Barron''''s How to Prepare for the SAT 23rd Edition (2008) _05 potx

Draw a Diagram. On any geometry question for which a figure is not pro- vided, 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. Example 2. 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 (ᐍ ) is 2 w . Now , write the required equation and solve it: 6 w = 4 ⇒ w = ⇒ 2 w = . The area of the rectangle = ᐍ w = = . Example 3. 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 60% , you have . Cross-multiplying gives: 5(10 + x ) = 3(40 + x ) ⇒ 50 + 5 x = 120 + 3 x ⇒ 2 x = 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 care- less mistake. Example 4. 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? (A) 2 (B) 2 (C) 4 (D) 4 (E) 8 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 A = d 2 (KEY FACT K8) to find that the area of square II is (4) 2 = (16) = 8, whereas the area of square I is 1. 1 2 1 2 1 2 2 I II 1 1111 1 1 1 1 2 2 10 40 3 5 + + = x x = ⎛ ⎝ ⎞ ⎠ 3 5 8 9 2 3 ⎛ ⎝ ⎞ ⎠ 4 3 ⎛ ⎝ ⎞ ⎠ 4 3 4 6 2 3 = 2w 2w P = 6w ww 1 1 11 P = 4 340 Math Strategies and Tactics Testing Tactics 1 Tactic x 30 10 Red Green Red Example 5. Tony drove 8 miles west, 6 miles north, 3 miles east, and 6 more miles north. How far was Tony from his 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 [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!] Example 6. 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 revo- lution, 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-con- suming to determine the angles between the hands at 1:27 and 1:28 (See TACTIC 10: Don’t do more than you have to).] II f a Diagram Is Drawn to Scale, Trust It, and Use Your Eyes. Remember that every diagram that appears on the SAT I 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, ques- tions of comparable difficulty would be answered cor- rectly 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. Example 7. In the figure at the right, 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? (A) 4 (B) 6 (C) 8 (D) 12 (E) 24 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 : A F B G BG Testing Tactics 341 C BA E D 6 3 6 6 35 8 F hour hand minute hand 12 1 30° 12 1 30° 30° in 1 hr. 0.5° in 1 min. 30° in 5 min. 6° in 1 min. A F B G CH D I E J 2 Tactic 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 mul- tiple-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. Example 8. In the figure at the right, square ABCD has been divided into four tri- angles by its diagonals. If the perime- ter of each triangle is 1, what is the perimeter of the square? (A) (B) 4( – 1) (C) 2 (D) 3 (E) 4 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 C, D, and E), but more than 1.5 (eliminate A). The answer must be B. 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 , which equals (see KEY FACT J8). Therefore, the perimeter of ᭝ BEC is + + s, which equals 1. Solving for s gives Finally, P = 4 s = 4( – 1). 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. Example 9. In the figure above, what is the value of x? (A) 55 (B) 95 (C) 125 (D) 135 (E) 145 Example 10. In the figure above, what is the value of x – y? (A) –30 (B) –15 (C) 0 (D) 15 (E) 30 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. 90° 50° 35° y° x° y° x° A C D B x° 35° 2 s = + =− 1 21 21. s 2 2 s 2 2 s 2 2 s 2 AE ED AP 1 1 s s s s P 1 21.5 AE ED AD 1 2 4 3 A F B C D E BG 342 Math Strategies and Tactics s A B D C E 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 30º or even 15º 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 + m∠ B + x = 90 + m∠ B + 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 prob- lems directly, practice TACTIC 2 and estimate the answers. The correct mathematical solutions without using this tactic are also given. Example 11. If, in the figure at the right, AB = AC, what is the value of x? (A) 135 (B) 125 (C) 115 (D) 65 (E) 50 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. Example 12. In the figure at the right, what is the sum of the mea- sures of all of the marked angles? (A) 360° (B) 540° (C) 720° (D) 900° (E) 1080° 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°. Example 13. In the diagram above, rays and are tangent to circle O. Which of the following is equal to z? (A) x (B) 180 – x (C) w + x + y (D) (E) 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 = 135 and = 90, 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). 270 3 270 2 wxy++ 3 wxy++ 2 PB  PA  • A P B O x° w° y° z° 105° 105° 110° 110° 100° 100° 30° 30° B C A x° 50° Testing Tactics 343 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 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 infor- mation you are given, and then to apply the technique of TACTIC 2. Helpful Hint 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. CAUTION: Redrawing a diagram, even roughly, takes time. Do this only when you do not see an easy direct solution to the problem. Example 14. In ᭝ACB, what is the value of x? Note : Figure not drawn to scale (A) 75 (B) 60 (C) 45 (D) 30 (E) 15 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 hypote- nuse of a right triangle is twice the length of one of the legs, you have a 30-60-90 triangle, and the angle formed by the hypotenuse and that leg is 60° (see Section 12-J). Example 15. In ᭝XYZ at the right, 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 Note : Figure not drawn to scale 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. Example 16. 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? (A) 15 (B) 25 (C) 30 (D) 45 (E) 60 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. O C B A x° 45° O x° C A B Z X z° x° y° Y Z Y X z° x° y° A C B x° 8 4 344 Math Strategies and Tactics 3 Tactic A C B x° 8 4 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. Example 17. In the figure at 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 — ? (A) r (B) r 2 (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 and PR — are equal to r (A). Note that you could also have made a “ruler” and seen that PR — is equal to r . Example 18. 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 (4)(4) = 8, and the area of ᭝ BAC is (6)(10) = 30. Then the area of ABCD is 30 + 8 = 38. Note that this problem could also have been solved by draw- ing 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 elimi- nate 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. Example 19. If the average (arithmetic mean) of 2, 7, and x is 12, what is the value of x? (A) 9 (B) 12 (C) 21 (D) 27 (E) 36 Solution. Use TACTIC 5. Test choice C: x = 21. • Is the average of 2, 7, and 21 equal to 12? • No: = = 10, 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: = = 12. The answer is D. Every problem that can be solved using TACTIC 5 can be solved directly, usually in less time. Therefore, we stress: if you are confident that you can solve a problem quickly and accurately, just do so. Here are two direct methods for solving Example 19, each of which is faster than backsolving. (See Section 12-E on averages.) If you know either method, you should use it and save TACTIC 5 for problems that you can’t easily solve directly. Direct Solution 1. If the average of three numbers is 12, their sum is 36. Then 2 + 7 + x = 36 ⇒ 9 + x = 36 ⇒ x = 27. Direct Solution 2. Since 2 is 10 less than 12 and 7 is 5 less than 12, to compensate, x must be 10 + 5 = 15 more than 12. Then x = 12 + 15 = 27. 36 3 2727 3 ++ 30 3 2721 3 ++ 1 2 1 2 (1,1) (5,1) (11,5) (1,7) A D C B E 4 4 10 6 F (1,1) (5,1) (11,5) (1,7) AD C B r 2 π r 2 π P Q O R Testing Tactics 345 4 Tactic P Q O R 5 Tactic A D C B E F 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. Example 20. If the sum of five consecutive even integers is 740, what is the largest of these integers? (A) 156 (B) 152 (C) 146 (D) 144 (E) 142 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. Example 21. 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 (C) $100,000 (D) $125,000 (E) $150,000 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 , 3 x , and 6 x. Therefore, x + 3 x + 6 x = $250,000 ⇒ 10 x = $250,000 ⇒ x = $25,000 ⇒ 6 x = $150,000. Helpful Hint 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. Example 22. If 2 + 5 = 8, then x = (A) (B) 0 (C) (D) 1 (E) 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 C . 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: 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 satis- fies a given condition. Examples 23 and 24 are two such problems. In Exam- ple 23, you are asked for the largest integer that satis- fies 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. Example 23. What is the largest integer, n, such that is an integer? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 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: = = 3.5. Eliminate E and try D. • Is an integer? Yes: 2 4 = 16, and = 7. • 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. 112 16 112 2 4 112 32 112 2 5 112 2 5 112 2 n 112 2 n 2+1= 9 4 2= 5 4 = 5 xxx⇒⇒ 88 22 +1+5=8 22 +1=3 2 +1= 3 2 xxx⇒⇒⇒ 5 8 2 2(1)+1+ 5 = 2 3 +5 8.46≈ 21+5 5 88 9 8 5 8 − 1 8 21x + 346 Math Strategies and Tactics 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 follow- ing ?” 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 eas- ier to test whole numbers than fractions and positive numbers than negative ones. Example 24. Which of the following is NOT equivalent to ? (A) (B) 60% (C) 0.6 (D) (E) Solution. Here, you have to test each of the choices until you find one that satisfies the condi- tion 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 immedi- ately eliminate choices B and C, since it is impossible that both are correct. • Test choice A. Reduce by dividing the numerator and denominator by 8: = . • 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. Example 19. 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 num- bers 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: = 13, 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. Example 20. 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. Example 25. If a is equal to b multiplied by c, which of the follow- ing is equal to b divided by c? (A) (B) (C) (D) (E) a bc 2 a c 2 a c ab c a bc 9 2 3 2730 3 39 3 ++ = = 9 2 3 2720 3 29 3 ++ = 3 5 15 49 3 7 5 7 × 3 7 7 5 ÷ 3 5 3 7 7 5 × 3 5 24 40 24 40 3 5 3 7 7 5 ÷ 3 7 7 5 × 24 40 3 5 Testing Tactics 347 6 Tactic Solution. • Pick three easy-to-use numbers that satisfy a = bc : for example, a = 6, b = 2, c = 3. • Solve the problem with these numbers: b ÷ c == . • Check each of the five choices to see which one is equal to : (A) = 1: NO. (B) = 6: NO. (C) = 2: NO. (D) = : YES! Still check (E): : NO. • The answer is D. Example 26. If the sum of four consecutive odd integers is s, then, in terms of s, what is the greatest of these integers? (A) (B) (C) (D) (E) Solution. • 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 25. a = bc ⇒ b = ⇒ b ÷ c = ÷ c = . Solution 26. Let n , n + 2, n + 4, and n + 6 be four con- secutive odd integers, and let s be their sum. Then: s = n + ( n + 2) + ( n + 4) + ( n + 6) = 4 n + 12. Therefore: n = ⇒ n + 6 = + 6 = + = . 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 alge- bra, you should use TACTIC 6 if you think you can solve the problem faster or will be less likely to make a mis- take. 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. 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; with- out 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. Helpful Hint 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. Example 27. 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? (A) (B) (C) (D) (E) csx Solution. • 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: , = 2, csx = 200. The answer is D. Example 28. If p painters can paint h houses in d days, how many houses can five painters, working at the same rate, paint in 2 days? (A) (B) (C) (D) (E) Solution. • Pick three easy-to-use numbers. Suppose that 1 painter can paint 1 house in 1 day. • Then, in 2 days each painter can paint 2 houses, and 5 painters can paint 10 houses. A quickly drawn chart can keep the numbers straight: 10dp h 10h dp 2 5 hp d 5 2 hp d dhp 10 x cs s cx = 1 8 xs c = 50, cx s = 8, x cs s cx xs c cx s s + 12 4 24 4 s − 12 4 s − 12 4 s − 12 4 a c 2 a c a c 28 4 s + = 16 4 32 4 . s + = 12 4 28 4 , s + = 6 4 22 4 , s − = 6 4 10 4 , s − = 12 4 4 4 , s + 16 4 s + 12 4 s + 6 4 s − 6 4 s − 12 4 a bc 22 6 23 6 18 1 3 === () 2 3 a c 2 == 6 3 6 9 2 a c = 6 3 1 63 3 1 ab c = ()() a bc = 6 23()() 2 3 2 3 b c 348 Math Strategies and Tactics Painters Houses Days 111 122 5102 • Evaluate the five choices when p = 1, h = 1, d = 1, and find the choice that equals 10: (A) = : NO. (B) = : NO. (C) = : NO. (D) = = 10: YES. (E) = = 10: YES. • Eliminate A, B, and C. But both D and E are 10. 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 1 day. Then 5 painters could paint lots of houses—certainly many more than 10. • 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: 3 x 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. The best choice in Example 28 would be to let p = 5 and d = 2, and let h be any number, say 4. Example 28 would then read, “If 5 painters can paint 4 houses in 2 days, how many houses can 5 painters, working at the same rate, paint in 2 days?” The answer is obviously 4, and only D is equal to 4 when p = 5 and d = 2. 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 d days, then 1 painter can paint houses in d days. In 1 day he can paint times the number of houses he can paint in d days; so, in 1 day, 1 painter can paint houses. Of course, in 1 day, 5 painters can paint 5 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. Example 29. 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 Example 30. If a = b + , b = 2c + , and c = 3d + , which of the following is an expression for d in terms of a? (A) (B) (C) (D) (E) Example 31. 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? (A) 20r (B) (C) 3rh (D) (E) 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 c = 3 , b = 7 , and a = 8. Which of the choices equals 1 when a = 8? Only (A). 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 20 of miles. Only (E) = 20 when r = 60 and h = 2. r 3 60 ⎞ ⎠ 1 3 ⎛ ⎝ 1 3 a − 2 6 1 2 1 2 r 3 hr 20 hr 3 43 24 a − 32 18 a − 23 12 a − 23 6 a − a − 2 6 1 2 1 2 1 2 10h dp 2 5h dp ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 5h dp 1 d h p h dp ×= 1 d h p 1 p x cs 3b a 3a b 3 x 10h dp 10 1 1 1 ()() 10dp h 10 1 11 () ()() 10h dp 21 1 51 2 5 ()() () = 2 5 hp d 51 1 21 5 2 ()() () = 5 2 hp d ()()()111 10 1 10 = dhp 10 Testing Tactics 349 [...]... in which the marbles could be drawn With 4 colors, there would ordinarily have been 24 orders, but since the first marble drawn was red, there are only 6 arrangements for the other 3 colors: BYG, BGY, YGB, YBG, GYB, GBY In 3 of these 6 the yellow comes before the blue, and in the other 3 the blue comes before the yellow Therefore, the probability that the yellow marble will be removed before the blue... closest corner Before picking B as the answer, however, ask yourself whether the point could possibly be closer to another corner of the table If the table were smaller, as in the diagram below, P would still be 5 feet from A, but could be closer to any of the other corners For example, now the distance from P to C is surely less than 5 A B 5 3 4 P D C The answer cannot be determined from the information... C 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 3 2 = 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... 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? (A) 4.0 (B) 5.7 (C) 6.0 (D) 6.5 (E) 6.7 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... find the difference of their absolute values and use the sign of the number with the larger absolute value 6+2=8 Example 5 What is the product of all the integers from –3 to 6? Solution Before reaching for your calculator, think You are asked for the product of 10 numbers, one of which is 0 Then, by KEY FACT A3, the product is 0 (–6) + (–2) = –8 To calculate either 6 + (–2) or (–6) + 2, take the difference,... elements or members of the set, and we say that the “thing” is in the set For example: • If A is the set of former presidents of the United States, then John Adams is an element of A • If B is the set of vowels in the English alphabet, then i is a member of B –3 –2 –1 0 – 1 1 2 2 3 4 2.53 2 Signed Numbers The numbers to the right of 0 on the number line are called positive, and those to the left of 0 are... “cents” you won’t forget to convert the units: 13 dollars is 1300 cents c cti Ta 12 Use Your Calculator You already know that you can use a calculator on the SAT (See Chapter 1 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,... 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 Example 55 1 1 If x2 = 2, what is the value of ⎛ x + ⎞ ⎛ x − ⎞ ? ⎝ x⎠ ⎝ x⎠ (A) 1 (B) 1.5 (C) 1 + (E) 1.5 + 2 2 2 (D) 2 + 2 Solution The College Board would consider this a hard question, and most students would either... not have to do long division, multiply three-digit numbers, or perform any other tedious calculations by hand If you use a calculator with fraction capability, you can even avoid finding least common denominators and reducing fractions The solutions to more than one-third of the mathematics questions on the SAT depend on your knowing the KEY FACTS in these sections Be sure to review them all The solution... is the set of all numbers that satisfy the equation Example 1 If A is the solution set of the equation x2 – 4 = 0 and B is the solution set of the equation x2 – 3x + 2 = 0, how many elements are in the union of the two sets? Solution Solving each equation (see Section 12-G if you need to review how to solve a quadratic equation), you get A = {–2, 2} and B = {1, 2} Therefore, A ∪ B = {–2, 1, 2} There . 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. 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. 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