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

Introduction to Part 4

Part 4 consists of five chapters. Chapter 10 presents several important strategies that can be used on any mathematics questions that appear on the GRE. In Chapters 11, 12, and 13 you will find tactics that are specific to one of the three different types of questions: discrete quantitative questions, quantitative comparison questions, and data interpretation questions, respectively. Chapter 14 contains a complete review of all the mathematics you need to know in order to do well on the GRE, as well as hundreds of sample problems patterned on actual test questions.

FIVE TYPES OF TACTICS

Five different types of tactics are discussed in this book.

1. In Chapters 1 and 2, you learned many basic tactics used by all good test-takers, such as read each question carefully, pace yourself, don’t get bogged down on any one question, and never waste time reading the directions. You also learned the specific tactics required to excel on a computer-adaptive test. These tactics apply to both the verbal and quantitative sections of the GRE.

2. In Chapters 4, 5, 6, and 7 you learned the important tactics needed for handling each of the four types of verbal questions.

3. In Chapter 9 you learned the strategies for planning and writing the two essays that constitute the analytical writing section of the GRE.

4. In Chapters 10–13 you will find all of the tactics that apply to the quantitative sections of the GRE. Chapter 10 contains those techniques that can be applied to all types of mathematics questions; Chapters 11, 12, and 13 present specific strategies to deal with each of the three kinds of quantitative questions found on the GRE: discrete quantitative questions, quantitative comparison questions, and data interpretation questions.

5. In Chapter 14 you will learn or review all of the mathematics that is needed for the GRE, and you will master the specific tactics and key facts that apply to each of the different mathematical topics.

Using these tactics will enable you to answer more quickly many questions that you already know how to do. But the greatest value of these tactics is that they will allow you to correctly answer or make educated guesses on problems that you do not know how to do.

AN IMPORTANT SYMBOL

Throughout the rest of this book, the symbol “⇒” is used to indicate that one step in the solution of a problem follows immediately from the preceding one, and no explanation is necessary. You should read

		3x = 12 ⇒ x = 4
	as	3x = 12 implies that x = 4
	or	3x = 12, which implies that x = 4
	or	since 3x = 12, then x = 4.

Here is a sample solution to the following problem using ⇒:

What is the value of 2x² – 5 when x = –4?

x = –4 ⇒ x² = (–4)² = 16 ⇒ 2x² = 2(16) = 32 ⇒ 2x² – 5 = 32 – 5 = 27

When the reason for a step is not obvious, ⇒ is not used: rather, an explanation is given, often including a reference to a KEY FACT from Chapter 14. In many solutions, some steps are explained, while others are linked by the ⇒symbol, as in the following example.

In the diagram below, if w = 10, what is the value of z?

·               By KEY FACT J1, w + x + y = 180.

·               Since ΔABC is isosceles, x = y (KEY FACT J5).

·               Therefore, w + 2y = 180 ⇒ 10 + 2y = 180 ⇒ 2y = 170 ⇒ y = 85.

·               Finally, since y + z = 180 (KEY FACT I3), 85 + z = 180 ⇒ z = 95.

Chapter 10. General Math Strategies

In Chapters 11 and 12, you will learn tactics that are specifically applicable to multiple-choice questions and quantitative comparison questions, respectively. In this chapter you will learn several important general math strategies that can be used on both of these types of questions.

The directions that appear on the screen at the beginning of the quantitative section include the following cautionary information:

Figures that accompany questions are intended to provide information useful in answering the questions.

However, unless a note states that a figure is drawn to scale, you should solve these problems NOT by estimating sizes by sight or measurement, but by using your knowledge of mathematics.

Despite the fact that they are telling you that you cannot totally rely on their diagrams, if you learn how to draw diagrams accurately, you can trust the ones you draw. Knowing the best ways of handling diagrams on the GRE is critically important. Consequently, the first five tactics all deal with diagrams.

	TACTIC 1.	Draw a diagram.
	TACTIC 2.	Trust a diagram that has been drawn to scale.
	TACTIC 3.	Exaggerate or change a diagram.
	TACTIC 4.	Add a line to a diagram.
	TACTIC 5.	Subtract to find shaded regions.

To implement these tactics, you need to be able to draw line segments and angles accurately, and you need to be able to look at segments and angles and accurately estimate their measures. Let’s look at three variations of the same problem.

1. If the diagonal of a rectangle is twice as long as the shorter side, what is the degree measure of the angle it makes with the longer side?

2. In the rectangle below, what is the value of x?

3. In the rectangle below, what is the value of x?

For the moment, let’s ignore the correct mathematical way of solving this problem. In the diagram in (3), the side labeled 2 appears to be half as long as the diagonal, which is labeled 4; consequently, you should assume that the diagram has been drawn to scale, and you should see that x is about 30, certainly between 25 and 35. In (1) you aren’t given a diagram, and in (2) the diagram is useless because you can see that it has not been drawn to scale (the side labeled 2 is nearly as long as the diagonal, which is labeled 4). However, if while taking the GRE, you see a question such as (1) or (2), you should be able to quickly draw on your scrap paper a diagram that looks just like the one in (3), and then look at your diagram and see that the measure of x is just about 30. If the answer choices for these questions were

(A) 15  (B) 30  (C) 45  (D) 60  (E) 75

you would, of course, choose 30, B. If the choices were

(A) 20  (B) 25  (C) 30  (D) 35  (E) 40

you might not be quite as confident, but you should still choose 30, here C.

When you take the GRE, even though you are not allowed to have rulers or protractors, you should be able to draw your diagrams very accurately. For example, in (1) above, you should draw a horizontal line, and then, either freehand or by tracing the corner of a piece of scrap paper, draw a right angle on the line. The vertical line segment will be the width of the rectangle; label it 2.

Mark off that distance twice on a piece of scrap paper and use that to draw the diagonal.

You should now have a diagram that is similar to that in (3), and you should be able to see that x is about 30.

By the way, x is exactly 30. A right triangle in which one leg is half the hypotenuse must be a 30-60-90 triangle, and that leg is opposite the 30° angle [see KEY FACT J11].

Having drawn an accurate diagram, are you still unsure as to how you should know that the value of x is 30 just by looking at the diagram? You will now learn not only how to look at any angle and know its measure within 5 or 10 degrees, but how to draw any angle that accurately.

You should easily recognize a 90° angle and can probably draw one freehand; but you can always just trace the corner of a piece of scrap paper. To draw a 45° angle, just bisect a 90° angle. Again, you can probably do this freehand. If not, or to be even more accurate, draw a right angle, mark off the same distance on each side, draw a square, and then draw in the diagonal.

To draw other acute angles, just divide the two 45° angles in the above diagram with as many lines as necessary.

Finally, to draw an obtuse angle, add an acute angle to a right angle.

Now, to estimate the measure of a given angle, just draw in some lines.

To test yourself, find the measure of each angle shown. The answers are found below.

Answers (a) 80° (b) 20° (c) 115° (d) 160°. Did you come within 10° on each one?

Testing Tactics

 Draw a Diagram

On any geometry question for which a figure is not provided, draw one (as accurately as possible) on your scrap paper — never attempt a geometry problem without first drawing a diagram.

What is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?

(A) 1 (B) 6 (C)  (D)  (E) 

SOLUTION. Don’t even think of answering this question until you have drawn a square and a rectangle and labeled each of them: each side of the square is 1, and if the width of the rectangle is w, its length () is 2w.

Now, write the required equation and solve it:

The area of the rectangle = w = , E.

Betty drove 8 miles west, 6 miles north, 3 miles east, and 6 more miles north. How many miles was Betty from her starting place?

(A) 13  (B) 17  (C) 19  (D) 21  (E) 23

SOLUTION. Draw a diagram. Now, extend line segment ED until it intersects AB at F. Then, AFE is a right triangle, whose legs are 5 and 12 and, therefore, whose hypotenuse is 13, A.

What is the difference in the degree measures of the angles formed by the hour hand and the minute hand of a clock at 12:35 and 12:36?

(A) 1°  (B) 5°  (C) 5.5°  (D) 6°  (E) 30°

SOLUTION. Draw a simple picture of a clock. The hour hand makes a complete revolution, 360°, once every 12 hours. So, in 1 hour it goes through 360° ÷ 12 = 30°, and in one minute it advances through 30° ÷ 60 = 0.5°. The minute hand moves through 30° every 5 minutes or 6° per minute. So, in the minute from 12:35 to 12:36 (or any other minute), the difference between the hands increased by 6° – 0.5° = 5.5°, C.

NOTE: It was not necessary, and would have been more time-consuming, to determine the angle between the hands at either 12:35 or 12:36. (See TACTIC 6: Don’t do more than you have to.)

Drawings should not be limited to geometry questions; there are many other questions on which drawings will help.

A jar contains 10 red marbles and 30 green ones. How many red marbles must be added to the jar so that 60% of the marbles will be red?

(A) 25  (B) 30  (C) 35  (D) 40  (E) 60

SOLUTION. First, draw a diagram and label it. From the diagram it is clear that there are now 40 + x marbles in the jar, of which 10 + x are red. Since we want the fraction of red marbles to be 60%, we have .

Cross-multiplying, we get:

50 + 5x = 120 + 3x ⇒ 2x = 70 ⇒ x = 35, C.

Of course, you could have set up the equation and solved it without the diagram, but the diagram makes the solution easier and you are less likely to make a careless mistake.

 Trust a Diagram That Has Been Drawn to Scale

Whenever diagrams have been drawn to scale, they can be trusted. This means that you can look at the diagram and use your eyes to accurately estimate the sizes of angles and line segments. For example, in the first problem discussed at the beginning of this chapter, you could “see” that the measure of the angle was about 30°.

To take advantage of this situation:

·               If a diagram is given that appears to be drawn to scale, trust it.

·               If a diagram is given that has not been drawn to scale, try to draw it to scale on your scrap paper, and then trust it.

·               When no diagram is provided, and you draw one on your scrap paper, try to draw it to scale.

In Example 5 (on page 311), we are told that ABCD is a square and that diagonal BD is 3. In the diagram provided, quadrilateral ABCD does indeed look like a square, and BD = 3 does not contradict any other information. We can, therefore, assume that the diagram has been drawn to scale.

In the figure at the right, diagonal BD of square ABCD is 3. What is the perimeter of the square?

(A) 4.5  (B) 12  (C) 3  (D) 6  (E) 12

SOLUTION. Since this diagram has been drawn to scale, you can trust it. The sides of the square appear to be about two thirds as long as the diagonal, so assume that each side is 2. Then the perimeter is 8. Which of the choices is approximately 8? Certainly not A or B. Since  ≈ 1.4, Choices C, D, and E are approximately 4.2, 8.4, and 12.6, respectively. Clearly, the answer must be D.

Direct mathematical solution. Let s be a side of the square. Then since ΔBCD is a 45-45-90 right triangle, s = , and the perimeter of the square is 4s = .

Remember the goal of this book is to help you get credit for all the problems you know how to do, and, by using the TACTICS, to get credit for many that you don’t know how to do. Example 5 is typical. Many students would miss this question. You, however, can now answer it correctly, even though you may not remember how to solve it directly.

In ΔABC, what is the value of x?

(A) 75  (B) 60  (C) 45  (D) 30  (E) 15

SOLUTION. If you don’t see the correct mathematical solution, you should use TACTIC 2 and trust the diagram; but to do that you must be careful that when you copy it onto your scrap paper you fix it. What’s wrong with the way it is drawn now? AB = 8 and BC = 4, but in the figure, AB and BC are almost the same length. Redraw it so that AB is twice as long as BC. Now, just look: x is about 60, B.

In fact, x is exactly 60. If the hypotenuse of a right triangle is twice the length of one of the legs, then it’s a 30-60-90 triangle, and the angle formed by the hypotenuse and that leg is 60° (see Section 14-J).

TACTIC 2 is equally effective on quantitative comparison questions that have diagrams. See pages 12–14 for directions on how to solve quantitative comparison questions.

SOLUTION. There are two things wrong with the given diagram: ∠C is labeled 40°, but looks much more like 60° or 70°, and AC and BC are each labeled 10, but BC is drawn much longer. When you copy the diagram onto your scrap paper, be sure to correct these two mistakes: draw a triangle that has a 40° angle and two sides of the same length.

Now, it’s clear: AB < 10. The answer is B.

SOLUTION. In the diagram on page 312, the value of x is at least 60, so if the diagram has been drawn to scale, the answer must be A. If, on the other hand, the diagram has not been drawn to scale, we can’t trust it. Which is it? The diagram is not OK — PQ is drawn almost as long as OR, even though OR is twice as long. Correct the diagram:

Now you can see that x is less than 45. The answer is B.

 Exaggerate or Otherwise Change a Diagram

Sometimes it is appropriate to take a diagram that appears to be drawn to scale and intentionally exaggerate it. Why would we do this? Consider the following example.

SOLUTION. In the diagram, which appears to be drawn correctly, AB and CD look as though they are the same length. However, there might be an imperceptible difference due to the fact that angle C is slightly smaller than angle A. So exaggerate the diagram: redraw it, making angle C much smaller than angle A. Now, it’s clear: CD is longer. The answer is B.

When you copy a diagram onto your scrap paper, you can change anything you like as long as your diagram is consistent with all the given data.

SOLUTION. You may redraw this diagram any way you like, as long as the two angles that are marked 45° remain 45°. If PQ and PR are equal, as they appear to be in the given diagram, then x would equal y. Since the given information doesn’t state that PQ = PR, draw a diagram in which PR and QR are clearly unequal. In the diagram below, PR is much longer than PQ, and x and y are clearly unequal. The answer is D.

 Add a Line to a Diagram

Occasionally, after staring at a diagram, you still have no idea how to solve the problem to which it applies. It looks as though not enough information has been given. When this happens, it often helps to draw another line in the diagram.

In the figure below, Q is a point on the circle whose center is O and whose radius is r, and OPQR is a rectangle. What is the length of diagonal PR?

(A) r  (B) r²  (C)   (D) 

(E) It cannot be determined from the information given.

SOLUTION. If after staring at the diagram and thinking about rectangles, circles, and the Pythagorean theorem, you’re still lost, don’t give up. Ask yourself, “Can I add another line to this diagram?” As soon as you think to draw in OQ, the other diagonal, the problem becomes easy: the two diagonals are equal and, since OQ is a radius, it is equal to r, A.

What is the area of quadrilateral ABCD?

(A) 24  (B) 28  (C) 30  (D) 38  (E) 60

SOLUTION. Since the quadrilateral is irregular, there isn’t a formula to find the area. However, if you draw in AC, you will divide ABCD into two triangles, each of whose areas can be determined. If you then draw in the height of each triangle, you see that the area of ΔACD is (4)(4) = 8, and the area of ΔBAC is (6)(10) = 30, so the area of ABCD is 30 + 8 = 38, D.

Note that this problem could also have been solved by drawing in lines to create rectangle ABEF, and subtracting the areas of ΔBEC and ΔCFD from the area of the rectangle.

 Subtract to Find Shaded Regions

Whenever part of a figure is shaded, the straightforward way to find the area of the shaded portion is to find the area of the entire figure and subtract from it the area of the unshaded region. Of course, if you are asked for the area of the unshaded region, you can, instead, subtract the shaded area from the total area. Occasionally, you may see an easy way to calculate the shaded area directly, but usually you should subtract.

In the figure below, ABCD is a rectangle, and BE and CF are arcs of circles centered at A and D. What is the area of the striped region?

(A) 10 – π  (B) 2(5 – π)  (C) 2(5 – 2π)  (D) 6 + 2π  (E) 5(2 – π)

SOLUTION. The entire region is a 2 × 5 rectangle whose area is 10. Since the white region consists of two quarter-circles of radius 2, the total white area is that of a semicircle of radius 2: π(2)² = 2π. Therefore, the area of the striped region is 10 – 2π = 2(5 – π), B.

In the figure below, square ABCD is inscribed in circle O. If the perimeter of ABCD is 24, what is the area of the shaded region?

(A) 18π – 36  (B) 18π – 24  (C) 12π – 36  (D) 9π – 36  (E) 9π – 24

SOLUTION. Since the perimeter of square ABCD is 24, each of its sides is 6, and its area is 6² = 36. Since diagonal AC is the hypotenuse of isosceles right triangle ABC, AC = 6. But AC is also a diameter of circle O, so the radius of the circle is 3, and its area is π(3)² = 18π. Finally, the area of the shaded region is 18π – 36, A.

 Don’t Do More Than You Have To

Very often a problem can be solved in more than one way. You should always try to do it in the easiest way possible. Consider the following examples.

If 5(3x – 7) = 20, what is 3x – 8?

(A)   (B) 0  (C) 3  (D) 14  (E) 19

It is not difficult to solve for x:

5(3x – 7) = 20 ⇒ 15x – 35 = 20 ⇒ 15x = 55 ⇒ x = .

But it’s too much work. Besides, once you find that x = , you still have to multiply to get 3x: 3 = 11, and then subtract to get 3x – 8: 11 – 8 = 3.

SOLUTION. The key is to recognize that you don’t need to find x. Finding 3x – 7 is easy (just divide the original equation by 5), and 3x – 8 is just 1 less:

5(3x – 7) = 20 ⇒ 3x – 7 = 4 ⇒ 3x – 8 = 3, C.

If 7x + 3y = 17 and 3x + 7y = 19, what is the average (arithmetic mean) of x and y?

(A)   (B)   (C) 1.8  (D) 3.6  (E) 36

The obvious way to do this is to first find x and y by solving the two equations simultaneously and then to take their average. If you know how to do this, try it now, before reading further. If you worked carefully, you should have found that x =  and y = , and their average is . This is not too difficult, but it is quite time-consuming, and questions on the GRE never require you to do that much work. Look for a shortcut. Is there a way to find the average without first finding x and y? Absolutely! Here’s the best way to do this.

Do not spend any time calculating how many minutes either of them worked. You only need to know which column is greater, and since Jeremy started earlier and finished later, he clearly worked longer. The answer is B.

 Pay Attention to Units

Often the answer to a question must be in units different from the data given in the question. As you read the question, write on your scratch paper exactly what you are being asked and circle it or put an asterisk next to it. Do they want hours or minutes or seconds, dollars or cents, feet or inches, meters or centimeters? On multiple-choice questions, an answer using the wrong units is almost always one of the choices.

Driving at 48 miles per hour, how many minutes will it take to drive 32 miles?

(A)   (B)   (C) 40  (D) 45  (E) 2400

SOLUTION. This is a relatively easy question. Just be attentive. Divide the distance, 32, by the rate, 48: , so it will take  of an hour to drive 32 miles. Choice A is , but that is not the correct answer, because you are asked how many minutes it will take. To convert hours to minutes, multiply by 60: it will take (60) = 40 minutes, C.

Note that you could have been asked how many seconds it would take, in which case the answer would be 40(60) = 2400, Choice E.

At Nat’s Nuts a 2-pound bag of pistachio nuts costs $6.00. At this rate, what is the cost in cents of a bag weighing 9 ounces?

(A) 1.5  (B) 24  (C) 150  (D) 1350  (E) 2400

SOLUTION. This is a relatively simple ratio, but make sure you get the units right. To do this you need to know that there are 100 cents in a dollar and 16 ounces in a pound.

Now cross-multiply and solve: 36x = 5400 ⇒ x = 150, C.

 Systematically Make Lists

When a question asks “how many,” often the best strategy is to make a list of all the possibilities. If you do this it is important that you make the list in a systematic fashion so that you don’t inadvertently leave something out. Usually, this means listing the possibilities in numerical or alphabetical order. Often, shortly after starting the list, you can see a pattern developing and you can figure out how many more entries there will be without writing them all down. Even if the question does not specifically ask “how many,” you may need to count something to answer it; in this case, as well, the best plan may be to write out a list.

A palindrome is a number, such as 93539, that reads the same forward and backward. How many palindromes are there between 100 and 1,000?

(A) 10  (B) 81  (C) 90  (D) 100  (E) 200

SOLUTION. First, write down the numbers that begin and end in 1:

101, 111, 121, 131, 141, 151, 161, 171, 181, 191

Next write the numbers that begin and end in a 2:

202, 212, 222, 232, 242, 252, 262, 272, 282, 292

By now you should see the pattern: there are 10 numbers beginning with 1, 10 beginning with 2, and there will be 10 beginning with 3, 4, …, 9 for a total of 9 × 10 = 90 palindromes, C.

The product of three positive integers is 300. If one of them is 5, what is the least possible value of the sum of the other two?

(A) 16  (B) 17  (C) 19  (D) 23  (E) 32

SOLUTION. Since one of the integers is 5, the product of the other two is 60. Systematically, list all possible pairs, (a, b), of positive integers whose product is 60 and check their sums. First let a = 1, then 2, and so on.

a	b	a + b
1	60	61
2	30	32
3	20	23
4	15	19
5	12	17
6	10	16

The answer is 16, A.

Practice Exercises

General Math Strategies

1. At Leo’s Lumberyard, an 8-foot long wooden pole costs $3.00. At this rate, what is the cost, in cents, of a pole that is 16 inches long?
(A) 0.5  (B) 48  (C) 50  (D) 64  (E) 96

2. In the figure below, vertex Q of square OPQR is on a circle with center O. If the area of the square is 8, what is the area of the circle?

(A) 8π
(B) 8π
(C) 16π
(D) 32π
(E) 64π

3. In 1999, Diana read 10 English books and 7 French books. In 2000, she read twice as many French books as English books. If 60% of the books that she read during the two years were French, how many books did she read in 2000?
(A) 16  (B) 26  (C) 32  (D) 39  (E) 48

4. In the figure below, if the radius of circle O is 10, what is the length of diagonal AC of rectangle OABC?

5. In writing all of the integers from 1 to 300, how many times is the digit 1 used?
(A) 60
(B) 120
(C) 150
(D) 160
(E) 180

6. In the figure below, ABCD is a square and AED is an equilateral triangle. If AB = 2, what is the area of the shaded region?

(A) 
(B) 2
(C) 3
(D) 4 – 2
(E) 4 – 

7. If 5x + 13 = 31, what is the value of 
(A) 
(B) 
(C) 7
(D) 13
(E) 169

8. If a + 2b = 14 and 5a + 4b = 16, what is the average (arithmetic mean) of a and b?
(A) 1.5  (B) 2  (C) 2.5  (D) 3  (E) 3.5

9. In the figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?

(A) 1
(B) 
(C) 2
(D) 2
(E) 4

10. In the figure below, three circles of radius 1 are tangent to one another. What is the area of the shaded region between them?

(A) 
(B) 1.5
(C) π – 
(D) 
(E) 2 – 

11. 

12. 

13. 

Questions 14–15 refer to the following definition.

{a, b} represents the remainder when a is divided by b.

14. 

c and d are positive integers with c < d.

15. 

Answer Key

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

Answer Explanations

Two asterisks (**) indicate an alternative method of solving.

1. (C) This is a relatively simple ratio problem, but use TACTIC 7 and make sure you get the units right. To do this you need to know that there are 100 cents in a dollar and 12 inches in a foot.

Now cross-multiply and solve:

96x = 4800 ⇒ x = 50.

2. (C) Use TACTICS 2 and 4. On your scrap paper, extend line segments OP and OR.

Square OPQR, whose area is 8, takes up most of quarter-circle OXY. So the area of the quarter-circle is certainly between 11 and 13. The area of the whole circle is 4 times as great: between 44 and 52. Check the five choices: they are approximately 25, 36, 50, 100, 200. The answer is clearly C.
**Another way to use TACTIC 4 is to draw in line segment OQ.

Since the area of the square is 8, each side is , and diagonal OQ is  ×  =  = 4. But OQ is also a radius, so the area of the circle is 16π.

3. (E) Use TACTIC 1: draw a picture representing a pile of books or a bookshelf.

In the two years the number of French books Diana read was 7 + 2x and the total number of books was 17 + 3x. Then 60% or . To solve, cross-multiply:

35 + 10x = 51 + 9x ⇒ x = 16.

In 2000, Diana read 16 English books and 32 French books, a total of 48 books.

4. (D) Use TACTIC 2. Trust the diagram: AC, which is clearly longer than OC, is approximately as long as radius OE.

Therefore, AC must be about 10. Check the choices. They are approximately 1.4, 3.1, 7, 10, and 14. The answer must be 10.
**The answer is 10. Use TACTIC 4: copy the diagram on your scrap paper and draw in diagonal OB.

Since the two diagonals of a rectangle are equal, and diagonal OB is a radius, OA = OB = 10.

5. (D) Use TACTIC 8. Systematically list the numbers that contain the digit 1, writing as many as you need to see the pattern. Between 1 and 99 the digit 1 is used 10 times as the units digit (1, 11, 21, …, 91) and 10 times as the tens digit (10, 11, 12, …, 19) for a total of 20 times. From 200 to 299, there are 20 more (the same 20 preceded by a 2). From 100 to 199 there are 20 more plus 100 numbers where the digit 1 is used in the hundreds place. So the total is 20 + 20 + 20 + 100 = 160.

6. (E) Use TACTIC 5: subtract to find the shaded area. The area of the square is 4. The area of the equilateral triangle (see Section 14-J) is . So the area of the shaded region is 4 – .

7. (C) Use TACTIC 6: don’t do more than you have to. In particular, don’t solve for x.

5x + 13 = 31 ⇒ 5x = 18 ⇒ 5x + 31 = 18 + 31 = 49 ⇒  = 7.

8. (C) Use TACTIC 6: don’t do more than is necessary. We don’t need to know the values of a and b, only their average. Adding the two equations, we get

6a + 6b = 30 ⇒ a + b = 5 ⇒  = 2.5.

9. (C) Use TACTIC 5: to get DE, subtract OD from radius OE, which is 4. Draw AO (TACTIC 4). Since ΔAOD is a 30-60-90 right triangle, OD is 2 (one half of OA). So, DE = 4 – 2 = 2.

10. (D) Use TACTIC 4 and add some lines: connect the centers of the three circles to form an equilateral triangle whose sides are 2.

Now use TACTIC 5 and find the shaded area by subtracting the area of the three sectors from the area of the triangle. The area of the triangle is  (see Section 14-J). Each sector is one sixth of a circle of radius 1. Together they form one half of such a circle, so their total area is π(1)² = . Finally, subtract: the shaded area is .

11.(B) If you don’t see how to do this, use TACTIC 2: trust the diagram. Estimate the measure of each angle: for example, a = 45, b = 70, c = 30, and d = 120. So c + d (150) is considerably greater than a + b (115). Choose B.

**In fact, d by itself is equal to a + b (an exterior angle of a triangle is equal to the sum of the opposite two interior angles). So c + d > a + b.

12. (B) From the figure, it appears that x and y are equal, or nearly so. However, the given information states that BC > CD, but this is not clear from the diagram. Use TACTIC 3: when you draw the figure on your scrap paper, exaggerate it. Draw it with BC much greater than CD. Now it is clear that y is greater.

**Since BC > CD, central angle 1 is greater than central angle 2, which means that x < y.

13. (C) Use TACTIC 8. Systematically list all the factors of 30, either individually or in pairs: 1, 30; 2, 15; 3, 10; 5, 6. Of the 8 factors, 4 are even and 4 are odd.

14. (A) Column A: When 10³ (1000) is divided by 3, the quotient is 333 and the remainder is 1. Column B: 10⁵ is divisible by 5, so the remainder is 0.
    Column A is greater.

15. (A) Column A: since c < d, the quotient when c is divided by d is 0, and the remainder is c. Column B: when d is divided by c the remainder must be less than c.
    So Column A is greater.