Barron's SAT, 26th edition (2012)
Part 3. TACTICS AND PRACTICE: CRITICAL READING
Chapter 8. Math Strategies and Tactics
Most of the questions in the mathematics sections of the SAT are multiple-choice questions. Immediately preceding each set of multiple-choice questions you will see the following directions:
In this section solve each problem, using any available space on the page for scratchwork. Then decide which is the best of the choices given and fill in the corresponding oval on the answer sheet. (Emphasis added.)
The directions are very simple. Basically, they tell you to ignore, at first, the fact that these are multiple-choice questions. Just solve each problem, and then look at the five choices to see which one is best. As you will learn in this chapter, however, that is not always the best strategy.
If you know how to solve a problem and are confident that you can do so accurately and reasonably quickly, JUST DO IT!
In this chapter you will learn all of the important strategies you need to help you answer these multiple-choice questions. As a bonus, almost every one of these tactics can also be used on the grid-in questions. However, as invaluable as these tactics are, use them only when you need them.
The first four tactics deal with the best ways of handling diagrams.
Draw a diagram.
If a diagram has been drawn to scale, trust it.
If a diagram has not been drawn to scale, redraw it.
Add a line to a diagram.
To implement these tactics, you need to be able to draw line segments and angles accurately, and also to be able to look at segments and angles and accurately estimate their measures. Let’s look at three variations of the same problem.
a. If the diagonal of a rectangle is twice as long as the shorter side, what is the degree measure of the angle the diagonal makes with the longer side?
b. In the rectangle below, what is the value of x?
c. In the rectangle below, what is the value of x?
Note: Figure not drawn to scale
For the moment, let’s ignore the correct mathematical way to solve this problem. You should be able to look at the diagram in (b) and “see” that x is about 30, certainly between 25 and 35. In (a), however, you aren’t given a diagram, and in (c) the diagram is useless because it hasn’t been drawn to scale. In each of these cases, you should be able to draw a diagram that looks just like the one in (b); then you can look at your diagram and “see” that the measure of the angle in question is about 30°.
If this were a multiple-choice question, and the choices were as follows:
(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 would be a little less confident, but you should still choose 30, here (C).
If this were a grid-in problem, you would be much less certain of your answer, but should surely bubble in 30, rather than guess a “strange” number such as 28 or 31.
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].
But how can you know the value of x just by looking at the diagram in (b)? In this section, you will learn not only how to look at any angle and know its measure within 5 or 10°, but also how to draw any angle with the same accuracy. You will also learn how to draw line segments of the correct lengths, so that your diagrams won’t be as bad as the one in (c). Do you see what is wrong with that diagram? The diagonal is labeled 4 and one of the sides is labeled 2, but the diagonal, as drawn, isn’t nearly twice as long as the side.
Consider the following example:
In the figure above, what is the value of d?
Solution. Since there is no note indicating that the diagram has not been drawn to scale, you can trust it [see TACTIC 2].
• Clearly, d is less than AC, which is 6; but all five choices are less than 6, so that doesn’t help.
• Actually, it looks as though d is less than half of AC, or 3.
• Assume it is, and eliminate choices C, D, and E.
You could now guess between choices A and B; but if you measure, you’ll know which is right. However, there’s a problem—on the SAT, you are not allowed to use a ruler, a compass, or a protractor. Then how can you measure anything? Use the back of your answer sheet! Here are two ways to do this, with the procedures illustrated below.
1. Turn your answer sheet over, place one corner of it on point A, and with your pencil make a small mark to indicate length d. Now use this “ruler” to measure AC. Put a dot on AC d units from A; slide the answer sheet, mark off a second segment of length d, and do this once more. The third mark is well past C, so 3d is more than 6; that is, d > 2. Eliminate A.
2. On the back of your answer sheet, measure AC and BC from the same point. The distance between them is 2. Compare 2 to d; d is longer. Eliminate A.
• The answer is 2.5 (B).
Finally, erase the dot or mark you made on the back of your answer sheet, so it won’t confuse you if you need to make a new “ruler” for another question. Also, there should be no stray pencil marks anywhere on the answer sheet when you hand it in.
To answer this question without TACTIC 2, use the Pythagorean theorem to obtain AB = 10 (or recognize that this is a 6-8-10 right triangle) and then solve the equation: d + 3d = 10 d = 2.5.
To take full advantage of TACTICS 1, 2, and 3, you need to be able to measure angles as well as line segments. Fortunately, this is very easy. In fact, you should be able to look at any angle and know its measure within 5–10°, and be able to draw any angle accurately within 10°. Let’s see how.
First, you should easily recognize a 90° angle and can probably draw one freehand, or you can always just trace the corner of your answer sheet.
Second, to draw a 45° angle, just bisect a 90° angle. Again, you can probably do this freehand. If not, or to be more accurate, draw a right angle, mark off the same distance on each side, draw a square, and then draw in the diagonal.
Third, to draw other acute angles, just divide the two 45° angles in the above diagram with as many lines as are 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 below. The answers follow.
Draw a Diagram.
On any geometry question for which a figure is not provided, draw one (as accurately as possible) in your test booklet. Drawings should not be limited, however, to geometry questions; there are many other questions on which drawings will help. Whether you intend to solve a problem directly or to use one of the tactics described in this chapter, drawing a diagram is the first step.
A good drawing requires no artistic ability. Usually, a few line segments are sufficient.
Let’s consider some examples.
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?
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
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?
Solution. 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 you want the fraction of red marbles to be , you have
5(10 + x) = 3(40 + x)
50 + 5x = 120 + 3x 2x = 70 x = 35.
Of course, you could have set up the equation and solved it without the diagram, but the drawing makes the solution easier and you are less likely to make a careless mistake.
The diagonal of square II is equal to the perimeter of square I. The area of square II is how many times the area of square I?
It is certainly possible to answer this question without drawing a diagram, but don’t. Get in the habit of always drawing a diagram for a geometry problem. Often a good drawing will lead you to the correct solution; other times, as you will see here, it prevents you from making a careless error or it allows you to get the right answer even if you don’t know how to solve the problem.
Solution. Draw a small square (square I), and next to it mark off a line segment equal in length to the perimeter of the square (4 times the side of the square). Then draw a second square (square II) whose diagonal is equal to the length of the line segment.
You can see how much larger square II is. In fact, if you draw four squares the size of square I inside square II, you can see that the answer to this question is certainly much more than 4, so eliminate choices A, B, and C. Then, if you don’t know how to proceed, just guess between D and E. In fact, 4 ≈ 5.6, and even that is too small, so you should choose 8 (E), the correct answer.
Mathematical Solution. If the side of square I is 1, its perimeter is 4. Then the diagonal of square II is 4. Now use the formula (KEY FACT K8) to find that the area of square II is , whereas the area of square I is 1.
Tony drove 8 miles west, 6 miles north, 3 miles east, and 6 more miles north. How far was Tony from his starting place?
Solution. Draw a diagram. Now, extend line segment ED until it intersects AB at F [see TACTIC 4]. Then, AFE is a right triangle whose legs are 5 and 12 and, therefore, whose hypotenuse is 13 (A).
[If you drew the diagram accurately, you could get the right answer by measuring!]
By how many degrees does the angle formed by the hour hand and the minute hand of a clock increase from 1:27 to 1:28?
Solution. Draw a simple picture of a clock. The hour hand makes a complete revolution, 360°, once every 12 hours. Therefore, in 1 hour it goes through 360°÷ 12 = 30°, and in 1 minute it advances through 30°÷ 60 = 0.5°. The minute hand moves through 30° every 5 minutes and 6° each 1 minute. Therefore, in the minute from 1:27 to 1:28 (or any other minute), the difference between the hands increases by 6 – 0.5 = 5.5 degrees. [Note that it was not necessary, and would have been more time-consuming to determine the angles between the hands at 1:27 and 1:28 (See TACTIC 10: Don’t do more than you have to).]
If a Diagram Is Drawn to Scale, Trust It, and Use Your Eyes.
Remember that every diagram that appears on the SAT has been drawn as accurately as possible unless you see “Note: Figure not drawn to scale” written below it.
For figures that are drawn to scale, the following are true: line segments that appear to be the same length are the same length; if an angle clearly looks obtuse, it is obtuse; and if one angle appears larger than another, you may assume that it is larger.
Try Examples 7 and 8, which have diagrams that have been drawn to scale. Both of these examples would be classified as hard questions. On an actual SAT, questions of comparable difficulty would be answered correctly by at most 20–35% of the students taking the exam. After you master TACTIC 2, you should have no trouble with problems like these.
In the figure above, EF, not shown, is a diagonal of rectangle AFJE and a diameter of the circle. D is the midpoint of AE, C is the midpoint of AD, and B is the midpoint of AC.
If AE is 8 and the radius of the circle is 5, what is the area of rectangle BGHC?
Solution. Since there is no note indicating that the diagram has not been drawn to scale, you can trust it.
• The area of rectangle BGHC is the product of its width, BC, and its length, BG.
• AE = 8 AD = 4 AC = 2 BC = 1.
• BG appears to be longer than AD, which is 4, and shorter than AE, which is 8. Therefore, BG is more than 4 and is less than 8.
• Then, the area of BGHC is more than 1 × 4 = 4 and less than 1 × 8 = 8.
• The only choice between 4 and 8 is 6. The answer is B.
Note that you never used the fact that the radius of the circle is 5, information that is necessary to actually solve the problem. You were able to answer this question merely by looking at the diagram. Were you just lucky? What if the five choices had been 4, 5, 6, 7, and 8, so that there were three choices between 4 and 8, not just one? Well, you could have eliminated 4 and 8 and guessed, or you could have looked at the diagram even more closely. BG appears to be about the same length as CE, which is 6. If BG is 6, then the area of BGHC is exactly 6. How can you be sure? Measure the lengths!
On the answer sheet make two small pencil marks to indicate length BG:
Now, use that length to measure CE:
The lengths are the same. BG is 6; the area is 6. It’s not a guess after all.
Mathematical Solution. Diameter EF, which is 10, is also the hypotenuse of right triangle EAF. Since leg AE is 8, AF, the other leg, is 6 (either you recognize this as a 6-8-10 triangle, or you use the Pythagorean theorem). Since BG = AF, BG is 6, and the area is 6.
If Example 7 had been a grid-in problem instead of a multiple-choice question, you could have used TACTIC 2 in exactly the same way, but you would have been less sure of your answer. If, based on a diagram, you know that the area of a rectangle is about 6 or the measure of an angle is about 30°, you can almost always pick the correct choice, but on a grid-in you can’t be certain that the area isn’t 6.2 or the angle 31°. Nevertheless, if you can’t solve a problem directly, you should always grid in a “simple” number that is consistent with the diagram.
In the figure above, square ABCD has been divided into four triangles by its diagonals. If the perimeter of each triangle is 1, what is the perimeter of the square?
(D) 4( + 1)
Solution Using TACTIC 2. Make a “ruler” and mark off the perimeter of BEC; label that 1.
Now, mark off the perimeter of square ABCD.
It should be clear that P is much less than 2 (eliminate B and C), but more than 1.5 (eliminate A). The answer must be D or E. If it is obvious to you that choice D is greater than 4, eliminate it and choose E; otherwise, use your calculator to evaluate choice D (9.66) and choice E (1.65).
Mathematical Solution. Let s be a side of the square. Then, since BEC is a 45-45-90 right triangle, BE and EC are each . Therefore, the perimeter of BEC is , which equals 1. Now solving for s:
Finally, . Even if you could do this (and most students can’t), it is far easier to use TACTIC 2.
Remember that the goal of this book is to help you get credit (i) for all the problems you know how to do, and (ii), by using the TACTICS, for many that you don’t know how to do. Example 8 is typical. Most students omit it because it is too hard. You, however, can now answer it correctly, even though you may not be able to solve it directly.
In the figure below, what is the value of x?
In the figure below, what is the value of x – y?
Solution 9 Using TACTIC 2. Since the diagram is drawn to scale, trust it. Look at x: it appears to be about 90 + 50 = 140. In this case, using TACTIC 2 did not get you the exact answer. It only enabled you to narrow down the choices to (D) or (E). At this point you would guess— unless, of course, you know the following mathematical solution.
Mathematical Solution 9. The sum of the measures of the four angles in any quadrilateral is 360° (KEY FACT K1). Then
360 = 90 + 90 + 35 + x = 215 + x
x = 360 – 215 = 145 (E).
Solution 10 Using TACTIC 2. In the diagram, x and y look about the same, so assume they are. Certainly, neither one is 30° or even 15° greater than the other. Therefore, x – y = 0 (C).
Mathematical Solution 10. The sums of the measures of the three angles in triangles ABC and CBD are equal (they are both 180°). Then
90 + mB + x = 90 + mB + y x = y x – y = 0 (C).
Now try Examples 11–13, in which the diagrams are drawn to scale, and you need to find the measures of angles. Even if you know that you can solve these problems directly, practice TACTIC 2 and estimate the answers. The correct mathematical solutions without using this tactic are also given.
If, in the figure above, AB = AC, what is the value of x?
Solution Using TACTIC 2. Ignore all the information in the question. Just “measure” x. Draw DC perpendicular to AB, and let EC divide right angle DCA into two 45° angles, DCE and ACE. Now, DCB is about half of DCE, say 20–25°. Therefore, your estimate for x should be about 110 (90 + 20) or 115 (90 + 25). Choose C.
Mathematical Solution. Since ABC is isosceles, with AB = AC, the other two angles in the triangle, B and C, each measure 65°. Therefore,
x + 65 = 180 x = 115.
In the figure above, what is the sum of the measures of all of the marked angles?
Solution Using TACTIC 2. Make your best estimate of each angle, and add up the values. The five choices are so far apart that, even if you’re off by 15° or more on some of the angles, you’ll get the right answer. The sum of the estimates shown below is 690°, so the correct answer must be 720° (C).
Mathematical Solution. Each of the eight marked angles is an exterior angle of the quadrilateral. If we take one angle from each pair, their sum is 360° (KEY FACT K3); so, taking both angles at each vertex, we find that the sum of the measures is 360° + 360° = 720°.
In the diagram above, rays PA and PB are tangent to circle O. Which of the following is equal to z?
(B) 180 – x
(C) w + x + y
Solution Using TACTIC 2. The diagram is drawn to scale, so trust it. In the figure, x is clearly greater than 90 and z is clearly less than 90, so choices A and C are surely wrong. Also, it appears that w and y are each about 90, so w + x + y is more than 270. Since and , neither could be equal to z.
Eliminate D and E. The answer must be B.
Mathematical Solution. Tangents to a circle are perpendicular to the radii drawn to the points of contact, so w = y = 90. The sum of the four angles in a quadrilateral is 360°, so w + x + y + z = 360. Then 90 + x + 90 + z = 360 x + z = 180 z = 180 – x (B).
If a Diagram Is Not Drawn to Scale, Redraw It to Scale, and Then Use Your Eyes.
For figures that have not been drawn to scale, you can make no assumptions. Lines that look parallel may not be; an angle that appears to be obtuse may, in fact, be acute; two line segments may have the same length even though one looks twice as long as the other.
In order to redraw a diagram to scale, you first have to ask yourself, “What is wrong with the original diagram?” If an angle is marked 45°, but in the figure it looks like a 75° angle, redraw it. If two line segments appear to be parallel, but you have not been told that they are, redraw them so that they are clearly not parallel. If two segments appear to have the same length, but one is marked 5 and the other 10, redraw them so that the second segment is twice as long as the first.
In the examples illustrating TACTIC 2, all of the diagrams were drawn to scale and could be used to advantage. When diagrams have not been drawn to scale, you must be much more careful. TACTIC 3 tells you to redraw the diagram as accurately as possible, based on the information you are given, and then to apply the technique of TACTIC 2.
CAUTION: Redrawing a diagram, even roughly, takes time. Do this only when you do not see an easy direct solution to the problem.
In ACB, what is the value of x?
Note: Figure not drawn to scale
Solution. In what way is this figure not drawn to scale? AB = 8 and BC = 4, but in the figure AB is not twice as long as BC. Redraw the triangle 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, the triangle is a 30-60-90 triangle, and the angle formed by the hypotenuse and that leg is 60° (see Section 9-J).
Note: Figure not drawn to scale
In XYZ above, if XY < YZ < ZX, then which of the following must be true?
(A) x < 60
(B) z < 60
(C) y < z
(D) x < z
(E) y < x
Solution. As drawn, the diagram is useless. The triangle looks like an equilateral triangle, even though the question states that XY < YZ < ZX. Redraw the figure so that the condition is satisfied (that is, ZX is clearly the longest side and XY the shortest).
From the redrawn figure, it is clear that y is the largest angle, so eliminate choices C and E. Also, since z < x, eliminate D as well. Both x and z appear to be less than 60, but only one answer can be correct. Since z < x, if only one of these angles is less than 60, it must be z. Therefore, z < 60 (B) must be true.
Note: Figure not drawn to scale
In the figure above, O is the center of the circle. If OA = 4 and BC = 2, what is the value of x?
Solution Using TACTIC 3. Do you see why the figure isn’t drawn to scale? BC, which is 2, looks almost as long as OA, which is 4. Redraw the diagram, making sure that BC is only one-half as long as OA. With the diagram drawn to scale, you can see that x is approximately 30 (C).
Mathematical Solution. Since OB is a radius, it has the same length as radius OA, which is 4. Then BCO is a right triangle in which the hypotenuse is twice as long as one leg. This can occur only in a 30-60-90 triangle, and the angle opposite that leg measures 30°. Therefore, x = 30.
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 there isn’t enough given information. In this case, it often helps to draw another line in the diagram.
In the figure to the right, 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?
(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 congruent, and, since OQ is a radius, OQ = PR = r (A).
Note that you could also have made a “ruler” and seen that PR is equal to r.
What is the area of quadrilateral ABCD?
Solution. Since the quadrilateral is irregular, you don’t know any formula to get the answer. 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 CE and CF, the heights of the two triangles, you see that the area of ACD is , and the area of BAC is . Then the area of ABCD is 30 + 8 = 38.
Note that this problem could also have been solved by drawing in lines to create rectangle ABEF, and subtracting the areas of BEC and CFD from the area of the rectangle.
Test the Choices, Starting with C.
TACTIC 5, often called backsolving, is useful when you are asked to solve for an unknown and you understand what needs to be done to answer the question, but you want to avoid doing the algebra. The idea is simple: test the various choices to see which one is correct.
NOTE: On the SAT the answers to virtually all numerical multiple-choice questions are listed in either increasing or decreasing order. Consequently, C is the middle value; and in applying TACTIC 5, you should always start with C. For example, assume that choices A, B, C, D, and E are given in increasing order. Try C. If it works, you’ve found the answer. If C doesn’t work, you should now know whether you need to test a larger number or a smaller one, and that information permits you to eliminate two more choices. If C is too small, you need a larger number, so A and B are out; if C is too large, you can eliminate D and E, which are even larger.
Example 19 illustrates the proper use of TACTIC 5.
If the average (arithmetic mean) of 2, 7, and x is 12, what is the value of x?
Solution. Use TACTIC 5. Test choice C: x = 21.
• Is the average of 2, 7, and 21 equal to 12?
• No: , which is too small.
• Eliminate C; also, since, for the average to be 12, x must be greater than 21, eliminate A and B.
• Try choice D: x = 27. Is the average of 2, 7, and 27 equal to 12?
• Yes: . The answer is D.
Every problem that can be solved using TACTIC 5 can be solved directly, often in less time. Therefore, we stress: if you are confident that you can solve a problem quickly and accurately, just do so.
Here is the direct method for solving Example 19, which is faster than backsolving. (See Section 9-E on averages.) If you know this method, you should use it and save TACTIC 5 for problems that you can’t easily solve directly.
Direct Solution. If the average of three numbers is 12, their sum is 36. Then
2 + 7 + x = 36 9 + x = 36 x = 27.
Some tactics allow you to eliminate a few choices so that you can make an educated guess. On problems where TACTIC 5 can be used, it always leads you to the right answer. The only reason not to use it on a particular problem is that you can easily solve the problem directly.
Now try applying TACTIC 5 to Examples 20 and 21.
If the sum of five consecutive even integers is 740, what is the largest of these integers?
Solution. Use TACTIC 5. Test choice C: 146.
• If 146 is the largest of the five integers, the integers are 146, 144, 142, 140, and 138. Quickly add them on your calculator. The sum is 710.
• Since 710 is too small, eliminate C, D, and E.
• If you noticed that the amount by which 710 is too small is 30, you should realize that each of the five numbers needs to be increased by 6; therefore, the largest is 152 (B).
• If you didn’t notice, just try 152, and see that it works.
This solution is easy, and it avoids having to set up and solve the required equation:
n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 740.
A competition offers a total of $250,000 in prize money to be shared by the top three contestants. If the money is to be divided among them in the ratio of 1:3:6, what is the value of the largest prize?
(A) $ 25,000
(B) $ 75,000
Solution. Use TACTIC 5. Test choice C: $100,000.
• If the largest prize is $100,000, the second largest is $50,000 (they are in the ratio of 6:3 = 2:1). The third prize is much less than $50,000, so all three add up to less than $200,000.
• Eliminate A, B, and C; and, since $100,000 is way too small, try E, not D.
• Test choice E. The prizes are $150,000, $75,000, and $25,000 (one-third of $75,000). Their total is $250,000. The answer is E.
Again, TACTIC 5 lets you avoid the algebra if you can’t do it or just don’t want to. Here is the correct solution. By TACTIC D1 the three prizes are x, 3x, and 6x. Therefore,
x + 3x + 6x = $250,000 10x = $250,000.
So, x = $25,000 and 6x = $150,000.
If 2 + 5 = 8, then x =
Solution. Since plugging in 0 is much easier than plugging in , start with B. If x = 0, the left-hand side of the equation is , which is equal to 7 and so is too small. Eliminate A and B, and try something bigger. Preferring whole numbers to fractions, try choice D. If x = 1, then .
Since that’s too big, eliminate D and E. The answer must be . Again, remember: no matter what the choices are, backsolve only if you can’t easily do the algebra. Some students would do this problem directly:
Don’t start with C if some other choice is much easier to work with. If you start with B and it is too small, you may be able to eliminate only two choices (A and B), instead of three, but you will save time if plugging in choice C would be messy.
and save backsolving for an even harder problem. You have to determine which method is better for you.
For some multiple-choice questions on the SAT, you have to test the various choices. On these problems you are not really backsolving (there is nothing to solve!); rather you are testing whether a particular choice satisfies a given condition.
Examples 23 and 24 are two such problems. In Example 23, you are asked for the largest integer that satisfies a certain condition. Usually, some of the smaller integers offered as choices also satisfy the condition, but your job is to find the largest one.
What is the largest integer, n, such that is an integer?
Solution. Since you want the largest value of n for which is an integer, start by testing 5, choice E, the largest of the choices.
• Is an integer? No: . Eliminate E and try D.
• Is an integer? Yes: 24 = 16, and .
• The answer is 4 (D).
It doesn’t matter whether any of the smaller choices works (you need the largest), although in this case they all do.
Surprisingly, on a problem that asks for the smallest number satisfying a property, you should also start with E, because the choices for these problems are usually given in decreasing order.
It is also better to start with E on questions such as Example 24, in which you are asked “which of the following...?” The right answer is rarely one of the first choices.
Sometimes a question asks which of the five choices satisfies a certain condition. Usually, in this situation there is no way to answer the question directly. Rather, you must look at the choices and test each of them until you find one that works. At that point, stop—none of the other choices could be correct. There is no particular order in which to test the choices, but it makes sense to test the easier ones first. For example, it is usually easier to test whole numbers than fractions and positive numbers than negative ones.
Which of the following is NOT equivalent to .
Solution. Here, you have to test each of the choices until you find one that satisfies the condition that it is not equal to . If, as you glance at the choices to see if any would be easier to test than the others, you happen to notice that 60% = 0.6, then you can immediately eliminate choices B and C, since it is impossible that both are correct.
• Test choice A. Reduce by dividing the numerator and denominator by .
• Test choice D. .
• You now know that E must be the correct answer. In fact, .
If you think of a grid-in problem as a multiple-choice question in which the choices accidentally got erased, you can still use TACTIC 5. To test choices, you just have to make them up. Let’s illustrate by looking again at Examples 19 and 20, except that now the choices are missing.
If the average (arithmetic mean) of 2, 7, and x is 12, what is the value of x?
Instead of starting with choice C, you have to pick a starting number. Any number will do, but when the numbers in the problem are 2, 7, and 12, it’s more likely that x is 10 or 20 than 100 or 1000.
Solution. You could start with 10; but if you immediately realize that the average of 2, 7, and 10 is less than 10 (so it can’t be 12), you’ll try a bigger number, say 20. The average of 2, 7, and 20 is
which is too small. Try x = 30:
just a bit too big. Since 12 is closer to 13 than it is to , your next choice should be closer to 30 than 20, surely more than 25. Your third try might well be 27, which works.
If the sum of five consecutive even integers is 740, what is the largest of these integers?
Solution. You can start with any number. If you realize that the sum of five numbers, each of which is near 100, is about 500, and that the sum of five numbers, each of which is near 200, is about 1000, you will immediately start with a number in between, say 150:
150 + 148 + 146 + 144 + 142 = 730.
Since 730 is too small but extremely close, try a number just slightly larger than 150, say 152, which works.
Replace Variables with Numbers.
Mastery of TACTIC 6 is critical for anyone developing good test-taking skills. This tactic can be used whenever the five choices in a multiple-choice math question involve the variables in the question. There are three steps:
1. Replace each variable with an easy-to-use number.
2. Solve the problem using those numbers.
3. Evaluate each of the five choices with the numbers you picked to see which choice is equal to the answer you obtained.
Examples 25 and 26 illustrate the proper use of TACTIC 6.
If a is equal to b multiplied by c, which of the following is equal to b divided by c?
• Pick three easy-to-use numbers that satisfy a = bc: for example, a = 6, b = 2, c = 3.
• Solve the problem with these numbers: .
• Check each of the five choices to see which one is equal to :
• The answer is D.
If the sum of four consecutive odd integers is s, then, in terms of s, what is the greatest of these integers?
• Pick four easy-to-use consecutive odd integers: say, 1, 3, 5, 7. Then s, their sum, is 16.
• Solve the problem with these numbers: the greatest of these integers is 7.
• When s = 16, the five choices are
• Only choice D, is equal to 7.
Of course, Examples 25 and 26 can be solved without using TACTIC 6 if your algebra skills are good. Here are the solutions.
Solution 26. Let n, n + 2, n + 4, and n + 6 be four consecutive odd integers, and let s be their sum. Then:
s = n + (n + 2) + (n + 4) + (n + 6) = 4n + 12.
The important point is that, if you are uncomfortable with the correct algebraic solution, you don’t have to omit these questions. You can use TACTIC 6 and always get the right answer. Of course, even if you can do the algebra, you should use TACTIC 6 if you think you can solve the problem faster or will be less likely to make a mistake. With the proper use of the tactics in this chapter, you can correctly answer many problems that you may not know how to solve mathematically.
Replace the variables with numbers that are easy to use, not necessarily ones that make sense. It is perfectly OK to ignore reality. A school can have five students, apples can cost $10 each, trains can go 5 miles per hour or 1000 miles per hour—it doesn’t matter.
Examples 27 and 28 are somewhat different. You are asked to reason through word problems involving only variables. Most students find problems like these mind-boggling. Here, the use of TACTIC 6 is essential; without it, Example 27 is difficult and Example 28 is nearly impossible. TACTIC 6 is not easy to master, but with practice you will catch on.
If a school cafeteria needs c cans of soup each week for each student, and if there are s students in the school, for how many weeks will x cans of soup last?
• Replace c, s, and x with three easy-to-use numbers. If a school cafeteria needs 2 cans of soup each week for each student, and if there are 5 students in the school, how many weeks will 20 cans of soup last?
• Since the cafeteria needs 2 × 5 = 10 cans of soup per week, 20 cans will last for 2 weeks.
• Which of the choices equals 2 when c = 2, s = 5, and x = 20?
• The five choices become:
The answer is D.
If p painters can paint h houses in 4 days, how many houses can 6 painters, working at the same rate, paint in 2 days?
• Pick two easy-to-use numbers, say p = 1 and h = 1. Then 1 painter can paint 1 house in 4 days.
• Then, 6 painters can paint 6 houses in 4 days, and so in 2 days 6 painters can paint 3 houses.
• Evaluate the five choices when p = 1 and h = 1, and find the choice that equals 3:
• Eliminate A, B, and C. But both D and E are 3. What now?
• Change one of the numbers, and test only D and E. Suppose that 1 painter could paint 100 houses, instead of just 1, in 4 days. Then 6 painters could paint lots of houses—certainly many more than 3.
• Of D and E, which will be bigger if you replace h by 100 instead of 1? In D, the numerator, and hence the whole fraction, which is will be much bigger. In E, the denominator will be larger and the value of the fraction smaller.
• The answer is D.
Example 28 illustrates that replacing a variable by 1 is not a good idea in this type of problem. The reason is that multiplying by 1 and dividing by 1 give the same result: 3x and are each equal to 3 when x = 1. It is also not a good idea to use the same number for different variables: and are each equal to 3 when a and b are equal.
A good choice in Example 28 would be to let p = 6 and h = 4. Example 28 would then read, “If 6 painters can paint 4 houses in 4 days, how many houses can 6 painters, working at the same rate, paint in 2 days?” The answer is obviously 2, and only choice D is equal to 2 when p = 6 and h = 4.
Even though Examples 27 and 28 are much more abstract than Examples 25 and 26, they too can be solved directly and more quickly if you can manipulate the variables.
Algebraic Solution 27. If each week the school needs c cans for each of the s students, then it will need cs cans per week. Dividing cs into x gives the number of weeks that x cans will last: .
Algebraic Solution 28. Since 1 painter can do times the amount of work of p painters, if p painters can paint h houses in 4 days, then 1 painter can paint houses in 4 days. In 1 day he can paint times the number of houses he can paint in 4 days; so, in 1 day, 1 painter can paint houses. Of course, in 1 day, 6 painters can paint 6 times as many houses: . Finally, in 2 days these painters can paint twice as many houses: . Even if you could carefully reason this out, why would you want to? Now, practice TACTIC 6 on the following problems.
Nadia will be x years old y years from now. How old was she z years ago?
(A) x + y + z
(B) x + y – z
(C) x – y – z
(D) y – x – z
(E) z – y – x
If , which of the following is an expression for d in terms of a?
Anne drove for h hours at a constant rate of r miles per hour. How many miles did she go during the final 20 minutes of her drive?
Solution 29. Assume Nadia will be 10 in 2 years. How old was she 3 years ago? If she will be 10 in 2 years, she is 8 now and 3 years ago was 5. Which of the choices equals 5 when x = 10, y = 2, and z = 3? Only x – y – z (C).
Solution 30. Let d = 1. Then , , and a = 8. Which of the choices equals 1 when a = 8? Only .
Solution 31. If Anne drove at 60 miles per hour for 2 hours, how far did she go in the last 20 minutes?
Since 20 minutes is of an hour, she went miles.
Only when r = 60 and h = 2.
Notice that h is irrelevant. Whether Anne had been driving for 2 hours or 20 hours, the distance she covered in her last 20 minutes would be the same.
Choose an Appropriate Number.
TACTIC 7 is similar to TACTIC 6 in that you pick a convenient number. However, here no variable is given in the problem. TACTIC 7 is especially useful in problems involving fractions, ratios, and percents.
At Central High School each student studies exactly one foreign language. Three-fifths of the students take Spanish, and one-fourth of the remaining students take Italian. If all of the others take French, what percent of the students take French?
Solution. The least common denominator of and is 20, so assume that there are 20 students at Central High. (Remember that the number you choose doesn’t have to be realistic.) Then the number of students taking Spanish is . Of the remaining 8 students, take Italian. The other 6 take French. Finally, 6 is 30% of 20. The answer is E.
In problems involving fractions, the best number to use is the least common denominator of all the fractions. In problems involving percents, the easiest number to use is 100. (See Sections 9-B and 9-C.)
From 2003 to 2004 the number of boys in the school chess club decreased by 20%, and the number of girls in the club increased by 20%. The ratio of girls to boys in the club in 2004 was how many times the ratio of girls to boys in the club in 2003?
Solution. This problem involves percents, so try to use 100. Assume that in 2003 there were 100 boys and 100 girls in the club. Since 20% of 100 is 20, in 2004 there were 120 girls (a 20% increase) and 80 boys (a 20% decrease). See the following chart:
The chart shows that the 2003 ratio of 1 was multiplied by . The answer is E. Here are two more problems where TACTIC 7 is useful.
In a particular triathlon the athletes cover of the total distance by swimming, of it by running, and the rest by bike. What is the ratio of the distance covered by bike to the distance covered by running?
From 2002 to 2003 the sales of a book decreased by 80%. If the sales in 2004 were the same as in 2002, by what percent did they increase from 2003 to 2004?
Solution 34. The least common denominator of the two fractions is 24, so assume that the total distance is 24 miles. Then, the athletes swim for 1 mile and run for miles. The remaining 15 miles they cover by bike. Therefore, the required ratio is 15:8 (B).
Solution 35. Use TACTIC 7, and assume that 100 copies were sold in 2002 (and 2004). Sales dropped by 80 (80% of 100) to 20 in 2003 and then increased by 80, from 20 back to 100, in 2004. The percent increase was
Eliminate Absurd Choices, and Guess.
When you have no idea how to solve a problem, eliminate all the absurd choices and guess from among the remaining ones.
In Part One, you read that only very infrequently should you omit a problem that you have time to work on. During the course of an SAT, you will probably find at least a few multiple-choice questions that you have no idea how to solve. Do not omit these questions! Often two or three of the answers are absurd. Eliminate them and guess. Occasionally, four of the choices are absurd. When this occurs, your answer is no longer a guess.
What makes a choice absurd? Lots of things. Even if you don’t know how to solve a problem, even with very hard ones, you may realize that:
• the answer must be positive, but some of the choices are negative;
• the answer must be even, but some of the choices are odd;
• the answer must be less than 100, but some choices exceed 100;
• a ratio must be less than 1, but some choices are greater than or equal to 1.
Let’s look at several examples. In a few of them the information given is intentionally insufficient to solve the problem, but you will still be able to determine that some of the answers are absurd. In each case the “solution” provided will indicate which choices you should have eliminated. At that point you would simply guess. [See Part One for a complete discussion of guessing.] Remember: on the SAT when you have to guess, don’t agonize. Just make your choice and then move on.
A region inside a semicircle of radius r is shaded. What is the area of the shaded region?
Solution. You may have no idea how to find the area of the shaded region, but you should know that, since the area of a circle is πr2, the area of a semicircle is . Therefore, the area of the shaded region must be less than , so eliminate C, D, and E. On an actual problem that includes a diagram, if the diagram is drawn to scale, you may be able to make an educated guess between A and B. If not, just choose one or the other.
The average of 5, 10, 15, and x is 20. What is x?
Solution. If the average of four numbers is 20, and three of them are less than 20, the other one must be greater than 20. Eliminate A and B and guess. If you further realize that, since 5 and 10 are a lot less than 20, x will probably be a lot more than 20, you can eliminate C, as well. Then guess either D or E.
If 25% of 220 equals 5.5% of w, what is w?
Solution. Since 5.5% of w equals 25% of 220, which is surely greater than 5.5% of 220, w must be greater than 220. Eliminate A, B, C, and D. The answer must be E!
Example 38 illustrates an important point. Even if you know how to solve a problem, if you immediately see that four of the five choices are absurd, just pick the remaining choice and move on.
A prize of $27,000 is to be divided in some ratio among three people. What is the largest share?
(C) $ 8100
(D) $ 5400
(E) $ 2700
Solution. If the prize were divided equally, each share would be worth $9000. If it is divided unequally, the largest share is surely more than $9000, so eliminate C, D, and E. In an actual question, you would be told what the ratio is, and that information should enable you to eliminate A or B. If not, you would just guess.
A jar contains only red and blue marbles. The ratio of the number of red marbles to the number of blue marbles is 5:3. What percent of the marbles are blue?
Solution. Since there are 5 red marbles for every 3 blue ones, there are fewer blue ones than red ones. Therefore, fewer than half (50%) of the marbles are blue. Eliminate B, C, D, and E. The answer is A.
In the figure above, four semicircles are drawn, each centered at the midpoint of one of the sides of square ABCD. Each of the four shaded “petals” is the intersection of two of the semicircles. If AB = 4, what is the total area of the shaded region?
(B) 32 – 8π
(C) 16 – 8π
(D) 8π – 32
(E) 8π – 16
Solution. The diagram is drawn to scale. Therefore, you can trust it in making your estimate (TACTIC 2).
• Since the shaded area appears to take up a little more than half of the square, it does.
• The area of the square is 16, so the area of the shaded region must be about 9.
• Using your calculator, but only when you need it, check each choice. Since π is slightly more than 3, 8π (which appears in each choice) is somewhat more than 24, approximately 25.
• (A) 8π ≈ 25. More than the whole square: way too big.
• (B) 32 – 8π ≈ 7. Too small.
• (C) 16 – 8π is negative. Clearly impossible!
• (D) 8π – 32 is also negative.
• (E) 8π – 16 ≈ 25 – 16 = 9. Finally! The answer is E.
Note: Three of the choices are absurd: A is more than the area of the entire square, and C and D are negative and so can be eliminated immediately. No matter what your estimate was, at worst you had to guess between two choices.
Now use TACTIC 8 on each of the following problems. Even if you know how to solve them, don’t. Practice this technique, and see how many choices you can eliminate without actually solving.
In the figure above, diagonal EG of square EFGH is one-half of diagonal AD of square ABCD. What is the ratio of the area of the shaded region to the area of ABCD?
Jim receives a commission of 25¢ for every $20.00 worth of merchandise he sells. What percent is his commission?
From 1990 to 2000, Michael’s weight increased by 25%. If his weight was W kilograms in 2000, what was it in 1990?
The average of 10 numbers is –10. If the sum of six of them is 100, what is the average of the other four?
What is 3% of 4%?
If f(x) = 4x2 + 2x4, what is the value of f(–2)?
Solution 42. Obviously, the shaded region is smaller than square ABCD, so the ratio must be less than 1. Eliminate A ( > 1.4). Also, from the diagram, it is clear that the shaded region is more than half of square ABCD, so the ratio is greater than 0.5. Eliminate D and E. Since 3:4 = 0.75 and :2 ≈ 0.71, B and C are too close to tell, just by looking, which is right, so guess. The answer is B.
Solution 43. Clearly, a commission of 25¢ on $20 is quite small. Eliminate D and E, and guess one of the small percents. If you realize that 1% of $20 is 20¢, then you know the answer is a little more than 1%, and you should guess A (maybe B, but definitely not C). The answer is A.
Solution 44. Since Michael’s weight increased, his weight in 1990 was less than W. Eliminate A, B, and C and guess. The answer is D.
Solution 45. Since the average of all 10 numbers is negative, so is their sum. However, the sum of the first six is positive, so the sum (and the average) of the others must be negative. Eliminate C, D, and E. The answer is B.
Solution 46. Since 3% of a number is just a small part of it, 3% of 4% must be much less than 4%. Eliminate D and E, and probably C. The answer is B.
Solution 47. Any nonzero number raised to an even power is positive, so 4x2 + 2x4 is positive. Eliminate A, B, and C. If you can’t evaluate f(–2), guess between D and E. If you have a hunch that E is too big, choose D. The answer is D.
Subtract to Find Shaded Regions.
Whenever part of a figure is white and part is shaded, the straightforward way to find the area of the shaded portion is to find the area of the entire figure and then subtract from it the area of the white region. Of course, if you are asked for the area of the white 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 above, ABCD is a rectangle, and and are arcs of circles centered at A and D. What is the area of the shaded 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 each white region is a quarter-circle of radius 2, the combined area of these regions is that of asemicircle of radius 2: . Therefore, the area of the shaded region is 10 – 2π = 2(5 – π) (B).
In the figure above, each side of square ABCD is divided into three equal parts. If a point is chosen at random inside the square, what is the probability it will be in the shaded region?
Solution. Since the answer doesn’t depend on the value of x (the probability will be the same no matter what x is), let x = 1. Then the area of the whole square is 32 = 9. The area of each shaded triangle or of all the white sections can be calculated, but there’s an easier way: notice that, if you slide the four shaded triangles together, they form a square of side 1. Therefore, the total shaded area is 1, and so the shaded area is of the total area and the probability that the chosen point is in the shaded area is .
The idea of subtracting a part from the whole works with line segments as well as areas.
In the figure above, the circle with center O is inscribed in square ABCD. Line segment AO intersects the circle at P. What is the length of AP?
(B) 2 –
(D) 2 – 2
(E) – 1
Solution. First use TACTIC 4 and draw some lines. Extend AO to form diagonal AC. Then, since ADC is an isosceles right triangle, AC = 2 (KEY FACT J8) and AO is half of that, or . Then draw in diameter EF parallel to AD. Since the diameter is 2 (EF = AD = 2), the radius is 1. Finally, subtract: AP = AO – PO = – 1(E).
Note: If you don’t realize which lines to add and/or you can’t reason a question like this one out, do not omit it. You can still use TACTIC 2: trust the diagram. Since AB = 2, then AE = 1, and AP is clearly less than 0.5.
With your calculator evaluate each choice. A, B, and D are all greater than 0.5. Eliminate them, and guess either C or E.
Don’t Do More Than You Have To.
In Example 6, you were asked, “By how many degrees does the angle formed by the hour hand and the minute hand of a clock increase from 1:27 to 1:28?” If you look at the solution, you will see that we didn’t have to calculate either angle. This is a common situation. Look for shortcuts. Since a problem can often be solved in more than one way, you should always look for the easiest method. Consider the following examples.
If 5(3x – 7) = 20, what is 3x – 8?
It’s not difficult to solve for x:
5(3x – 7) = 20 15x – 35 = 20 15x = 55
But it’s too much work. Besides, once you find that , you still have to multiply to get 3x: , and then subtract to get 3x – 8: 11 – 8 = 3.
Solution. The key is to recognize that you don’t need 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.
If 7x + 3y = 17 and 3x + 7y = 19, what is the average (arithmetic mean) of x and y?
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 are familiar with this method, try it now, before reading further. If you work carefully, you should find that and , and their average is or 1.8. This method is not too difficult; but it is quite time-consuming, and no problem on the SAT requires you to do so 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.
Solution. Add the two equations: 7x + 3y = 17
+ 3x + 7y = 19
10x + 10y = 36
Divide each side by 10: x + y = 3.6
Calculate the average:
When you learn TACTIC 17, you will see that adding two equations, as was done here, is the standard way to attack problems such as this on the SAT.
Pay Attention to Units.
Often the answer to a question must be in units different from those used in the given data. As you read the question, underline exactly what you are being asked. Do the examiners want hours or minutes or seconds, dollars or cents, feet or inches, meters or centimeters? On multiple-choice questions an answer with the wrong units is almost always one of the choices.
At a speed of 48 miles per hour, how many minutes will be required to drive 32 miles?
Solution. This is a relatively easy question. Just be attentive. Since , 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 will be required. (Did you underline the word “minutes” in the question?) The correct answer is .
Note that you could have been asked how many seconds would be needed, in which case the answer would be 40(60) = 2400 (E).
The wholesale price of potatoes is usually 3 pounds for $1.79. How much money, in cents, did a restaurant save when it was able to purchase 600 pounds of potatoes at 2 pounds for $1.15?
Solution. For 600 pounds the restaurant would normally have to buy 200 3-pound bags for 200 × $1.79 = $358. On sale, it bought 300 2-pound bags for 300 × $1.15 = $345. Therefore, the restaurant saved 13 dollars. Do notgrid in 13. If you underline the word “cents” you won’t forget to convert the units: 13 dollars is 1300 cents.
Use Your Calculator.
You already know that you can use a calculator on the SAT. (See Part One for a complete discussion of calculator usage.) The main reason to use a calculator is that it enables you to do arithmetic more quickly and more accurately than you can by hand on problems that you know how to solve. (For instance, in Example 54, you should use your calculator to multiply 200 × 1.79.) The purpose of TACTIC 12 is to show you how to use your calculator to get the right answer to questions that you do not know how to solve or you cannot solve.
If x2 = 2, what is the value of ?
(C) 1 +
(D) 2 +
(E) 1.5 + 2
Solution. The College Board would consider this a hard question, and most students would either omit it or, worse, miss it. The best approach is to realize that is a product of the form
(a + b)(a – b) = a2 – b2.
If you didn’t see this solution, you could still solve the problem by writing x = and then trying to multiply and simplify . It is likely, however, that you would make a mistake somewhere along the way.
The better method is to use your calculator: ≈ 1.414 and , so ≈ (1.414 + 0.707)(1.414 – 0.707) = (2.121)(0.707) = 1.499547.
Clearly, choose 1.5, the small difference being due to the rounding off of as 1.414. If this were a grid-in question, you should not grid in 1.49, since you know that that’s only an approximation. Rather, you should guess a simple number near 1.499, such as 1.5. In fact, if you don’t round off, and just use the value your calculator gives for (1.414213562, say), you will probably get 1.5 exactly (although you may get 1.49999999 or 1.50000001).
If a and b are positive numbers, with a3 = 3 and a5 = 12b2, what is the ratio of a to b?
Solution. This is another difficult question that most students would omit or miss. If you think to divide the second equation by the first, however, the problem is not too bad:
If you don’t see this, you can still solve the problem with your scientific calculator: .
Grid in 2.
What is the value of ?
Solution. There are two straightforward ways to do this: (i) multiply the numerator and denominator by 35, the LCM of 5 and 7, and (ii) simplify and divide:
However, if you don’t like working with fractions, you can easily do the arithmetic on any calculator. Be sure you know how your calculator works. Be sure that when you evaluate the given complex fraction you get 8.4.
If this had been a multiple-choice question, the five choices would probably have been fractions, in which case the correct answer would be . If you had solved this with your calculator, you would then have had to use the calculator to determine which of the fractions offered as choices was equal to 8.4.
Know When Not to Use Your Calculator.
Don’t get into the habit of using your calculator on every problem involving arithmetic. Since many problems can be solved more easily and faster without a calculator, learn to use your calculator only when you need it (see Part One).
John had $150. He used 85% of it to pay his electric bill and 5% of it on a gift for his mother. How much did he have left?
Solution. Many students would use their calculators on each step of this problem.
$150 × .85 = $127.50
Gift for mother:
$150 × .05 = $7.50
$127.50 + $7.50 = $135
$150 – $135 = $15
Good test-takers would have proceeded as follows, finishing the problem in less time than it takes to calculate the first percent: John used 90% of his money, so he had 10% left; and 10% of $150 is $15.
Systematically Make Lists.
When a question asks “how many,” often the best strategy is to make a list of all the possibilities. It is important that you make the list in a systematic fashion so that you don’t inadvertently leave something out. Often, shortly after starting the list, you can see a pattern developing and 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 some items to answer it; in this case, as well, the best plan may be to make a list.
Listing things systematically means writing them in numerical order (if the entries are numbers) or in alphabetical order (if the entries are letters). If the answer to “how many” is a small number (as in Example 59), just list all possibilities. If the answer is a large number (as in Example 60), start the list and write enough entries to enable you to see a pattern.
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?
Solution. Since one of the integers is 5, the product of the other two is 60 (5 × 60 = 300). 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
The answer is 16.
A palindrome is a number, such as 93539, that reads the same forward and backward. How many palindromes are there between 100 and 1000?
Solution. First, write down
101, 111, 121, 131, 141,
Now write the numbers
202, 212, 222, 232, 242,
By now you should see the pattern: there are 10 numbers beginning with 1, and 10 beginning with 2, and there will be 10 beginning with 3, 4, ..., 9 for a total of 9 × 10 = 90 palindromes.
In how many ways can Al, Bob, Charlie, Dan, and Ed stand in a line if Bob must be first and either Charlie or Dan must be last?
Solution. Represent the five boys as A, B, C, D, and E. Placing Charlie last, you see that the order is B __ __ __ C. Systematically fill in the blanks with A, D, and E. Write all the three-letter “words” you can in alphabetical order so you don’t accidentally skip one.
A D E
A E D
D A E
D E A
E A D
E D A
There are 6 possibilities when C is last. Clearly, there will be 6 more when D is last. Therefore, there are 12 ways in all to satisfy the conditions of the problem.
See Section O in Chapter 9 for additional examples and for another method of solving these problems using the counting principle.
Trust All Grids, Graphs, and Charts.
Figures that show the grid lines of a graph are always accurate, whether or not the coordinates of the points are given. For example, in the figure below, you can determine each of the following:
• the lengths of all three sides of the triangle;
• the perimeter of the triangle;
• the area of the triangle;
• the slope of each line segment.
In the grid above, what is the area of quadrilateral ABCD?
Solution. AB and CD are parallel (they’re both horizontal), so ABCD is a trapezoid. If you know that the formula for the area of a trapezoid is , use it. By counting boxes, you see that
b1 = CD = 9, b2 = AB = 4, and h = 3.
Therefore, the area is .
If you don’t know the formula, use AE and BF to divide ABCD into a rectangle (ABFE) and two right triangles (AED and BFC). Their areas are 12, 6, and 1.5, respectively, for a total area of 19.5.
For sample problems using grids, see Section 9-N on coordinate geometry.
SAT problems that use any kind of charts or graphs are always drawn accurately and can be trusted. For example, suppose that you are told that each of the 1000 students at Central High School studies exactly one foreign language. Then, from the circle graph that follows, you may conclude that fewer than half of the students study Spanish, but more students study Spanish than any other language; that approximately 250 students study French; that fewer students study German than any other language; and that approximately the same number of students are studying Latin and Italian.
FOREIGN LANGUAGES STUDIED BY
1000 STUDENTS AT CENTRAL HIGH SCHOOL
From the bar graph that follows, you know that in 2001 John won exactly three tournaments, and you can calculate that from 2000 to 2001 the number of tournaments he won decreased by 50% (from 6 to 3), whereas from 2001 to 2002 the number increased by 300% (from 3 to 12).
NUMBER OF TENNIS TOURNAMENTS
JOHN WON BY YEAR
The chart above depicts the number of electoral votes assigned to each of the six New England states. What is the average (arithmetic mean) number of electoral votes, to the nearest tenth, assigned to these states?
Solution. Since you can trust the chart to be accurate, the total number of electoral votes for the six states is
4 + 4 + 3 + 13 + 4 + 8 = 36
and the average is 36 ÷ 6 = 6 (C).
Several different types of questions concerning bar graphs, circle graphs, line graphs, and various charts and tables can be found in Section 9-Q.
Read the Definitions of Strange Symbols Very Carefully.
On almost every SAT a few questions use symbols, such as , that you have never before seen in a mathematics problem. How can you answer such a question? Don’t panic! It’s easy; you are always told exactly what the symbol means! All you have to do is follow the directions carefully.
For any numbers a and b, a b is defined as . What is the value of 25 15?
Solution. The definition of “” tells you that, whenever two numbers surround a “happy face,” you are to form a fraction in which the numerator is the sum of the numbers and the denominator is their difference. Here, 25 15 is the fraction whose numerator is 25 + 15 = 40 and whose denominator is .
Sometimes the same symbol is used in two (or even three) questions. In these cases, the first question is easy and involves only numbers; the second is a bit harder and usually contains variables.
Examples 65–67 refer to the following definition.
For any numbers x and y, let x y be defined as x y = xy – (x + y).
Follow the directions given in the definition and try to answer Examples 65–67 before reading the solutions.
What is the value of –2 3?
For what value of x does x 5 = x 10?
How many positive numbers are solutions of the equation y y = y?
(E) More than 3
– 2 3 = (–2)(3) – (–2 + 3) = –6 – (1) = –7 (B).
Solution 66. Since x 5 = x 10
5x – (5 + x) = 10x – (10 + x)
5x – 5 – x = 10x – 10 – x
4x – 5 = 9x –10
5x = 5 x = 1 (D).
y y = y y2 – 2y = y y2 – 3y = 0.
So, y(y – 3) = 0 y = 0 or y = 3.
There is only 1 positive solution, 3 (D).
For any real numbers c and d, c d is defined to be cd + dc. What is the value of 1 (2 3)?
Solution. Remember the correct order of operations: always do first what’s in the parentheses (see Section 9-A).
2 3 = 23 + 32 = 8 + 9 = 17
1 17 = 117 + 171 = 1 + 17 = 18.
When a question involves two equations that do not have exponents, either add the equations or subtract them. If there are three or more equations, add them.
Very often, answering a question that involves two or more equations does not require you to solve the equations. Remember TACTIC 10: Do not do any more than is necessary.
If 3x + 5y = 14 and x – y = 6, what is the average of x and y?
Solution. Add the equations:
Divide each side by 4: x + y = 5
The average of x and y is their sum divided by
The answer is B.
Note that you could have actually solved for x and y [x = 5.5, y = –0.5], and then taken their average. However, that would have been time-consuming and unnecessary.
Here are two more problems involving two or more equations.
If a – b + c = 7 and a + b – c = 11, which of the following statements MUST be true?
I. a is positive
II. b > c
III. bc < 0
(B) I only
(C) II only
(D) III only
(E) I and II only
If a – b = 1, b – c = 2, and c – a = d, what is the value of d?
(E) It cannot be determined from the information given.
Solution 70. Start by adding the two equations: a – b + c = 7
+ a + b – c = 11
2a = 18
Therefore, a = 9. (I is true.)
Replace a by 9 in each 9 – b + c = 7 – b + c = –2
equation to obtain two and
new equations: 9 + b – c = 11 b – c = 2
Since b – c = 2, then b > c. (II is true.)
As long as b = c + 2, however, there are no restrictions on b and c: if b = 2 and c = 0, bc = 0. (III is false.)
The answer is E.
Solution 71. Add the three equations:
The answer is A.
NOTE: On all the model tests in this book, the difficulty level of the math questions in each section proceeds from easy to medium to hard, just as they do on all real SATs. The questions in this exercise set, however, are not presented in any particular order of difficulty.
1. In 1995, Diana read 10 English books and 7 French books. In 1996, she read twice as many French books as English books. If 60% of the books that she read during the 2 years were French, how many English and French books did she read in 1996?
2. In the figure below, if the radius of circle O is 10, what is the length of diagonal AC of rectangle OABC?
3. 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?
4. In the figure below, AB and AC are two chords in a circle of radius 5. What is the sum of the lengths of the two chords?
Note: Figure not drawn to scale
(E) It cannot be determined from the information given.
5. 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?
(D) 4 – 2
(E) 4 –
6. If 5x + 13 = 31, what is the value of ?
7. At Nat’s Nuts a -pound bag of pistachio nuts costs $6.00. At this rate, what is the cost, in cents, of a bag weighing 9 ounces? (Note: 1 pound = 16 ounces)
8. If 12a + 3b = 1 and 7b – 2a = 9, what is the average (arithmetic mean) of a and b?
9. Jessica has 4 times as many books as John and 5 times as many as Karen. If Karen has more than 40 books, what is the least number of books that Jessica can have?
Questions 10 and 11 refer to the following definition.
For any numbers a and b, .
10. Which of the following is equal to (1 2) 3?
(A) 1 6
(B) 2 3
(C) 3 4
(D) 4 5
(E) 5 6
11. For how many ordered pairs of positive integers (x, y) is x y = 2?
(E) More than 3
12. What is the largest integer, n, that satisfies the inequality n2 + 8n – 3 < n2 + 7n + 8?
13. If a < b and c is the sum of a and b, which of the following is the positive difference between a and b?
(A) 2a – c
(B) 2b – c
(C) c – 2b
(D) c – a + b
(E) c – a – b
14. If w widgets cost c cents, how many widgets can you get for d dollars?
15. If 120% of a is equal to 80% of b, which of the following is equal to a + b?
16. In the figure below, WXYZ is a square whose sides are 12. AB, CD, EF, and GH are each 8, and are the diameters of the four semicircles. What is the area of the shaded region?
(A) 144 – 128π
(B) 144 – 64π
(C) 144 – 32π
(D) 144 – 16π
17. Which of the following numbers can be expressed as the product of three different integers greater than 1?
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
18. What is the average of 4y + 3 and 2y – 1?
(A) 3y + 1
(B) 3y + 2
(C) 3y + 4
(D) y + 1
(E) y + 2
19. If x and y are integers such that x3 = y2, which of the following CANNOT be the value of y?
20. What is a divided by a% of a?
21. If an object is moving at a speed of 36 kilometers per hour, how many meters does it travel in 1 second?
22. For what value of x is 82x – 4 = 16x?
23. On a certain Russian-American committee, of the members are men, and of the men are Americans. If of the committee members are Russians, what fraction of the members are American women?
24. If m is a positive integer, which of the following could be true?
I. m2 is a prime number.
II. is a prime number.
III. m2 =
(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III
25. If x% of y is 10, what is y?
26. In writing all of the integers from 1 to 300, how many times is the digit 1 used?
27. If a + 2b = 14 and 5a + 4b = 16, what is the average (arithmetic mean) of a and b?
28. In the figure below, the area of circle O is 12. What is the area of the shaded sector?
Note: Figure not drawn to scale
29. At a certain university, of the applicants failed to meet minimum standards and were rejected immediately. Of those who met the standards, were accepted. If 1200 applicants were accepted, how many applied?
30. How many integers between 1 and 1000 are the product of two consecutive integers?
Note: For many problems, an alternative solution, indicated by two asterisks (), follows the first solution. In this case, one of the solutions is the direct mathematical one and the other is based on one of the tactics discussed in this chapter.
1. (E). Use TACTIC 1: draw a diagram representing a pile of books or a bookshelf.
In the 2 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 1996, Diana read 16 English books and 32 French books, a total of 48 books.
2. (D). Even if you can’t solve this problem, don’t omit it. Use TACTIC 2: trust the diagram. AC is clearly longer than OC, and very close to radius OE (measure them).
Therefore, AC must be about 10. Either by inspection or with your calculator, check the choices. They are approximately as follows:
(A) = 1.4 (B) = 3.1
(C) 5 = 7 (D) 10 (E) 10= 14.
The answer must be 10.
The answer is 10. The two diagonals are equal, and diagonal OB is a radius.
3. (C). As in question 2, if you get stuck trying to answer this, use TACTIC 2: look at the diagram.
Square OPQR, whose area is 8, takes up most of the quarter circle, so the area of the quarter circle is certainly between 11 and 14. The area of the whole circle is 4 times as great: between 44 and 56. Check the choices. They are approximately as follows:
(A) 8π = 25 (B) 8π = 36 (C) 16π = 50
(D) 32π = 100 (E) 64π = 200.
The answer is clearly 16π.
Use TACTIC 4: 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 π(4)2= 16π.
4. (E). Use TACTIC 3. Since the diagram has not been drawn to scale, you are free to redraw it. AB and AC could each be very short, in which case the sum of their lengths could surely be less than 5. Therefore, none of choices A, B, C, and D could be the answer. The sum cannot be determined from the information given.
5. (E). Use TACTIC 9: subtract to find the shaded area. The area of square ABCD is 4. The area of AED is (see Section 9-J). Then the area of the shaded region is 4 – .
Use TACTIC 2: trust the diagram. The area of the square is 4 and the shaded area appears to take up more than half of the square. So the area should be more than 2 but definitely less than 3. Only choice E (4 – ) works.
6. (C). Use TACTIC 10: don’t do more than you have to. In particular, don’t solve for x. Here 5x + 13 = 31 5x = 18.
7. (C). This is a relatively simple ratio, but use TACTIC 11 and make sure you get the units right. 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.
8. (B). Use TACTIC 17, and add the two equations to get
10a + 10b = 10 a + b = 1
Remember TACTIC 10: don’t do more than you have to. In particular, do not solve for a and b.
9. (B). Use TACTIC 5: backsolve. Since you want the least number, start with the smallest answer, E. If Jessica had 200 books, Karen would have 40; but Karen has more than 40, so 200 is too small. Neither 205 (D) nor 210 (C), is a multiple of 4, so John wouldn’t have a whole number of books. Finally, 220 works. (So does 240, but you shouldn’t even test it since you want the smallest value.)
Since Karen has at least 41 books, Jessica has at least 205. But Jessica’s total must be a multiple of 4 and 5, hence of 20. The smallest multiple of 20 greater than 205 is 220.
So, x y = 2 = 2
x2 + y2 = 2xy x2 – 2xy + y2 = 0.
So, (x – y)2 = 0 x – y = 0 x = y.
So (1, 1), (2, 2), (3, 3), (4, 4), … are all ordered pairs of positive integers that satisfy the given condition.
12. (D). Use TACTIC 5: backsolve (using your calculator). Test the choices, starting with E (since you want the largest value):
112 + 8(11) – 3 = 121 + 88 – 3 = 206,
112 + 7(11) + 8 = 121 + 77 + 8 = 206.
The two sides are equal. When n = 10, however, the left-hand side is smaller: 100 + 80 – 3 = 177 and 100 + 70 + 8 = 178.
n2 + 8n – 3 < n2 + 7n + 8 n < 11.
13. (B). Use TACTIC 6. Pick simple values for a, b, and c. Let a = 1, b = 2, and c = 3. Then b – a = 1. Only choice B, 2b – c, is equal to 1 when b = 2 and c = 3.
c = a + b a = c – b.
So, b – a = b – (c – b) = 2b – c.
14. (A). Use TACTIC 6: replace variables with numbers. If 2 widgets cost 10 cents, then widgets cost 5 cents each; and for 3 dollars, you can get 60 widgets. Which of the choices equals 60 when w = 2, c = 10, and d = 3? Only .
Convert d dollars to 100d cents, and set the ratios equal: .
Multiply both sides by 100d: x = .
15. (C). Use TACTIC 7: choose appropriate numbers. Since 120% of 80 = 80% of 120, let a = 80 and b = 120. Then a + b = 200. Which of the choices equals 200 when a = 80? Only 2.5a.
16. (C). If you don’t know how to solve this, you must use TACTIC 8: eliminate the absurd choices and guess. Which choices are absurd? Certainly, A and B, both of which are negative. Also, using your calculator, you can see that choice D is about 94, which is much more than half the area of the square, and so is much too large. Guess between C (about 43) and E (about 50). If you remember that the way to find shaded areas is to subtract, guess C: 144 – 32π, the answer choice with a minus sign.
The area of the square is 122 = 144. The area of each semicircle is 8π, one-half the area of a circle of radius 4. Together the areas of the semicircles total 32π, and the area of the shaded region is 144 – 32π.
17. (B). Treat the number in each of the three Roman numeral choices as a separate true/false question.
• Could 25 be expressed as the product of three different integers greater than 1? No, 25 has only two positive factors greater than 1 (5 and 25), and so clearly cannot be the product of three different positive factors. (I is false.)
• Could 36 be expressed as the product of three different positive factors? Yes, 36 = 2 × 3 × 6.
(II is true.)
• Could 45 be expressed as the product of three different integers greater than 1? No, the factors of 45 that are greater than 1 are 3, 5, 9, 15, and 45; no three of them have a product equal to 45. (III is false.)
Only II is true.
18. (A). To find the average, add the two quantities and divide by 2:
Use TACTIC 6. Let y = 1. Then 4y + 3 = 7 and 2y – 1 = 1. The average of 7 and 1 is = 4. Of the five choices, only 3y + 1 is equal to 4 when y = 1.
19. (D). Use TACTIC 5: test the choices. When there is no advantage to starting with any particular choice, start with E. Could y = 27? Is there an integer x such that x3 = 272 = 729? Use your calculator to evaluate or test some numbers:103 = 1000—too large; 93 = 729 works.
Try choice D: 16. Is there an integer x such that x3 = 162 = 256? No: 53 = 125, 63 = 216, so 5 and 6 are too small; but 73 = 343, which is too large. Alternatively, use your calculator to see that is not an integer.
The answer is 16.
Use TACTICS 6 and 7: replace a by a number; use 100 since the problem involves percents.
100 ÷ (100% of 100) = 100 ÷ 100 = 1.
Test each choice; which one equals 1 when a = 100?
A and B: = 1.
Eliminate C, D, and E; and test A and B with another value, 50, for a:
50 ÷ (50% of 50) = 50 ÷ (25) = 2.
Now, only works: = 2.
21. (A). Set up a ratio:
= 10 meters/second
Use TACTIC 5: Test choices, starting with C:
100 meters/second = 6000 meters/minute = 360,000 meters/hour = 360 kilometers/hour.
Not only is that result too big, but it is too big by a factor of 10. The answer is 10.
22. (D). Use TACTIC 5: backsolve, using your calculator. Let x = 4: then 82(4) – 4 = 84 = 4096, whereas 164 = 65,536. Eliminate A, B, and C, and try a larger value. Let x = 6: then
82(6)–4 = 88 = 16,777,216
166 = 16,777,216.
82x – 4 = 16x (23)2x – 4 = (24)x.
Then 3(2x – 4) = 4x 6x – 12 = 4x and so 2x = 12 x = 6.
23. (A). Use TACTIC 7: choose appropriate numbers. The LCM of all the denominators is 120, so assume that the committee has 120 members. Then there are × 120 = 80 men and 40 women. Of the 80 men, 30 are Americans. Since there are 72 Russians, there are 120 – 72 = 48 Americans, of whom 30 are men, so the other 18 are women.
Finally, the fraction of American women is , as illustrated in the Venn diagram below.
24. (D). Check each statement separately.
• Could m2 be a prime number? No, 1 is not a prime, and for any integer m > 1, m2 is not a prime since it has at least three factors: 1, m, and m2. (I is false.)
• Could be a prime number? Yes, if m = 4, then = = 2, which is a prime. (II is true.)
• Could m2= ? Yes, if m = 1, then m2 = , since both are equal to 1. (III is true.)
II and III only are true.
25. (C). Use TACTICS 6 and 7. Since 100% of 10 is 10, let x = 100 and y = 10. When x = 100, choices C and E are each 10. Eliminate A, B, and D, and try some other numbers: 50% of 20 is 10.
Of C and E, only = 20 when x = 50.
26. (160) Use TACTIC 14. 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 times (the same 20 but preceded by 2). Finally, from 100 to 199 there are 20 more plus 100 numbers where the digit 1 is used in the hundreds place. The total is 20 + 20 + 20 + 100 = 160.
27. ( or 2.5) Use TACTIC 10: don’t do more than is necessary. You don’t need to solve this system of equations; you don’t need to know the values of a and b, only their average. Use TACTIC 17. Add the two equations:
6a + 6b = 30 a + b = 5.
Then, or 2.5.
28. ( or 1.5) The shaded sector is of the circle, so its area is of or 1.5. (Note that, since fits in the grid, it is not necessary to reduce it or to convert it to a decimal. See Chapter 8.)
If you didn’t see that, use TACTIC 3 and redraw the figure to scale by making the angle as close as possible to 45°. It is now clear that the sector is of the circle (or very close to it).
29. (4000) Use TACTIC 7: choose an appropriate number. The LCD of and is 20, so assume that there were 20 applicants. Then (20) = 5 failed to meet the minimum standards. Of the remaining 15 applicants, , or 6, were accepted, so 6 of every 20 applicants were accepted. Set up a proportion:
30. (31) Use TACTIC 14. List the integers systematically: 1 × 2, 2 × 3, ..., 24 × 25, .... You don’t have to multiply and list the products (2, 6, 12, ..., 600, ...); you just have to know when to stop. The largest product less than 1000 is 31 × 32 = 992, so there are 31 integers.