Math Workout for the GMAT
Part III Content and Strategy Review
Chapter 6 Facing Algebra
In the previous chapter, you learned how to avoid algebra by using Plugging In and Plugging In The Answers. Both are useful strategies if you are a bit algebraphobic. This chapter is for the proalgebra contingent of GMAT test takers (or at least the test takers who aren’t antialgebra). You’ll notice that many of the GMAT math problems involve algebra in some manner. Algebra is essentially arithmetic with one or more variables thrown into the mix. A variable is an unknown number and is usually indicated by an italicized letter, such as x or n.
SOLVING EQUATIONS
The golden rule for solving equations is “Do the same thing to both sides until you isolate the variable.” In other words, add, subtract, multiply, and divide both sides by whatever means necessary to get the variable all by itself on one side of the equals sign. You should usually (but not always) add and subtract before you multiply and divide.
For example, solve for x in the equation 3x + 7 = 19.
Subtract 7 from both sides to get 
3x 
= 12 
Divide both sides by 3 to get 
x 
= 4 
You should also know how to deal with numbers in parentheses, such as 5(x + 3). In this case, just multiply each number in the parentheses by 5. So 5(x + 3) = 5x + 15. You may also need to reverse this process by factoring. Just find the common factor of the numbers and put that outside the parentheses. For example, factor 6x + 12. Both numbers are divisible by 6, so factor out the 6 and put it in front of parentheses. So 6x + 12 = 6(x + 2).
Factoring is often required in problems with complicated fractions, such as . To simplify the fraction, you could try factoring out a number so that something on the top cancels with something on the bottom. This could let you reduce the fraction to a simpler form. If you factor out a 7 on the top to get = , then you can cancel the x + 2 on the top with the x + 2 on the bottom. So = = 7.
DRILL 1
Answers can be found on this page.
1.If x + 12 = 4, then x =
−16 

−8 

−4 

8 

6 
2.If x ≠ −2, then =
2 

3 

4 

5 

6 
3.If = , for x ≠ 1 and x ≠ −1, then x =
−3 

−2 

0 

2 

3 
4.If and y ≠ 1, then y =
4 

3 

2 

More Than One Variable (Simultaneous Equations)
You’ve seen how to solve an equation that contains one variable. What can you do if an equation has more than one variable, such as 2x + y = 7? Well, you can’t actually solve that equation without more information in the form of another equation. The general rule is that the number of different variables equals the number of different equations necessary to solve for those variables.
If you have two variables in two equations, such as 2x + y = 7 and x – y = −1, you can solve for the variables. The method is often referred to as simultaneous equations. Line up the two equations, as shown below, and then add the equations and eliminate one variable. This will leave you with an equation with one variable.
Solve that equation as you learned to do earlier in the chapter. 3x = 6 becomes x = 2. Now you can find y by plugging the value for x into one of the original equations (it doesn’t matter which). The first equation would become 2(2) + y = 7. That gives you a onevariable equation, which you can then solve for y. In this case, y = 3.
What if simply adding the equations doesn’t eliminate a variable, as with x + y = 9 and 2x + y = 12? In such cases, you need to multiply one or both of the equations to set up one of the variables for elimination. Multiply every term in the equation by the same number. In this example, you can multiply the first equation by −1. So x + y = 9 becomes –x – y = −9. Now you can add the equations and eliminate the y variable. The resulting equation is x = 3. Plug that back into one of the equations. Using the first equation, you get 3 + y = 9. Now you can determine that y = 6.
Look at a slightly more complex example: 3x + y = 11 and 2x + 3y = 19. In this case, you can multiply the first equation by −3 to eliminate the y variable. Combining −9x − 3y = −33 with 2x + 3y = 19, you get −7x = −14. So x = 2. Plug that into the first equation to get 3(2) + y = 11 and solve to get y = 5.
DRILL 2
Answers can be found on this page.
1.If x + 2y = 13 and 5x – y = 21, what is x + y ?
1 

4 

5 

8 

9 
Remember!
For Data Sufficiency
problems in this book, we
do not supply the answer
choices. The five possible
answer choices are the
same every time.
2.If 2a + 3b = 7, what is the value of b ?
(1) a − b = − 4
(2) 4a = 14 − 6b
INEQUALITIES
Inequalities are very similar to equations. Whereas equations use equals signs (=), inequalities use inequality signs (>, <, ≥, or ≤). Those symbols mean “greater than,” “less than,” “greater than or equal to,” and “less than or equal to,” respectively. Solving an inequality is very much like solving an equality. You add, subtract, multiply, and divide to isolate the variable. For example, solve 2x + 4 > 6.
Subtract 4 to get 
2x 
> 2 
Divide by 2 to get 
x 
> 1 
However, there is one critical difference between equalities and inequalities. When you multiply or divide an inequality by a negative number, you must reverse the inequality sign. For example, try −3x + 4 < −2.
Subtract 4 to get 
−3x < −6 
Divide by −3 (and reverse the inequality) to get 
x > 2 
Sometimes a problem provides two inequalities and ask you to combine them in some way. Plug in the highest and lowest values for each variable and list all of the results. You’ll have four numbers if there are two variables. From this list, choose the highest and lowest values and match the answer that includes those as the minimum and maximum values. Look at this next example.
1.If 2 < x < 5 and 3 < y < 10, which of the following expresses the possible range of values for x – y ?
−1 < x – y < 5 

−1 < x – y < 2 

−5 < x – y < 2 

−8 < x – y < 2 

−8 < x – y < −1 
Try all four combinations of the highest and lowest values for x and y. If x = 2 (well, obviously it couldn’t be 2, because 2 < x, but it could be very close, for instance 2.01; here we’ll use 2 as the closest approximation) and y = 3, then x – y = 2 − 3 = −1. If x = 2 and y = 10, then x – y = 2 − 10 = −8. If x = 5 and y = 3, then x – y = 2. Finally, when x = 5 and y = 10, then x – y = −5. The highest value for x – y is 2 and the lowest value is −8. Those are the numbers you want in the answer, so choose (D).
The following diagram shows this process visually. Perform the operation x – y with two vertical pairs and two diagonal pairs.
DRILL 3
Answers can be found on this page.
1.If 5 − 3x > 3, which of the following expresses the possible values of x ?
2.If −2 < p < 3 and −1 < q < 10, which of the following expresses the possible values of p – q ?
−12 < p – q < 4 

−12 < p – q < 7 

−3 < p – q < 4 

−3 < p – q < 7 

−1 < p – q < 4 
EXPONENTS
Exponents are just a shorthand expression for multiplication. For example, 4^{5} simply means 4 × 4 × 4 × 4 × 4. If you ever get confused by exponents, rewriting them as multiplication can demystify the problem. In the expression 4^{5}, the number 4 is the base and 5 is the exponent.
When two exponents have the same base, you can combine them in several ways. Here are the rules for manipulating exponents.
To multiply, add the exponents. 
3^{2} × 3^{5} = 3^{2+5} = 3^{7} 
To divide, subtract the exponents. 
= 7^{5−2}=7^{3} 
To raise to a power, multiply the exponents. 
(5^{2})^{3} = 5^{2×3}=5^{6} 
To find a root, divide the exponent by the root. 
= 5^{9÷3} = 5^{3} 
If you forget these rules, you can also rewrite the exponents as multiplication.
3^{2} × 3^{5} = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 3^{7}
= = 7 × 7 × 7 = 7^{3}
(5^{2})^{3} = (5×5)(5×5)(5×5)=5^{6}
= = 5^{3}
There are a few special cases for exponents that you should know.
Any number to the zero power equals 1. 
x^{0}=1 
Any number to the first power equals itself. 
x^{1}=x 
1 to any power equals 1. 
1^{x} = 1 
Negative Exponents
You may also see a negative exponent. The negative sign simply means the reciprocal of the positive exponent. For example
Also remember:
A Note About
Negative Exponents
People frequently make
the mistake of assuming
that negative exponents
have something to do with
negative numbers. That is
not the case. For example,
5^{−2} = = , not −25.
Exponents of Negative Numbers
A negative number raised to a power can become either positive or negative, depending on whether the exponent is even or odd. A negative number raised to an even exponent becomes positive; for example, (−3)^{2} = (−3)(−3) = 9. A negative number raised to an odd exponent becomes negative; for example, (−3)^{3} = (−3)(−3)(−3) = −27. Just remember that the exponent tells you how many times to multiply the number by itself. With an even exponent, the negative signs all pair up and cancel each other. With an odd exponent, most of the negative signs pair up, but there will be one left over that makes the resulting number negative.
· Negative number to an even exponent = positive
· Negative number to an odd exponent = negative
Scientific Notation
Scientific notation is a way to express very small and very large numbers (such as the number of inches between the Earth and the Sun) in a readable format. To express a number in scientific notation, move the decimal point so that there’s one digit to the left of the decimal point and then multiply by a power of 10. The exponent represents the number of places you moved the decimal point. A positive exponent means the original number was big (more than 1) and a negative exponent means the original number was small (less than 1). For example
1,200,000 = 1.2 × 10^{6}
0.0000567=5.67 × 10^{−5}
DRILL 4
Answers can be found on this page.
1. =
4^{2} 

4^{}+4^{} 

4^{3}+4 

4^{4} 

4^{6} 
2.If x = −3, then =
− 

− 

3.The number of people surfing the Internet doubles every three months. If the number of people surfing the Internet today is 10^{3}, how many people will be surfing the Internet in one year?
4 × 10^{3} 

10^{3} × 2^{4} 

10^{7} 

10^{7} 

10^{7} × 2^{4} 

10^{11} 
ROOTS
A root is like an exponent in reverse. For example, 5^{3} = 125 and = 5. So the square root of x is the number that you would square to get x. Almost all of the roots on the GMAT are square roots (as opposed to cube roots and so forth). Square roots are generally indicated by a radical sign (); other roots will include a small number over the left side of the radical sign that tells you what kind of root it is. For example, the 3 in tells you to find the third root (or cube root).
Adding and subtracting roots is just like adding and subtracting variables. If the roots are the same, you can add and subtract. If they’re different, you can’t add or subtract. Just as you can add x + 4x = 5x, you can add + 4 = 5. Just as you can’t do anything with x + y, you can’t combine + . You can add or subtract only if the roots are the same.
If you’re multiplying or dividing several roots, it’s usually easiest to combine them under one radical sign and then do the calculation. Look at these examples.
2 × = 2 × = 2 = 2 × 4 = 8
= = = 2
Sometimes you need to take a single root and split it up into several roots. Try to factor out perfect squares so that you can simplify them to integers. Look at these examples.
= =
= = × = 4
Sometimes your answer will have a root in the denominator of a fraction. If it does, that answer won’t be in the answer choices. On the GMAT, you can’t have a radical in the denominator of a fraction, so you’ll need to convert it to something that’s acceptable. Just multiply both the numerator and denominator of the fraction by that root. It will get rid of the root in the denominator. You haven’t changed the value of the number, just what it looks like. Look at this example.
= × =
Principal Square Roots
At The Princeton Review, we teach you the math that you need to know to crack the test. We don’t review every facet of theoretical math—just what you must know to crack the GMAT. To that end, let’s review a seeming contradiction in exponents and square roots.
For the purposes of the GMAT, = 3 means that the square root of 9 is 3. Technically, a square root can be positive or negative and the square root that’s positive is called the principal square root. For testtaking purposes, though, we’re always talking about the principal square root, so = 3, never −3. Things on the flip side of that aren’t quite as cutanddry, though.
On the GMAT, you might also see x^{2} = 9, but in this case, unlike what we saw above, x = 3 or −3. Confusing, huh? Just remember: For the purposes of test taking, a radical sign always indicates a positive number, but a variable that is raised to an exponent can be reduced to positive or negative.
Estimating Roots
In some calculations involving square roots, your final answer will have a root in it. In these cases, you may want to know the approximate value of the number, so that you can use Ballparking or other elimination strategies.
You should memorize the easier perfect squares and their roots, such as = 1, = 2, and = 10. Learn the perfect squares at least up to 10^{2} = 100; preferably, you should know them up to 15^{2} = 225 or 20^{2} = 400. You can use these perfect squares as guideposts for estimating the value of other roots. For example, you know that = 4 and = 5. Since is between and , its value should be somewhere between 4 and 5.
Two particular roots that you will see frequently are and . They often crop up in geometry problems, especially those involving right triangles, so you should learn their approximate values: ≈ 1.4 and ≈ 1.7.
Fractional Exponents
It’s possible that you’ll see a fractional exponent on the GMAT. While they appear strange, they’re not really that tough. A fractional exponent is just a way of expressing both an exponent and a root at the same time. Use the numerator as the exponent and the denominator as the root. Here are some examples.
x^{} = or
x^{} = or
x^{} =
This ties in with the rule you learned earlier: To find a root, divide the exponent by the root.
= 5^{} = 5^{4}
DRILL 5
Answers can be found on this page.
1. × +
+ 

2 

3 

4 

3 
Remember!
For Data Sufficiency
problems in this book, we
do not supply the answer
choices. The five possible
answer choices are the
same every time.
2.What is n ?
(1) n =
(2) n^{2} = 25
QUADRATIC EQUATIONS
On the GMAT, you will most likely see some problems that involve quadratic equations and polynomials. The good news is you don’t really have to know what either of those terms mean (or even how to spell them). Just follow a few guidelines to get you through these problems.
One issue that will probably come up is multiplying two sets of parentheses, such as (x + 2)(x + 3). To do this, remember FOIL. It stands for First, Outer, Inner, Last. Those are the pairs of numbers you need to multiply. Check out the diagram below.
Notice that you can usually combine the inner and outer terms into one term.
Sometimes you’ll be given an equation that you need to factor out. Start by looking at the first number. That will tell you the first number in each set of parentheses. Then look at the signs (positive or negative) of the middle number and last number of the original equation. These tell you the signs of the second numbers inside the parentheses. Then find the factors of that last number because those are the potential choices for the second numbers in each set of parentheses. Last, look at the middle number of the original equation to help choose which pair of factors is appropriate. Choose the numbers that add up to this middle number. Be very careful with positive and negative signs.
Suppose you need to factor x^{2} − 4x + 4. From the x^{2}, you can tell that the first numbers in the parentheses are x and x. From the sign of the last number (+4), you can tell that you need two positives or two negatives because the product must be positive. The negative sign of −4x tells you it must be two negatives. The factors of the 4 are −1 and −4 or −2 and −2 (you already know that they’re negative). To add up to −4x, you need −2 and −2. So the equation x^{2} − 4x + 4 factors out to (x − 2)(x − 2).
x^{2} − 4x + 4 = (x )(x )
= (x − )(x − )
= (x − 2)(x − 2)
There are three patterns of quadratic equations that the GMAT writers love to use. You should memorize these and look for them whenever you see a problem with quadratic equations. You’ll be able to avoid the tedious calculations if you recognize the patterns. Here are the three patterns:
(x + y)(x + y) = x^{2} + 2xy + y^{2}
(x – y)(x – y) = x^{2} − 2xy + y^{2}
(x + y)(x – y) = x^{2} – y^{2}
The final thing you need to know about quadratic equations is how to solve one for the value(s) of the variable. Suppose you’re told to find the value of x in the equation x^{2} − 2x = 35. First, move everything to one side of the equation, so that you have a zero on the other side. In the example, you should get x^{2} − 2x − 35 = 0. Next, factor the nonzero side using the method you learned. In the example, you get (x − 7)(x + 5) = 0.
Now for the tricky part. The only way to multiply two things together and get zero is for one of the things to equal zero. So if (x − 7)(x + 5) = 0, then either x − 7 = 0 or x + 5 = 0. Solve both of these miniequations to get x = 7 and x = −5. Note that you will get two answers (unless the two miniequations are the same). You can tell that x is either 7 or −5, but you can’t pin it down to one value. That’s the best you can do with a quadratic equation. The possible values are often referred to as the solutions or roots of the equation, so learn to recognize those terms.
Note that the miniequations are pretty easy to solve. If there is no coefficient in front of the x, simply take the number in each parentheses and reverse the positive/negative sign. Suppose you have x^{2} − 4x + 7 = 4. First, subtract 4 from each side to get x^{2} − 4x + 3 = 0. Then, factor the left side to get (x − 3)(x − 1) = 0. Take the −3 and the −1 and reverse the signs, so x = 3 or x = 1. Note that this method works only if there is no coefficient in front of the x^{2} in the original equation.
DRILL 6
Answers can be found on this page.
1.If = 5, what is the value of x + 5 ?
2 

3 

5 

7 

10 
2.If x^{2} − 5x − 6 = 0, which of the following could be x ?
−2 

−1 

1 

2 

3 
Comprehensive Facing Algebra Drill
Answers can be found on this page.
Remember!
For Data Sufficiency problems in this book, we do not supply the answer choices. The five possible answer choices are the same every time.
1.Which one of the following equations has a root in common with m^{2} + 4m + 4 = 0 ?
m^{2} − 4m + 4 = 0 

m^{2} + 4m + 3 = 0 

m^{2} − 4 = 0 

m^{2} + m − 6 =0 

m^{2} + 5m + 4 = 0 
2.If = 2, then x =
−6 

− 





6 
3.If 3^{n}^{ + 1} = 9^{n}^{ − 1}, then n =
−1 

0 

1 

2 

3 
4.If 7 < m < 11 and −2 < n < 5, then which of the following expresses the possible values of mn ?
−14 < mn < 35 

−14 < mn < 55 

−22 < mn < 35 

−22 < mn < 55 

5 < mn < 16 
5.Which of the following most closely approximates ?
1 

5 

10 

20 

100 
6.If x + 3y = 15 and y – x = 5, then y =
−5 

0 

3 

5 

15 
7.If n = , what is the value of n^{2} + 5n + 6 ?
−3 

−2 

0 

2 

3 
8.If = 1.5 × 10^{3}, then a – b =
1 

2 

3 

4 

5 
9. =
−2 

6 − 2 

2 

9 − 2 

8 
10.If (x − 2)^{2} = 100, which of the following could be the value of x + 2 ?
−10 

−8 

−6 

10 

12 
11.What is the value of z ?
(1) 6z = z^{2} + 9
(2) z^{2} = –z + 12
12.If y^{c} = y^{d}^{ + 1}, what is the value of y ?
(1) y < 1
(2) d = c
13.If 3 is one value of x for the equation x^{2} − 7x + k = −5, where k is a constant, what is the other solution?
2 

4 

5 

6 

12 
14. =
19 

38 

361 
15. =
0.04 

0.25 

0.4 

4.0 

25.0 
16. =
2 

5 

10 

20 

25 
17.What is the value of t ?
(1) 2t − 1 ≥ 5
(2) 3t − 2 ≤ 11
18. Which of the following expresses 3^{4} × 25^{6} × 2^{12} in scientific notation?
1.2 × 10^{12} 

8.1 × 10^{12} 

8.1 × 10^{13} 

1.5 × 10^{22} 

1.2 × 10^{24} 
19.What is the value of x – y ?
(1) x = y + 3
(2) x^{2} − 2xy + y^{2} = 9
Challenge!
Take a crack at this highlevel GMAT question.
20.If is x times 4^{7}, what is the value of x ?
5 

13 

17 
ANSWERS AND EXPLANATIONS
Drill 1
1.ASubtract 12 from both sides to get x = −8. Then multiply both sides by 2 to get x = −16. Choose (A).
2.DFirst, factor out 5 in the top of the fraction. That gives you . Now you can cancel the x + 2 on the top with the one on the bottom to get = 5. Choose (D).
3.EThe variable in the denominator of the fractions is hard to work with, unless you move it to the top. Multiply both sides by x + 1 to get 2 = . Next, multiply both sides by x − 1 to get 2(x − 1) = x + 1 or 2x − 2 = x + 1. You can also get this by crossmultiplying the original two fractions. Subtract x from both sides to get x − 2 = 1. Finally, add 2 to both sides to get x = 3.
4.AAgain, move the variable to the numerator by multiplying by the denominator. Multiply both sides by 1 + to get 1 = (1 + ), which becomes 1 = + . Subtract from both sides to get = . Multiply both sides by 4 to simplify some of the fractions: 1 = . Now multiply both sides by y − 1 to get y − 1 = 3. Add 1 to each side to get y = 4. Note: You might also try PITA on this question.
Drill 2
1.EYou can’t eliminate a variable by adding the equations as they are. So multiply the second equation by 2 to eliminate the y variable. Adding the equations, you get 11x = 55. Divide both sides by 11 to get x = 5. Plug that back into the first equation to get 5 + 2y = 13. Subtract 5 from each side to get 2y = 8. Now divide by 2 to get y = 4. So x + y = 5 + 4 = 9. Choose (E).
2.AStart with Statement (1). This gives you two different equations, so you can solve for the two variables. You can answer the question, so eliminate (B), (C), and (E). Look at Statement (2). This does not give you two different equations. The equation from the statement is the same as the equation from the question. Take the question equation and multiply by 2 to get 4a + 6b = 14. Then subtract 6b to get 4a = 14 − 6b. If you tried to solve for the variables by adding the equations, you would find that everything was eliminated, leaving you with 0 = 0. The key is that you need two different equations. You can’t answer the question, so choose (A).
Drill 3
1.CSubtract 5 from each side of the inequality to get −3x > −2. Then, divide each side by −3 to get x < . Remember: You need to flip the inequality sign when you divide (or multiply) by a negative number. Choose (C).
2.ATo combine the inequalities, calculate p – q using all four numbers. When p = −2 and q = −1, then p – q = −2 – (−1) = −2 + 1 = −1. When p = −2 and q = 10, then p – q = −2 − 10 = −12. When p = 3 and q = −1, then p – q = 3 – (−1) = 3 + 1 = 4. Finally, when p = 3 and q = 10, then p– q = 3 − 10 = −7. The greatest value of p – q is 4 and the least value is −12, so −12 < p – q < 4. Choose (A).
Drill 4
1.CFirst, factor out 4^{2} in the numerator of the fraction. That gives you . Cancel the 4^{2} in the numerator with the one in the denominator. That gives you 4^{3} + 4. Choose (C).
2.DPlug x = −3 into the fraction to get = = = . Don’t forget about the positive/negative rules for multiplication. A negative number to an even exponent will be positive and a negative number to an odd exponent will be negative. Choose (D).
3.BEvery time the number of Internet surfers doubles, you’re just multiplying by 2. It will double four times in a year, so that’s 2 × 2 × 2 × 2, or 2^{4}. You need to multiply that by the original 10^{3} to get 10^{3} × 2^{4}. Choose (B).
Drill 5
1.CWhen you multiply and divide roots, you can combine them under one radical sign. But you can’t do that with adding or subtracting roots. So you can combine × = = . You can also combine = = . Putting this together, you get + . You can factor the perfect square 4 out of 20 to get + = + = × + = 2 + . Since the base is the same, you can add 2 + = 3. Choose (C).
2.AStart with Statement (1). You know that the radical sign refers only to the positive root. So n = 5. You can answer the question. Eliminate (B), (C), and (E). Try Statement (2). In this case, n could be either the positive root (5) or the negative root (−5). You have more than one value, so you can’t answer the question. Choose (A).
Drill 6
1.DFirst, factor the numerator of the fraction. The x^{2} tells that you have x and x for the first numbers inside the parentheses. The signs of the middle (+ 5x) and last numbers (+ 6) tell you that you’re adding in both of the parentheses. The factors of 6 are 1 and 6 or 2 and 3. To add up to 5, you need 2 and 3. So x^{2} + 5x + 6 factors into (x + 2)(x + 3) to give you = 5. You can cancel the x + 2 in the numerator and the denominator of the fraction to get x + 3 = 5, so x = 2. However, the question asks for x + 5. So x + 5 = 2 + 5 = 7. Choose (D).
2.BIf you’re a little nervous about factoring, this is an easy place to Plug In The Answers. Start with (C): −1^{2} − 5(1) − 6 = 0, or −10 = 0. That obviously doesn’t work, so now you need to decide whether to try a bigger or smaller number. It doesn’t take much time to substitute the values into the equation, so if you don’t know whether you need a bigger or smaller number, experiment. Eventually, when you test (B), you’ll get: −1^{2} − 5(−1) − 6 = 0, which can be simplified to 1 + 5 − 6 = 0, or 0 = 0. (B) is the right answer.
Comprehensive Facing Algebra Drill
1.CFind the factors of the equation in the question and then factor the answer choices until you find one that matches. The equation in the question factors to (m + 2)(m + 2) = 0. Answer (A) factors to (m − 2)(m − 2) = 0. Answer (B) factors to (m + 3)(m + 1) = 0. Answer (C) factors to (m + 2)(m − 2) = 0. Answer (D) factors to (m + 3)(m − 2) = 0. Answer (E) factors to (m + 4)(m + 1) = 0. The only one that matches is the (m + 2) from (C).
2.EFirst, multiply both sides by 1 + . This gives you 3 = 2 + . Next, multiply both sides by x to get 3x = 2x + 6. Subtract 2x from both sides to get x = 6. Choose (E).
3.EYou can’t combine exponents unless the bases are the same. So translate the 9 into 3^{2}. That gives you 3^{n}^{+1} = (3^{2})^{n}^{−1}. To raise an exponent to a power, just multiply the exponents. That gives you 3^{n}^{+1} = 3^{2n−2}. So n + 1 = 2n − 2. Add 2 to both sides to get n + 3 = 2n. Subtract n from each side to get 3 = n. Choose (E). Note: You might also try PITA on this question.
4.DPlug in both of the values for m and both of the values for n to get the four possible numbers. They are −14, 35, −22, and 55. The greatest value is 55 and the least value is −22, so −22 < mn < 55. Choose (D).
5.CThe problem says “approximates,” so round off the numbers and Ballpark. The problem becomes = = = 10. Choose (C).
6.DIf you didn’t see the opportunity to solve the simultaneous equations here, you could also have solved by Plugging In The Answers. By plugging in values for y, you can derive a value for x. Then you simply substitute those numbers into the original equation to test them. Start with (C). If y = 3, then 3 − x = 5. So, x would equal −2. Next, plug those values for x and y into the original equation: −2 + 3(3) = 15. Since 7 is obviously not equal to 15, (C) isn’t your answer. If you can’t tell whether you need a bigger or smaller value for y, take a guess and find out. If you choose (D), y = 5, which means that x = 0. Next plug those values into the original equation: 0 + 3(5) = 15.
7.CMultiply both sides of the equation by n + 6 to get n^{2} + 6n = n − 6. Subtract n and add 6 to both sides to get n^{2} + 5n + 6 = 0. Surprisingly enough, that’s the value you need. Choose (C).
8.BSplit the decimals and the powers of 10 into separate fractions to get × = 1.5×10^{3}. When you do the division, you get 15 × = 1.5×10^{3}. When you divide numbers with exponents, just subtract the exponents. So the equation becomes 15 × 10^{a}^{–b} = 1.5 × 10^{3}. Divide both sides by 15 to get 10^{a}^{–b} = 0.1 × 10^{3}, which becomes 10^{a}^{−b} = 10^{2}. So a − b = 2. Choose (B).
9.AUse FOIL to multiply these terms. You get 9 − 3 + 3 − 11 = 9 − 11 = −2. Notice that this is one of the three common patterns for quadratic equations. Choose (A).
10.CIf you take the square root of both sides, you get x − 2 = 10 or x − 2 = −10. Solving both of these equations, you get x = 12 and x = −8. However, the question asks for x + 2, which will be 12 + 2 = 14 or −8 + 2 = −6. The only answer that matches is (C).
11.AIn Statement (1), rearrange the equation to z^{2} − 6z + 9 = 0. You can factor that into (z − 3)(z − 3) = 0. Although many quadratic equations have two solutions, this is one with only one solution, z = 3. The tipoff is that the two factors (the parts in parentheses) are the same. Narrow the choices to (A) and (D). With Statement (2), you can rearrange the equation to z^{2} + z − 12 = 0, which factors into (z + 4)(z − 3) = 0. With this equation, there are two possible solutions, z = 3 and z = −4. Since you can’t determine just one value, Statement (2) is insufficient. Choose (A).
12.CWith Statement (1), you don’t know anything about the values of c and d. If c = d + 1, so that the two exponents are equal, y could be almost anything (limited by y < 1, of course), so Statement (1) is insufficient. Narrow the choices to (B), (C), and (E). With Statement (2), you can substitute c = d into the equation to get y^{c} = y^{c}^{ + 1}. It may be helpful to make up some values for c to get y^{2} = y^{3} or y^{5} = y^{6}. Most potential numbers for y won’t fit those equations, but two will: y = 0 and y = 1. With two possible solutions, Statement (2) is insufficient. Eliminate (B). With both statements, you can find a single value for y. From Statement (2) you know that y is either 0 or 1, and Statement (1) eliminates 1 as a possible value, so y = 0. Choose (C).
13.BThe question tells you that x can equal 3. So plug that into the equation to solve for k. You get 3^{2} − 7(3) + k = −5, which becomes 9 − 21 + k = −5. So −12 + k = −5, or k = 7. Now you can plug the value of k into the original equation to solve for the other possible value of x. You getx^{2} − 7x + 7 = −5. Add 5 to each side to get x^{2} − 7x + 12 = 0. That factors into (x − 4)(x − 3) = 0. Set up the two miniequations x − 4 = 0 and x − 3 = 0. Solving them, you get x = 4 or x = 3. So the two possible values for x are 4 (the new one) and 3 (which you already had). Choose (B).
14.CFactor out a 19 in the numerator of the fraction to get = = 19. Choose (C).
15.EYou can cancel a (0.2)^{2} from the numerator and denominator of the fraction, leaving you with = = 25. Choose (E).
16.BDon’t fall for the Joe Bloggs trap of 25. Add the numbers to get = . Then factor out the perfect square to get = = × = 5. Choose (B).
17.CIf you solve the inequality in Statement (1), you find that t ≥ 3. Since there are many possible values for t, that is insufficient. Narrow the choices to (B), (C), and (E). With Statement (2), you can find that t ≤ 3. Again, there are many possible values for t. Eliminate (B). With both statements, you can find the value of t. Only t = 3 fits both inequalities. Choose (C).
18.CThe best way to solve this problem is to work with the factors, rather than combine them into a single number and convert that to scientific notation. You know that you need some factors of 10, because it’s in all of the answer choices. Split 25^{6} into (5 × 5)^{6} = 5^{6} × 5^{6} = 5^{12}. Next, combine that with the 2^{12}: The result is 5^{12} × 2^{12} = (5 × 2)^{12} = 10^{12}. Since 3^{4} = 81, the overall expression is 81 × 10^{12} or 8.1 × 10^{13}, so the answer is (C).
19.AWith Statement (1), you can subtract y from both sides to get x – y = 3. Even though you don’t know the individual values of x and y, you can answer the question. Narrow the choices to (A) and (D). With Statement (2), you can factor the quadratic term to get (x – y) (x – y) = 9 or (x– y)^{2} = 9. However, if you take the square root of both sides, you get two possible solutions: x – y = 3 and x – y = −3. That’s not sufficient, so choose (A).
20.EFirst, factor the expression in the numerator. You can take the common factor, 4^{7}, out of the expression, leaving . Next, simplify the expression in the parenthesis to 1 + 4 + 16 + 64 = 85. So, now the expression reads , which reduces to 4^{7}(17). So, x must equal 17. Choose (E).