT K5

Parallelograms have the following properties:

· Opposite sides are equal: AB = CD and AD = BC.

· Opposite angles are equal: a = c and b = d.

· Consecutive angles add up to 180°: a + b = 180, b + c = 180, c + d = 180, and a + d = 180.

· The two diagonals bisect each other: AE = EC and BE = ED.

· A diagonal divides the parallelogram into two triangles that have the exact same size and shape. (The triangles are congruent.)

EXAMPLE 3

SOLUTION. In ΔABD the larger angle is opposite the larger side (KEY FACT J2); so x > m∠ABD. However, since AB and CD are parallel sides cut by transversal BD, y = m∠ABD. Therefore, x > y. Column A is greater.

A rectangle is a parallelogram in which all four angles are right angles. Two adjacent sides of a rectangle are usually called the length () and the width (w). Note in the right-hand figure that the length is not necessarily greater than the width.

KEY FACT K6

Since a rectangle is a parallelogram, all of the properties listed in KEY FACT K5 hold for rectangles. In addition:

· The measure of each angle in a rectangle is 90°.

· The diagonals of a rectangle have the same length: AC = BD.

A square is a rectangle in which all four sides have the same length.

KEY FACT K7

Since a square is a rectangle, all of the properties listed in KEY FACTS K5 and K6 hold for squares. In addition:

· All four sides have the same length.

· Each diagonal divides the square into two 45-45-90 right triangles.

· The diagonals are perpendicular to each other: AC ⊥ BD.

EXAMPLE 4

What is the length of each side of a square if its diagonals are 10?
(A) 5 (B) 7 (C) 5 (D) 10 (E) 10

SOLUTION. Draw a diagram. In square ABCD, diagonal AC is the hypotenuse of a 45-45-90 right triangle, and side AB is a leg of that triangle. By KEY FACT J7,

A trapezoid is a quadrilateral in which one pair of sides is parallel and the other pair of sides is not parallel. The parallel sides are called the bases of the trapezoid. The two bases are never equal. In general, the two nonparallel sides are not equal; if they are the trapezoid is called an isosceles trapezoid.

Trapezoid

Isoceles trapezoid

The perimeter (P) of any polygon is the sum of the lengths of all of its sides.

P = 3 + 4 + 5 = 12

P = 13 + 17 + 20 + 38 = 88

KEY FACT K8

In a rectangle, P = 2( + w); in a square, P = 4s.

P = + w + + w = 2( + w)

P = s + s + s + s = 4s

EXAMPLE 5

The length of a rectangle is 7 more than its width. If the perimeter of the rectangle is the same as the perimeter of a square of side 8.5, what is the length of a diagonal of the rectangle?
(A) 12 (B) 13 (C) 17 (D) 34 (E) 169

SOLUTION. Don’t do anything until you have drawn a diagram.

Since the perimeter of the square = 4 × 8.5 = 34, the perimeter of the rectangle is also 34: 2( + w) = 34 ⇒ + w = 17. Replacing by w + 7, we get:

w + 7 + w = 17 ⇒ 2w + 7= 17 ⇒ 2w = 10 ⇒w = 5

Then = 5 + 7 = 12. Finally, realize that the diagonal is the hypotenuse of a 5-12-13 triangle, or use the Pythagorean theorem:

d² = 5² + 12² = 25 + 144 = 169 ⇒d = 13 (B).

In Section 14-J we reviewed the formula for the area of a triangle. The only other figures for which you need to know area formulas are the parallelogram, rectangle, square, and trapezoid.

KEY FACT K9

Here are the area formulas you need to know:

· For a parallelogram: A = bh.

· For a rectangle: A = w.

· For a square: A = s² or A = d².

· For a trapezoid: A = (b₁ + b₂)h.

EXAMPLE 6

In the figure below, the area of parallelogram ABCD is 40. What is the area of rectangle AFCE?

(A) 20 (B) 24 (C) 28 (D) 32 (E) 36

SOLUTION. Since the base, CD, is 10 and the area is 40, the height, AE, must be 4. Then ΔAED must be a 3-4-5 right triangle with DE = 3, which implies that EC = 7. So the area of the rectangle is 7 × 4 = 28 (C).

Two rectangles with the same perimeter can have different areas, and two rectangles with the same area can have different perimeters. These facts are a common source of questions on the GRE.

Rectangles Whose Perimeters Are 100

Rectangles Whose Areas Are 100

KEY FACT K10

For a given perimeter, the rectangle with the largest area is a square. For a given area, the rectangle with the smallest perimeter is a square.

EXAMPLE 7

EXAMPLE 8

SOLUTION 7. Draw any rectangles whose perimeters are 12 and 14 and compute their areas. As drawn below, Column A = 8 and Column B = 12.

This time Column B is greater. Is it always? Draw a different rectangle whose perimeter is 14.

The one drawn here has an area of 6. Now Column B isn’t greater. The answer is D.

SOLUTION 8. There are many rectangles of different areas whose perimeters are 12. But the largest area is 9, when the rectangle is a 3 × 3 square. Column B is greater.

Practice Exercises — Quadrilaterals

Multiple-Choice Questions

1. If the length of a rectangle is 4 times its width, and if its area is 144, what is its perimeter?
(A) 6 (B) 24 (C) 30 (D) 60 (E) 96

Questions 2–3 refer to the diagram below in which the diagonals of square ABCD intersect at E.

2. What is the area of ΔDEC?
(A) (B) 1 (C) (D) 2 (E) 2

3. What is the perimeter of ΔDEC?
(A) 1 +
(B) 2 +
(C) 4
(D) 2 + 2
(E) 6

4. If the angles of a five-sided polygon are in the ratio of 2:3:3:5:5, what is the measure of the smallest angle?
(A) 20 (B) 40 (C) 60 (D) 80 (E) 90

5. If in the figures below, the area of rectangle ABCD is 100, what is the area of rectangle EFGH?

(A)98
(B)100
(C)102
(D)104
(E)106

Questions 6–7 refer to a rectangle in which the length of each diagonal is 12, and one of the angles formed by the diagonal and a side measures 30°.

6. What is the area of the rectangle?
(A) 18
(B) 72
(C) 18
(D) 36
(E) 36

7. What is the perimeter of the rectangle?
(A) 18
(B) 24
(C) 12 + 12
(D) 18 + 6
(E) 24

8. How many sides does a polygon have if the measure of each interior angle is 8 times the measure of each exterior angle?
(A) 8 (B) 9 (C) 10 (D) 12 (E) 18

9. The length of a rectangle is 5 more than the side of a square, and the width of the rectangle is 5 less than the side of the square. If the area of the square is 45, what is the area of the rectangle?
(A) 20 (B) 25 (C) 45 (D) 50 (E) 70

Questions 10–11 refer to the following figure, in which M, N, O, and P are the midpoints of the sides of rectangle ABCD.

10. What is the perimeter of quadrilateral MNOP?
(A) 24 (B) 32 (C) 40 (D) 48 (E) 60

11. What is the area of quadrilateral MNOP?
(A) 48 (B) 60 (C) 72 (D) 96 (E) 108

12. In the figure at the right, what is the sum of the measures of all of the marked angles?
(A) 360
(B) 540
(C) 720
(D) 900
(E) 1080

13. In quadrilateral WXYZ, the measure of angle Z is 10 more than twice the average of the measures of the other three angles. What is the measure of angle Z?
(A) 100
(B) 105
(C) 120
(D) 135
(E) 150

Questions 14–15 refer to the following figure, in which M and N are the midpoints of two of the sides of square ABCD.

14. What is the perimeter of the shaded region?
(A) 3
(B) 2 + 3
(C) 3 + 2
(D) 5
(E) 8

15. What is the area of the shaded region?
(A) 1.5
(B) 1.75
(C) 3
(D) 2
(E) 3

Quantitative Comparison Questions

ANSWER KEY

1. D	5. C	9. A	13. E	17. B
2. B	6. D	10. C	14. B	18. A
3. D	7. C	11. D	15. A	19. A
4. C	8. E	12. C	16. C	20. C

Answer Explanations

1. (D) Draw a diagram and label it.

Since the area is 144, then
144 = (4x)(x) = 4x² ⇒ x² = 36 ⇒ x = 6.
So the width is 6, the length is 24, and the perimeter is 60.

2. (B) The area of the square is 2² = 4, and each triangle is one-fourth of the square. So the area of ΔDEC is 1.

3. (D) ΔDEC is a 45-45-90 right triangle whose hypotenuse, DC, is 2. Therefore, each of the legs is . So the perimeter is 2 + 2.

4. (C) The sum of the angles of a five-sided polygon is (5 – 2) × 180 = 3 × 180 = 540. Therefore,
         540 = 2x + 3x + 3x + 5x + 5x = 18x ×
                 x = 540 ÷ 18 = 30.
The measure of the smallest angle is
                 2x = 2 × 30 = 60.

5. (C) The area of rectangle ABCD is
(x + 1)(x + 4) = x² + 5x + 4.
The area of rectangle EFGH is
(x + 2)( x + 3) = x² + 5x + 6,
which is exactly 2 more than the area of rectangle ABCD: 100 + 2 = 102.

6. (D) Draw a picture and label it.

Since ΔBCD is a 30-60-90 right triangle, BC is 6 (half the hypotenuse) and CD is 6. So the area is w = 6(6) = 36.

7. (C) The perimeter of the rectangle is
2( + w) = 2(6 + 6) = 12 + 12.

8. (E) The sum of the degree measures of an interior and exterior angle is 180, so
180 = 8x + x = 9x ⇒ x = 20.
Since the sum of the measures of all the exterior angles is 360, there are 360 ÷ 20 = 18 angles and 18 sides.

9. (A) Let x represent the side of the square. Then the dimensions of the rectangle are (x + 5) and (x – 5), and its area is (x + 5)(x – 5) = x² – 25. Since the area of the square is 45, x² = 45 ⇒ x² – 25 = 45 – 25 = 20.

10. (C) Each triangle surrounding quadrilateral MNOP is a 6-8-10 right triangle. So each side of MNOP is 10, and its perimeter is 40.

11. (D) The area of each of the triangles is (6)(8) = 24, so together the four triangles have an area of 96. The area of the rectangle is 16 × 12 = 192. Therefore, the area of quadrilateral MNOP is 192 – 96 = 96.
Note: Joining the midpoints of the four sides of any quadrilateral creates a parallelogram whose area is one-half the area of the original quadrilateral.

12. (C) Each of the 10 marked angles is an exterior angle of the pentagon. If we take one angle at each vertex, the sum of those five angles is 360; the sum of the other five is also 360: 360 + 360 = 720.

13. (E) Let W, X, Y, and Z represent the measures of the four angles. Since W + X + Y + Z = 360, W + X + Y = 360 – Z. Also,
.

14. (B) Since M and N are midpoints of sides of length 2, AM, MB, AN, and ND are all 1.
MN = , since it’s the hypotenuse of an isosceles right triangle whose legs are 1; and BD = 2, since it’s the hypotenuse of an isosceles right triangle whose legs are 2. So the perimeter of the shaded region is
1 + + 1 + 2 = 2 + 3.

15. (A) The area of ΔABD = (2)(2) = 2, and the area of ΔAMN is (1)(1) = 0.5. So the area of the shaded region is 2 – 0.5 = 1.5.

16. (C) The area of ΔAED is .
The area of ΔEDC is .
Note: Each of the four small triangles has the same area.

17. (B) By KEY FACT J5, since ∠ Z is acute and ∠ Y is obtuse, (WY)² < a² + b², whereas (XZ)² > a² + b².

18. (A) Since an octagon has eight sides, Column B is 8x.
Column A: By KEY FACT J10, the hypotenuse of the triangle is 4x, and the longer leg is 2x. So the perimeter is 2x + 4x + 2x. Since > 1, then 2x + 4x + 2x > 2x + 4x + 2x = 8x.

19. (A) The perimeter of a rectangle of area 50 can be as large as we like, but the least it can be is when the rectangle is a square. In that case, each side is , which is greater than 7, and so the perimeter is greater than 28.

20. (C) TR is a transversal cutting the parallel sides PQ and RS. So b = x and 2b = 2x. But since the opposite angles of a parallelogram are equal, a = 2x. So a = 2b.

14-L. CIRCLES

A circle consists of all the points that are the same distance from one fixed point called the center. That distance is called the radius of the circle. The figure below is a circle of radius 1 unit whose center is at the point O. A, B, C, D, and E, which are each 1 unit from O, are all points on circle O. The word radius is also used to represent any of the line segments joining the center and a point on the circle. The plural of radius is radii. In circle O, below, OA, OB, OC, OD, and OE are all radii. If a circle has radius r, each of the radii is r units long.

KEY FACT L1

Any triangle, such as ΔCOD in the figure on page 504, formed by connecting the endpoints of two radii, is isosceles.

EXAMPLE 1

If P and Q are points on circle O, what is the value of x?
(A) 35 (B) 45 (C) 55 (D) 65 (E) 70

SOLUTION. Since ΔPOQ is isosceles, angles P and Q have the same measure. Then, 70 + x + x = 180 ⇒ 2x = 110 ⇒ x = 55 (C).

A line segment, such as CD in circle O at the beginning of this section, both of whose endpoints are on a circle is called a chord. A chord such as BE, which passes through the center of the circle, is called a diameter. Since BE is the sum of two radii, OB and OE, it is twice as long as a radius.

KEY FACT L2

If d is the diameter and r the radius of a circle, d = 2r.

KEY FACT L3

A diameter is the longest chord that can be drawn in a circle.

EXAMPLE 2

SOLUTION. Since the radius of the circle is 0.1, the diameter is 0.2. Therefore, the length of each of the five line segments is less than 0.2, and the sum of their lengths is less than 5 × 0.2 = 1. The answer is B.

The total length around a circle, from A to B to C to D to E and back to A, is called the circumference of the circle. In every circle the ratio of the circumference to the diameter is exactly the same and is denoted by the symbol π (the Greek letter “pi”).

KEY FACT L4

KEY FACT L5

The value of π is approximately 3.14.

On GRE questions that involve circles, you are expected to leave your answer in terms of π. So never multiply by 3.14. If you are ever stuck on a problem whose answers involve π, try to estimate the answer, and then use 3 as an approximation of π to test the answers. For example, assume that you think that an answer is about 50, and the answer choices are 4π, 6π, 12π, 16π, and 24π. Since π is slightly greater than 3, these choices are a little greater than 12, 18, 36, 48, and 72. The answer must be 16π. (To the nearest hundredth, 16π is actually 50.27, but approximating it by 48 was close enough.)

EXAMPLE 3

SOLUTION. Column A: C = πd = π(12). Column B: P = 4s = 4(12).

Since 4 > π, Column B is greater. (Note: 12π = 12(3.14) = 37.68, but you should not have wasted any time calculating this.)

An arc consists of two points on a circle and all the points between them. On the GRE, arc AB always refers to the small arc joining A and B. If we wanted to refer to the large arc going from A to B through P and Q, we would refer to it as arc APB or arc AQB. If two points, such as P and Q in circle O, are the endpoints of a diameter, they divide the circle into two arcs called semicircles.

An angle whose vertex is at the center of a circle is called a central angle.

KEY FACT L6

The degree measure of a complete circle is 360°.

KEY FACT L7

The degree measure of an arc equals the degree measure of the central angle that intercepts it.

CAUTION

Degree measure is not a measure of length. In the circles above, arc AB and arc CD each measure 72°, even though arc CD is much longer.

How long is arc CD? Since the radius of Circle P is 10, its diameter is 20, and its circumference is 20π. Since there are 360° in a circle, arc CD is , of the circumference: (20π) = 4π.

KEY FACT L8

The formula for the area of a circle of radius r is A = πr².

The area of Circle P, in KEY FACT L7, is π(10)² = 100π square units. The area of sector CPD is of the area of the circle: (100π) = 20π.

KEY FACT L9

If an arc measures x°, the length of the arc is , and the area of the sector formed by the arc and 2 radii is .

Examples 4 and 5 refer to the circle below.

EXAMPLE 4

What is the area of the shaded region?

· (A) 144π − 144

· (B) 144π − 36

· (C) 144 − 72

· (D) 24π − 36

· (E) 24π − 72

EXAMPLE 5

What is the perimeter of the shaded region?

· (A) 12 + 4π

· (B) 12 + 12π

· (C) 12 + 24π

· (D) 12 + 4π

· (E) 12 + 24π

SOLUTION 4. The area of the shaded region is equal to the area of sector COD minus the area of ΔCOD. The area of the circle is π(12)² = 144π.

Since , the area of sector COD is . Since m∠O = 60°, m∠C + ∠D = 120°; but ΔCOD is isosceles, so m∠C = m∠D. Therefore, they each measure 60°, and the triangle is equilateral. By KEY FACT J15, area of ΔCOD = , so the area of the shaded region is 24π − 36 (D).

SOLUTION 5. Since ΔCOD is equilateral, CD = 12. Since the circumference of the circle = 2π(12) = 24π, arc . So the perimeter is 12 + 4π (A).

Suppose that in Example 5 you see that CD = 12, but you don’t remember how to find the length of arc CD. From the diagram, it is clear that it is slightly longer than CD, say 13. So you know that the perimeter is about 25. Now, approximate the value of each of the choices and see which one is closest to 25. Only Choice A is even close.

A line and a circle or two circles are tangent if they have only one point of intersection. A circle is inscribed in a triangle or square if it is tangent to each side. A polygon is inscribed in a circle if each vertex is on the circle.

KEY FACT L10

If a line is tangent to a circle, a radius (or diameter) drawn to the point where the tangent touches the circle is perpendicular to the tangent line.

EXAMPLE 6

A is the center of a circle whose radius is 8, and B is the center of a circle whose diameter is 8. If these two circles are tangent to one another, what is the area of the circle whose diameter is AB?

(A) 12π (B) 36π (C) 64π (D) 144π (E) 256π

SOLUTION. Draw a diagram. Since the diameter, AB, of the dotted circle is 12, its radius is 6 and the area is π(6)² = 36π (B).

Practice Exercises — Circles

Multiple-Choice Questions

1. What is the circumference of a circle whose area is 100π?
(A) 10 (B) 20 (C) 10π (D) 20π (E) 25π

2. What is the area of a circle whose circumference is π?
(A) (B) (C) π (D) 2π (E) 4π

3. What is the area of a circle that is inscribed in a square of area 2?
(A) (B) (C) π (D) (E) 2π

4. A square of area 2 is inscribed in a circle. What is the area of the circle?
(A) (B) (C) π (D) (E) 2π

5. A 5 × 12 rectangle is inscribed in a circle. What is the radius of the circle?
(A) 6.5 (B) 7 (C) 8.5 (D) 13 (E) 17

6. If, in the figure below, the area of the shaded sector is 85% of the area of the entire circle, what is the value of w?

(A) 15 (B) 30 (C) 45 (D) 54 (E) 60

7. The circumference of a circle is aπ units, and the area of the circle is bπ square units. If a = b, what is the radius of the circle?
(A) 1 (B) 2 (C) 3 (D) π (E) 2π

Questions 8−9 refer to the following figure.

8. What is the length of arc RS?
(A) 8 (B) 20 (C) 8π (D) 20π (E) 40π

9. What is the area of the shaded sector?
(A) 8 (B) 20 (C) 8π (D) 20π (E) 40π

10. In the figure above, what is the value of x?
(A) 30 (B) 36 (C) 45 (D) 54 (E) 60

11. If A is the area and C the circumference of a circle, which of the following is an expression for A in terms of C?

12. What is the area of a circle whose radius is the diagonal of a square whose area is 4?
(A) 2π
(B) 2π
(C) 4π
D) 8π
(E) 16π

Quantitative Comparison Questions

	Column A	Column B

13.	The perimeter of the pentagon	The circumference of the circle

	The circumference of a circle is C inches.
	The area of the same circle is A square inches.
14.

15.

The area of a circle
of radius 2

The area of a semicircle
of radius 3

	C is the circumference of a circle of radius r
16.		6


17.	The area of sector A	The area of sector B


	Each of the triangles is equilateral.
18.	The area of the shaded region	6π

	Column A	Column B

	ABCD and EFGH are squares, and all the circles are tangent to one another and to the sides of the squares.
19.	The area of the shaded region in Figure 1	The area of the shaded region in Figure 2

	A square and a circle have equal areas.
20.	The perimeter of the square	The circumference of the circle

Answer Key

1. D	5. A	9. E	13. B	17. B
2. A	6. D	10. D	14. D	18. C
3. B	7. B	11. A	15. B	19. C
4. C	8. C	12. D	16. A	20. A

Answer Explanations

1. (D) A = πr² = 100π⇒r² = 100 ⇒r = 10 ⇒
C = 2πr = 2π(10) = 20π.

2. (A) C = 2πr = π ⇒ 2r = 1 ⇒ r = ⇒
.

3. (B) Draw a diagram.

Since the area of square ABCD is 2, AD = .

Then diameter EF = and radius

OE = . Then the area of the circle =

4. (C) Draw a diagram.

Since the area of square ABCD is 2, AD = . Then, since ΔABD is an isosceles right triangle, diagonal BD = × = 2.

But BD is also a diameter of the circle. So the diameter is 2 and the radius is 1. Therefore, the area is π(1)² = π.

5. (A) Draw a diagram.

By the Pythagorean theorem (or by recognizing a 5-12-13 triangle), we see that diagonal AC is 13. But AC is also a diameter of the circle, so the diameter is 13 and the radius is 6.5.

6. (D) Since the shaded area is 85% of the circle, the white area is 15% of the circle. So, w is 15% of 360°: 0.15 × 360 = 54.

7. (B) Since C = aπ = bπ = A, we have 2πr = πr² ⇒ 2r = r² ⇒r = 2.

8. (C) The length of arc RS is of the circumference: .

9. (E) The area of the shaded sector is of the area of the circle:

10. (D) Since two of the sides are radii of the circles, the triangle is isosceles. So the unmarked angle is also x:

180 = 72 + 2x ⇒ 2x = 108 ⇒ x = 54.

11. (A) C = 2πr ⇒ r = ⇒

12.(D) If the area of the square is 4, each side is 2, and the length of a diagonal is 2. The area of a circle whose radius is 2 is π(2)² = 8π.

13.(B) There’s nothing to calculate here. Each arc of the circle is clearly longer than the corresponding chord, which is a side of the pentagon. So the circumference, which is the sum of all the arcs, is greater than the perimeter, which is the sum of all the chords.

14. (D) Column A: .

Column B: .

If r = 2, the columns are equal; otherwise, they’re not.

15.(B) Column A: A = π(2)² = 4π.
Column B: The area of a semicircle of radius
.

16. (A) By KEY FACT L4,

, which is greater than 6.

17. (B) The area of sector A is .

The area of sector B is .

So sector B is twice as big as sector A.

18.(C) Since the triangles are equilateral, the two white central angles each measure 60°, and their sum is 120°. So the white area is of the circle, and the shaded area is of the circle. The area of the circle is π(3)² = 9π, so the shaded area is .

19.(C) In Figure 1, since the radius of each circle is 3, the area of each circle is 9π, and the total area of the 4 circles is 36π. In Figure 2, the radius of each circle is 2, and so the area of each circle is 4π, and the total area of the 9 circles is 36π. In the two figures, the white areas are equal, as are the shaded areas.

20. (A) Let A represent the area of the square and the circle.

14-M. SOLID GEOMETRY

There are very few solid geometry questions on the GRE, and they cover only a few elementary topics. Basically, all you need to know are the formulas for the volume and surface areas of rectangular solids (including cubes) and cylinders.

A rectangular solid or box is a solid formed by six rectangles, called faces. The sides of the rectangles are called edges. As shown in the diagram below (left), the edges are called the length, width, and height. A cube is a rectangular solid in which the length, width, and height are equal; so all the edges are the same length.

The volume of a solid is the amount of space it takes up and is measured in cubic units. One cubic unit is the amount of space occupied by a cube all of whose edges are one unit long. In the figure above (right), if each edge of the cube is 1 inch long, then the area of each face is 1 square inch, and the volume of the cube is 1 cubic inch.

KEY FACT M1

• The formula for the volume of a rectangular solid is V = wh.

• The formula for the volume of a cube is V = e·e·e = e³.

EXAMPLE 1

The base of a rectangular tank is 12 feet long and 8 feet wide; the height of the tank is 30 inches. If water is pouring into the tank at the rate of 2 cubic feet per second, how many minutes will be required to fill the tank?

(A) 1 (B) 2 (C) 10 (D) 120 (E) 240

SOLUTION. Draw a diagram. In order to express all of the dimensions of the tank in the same units, convert 30 inches to 2.5 feet. Then the volume of the tank is 12 × 8 × 2.5 = 240 cubic feet. At 2 cubic feet per second, it will take 240 ÷ 2 = 120 seconds = 2 minutes to fill the tank (B).

The surface area of a rectangular solid is the sum of the areas of the six faces. Since the top and bottom faces are equal, the front and back faces are equal, and the left and right faces are equal, we can calculate the area of one from each pair and then double the sum. In a cube, each of the six faces has the same area.

KEY FACT M2

• The formula for the surface area of a rectangular solid is A = 2(w + h + wh).

• The formula for the surface area of a cube is A = 6e².

EXAMPLE 2

The volume of a cube is v cubic yards, and its surface area is a square feet.
If v = a, what is the length in inches of each edge?

(A) 12 (B) 36 (C) 144 (D) 648 (E) 1944

SOLUTION. Draw a diagram. If e is the length of the edge in yards, 3e is the length in feet, and 36e the length in inches. Therefore, v = e³ and a = 6(3e)² = 6(9e²) = 54e². Since v = a, e³ = 54e² ⇒ e = 54, so the length is 36(54) = 1944 inches (E).

A diagonal of a box is a line segment joining a vertex on the top of the box to the opposite vertex on the bottom. A box has 4 diagonals, all the same length. In the box below they are line segments AG, BH, CE, and DF.

KEY FACT M3

A diagonal of a box is the longest line segment that can be drawn between two points on the box.

KEY FACT M4

If the dimensions of a box are , w, and h, and if d is the length of a diagonal, then d² = ² + w² + h².

For example, in the box below: d² = 3² + 4² + 12² = 9 + 16 + 144 = 169 ⇒d = 13.

This formula is really just an extended Pythagorean theorem. EG is the diagonal of rectangular base EFGH. Since the sides of the base are 3 and 4, EG is 5. Now, ΔCGE is a right triangle whose legs are 12 and 5, so diagonal CEis 13.

EXAMPLE 3

What is the length of a diagonal of a cube whose edges are 1?

(A) 1 (B) 2 (C) 3 (D) (E)

SOLUTION. Use the formula:

d² = 1² + 1² + 1² = 3 ⇒ d = (E).

Without the formula you would draw a diagram and label it. Since the base is a 1 × 1 square, its diagonal is . Then the diagonal of the cube is the hypotenuse of a right triangle whose legs are 1 and , so

d² = 1² + ()² = 1 + 2 = 3, and d = .

A cylinder is similar to a rectangular solid except that the base is a circle instead of a rectangle. The volume of a cylinder is the area of its circular base (πr²) times its height (h). The surface area of a cylinder depends on whether you are envisioning a tube, such as a straw, without a top or bottom, or a can, which has both a top and a bottom.

KEY FACT M5

• The formula for the volume, V, of a cylinder whose circular base has radius r and whose height is h is V = πr²h.

• The surface area, A, of the side of the cylinder is the circumference of the circular base times the height: A = 2πrh.

• The area of the top and bottom are each πr², so the total area of a can is 2πrh + 2πr².

EXAMPLE 4

Column A		Column B
	The radius of cylinder II equals the height of cylinder I.
	The height of cylinder II equals the radius of cylinder I.
The volume of cylinder I		The volume of cylinder II

SOLUTION. Let r and h be the radius and height, respectively, of cylinder I. Then

	Column A	Column B
	πr²h	πh²r
Divide each column by πrh:	r	h

Either r or h could be greater, or the two could be equal. The answer is D.

These are the only formulas you need to know. Any other solid geometry questions that might appear on the GRE would require you to visualize a situation and reason it out, rather than to apply a formula.

EXAMPLE 5

How many small cubes are needed to construct the tower in the figure below?

(A) 25 (B) 28 (C) 35 (D) 44 (E) 67

SOLUTION. You need to “see” the answer. The top level consists of 1 cube, the second and third levels consist of 9 cubes each, and the bottom layer consists of 25 cubes. The total is 1 + 9 + 9 + 25 = 44 (D).

Practice Exercises — Solid Geometry

Multiple-Choice Questions

1. The sum of the lengths of all the edges of a cube is 6 centimeters. What is the volume, in cubic centimeters, of the cube?
(A) (B) (C) (D) 1 (E) 8

2. What is the volume of a cube whose surface area is 150?
(A) 25
(B) 100
(C) 125
(D) 1000
(E) 15,625

3. What is the surface area of a cube whose volume is 64?
(A) 16 (B) 64 (C) 96 (D) 128 (E) 384

4. What is the number of cubic inches in one cubic foot?
(A) 12
(B) 24
(C) 144
(D) 684
(E) 1728

5. A solid metal cube of edge 3 feet is placed in a rectangular tank whose length, width, and height are 3, 4, and 5 feet, respectively. What is the volume, in cubic feet, of water that the tank can now hold?
(A) 20 (B) 27 (C) 33 (D) 48 (E) 60

6. A 5-foot-long cylindrical pipe has an inner diameter of 6 feet and an outer diameter of 8 feet. If the total surface area (inside and out, including the ends) is kπ, what is the value of k?
(A) 7 (B) 40 (C) 48 (D) 70 (E) 84

7. The height, h, of a cylinder is equal to the edge of a cube. If the cylinder and cube have the same volume, what is the radius of the cylinder?

8. A rectangular tank has a base that is 10 centimeters by 5 centimeters and a height of 20 centimeters. If the tank is half full of water, by how many centimeters will the water level rise if 325 cubic centimeters of water are poured into the tank?
(A) 3.25
(B) 6.5
(C) 16.25
(D) 32.5
(E) 65

9. If the height of a cylinder is 4 times its circumference, what is the volume of the cylinder in terms of its circumference, C?

10. Three identical balls fit snugly into a cylindrical can: the radius of the spheres equals the radius of the can, and the balls just touch the bottom and the top of the can. If the formula for the volume of a sphere is V = πr³, what fraction of the volume of the can is taken up by the balls?

Quantitative Comparison Questions

	Column A	Column B
	Jack and Jill each roll a sheet of 9 × 12 paper to form a cylinder.
	Jack tapes the two 9-inch edges together.
	Jill tapes the two 12-inch edges together.
11.	The volume of Jack’s cylinder	The volume of Jill’s cylinder

12.

The volume of
a cube whose edges
are 6

The volume of a box
whose dimensions
are 5, 6, and 7

	A is the surface area of a rectangular box in square units.
	V is the volume of the same box in cubic units.
13.	A	V


	P is a point on edge GH of cube ABCDEFGH.
	Each edge of the cube is 1.
14.	The area of ΔABP	1

15.

The volume of a
sphere whose radius
is 1

The volume of a
cube whose edge is 1

ANSWER KEY

1. A	4. E	7. A	10. D	13. D
2. C	5. C	8. B	11. A	14. B
3. C	6. E	9. A	12. A	15. A

Answer Explanations

1. (A) Since a cube has 12 edges, we have 12e = 6 ⇒e = . Therefore, .

2. (C) Since the surface area is 150, each of the 6 faces is a square whose area is 150 ÷ 6 = 25. So the edges are all 5, and the volume is 5³ = 125.

3. (C) Since the volume of the cube is 64, we have e³ = 64 ⇒e = 4. The surface area is 6e² = 6 × 16 = 96.

4. (E) The volume of a cube whose edges are 1 foot can be expressed in either of two ways:

5. (C) The volume of the tank is 3 × 4 × 5 = 60 cubic units, but the solid cube is taking up 3³ = 27 cubic units. Therefore, the tank can hold 60 − 27 = 33 cubic units of water.

6. (E) Draw a diagram and label it. Since the surface of a cylinder is given by A = 2πrh, the area of the exterior is 2π(4)(5) = 40π, and the area of the interior is 2π(3)(5) = 30π. The area of each shaded end is the area of the outer circle minus the area of the inner circle: 16π − 9π = 7π, so total surface area = 40π + 30π + 7π + 7π = 84π⇒k = 84.

7. (A) Since the volumes are equal, πr²h = e³ = h³. Therefore,

8. (B) Draw a diagram. Since the area of the base is 5 × 10 = 50 square centimeters, each 1 centimeter of depth has a volume of 50 cubic centimeters. Therefore, 325 cubic centimeters will raise the water level 325 ÷ 50 = 6.5 centimeters.

(Note that we didn’t use the fact that the tank was half full, except to be sure that the tank didn’t overflow. Since the tank was half full, the water was 10 centimeters deep, and the water level could rise by 6.5 centimeters. Had the tank been three-fourths full, the water would have been 15 centimeters deep and the extra water would have caused the level to rise 5 centimeters, filling the tank; the rest of the water would have spilled out.)

9. (A) Since V = πr²h, we need to express r and h in terms of C. It is given that h = 4C and since C = 2πr, then . Therefore, .

10. (D) To avoid using r, assume that the radii of the spheres and the can are 1. Then the volume of each ball is , and the total volume of the 3 balls is .

Since the height of the can is 6 (the diameter of each sphere is 2), the volume of the can is π(1)²(6) = 6π. So the balls take up of the can.

11. (A) Drawing a diagram makes it easier to visualize the problem. The volume of a cylinder is πr²h. In each case, we know the height but have to determine the radius in order to calculate the volume.

Jack’s cylinder has a circumference of 12:

Jill’s cylinder has a circumference of 9:

12.(A) Column A: V = 6³ = 216.
Column B: V = 5 × 6 × 7 = 210.

13.(D) There is no relationship between the two columns. If the box is a cube of edge 1, A = 6 and V = 1. If the box is a cube of edge 10, A = 600 and V = 1000.

14. (B) The base, AB, of ΔABP is 1. Since the diagonal is the longest line segment in the cube, the height, h, of the triangle is definitely less than the diagonal, which is .

So the area of the triangle is less than , which is less than 1. (You could also have just calculated the area: h = BG = , so the area is

15. (A) You probably don’t know how to find the volume of a sphere; fortunately, you don’t need to. You should be able to visualize that the sphere is much bigger than the cube. (In fact, it is more than 4 times the size.)

14-N. COORDINATE GEOMETRY

The GRE has very few questions on coordinate geometry. Most often they deal with the coordinates of points and occasionally with the slope of a line. You will never have to draw a graph; nor will you have to provide the equation for a given graph.

The coordinate plane is formed by two perpendicular number lines called the x-axis and y-axis, which intersect at the origin. The axes divide the plane into four quadrants, labeled I, II, III, and IV.

Each point in the plane is assigned two numbers, an x-coordinate and a y-coordinate, which are written as an ordered pair, (x, y).

• Points to the right of the y-axis have positive x-coordinates, and those to the left have negative x-coordinates.

• Points above the x-axis have positive y-coordinates, and those below it have negative y-coordinates.

• If a point is on the x-axis, its y-coordinate is 0.

• If a point is on the y-axis, its x-coordinate is 0.

For example, point A in the following figure is labeled (2, 3), since it is 2 units to the right of the y-axis and 3 units above the x-axis. Similarly, B(−3, −5) is in Quadrant III, 3 units to the left of the y-axis and 5 units below the x-axis.

EXAMPLE 1

Column A		Column B

b		c

EXAMPLE 2


r		s

SOLUTION 1. Since (a, b) lies on the x-axis, b = 0. Since (c, d) lies on the y-axis, c = 0. The answer is C.

SOLUTION 2. Since (r, s) is in Quadrant II, r is negative and s is positive. The answer is B.

Often a question requires you to calculate the distance between two points. This is easiest when the points lie on the same horizontal or vertical line.

KEY FACT N1

• All the points on a horizontal line have the same y-coordinate. To find the distance between them, subtract their x-coordinates.

• All the points on a vertical line have the same x-coordinate. To find the distance between them, subtract their y-coordinates.

HELPFUL HINT

If the points have been plotted on a graph, you can find the distance between them by counting boxes.

The distance from A to C is 6 − 1 = 5. The distance from B to C is 4 − 1 = 3.

It is a little more difficult to find the distance between two points that are not on the same horizontal or vertical line. In this case, use the Pythagorean theorem. For example, in the previous figure, if d represents the distance from A to B, d² = 5² + 3² = 25 + 9 = 34 ⇒ d = .

CAUTION

You cannot count boxes unless the points are on the same horizontal or vertical line. The distance between A and B is 5, not 4.

KEY FACT N2

The distance, d, between two points, A(x₁, y₁) and B(x₂, y₂), can be calculated using the distance formula:

HELPFUL HINT

The “distance formula” is nothing more than the Pythagorean theorem. If you ever forget the formula, and you need the distance between two points that do not lie on the same horizontal or vertical line, do as follows: create a right triangle by drawing a horizontal line through one of the points and a vertical line through the other, and then use the Pythagorean theorem.

Examples 3−4 refer to the triangle in the following figure.

EXAMPLE 3

What is the area of ΔRST?

(A) 6 (B) 9 (C) 12 (D) 15 (E) 18

SOLUTION. R(−2, 1) and S(4, 1) lie on the same horizontal line, so RS = 4 − (−2) = 6. Let that be the base of the triangle. Then the height is the distance along the vertical line from T to RS: 4 − 1 = 3. The area is

EXAMPLE 4

What is the perimeter of ΔRST?

(A) 13 (B) 14 (C) 16 (D) 11 + (E) 11 +

SOLUTION. The perimeter is RS + ST + RT. From the solution to Example 3, you know that RS = 6. Also, ST = 5, since it is the hypotenuse of a 3-4-5 right triangle. To calculate RT, either use the distance formula:

or the Pythagorean theorem:

So the perimeter is: .

The slope of a line is a number that indicates how steep the line is.

KEY FACT N3

• Vertical lines do not have slopes.

• To find the slope of any other line proceed as follows:

1. Choose any two points A(x₁, y₁) and B(x₂, y₂) on the line.

2. Take the differences of the y-coordinates, y₂ − y₁, and the x-coordinates, x₂ − x₁.

3. Divide: slope = .

We will illustrate the next KEY FACT by using this formula to calculate the slopes of RS, RT, and ST from Example 3: R(−2, 1), S(4, 1), T(0, 4).

KEY FACT N4

• The slope of any horizontal line is 0: slope of

• The slope of any line that goes up as you move from left to right is positive: slope of

• The slope of any line that goes down as you move from left to right is negative: slope of

EXAMPLE 5

Column A		Column B
	Line , passes through (1, 2) and (3, 5)
	Line m is perpendicular to ,
The slope of ,		The slope of m

SOLUTION. First, make a quick sketch. Do not use the formula to calculate the slope of . Simply notice that slopes upward, so its slope is positive, whereas m slopes downward, so its slope is negative. Column A is greater.

Practice Exercises — Coordinate Geometry

Multiple-Choice Questions

1. What is the slope of the line that passes through points (0, −2) and (3, 0)?

2. If the coordinates of ΔRST are R(0, 0), S(7, 0), and T(2, 5), what is the sum of the slopes of the three sides of the triangle?
(A) −1.5 (B) 0 (C) 1.5 (D) 2.5 (E) 3.5

3. If A(−1, 1) and B(3, −1) are the endpoints of one side of square ABCD, what is the area of the square?
(A) 12 (B) 16 (C) 20 (D) 25 (E) 36

4. If the area of circle O above is kπ, what is the value of k?
(A) 3 (B) 6 (C) 9 (D) 18 (E) 27

5. If P(2, 1) and Q(8, 1) are two of the vertices of a rectangle, which of the following could not be another of the vertices?
(A) (2, 8)
(B) (8, 2)
(C) (2, −8)
(D) (−2, 8)
(E) (8, 8)

6. A circle whose center is at (6, 8) passes through the origin. Which of the following points is not on the circle?
(A) (12, 0)
(B) (6, −2)
(C) (16, 8)
(D) (−2, 12)
(E) (−4, 8)

Questions 7−8 concern parallelogram JKLM, whose coordinates are J(−5, 2), K(−2, 6), L(5, 6), M(2, 2).

7. What is the area of parallelogram JKLM?
(A) 35 (B) 28 (C) 24 (D) 20 (E) 12

8. What is the perimeter of parallelogram JKLM?
(A) 35 (B) 28 (C) 24 (D) 20 (E) 12

8. If (a, b) and are two distinct points, what is the slope of the line that passes through them?

9. If c ≠ 0 and the slope of the line passing through (−c, c) and (3c, a) is 1, which of the following is an expression for a in terms of c?
(A) −3c (B) (C) 2c (D) 3c (E) 5c

Quantitative Comparison Questions

	Column A	Column B
	m is the slope of one of the diagonals of a square.
11.	m²	1


12.	a − b	0

	Column A	Column B

13.	The slope of line k	The slope of line


	The slope of line is −0.8.
14.	c	b

	The distance from (b, 5) to (c, −3) is 10. b < c
15.	c − b	6

ANSWER KEY

1. D	4. D	7. B	10. E	13. A
2. C	5. D	8. C	11. D	14. A
3. C	6. D	9. A	12. A	15. C

Answer Explanations

1. (D) If you sketch the line, you see immediately that the slope of the line is positive. Without even knowing the slope formula, therefore, you can eliminate Choices A, B, and C. To determine the actual slope, use the formula: .

2. (C) Sketch the triangle, and then calculate the slopes.

Since RS is horizontal, its slope is 0.
The slope of RT = = 2.5.
The slope of ST = .
Now add: 0 + 2.5 + (−1) = 1.5

3. (C) Draw a diagram and label it. The area of square ABCD is s², where s = AB = length of a side. By the Pythagorean theorem:

s² = 2² + 4² = 4 + 16 = 20.

4. (D) Since the line segment joining (3, 3) and (0, 0) is a radius of the circle, r² = 3² + 3² = 18. Therefore, area = πr² = 18π⇒k = 18. Note that you do not actually have to find that the value of r is 3.

5. (D) Draw a diagram. Any point whose x-coordinate is 2 or 8 could be another vertex. Of the choices, only (−2, 8) is not possible.

6. (D) Draw a diagram. The radius of the circle is 10 (since it’s the hypotenuse of a 6-8-10 right triangle). Which of the choices are 10 units from (6, 8)? First, check the easy ones: (−4, 8) and (16, 8) are 10 units to the left and right of (6, 8), and (6, −2) is 10 units below. What remains is to check (12, 0), which works, and (−2, 12), which doesn’t.

Here is the diagram for solutions 7 and 8.

7. (B) The base is 7 and the height is 4. So, the area is 7 × 4 = 28.

8. (C) Sides JM and KL are each 7, and sides JK and LM are each the hypotenuse of a 3-4-5 right triangle, so they are 5. The perimeter is 2(7 + 5) = 24.

9. (A) The formula for the slope is , but before using it, look. Since the y-coordinates are equal and the x-coordinates are not equal, the numerator is 0 and the denominator is not 0. So the value of the fraction is 0.

10. (E) The slope is equal to

a − c = 4c⇒a = 5c.

11. (D) If the sides of the square are horizontal and vertical, then m is 1 or −1, and m² is 1. But the square could be positioned any place, and the slope of a diagonal could be any number.

12. (A) Line , which goes through (0, 0) and (1, 1), also goes through (a, a), and since (a, b) is below (a, a), b < a. Therefore, a − b is positive.

13. (A) The line going through (−3, 3) and (0, 0) has slope −1. Since, is steeper, its slope is a number such as −2 or −3; since k is less steep, its slope is a number such as −0.5 or −0.3. Therefore, the slope of k is greater.

14. (A) Since (a, b) is on the y-axis, a = 0; and since (c, d) is on the x-axis, d = 0. Then by the slope formula,
.

15. (C) Draw a diagram.

Since the distance between the two points is 10, by the distance formula:

Squaring both sides gives
100 = (c − b)² + 64 ⇒ (c − b)² = 36 ⇒ c − b = 6.

14-O. COUNTING AND PROBABILITY

Some questions on the GRE begin, “How many ….” In these problems you are being asked to count something: how many apples can she buy, how many dollars did he spend, how many pages did she read, how many numbers satisfy a certain property, or how many ways are there to complete a particular task. Sometimes these problems can be handled by simple arithmetic. Other times it helps to use TACTIC 8 from Chapter 10 and systematically make a list. Occasionally it helps to know the counting principle and other strategies that we will review in this section.

COUNTING

USING ARITHMETIC TO COUNT

The following three examples require only arithmetic. But be careful; they are not the same.

EXAMPLE 1

Brian bought some apples. If he entered the store with $113 and left with $109, how much did the apples cost?

EXAMPLE 2

Scott was selling tickets for the school play. One day he sold tickets numbered 109 through 113. How many tickets did he sell that day?

EXAMPLE 3

Brian is the 109th person in a line and Scott is the 113th person. How many people are there between Brian and Scott?

SOLUTIONS 1–3.

· It may seem that each of these examples requires a simple subtraction: 113 − 109 = 4. In Example 1, Brian did spend $4 on apples; in Example 2, however, Scott sold 5 tickets; and in Example 3, only 3 people are on line between Brian and Scott!

· Assume that Brian went into the store with 113 one-dollar bills, numbered 1 through 113; he spent the 4 dollars numbered 113, 112, 111, and 110, and still had the dollars numbered 1 through 109; Scott sold the 5 tickets numbered 109, 110, 111, 112, and 113; and between Brian and Scott the 110th, 111th, and 112th persons — 3 people — were on line.

In Example 1, you just needed to subtract: 113 − 109 = 4. In Example 2, you need to subtract and then add1: 113 − 109 + 1 = 4 + 1 = 5. And in Example 3, you need to subtract and then subtract 1 more: 113 − 109 − 1 = 3. Although Example 1 is too easy for the GRE, questions such as Examples 2 and 3 do appear, because they’re not as obvious and they require that little extra thought. When do you have to add or subtract 1?

The issue is whether or not the first and last numbers are included. In Example 1, Brian spent dollar number 113, but he still had dollar number 109 when he left the store. In Example 2, Scott sold both ticket number 109 and ticket 113. In Example 3, neither Scott (the 113th person) nor Brian (the 109th person) was to be counted.

KEY FACT O1

To count how many integers there are between two integers, follow these rules:

· If exactly one of the endpoints is included, subtract.

· If both endpoints are included, subtract and add 1.

· If neither endpoint is included, subtract and subtract 1 more.

EXAMPLE 4

From 1:09 to 1:13, Adam read pages 109 through 113 in his English book. What was his rate of reading, in pages per minute?

SOLUTION. Since Adam read both pages 109 and 113, he read 113 − 109 + 1 = 5 pages. He started reading during the minute that started at 1:09 (and ended at 1:10). Since he stopped reading at 1:13, he did not read during the minute that began at 1:13 (and ended at 1:14). So he read for 1:13 − 1:09 = 4 minutes. He read at the rate of pages per minute.

SYSTEMATICALLY MAKING A LIST

When a question asks “How many…?” and the numbers in the problem are small, just systematically list all of the possibilities.

Proper use of TACTIC O1 eliminates the risk of making an error in arithmetic. In Example 4, rather than even thinking about whether or not to add 1 or subtract 1 after subtracting the number of pages, you could have just quickly jotted down the numbers of the pages Adam read (109, 110, 111, 112, 113), and then counted them.

EXAMPLE 5

Ariel has 4 paintings in the basement. She is going to bring up 2 of them and hang 1 in her den and 1 in her bedroom. In how many ways can she choose which paintings go in each room?

(A) 4 (B) 6 (C) 12 (D) 16 (E) 24

SOLUTION. Label the paintings 1, 2, 3, and 4, write B for bedroom and D for den, and make a list.

B-D	B-D	B-D	B-D
1-2	2-2	3-1	4-1
1-3	2-3	3-2	4-2
1-4	2-4	3-4	4-3

There are 12 ways to choose which paintings go in each room (C).

In Example 5, making a list was feasible, but if Ariel had 10 paintings and needed to hang 4 of them, it would be impossible to list all the different ways of hanging them. In such cases, we need the counting principle.

THE COUNTING PRINCIPLE

KEY FACT O2

If two jobs need to be completed and there are m ways to do the first job and n ways to do the second job, then there are m × n ways to do one job followed by the other. This principle can be extended to any number of jobs.

In Example 5, the first job was to pick 1 of the 4 paintings and hang it in the bedroom. That could be done in 4 ways. The second job was to pick a second painting to hang in the den. That job could be accomplished by choosing any of the remaining 3 paintings. So there are 4 × 3 = 12 ways to hang 2 of the paintings.

Now, assume that there are 10 paintings to be hung in 4 rooms. The first job is to choose one of the 10 paintings for the bedroom. The second job is to choose one of the 9 remaining paintings to hang in the den. The third job is to choose one of the 8 remaining paintings for the living room. Finally, the fourth job is to pick one of the 7 remaining paintings for the dining room. These 4 jobs can be completed in 10 × 9 × 8 × 7 = 5040 ways.

EXAMPLE 6

How many integers are there between 100 and 1000 all of whose digits are odd?

SOLUTION. We’re looking for three-digit numbers, such as 135, 711, 353, and 999, in which all three digits are odd. Note that we are not required to use three different digits. Although you certainly wouldn’t want to list all of them, you could count them by listing some of them and seeing if a pattern develops. In the 100s there are 5 numbers that begin with 11: 111, 113, 115, 117, 119. Similarly, there are 5 numbers that begin with 13: 131, 133, 135, 137, 139; 5 that begin with 15; 5 that begin with 17; and 5 that begin with 19, for a total of 5 × 5 = 25 in the 100s. In the same way there are 25 in the 300s, 25 in the 500s, 25 in the 700s, and 25 in the 900s, for a grand total of 5 × 25 = 125. You can actually do this in less time than it takes to read this paragraph.

The best way to solve Example 6, however, is to use the counting principle. Think of writing a three-digit number as three jobs that need to be done. The first job is to select one of the five odd digits and use it as the digit in the hundreds place. The second job is to select one of the five odd digits to be the digit that goes in the tens place. Finally, the third job is to select one of the five odd digits to be the digit in the units place. Each of these jobs can be done in 5 ways. So the total number of ways is 5 × 5 × 5 = 125.

EXAMPLE 7

How many different arrangements are there of the letters A, B, C, and D?

(A) 4 (B) 6 (C) 8 (D) 12 (E) 24

Since from the choices given, we know that the answer is a relatively small number, we could just use TACTIC O1 and systematically list them: ABCD, ABDC, ACBD, …. However, this method would not be suitable if you had to arrange as few as 5 or 6 letters and would be practically impossible if you had to arrange 10 or 20 letters.

SOLUTION. Think of the act of arranging the four letters as four jobs that need to be done, and use the counting principle. The first job is to choose one of the four letters to write in the first position; there are 4 ways to complete that job. The second job is to choose one of the remaining three letters to write in the second position; there are 3 ways to complete that job. The third job is to choose one of the two remaining letters to write in the third position; there are 2 ways to complete that job. Finally, the fourth job is to choose the only remaining letter and to write it: 4 × 3 × 2 × 1 = 24.

VENN DIAGRAMS

A Venn diagram is a figure with two or three overlapping circles, usually enclosed in a rectangle, which is used to solve certain counting problems. To illustrate this, assume that a school has 100 seniors. The following Venn diagram, which divides the rectangle into four regions, shows the distribution of those students in the band and the orchestra.

The 32 written in the part of the diagram where the two circles overlap represents the 32 seniors who are in both band and orchestra. The 18 written in the circle on the right represents the 18 seniors who are in band but not in orchestra, while the 37 written in the left circle represents the 37 seniors who are in orchestra but not in band. Finally, the 13 written in the rectangle outside of the circles represents the 13 seniors who are in neither band nor orchestra. The numbers in all four regions must add up to the total number of seniors: 32 + 18 + 37 + 13 = 100. Note that there are 50 seniors in the band — 32 who are also in the orchestra and 18 who are not in the orchestra. Similarly, there are 32 + 37 = 69 seniors in the orchestra. Be careful: the 50 names on the band roster and the 69 names on the orchestra roster add up to 119 names — more than the number of seniors. That’s because 32 names are on both lists and so have been counted twice. The number of seniors who are in band or orchestra is only 119 − 32 = 87. Those 87 together with the 13 seniors who are in neither make up the total of 100.

On the GRE, Venn diagrams are used in two ways. It is possible to be given a Venn diagram and asked a question about it, as in Example 7. More often, you will come across a problem, such as Example 8, that you will be able to solve more easily if you think to draw a Venn diagram.

EXAMPLE 7

If the integers from 1 through 15 are each placed in the diagram at the right, which regions are empty?

(A) D only
(B) F only
(C) G only
(D) F and G only
(E) D and G only

SOLUTION. The easiest way is just to put each of the numbers from 1 through 15 in the appropriate region.

The empty regions are F and G only (D).

EXAMPLE 8

Of the 410 students at H. S. Truman High School, 240 study Spanish and 180 study French. If 25 students study neither language, how many study both?

(A) 25 (B) 35 (C) 60 (D) 170 (E) 230

SOLUTION. Draw a Venn diagram. Let x represent the number of students who study both languages, and write x in the part of the diagram where the two circles overlap. Then the number who study only Spanish is 240 − x, and the number who study only French is 180 − x. The number who study at least one of the languages is 410 − 25 = 385, so we have

385 = (240 − x) + x + (180 − x) = 420 − x ⇒x = 420 − 385 = 35

students who study both (B).

Note: No problem requires the use of a Venn diagram. On some problems you might even find it easier not to use one. In Example 8, you could have reasoned that if there were 410 students in the school and 25 didn’t study either language, then there were 410 − 25 = 385 students who studied at least one language. There are 240 names on the Spanish class lists and 180 on the French class lists, a total of 240 + 180 = 420 names. But those 420 names belong to only 385 students. It must be that 420 − 385 = 35 names were repeated. In other words, 35 students are in both French and Spanish classes.

PROBABILITY

The probability that an event will occur is a number between 0 and 1, usually written as a fraction, which indicates how likely it is that the event will happen. For example, if you spin the spinner on the next page, there are 4 possible outcomes. It is equally likely that the spinner will stop in any of the 4 regions. There is 1 chance in 4 that it will stop in the region marked 2. So we say that the probability of spinning a 2 is one-fourth and write P(2) = . Since 2 is the only even number on the spinner we could also say P(even) = . There are 3 chances in 4 that the spinner will land in a region with an odd number in it, so P(odd) = .

KEY FACT O3

If E is any event, the probability that E will occur is given by

assuming that the possible outcomes are all equally likely.

In the preceding example, each of the 4 regions is the same size, so it is equally likely that the spinner will land on the 2, 3, 5, or 7. Therefore,

Note that the probability of not getting an odd number is 1 minus the probability of getting an odd number: 1 − . Let’s look at some other probabilities associated with spinning this spinner once.

KEY FACT O4

Let E be an event, and P(E) the probability it will occur.

· If E is impossible (such as getting a number greater than 10), P(E) = 0.

· If it is certain that E will occur (such as getting a prime number), P(E) = 1.

· In all cases 0 ≤ P(E) ≤ 1.

· The probability that event E will not occur is 1 − P(E).

· If 2 or more events constitute all the outcomes, the sum of their probabilities is 1.
[For example, P(even) + P(odd) = = 1.]

· The more likely it is that an event will occur, the higher its probability (the closer to 1 it is); the less likely it is that an event will occur, the lower its probability (the closer to 0 it is).

Even though probability is defined as a fraction, we can write probabilities as decimals or percents, as well.

Instead of writing P(E) = , we can write P(E) = .50 or P(E) = 50%.

EXAMPLE 9

An integer between 100 and 999, inclusive, is chosen at random. What is the probability that all the digits of the number are odd?

SOLUTION. By KEY FACT O1, since both endpoints are included, there are 999 − 100 + 1 = 900 integers between 100 and 999. In Example 6, we saw that there are 125 three-digit numbers all of whose digits are odd. So the probability is

KEY FACT O5

If an experiment is done two (or more) times, the probability that first one event will occur and then a second event will occur is the product of the probabilities.

EXAMPLE 10

A fair coin is flipped three times. What is the probability that the coin lands heads each time?

SOLUTION. When a fair coin is flipped:

P(head) = and P(tail) = .

By KEY FACT O5, P(3 heads) =

P(head 1st time) ×P(head 2nd time) ×P(head 3rd time) = .

Another way to handle problems such as Example 10 is to make a list of all the possible outcomes. For example, if a coin is tossed three times, the possible outcomes are

	head, head, head	head, head, tail
	head, tail, head	head, tail, tail
	tail, head, head	tail, head, tail
	tail, tail, head	tail, tail, tail

On the GRE, of course, if you choose to list the outcomes on your scrap paper, you should abbreviate and just write HHH, HHT, and so on. In any event, there are eight possible outcomes, and only one of them (HHH) is favorable. So the probability is .

EXAMPLE 11

Column A	Column B
Three fair coins are flipped.
The probability of getting more heads than tails	The probability of getting more tails than heads

SOLUTION. From the list of the 8 possible outcomes mentioned, you can see that in 4 of them (HHH, HHT, HTH, THH) there are more heads than tails, and that in 4 of them (TTT, TTH, THT, HTT) there are more tails than heads. Each probability is . The answer is C.

In Example 11, it wasn’t even necessary to calculate the two probabilities. Since heads and tails are equally likely, when several coins are flipped, it is just as likely to have more heads as it is to have more tails. This is typical of quantitative comparison questions on probability; you usually can tell which of the two probabilities is greater without having to calculate either one. This is another instance where TACTIC 5 from Chapter 13 (don’t calculate, compare) is useful.

EXAMPLE 12

Column A	Column B
The numbers from 1 to 1000 are each written on a slip of paper and placed in a box. Then 1 slip is removed.
The probability that the number drawn is a multiple of 5	The probability that the number drawn is a multiple of 7

SOLUTION. Since there are many more multiples of 5 than there are of 7, it is more likely that a multiple of 5 will be drawn. Column A is greater.

Practice Exercises — Counting and Probability

Multiple-Choice Questions

1. Alyssa completed exercises 6−20 on her math review sheet in 30 minutes. At this rate, how long, in minutes, will it take her to complete exercises 29−57?
(A) 56 (B) 57 (C) 58 (D) 60 (E) 65

2. A diner serves a lunch special, consisting of soup or salad, a sandwich, coffee or tea, and a dessert. If the menu lists 3 soups, 2 salads, 7 sandwiches, and 8 desserts, how many different lunches can you choose? (Note: Two lunches are different if they differ in any aspect.)
(A) 22
(B) 280
(C) 336
(D) 560
(E) 672

3. Dwight Eisenhower was born on October 14, 1890 and died on March 28, 1969. What was his age, in years, at the time of his death?
(A) 77 (B) 78 (C) 79 (D) 80 (E) 81

4. How many four-digit numbers have only even digits?
(A) 96
(B) 128
(C) 256
(D) 500
(E) 625

5. There are 27 students on the college debate team. What is the probability that at least 3 of them have their birthdays in the same month?

6. Let A be the set of primes less than 6, and B be the set of positive odd numbers less than 6. How many different sums of the form a + b are possible, if a is in A and b is in B?
(A) 6 (B) 7 (C) 8 (D) 9 (E) 10

7. There are 100 people on a line. Aviva is the 37th person and Naomi is the 67th person. If a person on line is chosen at random, what is the probability that the person is standing between Aviva and Naomi?

8. A jar has 5 marbles, 1 of each of the colors red, white, blue, green, and yellow. If 4 marbles are removed from the jar, what is the probability that the yellow one was removed?

9. Josh works on the second floor of a building. There are 10 doors to the building and 8 staircases from the first to the second floor. Josh decided that each day he would enter by one door and leave by a different one, and go up one staircase and down another. How many days could Josh do this before he had to repeat a path he had previously taken?
(A) 80
(B) 640
(C) 800
(D) 5040
(E) 6400

10. A jar contains 20 marbles: 4 red, 6 white, and 10 blue. If you remove marbles one at a time, randomly, what is the minimum number that must be removed to be certain that you have at least 2 marbles of each color?
(A) 6 (B) 10 (C) 12 (D) 16 (E) 18

11. At the audition for the school play, n people tried out. If k people went before Judy, who went before Liz, and m people went after Liz, how many people tried out between Judy and Liz?
(A) n − m − k − 2
(B) n − m − k − 1
(C) n − m − k
(D) n − m − k + 1
(E) n − m − k + 2

12. In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 are in the club and 20 are neither on the team nor in the club, what is the minimum number of students who could be both on the team and in the club?
(A) 10 (B) 49 (C) 50 (D) 60 (E) 61

13. In a singles tennis tournament that has 125 entrants, a player is eliminated whenever he loses a match. How many matches are played in the entire tournament?
(A) 62 (B) 63 (C) 124 (D) 125 (E) 246

Questions 14−15 refer to the following diagram. A is the set of positive integers less than 20; B is the set of positive integers that contain the digit 7; and C is the set of primes.

14. How many numbers are in the region labeled x?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8

15. What is the sum of all the numbers less than 50 that are in the region labeled y?
(A) 24 (B) 37 (C) 47 (D) 84 (E) 108

Quantitative Comparison Questions

	Column A	Column B
16.	The probability of getting no heads when a fair coin is flipped 7 times	The probability of getting 7 heads when a fair coin is flipped 7 times

	A jar contains 4 marbles: 2 red and 2 white. 2 marbles are chosen at random.
17.	The probability that the marbles chosen are the same color	The probability that the marbles chosen are different colors

18.

The number of ways to assign a number from 1 to 5 to each of 4 people

The number of ways to assign a number from 1 to 5 to each of 5 people

19.

The probability that 2 people chosen at random were born on the same day of the week

The probability that 2 people chosen at random were born in the same month

20.

The probability that a number chosen at random from the primes between 100 and 199 is odd.

.99

Answer Key

1. C	5. E	9. D	13. C	17. B
2. D	6. B	10. E	14. C	18. C
3. B	7. B	11. A	15. D	19. A
4. D	8. D	12. E	16. C	20. A

Answer Explanations

1. (C) Alyssa completed 20 − 6 + 1 = 15 exercises in 30 minutes, which is a rate of 1 exercise every 2 minutes. Therefore, to complete 57 − 29 + 1 = 29 exercises would take her 58 minutes.

2. (D) You can choose your soup or salad in any of 5 ways, your beverage in any of 2 ways, your sandwich in 7 ways, and your dessert in 8 ways. The counting principle says to multiply: 5 × 2 × 7 × 8 = 560. (Note that if you got soup and a salad, then instead of 5 choices for the first course there would have been 2 × 3 = 6 choices for the first two courses.)

3. (B) His last birthday was in October 1968, when he turned 78: 1968 − 1890 = 78

4. (D) The easiest way to solve this problem is to use the counting principle. The first digit can be chosen in any of 4 ways (2, 4, 6, 8), whereas the second, third, and fourth digits can be chosen in any of 5 ways (0, 2, 4, 6, 8). Therefore, the total number of four-digit numbers all of whose digits are even is 4 × 5 × 5 × 5 = 500.

5. (E) If there were no month in which at least 3 students had a birthday, then each month would have the birthdays of at most 2 students. But that’s not possible. Even if there were 2 birthdays in January, 2 in February, …, and 2 in December, that would account for only 24 students. It is guaranteed that with more than 24 students, at least one month will have 3 or more birthdays. The probability is 1.

6. (B) A = {2, 3, 5} and B = {1, 3, 5}. Any of the 3 numbers in A could be added to any of the 3 numbers in B, so there are 9 sums that could be formed. However, there could be some duplication. List the sums systematically; first add 1 to each number in A, then 3, and then 5: 3, 4, 6; 5, 6, 10; 7, 8, 10. There are 7 different sums.

7. (B) There are 67 − 37 − 1 = 29 people between Aviva and Naomi, so, the probability that one of them is chosen is .

8. (D) It is equally likely that any one of the 5 marbles will be the one that is not removed. So, the probability that the yellow one is left is and the probability that it is removed is .

9. (D) This is the counting principle at work. Each day Josh has four jobs to do: choose 1 of the 10 doors to enter and 1 of the 9 other doors to exit; choose 1 of the 8 staircases to go up and 1 of the other 7 to come down. This can be done in 10 × 9 × 8 × 7 = 5040 ways. So on each of 5040 days Josh could choose a different path.

10. (E) In a problem like this the easiest thing to do is to see what could go wrong in your attempt to get 2 marbles of each color. If you were really unlucky, you might remove 10 blue ones in a row, followed by all 6 white ones. At that point you would have 16 marbles, and you still wouldn’t have even 1 red one. The next 2 marbles, however, must both be red. The answer is 18.

11. (A) It may help to draw a line and label it:

Since k people went before Judy, she was number k + 1 to try out; and since m people went after Liz, she was number n − m to try out. So the number of people to try out between them was

(n − m) − (k + 1) − 1 = n − m − k − 2.

12. (E) Draw a Venn diagram, letting x be the number of students who are on the team and in the club. Of the 100 students, 70 are in the club, so 30 are not in the club. Of these 30, 20 are also not on the team, so 10 are on the team but not in the club.

Since more students are on the team than in the club, 10 + x > 70 ⇒ x > 60. Since x must be an integer, the least it can be is 61.

13. (C) You could try to break it down by saying that first 124 of the 125 players would be paired off and play 62 matches. The 62 losers would be eliminated and there would still be 63 people left, the 62 winners and the 1 person who didn’t play yet. Then continue until only 1 person was left. This is too time-consuming. An easier way is to observe that the winner never loses and the other 124 players each lose once. Since each match has exactly one loser, there must be 124 matches.

14. (C) The region labeled x contains all of the primes less than 20 that do not contain the digit 7. They are 2, 3, 5, 11, 13, 19.

15. (D) Region y consists of primes that contain the digit 7 and that are greater than 20. There are two of them that are less than 50: 37 and 47. Their sum is 84.

16. (C) Don’t calculate the probabilities. The probability of no heads is equal to the probability of no tails; but no tails means all heads.

17. (B) The simplest solution is to notice that whatever color the first marble is, there is only 1 more marble of that color, but there are 2 of the other color, so it is twice as likely that the marbles will be of different colors.

18. (C) By the counting principle, Column A is 5·4·3·2 and Column B is 5·4·3·2·1. Clearly, the columns are equal.

19. (A) Column A: the probability is .
Column B: the probability is .

20. (A) Every prime between 100 and 199 is odd (the only even prime is 2). So the probability in Column A is 1, which is greater than .99.