Answers and explanations for ISEE practice drills - The ISEE

Cracking the SSAT & ISEE - The Princeton Review 2019

Answers and explanations for ISEE practice drills
The ISEE

ISEE MATH

Practice Drill 1—Lower Level

1. DList the factors of 24: 1 and 24, 2 and 12, 3 and 8, and 4 and 6. This totals 8 different factors of 24. The correct answer is (D).

2. DSince 12 is a factor of a certain number, Plug In for that certain number. For instance, Plug In 36 since 12 is a factor, and use POE. 2 and 6 are factors of 36, but they are not the only factors of 36 listed in the choices. Eliminate (A). Similarly, 3 and 4 are not the only factors of 36 listed, so eliminate (B). 12 is a factor, but it is not the only factor listed, so eliminate (C). Choice (D) contains all the other factors listed in previous choices. The correct answer is (D).

3. BA multiple of 3 will be 3 times a number. 2 is not a multiple of 3, so eliminate (A). 3 × 2 = 6, so keep (B). 10 and 14 are not divisible by 3, and therefore cannot be multiples of 3. The correct answer is (B).

4. CThe question asks which number is NOT a multiple of 6. 6 × 2 = 12 and 6 × 3 = 18, so (A) and (B) are multiples. 6 does not divide evenly into 23, so keep (C). 6 × 4 = 24 and 6 × 7 = 42 so (D) can be eliminated. The correct answer is (C).

5. DThe question is essentially asking for a number that is divisible by both 3 and 5. 10, 20, and 25 are all divisible by 5, but not by 3. Eliminate (A), (B), and (C). 45 ÷ 3 = 15 and 45 ÷ 5 = 9, so 45 is divisible by both 3 and 5. The correct answer is (D). Remember, you can also use the divisibility rules for 3 and 5 to help with this question! All the numbers end in either 5 or 0, so they are all divisible by 5. However, only (D) has digits that add up to a number divisible by 3.

6. BUse long division to find the remainder of 1,024 divided by 9. The remainder is 7, so add 2 to the total to make the number be divisible by 9. The correct answer is (B). Alternatively, you can use the divisibility rule for 9—the sum of the digits is divisible by 9. 1 + 0 + 2 + 4 = 7, so if 2 is added to that, the sum is 9, which is divisible by 9. Remember, the question asks for the smallest number that can be added to 1,024.

7. DUse PITA to solve this question. Since the question asks for the square of the largest of the five consecutive integers, label the choices as such. Then, start with (C) to get rid of choices more effectively. If 49 were the square of the largest integer, the integer would be 7, and other consecutive integers would be 6, 5, 4, and 3. The sum of these numbers 3 + 4 + 5 + 6 + 7 = 25, which is less than 30. Therefore, eliminate (A), (B), and (C), since these are too small. Try (D): the square root of 64 is 8, and the other integers would therefore be 7, 6, 5, and 4. 4 + 5 + 6 + 7 + 8 = 30, so the correct answer is (D).

8. CSince the question is asking for a specific amount and has real numbers in the choices, one way to solve this problem is to use PITA to test the answers, starting with (C). 75,000 × 6 = 450,000, which works. The correct answer is (C). Alternatively, you can translate the words. How many times as great as means divide the two numbers, so Image = Image = 6.

9. BUse PITA since the question is asking for a specific value and there are real numbers in the choices. The question asks for Joanne’s portion of the furniture, which is one-third of the total. Start with (C): 6 × 3 = 18, which is too large. Eliminate (C) and (D). Try (B): 4 × 3 = 12, which works. The correct answer is (B).

10. CSince the question asks for a specific value, use PITA to answer the question, starting with (C). The choices represent the amount of oil in the tank now, which is one-third of the total amount. Choice (C) is 30: is 30 one-third of the total amount (90 gallons) the tank holds? 30 = Image(90) is true, so the correct answer is (C).

11. DGinger sleeps Image of each day. To find how many days she sleeps over the course of four days, multiply: Image × 4. Simplify to solve: Image = Image = 3. The correct answer is (D).

12. ATo find the greatest value, use the choices and ballpark wherever possible to help. Choice (A) is close to 1, so use this as a comparison point. Choice (B) is smaller, since it is less than Image. Pay attention to the division sign in (C): Image ÷ Image = Image × Image = Image, which is also less than Image. In (D), multiply the fractions: Image × Image = Image = Image, which is equal to (C) and therefore less than Image. The greatest value is (A).

13. CRearrange these values by grouping together fractions with like denominators: Image + Image = 1, Image + Image = Image = 1, and Image + Image = Image = 1. Then add the whole numbers: 1 + 1 + 1 = 3. The correct answer is (C).

14. DWhen multiplying by a factor of 10, simply move the decimal point to the right for each zero. In this case, you are multiplying by 1,000, so move the decimal point to the right 3 places for the three zeros in 1,000. The decimal 0.34 becomes 340, which is closest to 350. The correct answer is (D).

15. CThe question is testing knowledge of decimal places. The answer should not have multiplication in it, so eliminate (A). Eliminate (D) as well since it does not have 2 included. In the number 2.398, 0.3 is equivalent to Image, 0.09 to Image, and 0.008 to Image. This correlates to (C), which is the correct answer.

Practice Drill 2—Middle and Upper Levels

1. ASince 4 is not a prime number and no multiple of 4 will be prime either, there will not be any numbers in common. Therefore, the correct answer is (A).

2. BFirst, list all the factors of 24: 1 and 24, 2 and 12, 3 and 8, 4 and 6. Next, list all the factors of 81: 1 and 81, 3 and 27, 9 and 9. The only factors that 24 and 81 have in common are 1 and 3, so there are two factors in common. The correct answer is (B).

3. DUse PITA here to find the correct answer. Usually, you would start with (B) or (C), but notice that finding Image (i.e., 25%) of either of these numbers will create a fraction. Try (D): Find Image of $40: Image (40) = Image = 10. If the tip was $10 and the price of the dinner was $40, then the total was $50. Since this matches the total in the problem, stop here. The correct answer is (D).

4. BFirst, since there are fewer multiples of 7, list the multiples of 7 from 1—99. The multiples are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, and 98. The multiples that would also be multiples of 2 would be the even numbers: 14, 28, 42, 56, 70, 84, and 98. This is a total of 7 numbers. The correct answer is (B).

5. BRemember, with exponents, you can write it out! 23 = 2 × 2 × 2 and 22 = 2 × 2, so you have (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2). Count up the number of 2s that you have, which is 8, and make that number the new exponent: 28. The correct choice is (B). Alternatively, you can use MADSPM: when multiplying the same base, add the exponents. Simply add 3 + 3 + 2 = 8. The answer will be 28.

6. BUse PITA to plug in the choices for m, starting with (C). Plug 3 in for m: 2(3) + 4 = 10 and 33 = 27, which are not equal. Since 3 is too large, eliminate (C) and (D). Next try (B): 2(2) + 4 = 8 and 23 = 8. Since this works, stop here. The correct answer is (B).

7. DFirst, solve for x and then find what the question is asking: x + 10. Start with 6x — 4 = 38. Add 4 to each side, and 6x = 42. Divide by 6 on each side to find that x = 7. Now plug 7 into x + 10 to find that 7 + 10 = 17. The correct answer is (D).

8. DLet the choices help here. Since the number must be greater than 50, eliminate (A) and (B). 51 is not a multiple of 7, so eliminate (C). 56 is a multiple of 7 and is the only remaining choice. Therefore, the correct answer is (D).

9. DFor word problem questions, translate the words to their math equivalents: if 4 (the students who chose recycling) is equal to one-fifth of the students in the class, then 4 = Image × n, where n is the number of students in the class. To solve, multiply both sides by 5, and n = 20. There are 20 students in the class. The correct choice is (D).

10. ATo find a percentage, find the portion the question asks for out of the total. First, find the total of all the grains: 60 bushels of corn + 20 bushels of wheat + 40 bushels of soybeans = 120 total bushels. The question asks for the percent of corn, so Image = Image = Image, which is equal to Image, or 50%. The correct answer is (A).

11. CTo find percent change, use this formula: % change = Image × 100. The difference here is $45 — $30 = $15, and the item was originally $45, so Image × 100. This reduces to Image × 100. To solve, Image = Image = 33 Image%. The correct answer is (C).

12. CUse Ballparking! 19.95 is roughly 20, and 35% is close to Image, so Image of 20 is between 6 and 7. Eliminate (D) since it is too big. Choices (A) and (B) are too small, so that leaves (C) as the closest. The correct answer is (C).

13. ATo find percent change, use this formula: % change = Image × 100. The original value is $50 and the final value is $20, so the difference is $30. Image × 100 reduces to Image × 100 = Image = Image = 60. The correct answer is (A).

14. BTake this question in bite-sized pieces. If four friends each pay $5 for a pizza, the pizza costs 4 × 5 = $20. Therefore, if a fifth friend joins, then 5 × p = $20. Divide both sides by 5, and each friend pays $4. The correct answer is (B).

15. CThe perimeter is all the sides added together. The sides of a square are all equal, so divide 56 by 4 to find that each side has a length of 14. The correct answer is (C).

16. AAn equilateral triangle has equal sides. Therefore, if one side has a length of 4, all sides have a length of 4. Add all the sides to find the perimeter: 4 + 4 + 4 = 12. The correct answer is (A).

More Practice: Upper Level Only

17. AIf bº = 45º, the other angle must also be 45º since 180º — 90º — 45º = 45º, which makes this an isosceles right triangle. Therefore, the other leg of the triangle is also 4. From here, use the Pythagorean Theorem to find v2: 42 + 42 = v2. Simplify the left side of the equation to get 16 + 16 = 32. The correct answer is (A). Be careful here. Notice that the question is asking for v2, not just v.

18. BTranslate this question into math: one-half of something means to multiply by Image, difference between means to subtract, degrees in a square is 360º, and degrees in a triangle is 180º. Thus, the equation will be Image(360 − 180). Simplify to get Image(180) = 90. The correct answer is (B).

19. CSince the question asks for a specific value, use PITA to answer this question, starting with (C). If the side length of a square is 4, its perimeter is 16 because 4 + 4 + 4 + 4 = 16 and its area is also 16 because 42 = 16. Since those are equal, stop here. The correct answer is (C).

20. DFirst find the area of the rectangle with a width of 4 and length of 3: A = l × w = 3 × 4 = 12. The area of the triangle is also equal to 12, so A = Image bh = 12. Plug in the given value for the base: Image(6)h = 12. Simplify to find that 3h = 12, and then divide both sides by 3. The height must be 4, so the correct answer is (D).

21. BFirst, find the volume of the box that has all dimensions known. V = lwh, so V = 3 × 4 × 10 = 120. Since the other box has the same volume, 120 = 6 × 4 × h. 120 = 24h, so h = 5. The correct answer is (B).

22. DUse the formula for the area of a square: A = s2. If A = 64p2, then to find the side length of the square, take the square root: Image = 8. Eliminate (A) and (C) since both choices have 64. For the square root of p2, you can plug in for p. Pick an easy number like 2. If p = 2, then p2 = 4. So Image = Image = 2, which means your answer should equal 2 when you plug in for p. Choice (B) has p2, which would be 4, so eliminate (B). Choice (D) has p, which is 2. This matches, so the correct answer is (D).

23. DThe length of AB is the same as all the different heights added together on the right-hand side of the figure. Therefore, the perimeter will contain two lengths of 10. Similarly, the length of AC is the same as all the different lengths added together that are across the figure (in this case, above AC), so there will be two lengths of 15. To find the perimeter, add all the sides: P = 10 + 10 + 15 + 15 = 50. The correct answer is (D).

24. CNotice the three triangles that have been created within the rectangle. Look at the two right triangles that surround the larger (possibly) equilateral triangle in the middle. Since each triangle has a right angle, the other two angles must equal 90º since 180º — 90º = 90º. Thus, in the triangle on the left side that includes side AB, w + x = 90º, and in the triangle on the right side that includes side CD, y + z = 90º. Add all these together to find that 90º + 90º = 180º. The correct answer is (C).

25. DNotice that the part that juts out on the left side of the shape would fit into the indented part on the right side of the shape. Filling in the hole would make a rectangle with a length of 8 and a width of 4 + 3 + 4 = 11. To find the area of a rectangle, use the formula: A = l × w. Therefore, A = 8 × 11 = 88. The correct answer is (D).

26. BThe length of police tape wrapping around a rectangle is the same as the perimeter. Draw a rectangle and label the length as 28 and the width as 6. Remember, in a rectangle, opposite sides are equal to each other. Calculate the perimeter by adding all the sides: 6 + 6 + 28 + 28 = 68. The correct answer is (B).

27. DTo find the distance between two points, draw a right triangle and use the Pythagorean Theorem. Draw a line straight down from point B and directly right from point A. That point will be (7,1), which you can label C. The distance from A to C is 6, and the distance from C to B is 8. Use the Pythagorean Theorem to find the missing side: 62 + 82 = c2. Simplify the left side to get 36 + 64 = 100. Take the square root of both sides to get c = 10. The correct answer is (D).

28. BEven though the length of the radius is unknown, it is still possible to find the angle measurements. There is a 90º angle in the center of the circle, and both OQ and OP are radii of the circle, which means they are the same length. Therefore, this is an isosceles right triangle, meaning the two smaller angles are congruent. All triangles have 180º, so 180º — 90º = 90º. The two smaller angles add up to 90º, so Image. The correct answer is (B).

29. DNotice that the four intersecting lines form a quadrilateral. All quadrilaterals contain 360º, so keep a tally of the vertices and find the missing angle. 80º is already provided, so 360º — 80º = 280º. All straight lines add up to 180º, so use the exterior angles to find the interior angles. If one of the exterior angles is 75º, the supplementary angle must be 105º. Subtract this from 280º to find that 280º — 105º = 175º. The other exterior angle, 108º, is opposite the interior vertex. Since opposite angles are equal, the interior vertex must also be 108º. Subtract this from the current total to find that 175º — 108º = 67º. The missing angle is 67º. The correct answer is (D).

30. CThe question states that triangle ABC is equilateral, so all 3 sides are equal to 4. Label AC as 4 and BC as 4. The question asks for perimeter, not area. So, there are two sides of the triangle that are part of the perimeter, so add them together: AB + BC = 4 + 4 = 8. Eliminate (A) and (B) since the choice must have an 8 in it. Now, find the rounded portion. The rounded portion is half of the circumference (i.e., a semicircle). Since you labeled BC as 4, you should see that the diameter of the circle must also be 4. If C = πd, then half of the circumference is Imageπd. Plug in the value for the diameter and simplify: Imageπ(4) = 2π. The full expression for the perimeter will then read 8 + 2π. The correct answer is (C).

31. DBreak the trapezoid into two triangles and a rectangle. Next, figure out the missing segment lengths. If MN = 8, then QP = 8. That means that 20 — 8 = 12, which is what LQ and PO must total. If each segment were the same, then LQ = 6 and PO = 6 since Image = 6. Next, find out QM and PN. You may notice that these are 6-8-10 right triangles. Otherwise, use the Pythagorean Theorem to find the height of the triangles: 62 + b2 = 102. Solve for b: 36 + b2 = 100; subtract 36 from both sides to get b2 = 64. Then take the square root of both sides, and b = 8. To find the area of the triangles, plug in the base and height into the formula for the area of a triangle: A = Image bh = Image(6)(8) = 24. There are two triangles, so 24 + 24 = 48. Eliminate (A) since you haven’t finished finding the area of the whole trapezoid yet. The rectangle QMNP is actually a square since all sides equal 8. To find the area of a square, use the formula: A = s2 = 82 = 64. Finally, add the areas: (2 triangles) + (1 square) = 48 + 64 = 112. The correct answer is (D).

Practice Drill 3—Ratios

1. CWhen the question asks about ratios, make a Ratio Box. The ratio of single rooms to doubles to triples is 3:4:5, so label the boxes and place this ratio in the top row. Add 3 + 4 + 5 to find the total for the room types is 12. Put 12 in the total column for the ratio row. The question states that there are 36 total rooms in the hotel, so this number goes in the total column for the actual number row. 12 times what equals 36? Since 12 × 3 = 36, the multiplier is 3. Find the number of doubles by multiplying: 4 × 3 = 12. The correct answer is (C).

Image

2. CThis question gives information about the total number of players first, so place this information in the bottom row and add 8 and 4 to find the total for the actual number row. Be careful to order the ratio in the way the question asks. There are more right-handed players, so the first number should be the bigger of the two numbers. Eliminate (A) and (B)! Next, divide out the largest possible common denominator, in this case 4, to find the most reduced form of the ratio. That means the ratio of right-handed players to left-handed players is 2:1. The correct answer is (C). Note: if you chose (A), you set up the ratio box backwards, showing left-handed players to right-handed players. Read carefully!

Image

3. CThe ratio of goat food to grass is 2:1, so place this in the ratio row of the box. The question also states that the goat eats 15 total pounds per day, so place this number in total column of the actual number row. To find the multiplier, find the total of the ratio, 2 + 1 = 3, and find what times 3 equals 15. 3 × 5 = 15, so the multiplier is 5. The question asks for the total amount of grass the goat eats, so 1 × 5 = 5. The correct answer is (C).

Image

Practice Drill 4—Averages

1. AUse an Average Pie to solve this question: Image. Place 3 in the # of items and 18 in for the average. Multiply these numbers to find the total, which is 54. The question asks for twice the sum, which is the same as twice the total, so 2 × 54 = 108. The correct answer is (A).

2. BUse two Average Pies to organize the information in this question—every time you see the word average draw an Average Pie Image. The first pie represents the information about the boys: 4 boys average 2 projects each, so place 4 in the # of items place and 2 in the average place. To find the total number of projects the boys complete, multiply 4 and 2 to find a total of 8 projects. Repeat this same process with the girls in the second pie. The 5 girls average 3 projects each, so place these numbers in their respective places in the Average Pie, and multiply to find a total of 15 projects. The question asks for the total number of projects in the class, so 8 + 15 = 23. The correct answer is (B).

3. AFirst, add the three scores to find Catherine’s current point total. 84 + 85 + 88 = 257. Next, make an Average Pie Image with 4 in the # of items place since there will be a fourth test, and 89 as the desired average. Multiply 4 × 89 to find a total of 356. Subtract the totals to find that 356 — 257 = 99. This means that she must score a 99 on the fourth test to raise her average to an 89. The correct answer is (A).

4. AUse two Average Pies to organize the information in this question—every time you see the word average draw an Average Pie Image. There are 6 students with an average test score of 72. Place 6 in the # of items place and 72 in the average place. Find the total number of points by multiplying 6 × 72 = 432. Make a separate Average Pie for the next portion of the question. If a seventh student joins the class, the # of items place now contains 7, and the desired average is 76. Multiply these together to find that 7 × 76 = 532. The difference between 532 and 432 is 100, so the seventh student must score 100 to change the average to 76. The correct answer is (A).

More Practice: Lower Level

5. AUse an Average Pie to solve this question: Image. List the number of donuts Anna ate each day: Monday = 2, Tuesday = 4, Wednesday = 2, Thursday = 4, Friday = 2, Saturday = 0, and Sunday = 0. The question asks for the average number of donuts she ate over the course of all 7 days, so add all the donuts she ate: 2 + 4 + 2 + 4 + 2 + 0 + 0 = 14. Put 14 in the total spot in your Average Pie. The # of items is 7 since the question asks about the whole week. Finally, divide these two numbers to get the average: Image = 2. The correct answer is (A).

6. BWhen you see the word average, draw an Average Pie Image. To find the average for the whole trip, find the total number of miles: 240 mi + 350 mi = 590 mi. Put 590 in the total part of the Average Pie. Then find the total number of hours Merry drove: 4 hrs + 7 hrs = 11 hrs. 11 will go in the # of items spot. Next, divide to find the average: Image. To save yourself some time, remember to ballpark! Notice that the result will not be a whole number, so the correct answer must be (B).

7. DTo find the profit, find the total amount of ticket sales and then subtract the expenses the school had for stage production and advertising. Since the show sold out, all 300 seats were purchased. If each seat cost $6, then the sales total was 300 × 6 = 1,800. Subtract the expenses from the total sales to find the net profit: 1,800 — (550 + 250) = 1,800 — 800 = 1,000. The correct answer is (D).

More Practice: Middle and Upper Levels

8. CUse two Average Pies to organize the information in this question—every time you see the word average draw an Average Pie Image.The question states that Michael scored an average of 24 points over his first 5 basketball games. Therefore, place 24 in the average place of the pie, and 5 in the # of items place. Multiply these numbers together to find that Michael scored a total of 120 points over the five games. To find how many points he must score on his sixth game to bring his average up to 25, use the second Average Pie to plug in the given information. Write 6 in the # of items place of to account for all six games, and 25 in the average place since that’s the desired average. Multiply these numbers to find he must score a total of 150 points over the entire 6 games. The difference between 150 and 120 is 30, so Michael must score 30 points in the sixth game to raise his average to 25. The correct answer is (C).

9. BEven though this problem doesn’t use the word average, you can still use Average Pies to help solve this question. The problem gives information about the weekly amount of rain, but the question asks about the daily amount instead. The daily amount will be the average (i.e., the amount of rain per day). Place 245 in the total spot of the first Average Pie and 7 in the # of items place. That gives you an average of Image = 35, which is the average daily amount for the current year. Do the same for the previous year in a second Average Pie. This time, 196 goes in the total spot and 7 goes in the # of items place. That equals an average of Image = 28. The question asks for how many more inches, so you will need to subtract the two daily amounts of rain: 35 — 28 = 7. The correct answer is (B).

10. CSince the question mentions the mean, create an Average Pie. Image. Joe wants to have an average of 230 or more, so place 230 in the average spot of the pie. In the # of items place, write in 5 because he has already read 4 books that were 200, 200, 220, and 260 pages long, and he is going to read one more. Multiply to find the total number of pages he must read: 5 × 230 = 1,150. He has already read 200 + 200 + 220 + 260 = 880 pages, so find the difference between these two totals to see how many pages long the fifth book must at least be: 1,150 — 880 = 270. The correct answer is (C).

Practice Drill 5—Percent Change

1. DThe question is testing percent change since it asks by what percent did the temperature drop? To find percent change, use this formula: % change = Image × 100. The change in temperature was 20°: 10° — (—10°) = 20°. Since the question asks for the percent the temperature dropped, the larger number will be the original number. Thus, the equation should read Image × 100, which reduces to 2 × 100 = 200. The correct answer is (D).

2. CThe question is testing percent change since it asks by what percent did the patty increase? To find percent change, use this formula: % change = Image × 100. The change in patty size is 4, which is given in the question. The new patty size is 16 oz, so the original patty size must have been 12 oz since 16 — 4 = 12. The equation will read Image × 100, which reduces to Image × 100 = Image = 33Image. The correct answer is (C).

Practice Drill 6—Plugging In

1. BThis is a Plugging In question because there are variables in the choices and the question stem contains the phrase in terms of. Plug In a value, work through the problem to find a target answer, and then check each of the choices to see which yields the target answer. For instance, Plug In x = $3. The question asks for the total amount of money donated, so 3 × 200 = 600. $600 is the target answer. Now, plug 3 into the choices for x to see which choice matches your target answer (600). Eliminate (A) because Image is way too small. Choice (B) works because 200(3) = 600. Eliminate (C) because Image is too small. Eliminate (D) because 200 + 3 or 203 ≠ 600. The correct answer is (B).

2. DThis is a Plugging In question because there are variables in the choices. Plug In a value, work through the problem to find a target answer, and then check each of the choices to see which yields the target answer. For instance, Plug In 6 for d dollars. If 10 magazines cost $6, then $3 would buy 5 magazines—you spend half as much money, so you can get only half as many magazines. So 5 is the target answer. Now, plug 6 into the choices to see which answer yields 5, the target answer. Eliminate (A) because Image = Image = 1.8 does not equal 5. Eliminate (B) because 30(6) is way too large. Choice (C) is a fraction, Image = Image, so it will not equal 5. Choice (D) works, as Image = 5, so keep this choice. The correct answer is (D).

3. CThis is a Plugging In question because there are variables in the choices and the question stem contains the phrase in terms of. Plug In a value, work through the problem to find a target answer, and then check each of the choices to see which yields the target answer. The zoo has four times as many monkeys as lions, so, for instance, Plug In 40 for the monkeys, which translates to 4 × lions = 40, so there are 10 lions. There are four more lions than zebras, which means that 10 — 4 = 6 zebras, so z = 6. The question asks how many monkeys are there in the zoo, so the target answer is 40. Now, plug 6 into the choices for z to see which choice matches your target answer (40). Eliminate (A) because 4 × 6 = 24 is not equal to 40. Eliminate (B) because 6 + 8 = 14 is still too small. Since 4(6) + 16 = 40, keep (C). Remember to try all four choices when Plugging In, so check (D) as well. 4(6) + 4 = 28, which is too small, so eliminate (D). The correct answer is (C).

More Practice: Lower Level

4. BThis is a Plugging In question because there are variables in the choices and the question stem contains the phrase in terms of. Plug In a value, work through the problem to find a target answer, and then check each of the choices to see which yields the target answer. For instance, say that y = 10 pairs of earrings. The total amount of money for 6 pairs of earrings would be 6 × 10 = 60, which is the target answer. Now, plug 10 into the choices for y to see which choice matches your target answer (60). Eliminate (A) because 6 + 10 = 16, which does not equal 60. Keep (B) because 6(10) = 60. Remember to check the remaining choices when Plugging In. 610 is a very large number, much greater than 60, so eliminate (C). Eliminate (D) as well because 6 + 6(10) = 66, which does not equal 60. The correct answer is (B).

5. BThis is a Plugging In question because there are variables in the choices. Plug In a value, work through the problem to find a target answer, and then check each of the choices to see which yields the target answer. For instance, say that p pieces of candy is equal to 5 pieces, and c cents is 10 cents. Therefore, 10 pieces of candy will cost 20 cents—you have twice as many pieces, so it will cost twice as much money. So, the target answer is 20. Now, Plug In your values for p and c into the choices to find the choice that equals your target answer (20). Eliminate (A) because Image = Image = 5, which is too small. Image = Image = 20, so keep (B). Remember to check the remaining choices when Plugging In. Image = Image = 5, so eliminate (C) as well. Cross off (D) because 10(5)(10) = 500, which is way too large. The correct answer is (B).

More Practice: Middle Level

6. CIn this question, J is an odd integer, so Plug In an odd integer for J. Since this is a must be question, see if there is a number that would make the answer untrue. Plug In 1 for J to make (A) untrue, since Image is not greater than 1. This number for J will also eliminate (B) since 1 — 2 = −1, which is not a positive integer. Choice (C) is true since 2 × 1 = 2, which is an even integer. Eliminate (D) since J could be negative. For example, if J = −3, −3 is not greater than 0. Check that value for (C) to be sure it always works. Again, if J = −3, then 2 × −3 = −6, which is still an even integer. Since it always works, the correct answer is (C).

7. CWhen there are percents or fractions without a starting or ending value in the question stem, feel free to Plug In. What number would make the math easy? 8 is a common denominator for Image and Image, so draw a picture of a fruit tart and divide it into 8 equal parts. Shade in the number of pieces she has eaten. On Monday, she ate Image of the pie, so Image of 8 is Image or 4 slices, leaving 4 slices for later. The next day, she ate Image of what was left. Image of 4 slices is 1, so she ate 1 slice. There are now 3 out of 8 slices left. Beware of choosing (A), however! The question asks how much she ate, so add up the slices she consumed. There should be 5 slices shaded (4 + 1 = 5), so the correct answer is (C).

More Practice: Middle and Upper Levels

8. CWhen there are percents or fractions without a starting or ending value in the question stem, feel free to Plug In. For instance, Plug In $100 for the starting price of the suit. It is reduced by 20%, so 20% of $100 is equal to Image(100) = Image = 20. That is the amount the suit is reduced. Subtract that from $100 to find the resulting price: $100 — $20 = $80. The suit is then reduced by 10%, so 10% of $80 is Image(80) = Image(80) = Image = 8. Subtract this from $80 to find the final price of the suit: $80 — $8 = $72. The final price is $72. The final price is what percent off of the original is another way of asking the final price is what percent less than the original. So, use the percent change formula: % change = Image × 100. The difference is $100 — $72 = 28. The original price was $100. Therefore, Image(100) = 28. The correct answer is (C).

9. CTry Plugging In values that satisfy the question stem, and eliminate choices. It may be necessary to Plug In twice on must be true or always true questions. If m is an even number, let m = 2, and let n = 3 since it must be an odd integer. If p is the product of m and n, then p = (2)(3) = 6. Now check the choices. Eliminate (A) because p is not a fraction. Eliminate (B) as well since p is not an odd integer. Keep (C) because 6 is divisible by 2. Finally, keep (D) because 6 is greater than zero. Plug In again to compare the remaining choices. Perhaps keep one number the same, so n = 3, but make m = −2 instead of 2. Now p = (−2)(3) = −6. Choice (C) still works since −6 is divisible by 2, but (D) no longer works since p is less than zero. Since it is always true, the correct answer is (C).

More Practice: Upper Level

10. BSince there are variables in the choices, Plug In a value for p, paying attention to the restrictions in the question. If p is an odd integer, make sure to Plug In an odd integer, for instance p = 3. Now, test the choices to see which one can be eliminated. Cross off (A) because (3)2 + 3 = 9 + 3 = 12, which is not odd. Choice (B) works since 2(3) + 1 = 6 + 1 = 7, which is odd. Choice (C) works since Image = 1. Choice (D) does not work since 3 — 3 = 0. Remember 0 is even, not odd. Plug In a second time for the remaining choices. Try p = 5. Choice (B) still works because 2(5) + 1 = 10 + 1 = 11, but eliminate (C) because Image is no longer an integer. The correct answer is (B).

11. BThe wording on this problem is tricky: it asks for which CANNOT be true, so try to find examples that COULD be true to eliminate choices. Pay attention to the restrictions in the problem, and Plug In two positive even integers: say 4 and 6. Thus, 4 + 6 = 10 = m. Next, eliminate choices that WORK. Choice (A) does not work since 10 is greater than 5. Keep it. Choice (B) does not work because 3(10) = 30, which is even, not odd. Keep it. Eliminate (C) because m = 10, which is even, so it works. Eliminate (D) as well because 103 ends in a zero, which is also even, so this statement works. Now, Plug In a second time for the remaining choices. Try new numbers, and remember that the numbers do not have to be distinct from one another. Try Plugging In 2 for both positive even integers. Thus, 2 + 2 = 4 = m. Check the remaining answers and eliminate the choices that WORK. For (A), 4 is less than 5. That works, so eliminate (A). For (B), 3(4) = 12, which does not work since it’s even, so keep it. The only choice left is (B), which is the correct answer.

12. DThis is a Plugging In question because there are variables in the choices and the question stem contains the phrase in terms of. Plug In a value, work through the problem to find a target answer, and then check each of the choices to see which yields the target answer. For instance, Plug In 20 for Anthony. Since Anthony has twice as many baseball cards as Keith, Keith has Image the number of cards that Anthony has. Therefore, Keith must have 10 cards, and k = 10. Keith has one-third as many baseball cards as Ian, so Ian has 3 times as many as Keith has: 10 × 3 = 30, or 30 cards. Together, Anthony and Ian have 20 + 30 = 50. So, 50 is the target answer. Now, Plug In 10 for k to find which choice yields 50, your target answer. Eliminate (A) because Image = Image = 15, not 50. Eliminate (B) because Image = Image = 30 is still too small. Choice (C) is still too small, as Image = Image = 40. Choice (D) works because Image = Image = 50. The correct answer is (D).

13. DThis is a Plugging In question because there are variables in the choices. Plug In a value, work through the problem to find a target answer, and then check each of the choices to see which yields the target answer. Let b = 4 and a = 3. Finding the product means multiply, so Image(4) × 32 = 2 × 9 = 18. The target answer is 18. Now, Plug In your values for b and a into the choices to find the choice that equals your target answer (18). Eliminate (A) since (3 × 4)2 = (12)2 = 144, which is too big. Eliminate (B) since Image = Image and is not equal to 18. Also eliminate (C) since 2(3) × Image(4) 6 × 2 = 12, which does not equal 18. Choice (D) works: Image = Image = Image = 18. Keep it. The correct answer is (D).

14. DUse MADSPM to simplify the exponents in the equations first. When raising a power to a power, multiply the exponents together. For the first equation, (x3)3 = x3×3 = x9, so a = 9. When dividing by the same base, subtract the exponents. For the second equation, Image = y10−2 = y8, so b = 8. The question asks to find a × b, so 9 × 8 = 72. The correct answer is (D).

15. BSince the question involves averages, use the Average Pie Image. However, save yourself some time by reading carefully! Notice that the classes have an equal number of students donating money. Because of this, simply Plug In values for the averages since it doesn’t matter how many actual students are donating money from each class. You only need values for the average of each class, so start there. Those two averages will become the numbers for your total in the next part of the problem. For instance, Plug In $3 as the average, y, for Mr. Greenwood’s class, and $5 for z, the average for Ms. Norris’s class. Add these two numbers to find the total amount of money donated (3 + 5 = 8), and put 8 in the total spot. There are 2 classes donating money, so the # of items is 2. Find the average by dividing: Image = 4. The target answer is 4. Now, Plug In 3 for y and 5 for z to find which choice yields 4, the target answer. Eliminate (A) because Image ≠ 4. Choice (B) works because Image = Image = 4. Eliminate (C) because 3 + 5 ≠ 4. Finally, eliminate (D) because 2(3 + 5) = 2(8) = 16, which is way too large. The correct answer is (B).

16. CFirst, simplify the first expression: (3xy)3 = 33x3y3 = 27x3y3. While comparing it to the other expression, 3x2y5, you can work in bite-sized pieces. Start with the coefficients: the greatest common factor of 3 and 27 is 3. Eliminate (A) and (D) since those don’t contain 3. Both of the remaining answers contain x2, so compare y in the two expressions. One has y3 = y × y × y and the other has y5 = y × y × y × y × y. The greatest common factor is y3 since both expressions have at least 3 y’s. Eliminate (B). The correct answer is (C).

Practice Drill 7—Plugging In The Answers

1. BThe question is asking for a specific value and there are real numbers in the choices, so use PITA to solve. Ted can read 60 pages per hour, which is 60 pages in 60 minutes, and Naomi can read 45 pages in 60 minutes. Combined, they can read 105 pages (60 + 45) in 60 minutes. Now, start with one of the middle choices to see which answer will yield a total of 210 pages. Try (B) since if they read for 120 minutes, they will read double the amount they did in 60 minutes. That will make the math easy. In 60 minutes they read 105 pages, so 105 × 2 = 210. This satisfies the question, so (B) is correct.

2. BThe question is asking for a specific value and there are real numbers in the choices, so use PITA to solve, starting with (C). The choices represent how much Sara pays. If Sara pays $30 and she pays twice as much as John, then John would have paid $15 since Image × 30 = 15. Paul paid three times as much as Sara, so he would have paid 3 × 30 = 90. This added together is more than $90, so eliminate (C) and (D), as these will amount to a total that is too much. Try (B): if Sara paid $20, John would have paid $10 since Image × 20 = 10. Paul paid 3 × 20 = 60. Add these amounts together to find that $20 + $10 + $60 = $90, which satisfies the question. The correct answer is (B).

3. DFirst, translate the English into math and then use PITA to test the choices. Four less than a certain number translates to n — 4, and two-thirds of a number translates to Image × n. So the equation is n − 4 = Image × n. Now, Plug In the Answers to find the one that satisfies the equation, starting with (C). If n = 8, then the equation will read 8 − 4 = Image(8). Since 4 ≠ Image, eliminate (C) and try another choice. Try (D): if n = 12, then 12 − 4 = Image(12), which is 8 = Image or 8 = 8. Since 12 works, stop here. The correct answer is (D).

More Practice: Lower Level

4. DThe question is asking for a specific value and there are real numbers in the choices, so use PITA to solve. The question asks for how many girls are in the class, so label the choices “girls” and create another label next to it marked “boys.” Test the choices, starting with (C). If there are 20 girls in the classroom, then subtract 12 to find the number of boys: 20 — 12 = 8 boys. 20 + 8 = 28, which means that (C) is too small. Eliminate (A), (B), and (C). Choice (D) must be the correct answer, so be aggressive if you’re worried about the time. If you have time to check, you will see that it works because if there are 21 girls, then 21 — 12 = 9, so there are 9 boys. 21 + 9 = 30, which matches the total given in the problem. The correct answer is (D).

5. BThe question is asking for a specific value and there are real numbers in the choices, so use PITA to solve. The question asks for how much Jonathan paid, so label the choices as such and create 2 additional columns next to it, one labeled as Victor and the other as Russell. Now, test the choices and follow instructions from the question stem, starting with (C). Choice (C) means that Jonathan paid $100. The question states that Victor paid twice as much as Jonathan, so Victor paid 100 × 2 = 200. The question then states that Victor paid half as much as Russell, so Russell paid twice as much as Victor: 200 × 2 = 400. Find the total: $100 + $200 + $400 = $700, which is too much, so eliminate (C) and (D). Try (B): if Jonathan paid $80, then Victor paid 80 × 2 = 160, and Russell paid 160 × 2 = 320. Find the total: $80 + $160 + $320 = $560, which is the amount stated in the question. The correct answer is (B).

More Practice: Middle and Upper Levels

6. DThe question is asking for a specific value and there are real numbers in the choices, so use PITA to solve. The question asks for Bob’s age, so label the choices “Bob” and create 2 additional columns next to it, one labeled as Adam and the other as Cindy. Now, follow the steps of the question and test the choices, starting with (C). If Bob is 18, then Adam must be 9 because Adam is half as old as Bob. It is also stated that [Adam] is three times as old as Cindy (remember, Adam is the subject of the sentence), so Cindy must be 3 since Image = 3. Find the total of their ages: 18 + 9 + 3 = 30, which is too small. Eliminate (A), (B), and (C). Choice (D) is the only answer left, so it must be correct. Stop work and move on! If you have time later to check, you’ll see that if Bob is 24, then Adam must be 12 and Cindy is 4, which makes the total 24 + 12 + 4 = 40. The correct answer is (D).

7. DThe question is asking for a specific value and there are real numbers in the choices, so use PITA to solve. The question asks for a possible value of x. So, Plug In for x and see if there is an integer that would work to make the rest of the equation balance. Start with one of the middle choices. Try (C): if x = 55, then 70(55) = 3,850. Solve the rest of the equation: 3,850 + 33y = 4,233 to see if y is an integer. Subtract 3,850 from both sides to find that 33y = 383. You can try dividing, but you could also ballpark. 33 × 11 = 363 and 33 × 12 = 396, so 383 is not divisible by 33. Therefore, eliminate (C) and try another choice. Try (D): 70(60) + 33y = 4,233 simplifies to 4,200 + 33y = 4,233. Subtract 4,200 from both sides to find that 33y = 33. Divide each side by 33 to find that y = 1. 1 is an integer, so this satisfies the question. The correct answer is (D).

8. AThis question gives a fair amount of information in the question, that the smallest of the three integers is 2, the sum of 2 + x + y = 9, and the product of 2xy = 24. Simplify these equations first and then use PITA to solve. Subtract 2 from either side of the sum to find that x + y = 7, and divide by 2 in the second equation to find that xy = 12. Now Plug In, starting with the largest number to find the largest number as efficiently as possible. Eliminate (C) and (D) right away since 8 or 9 added to another positive integer cannot equal 7. Try (B): Plug In 6 to get 6 + y = 7, so y would have to equal 1. This cannot be true, however, since the smallest integer has to be 2. Eliminate (B). Choice (A) is the only answer left, so it must be correct. Stop work and move on! If you have time later to check, you’ll see that if you Plug In 4 that 4 + y = 7 means that y = 3. These are both larger than 2, the smallest number. Do these numbers work in the second equation? Yes! 3 × 4 = 12. The correct answer is (A).

9. CBe sure to label the choices very carefully to stay visually organized. The question asks for Lori’s age now, so label the choices “L,” and create another column to the right and label it “C” for Carol’s age. Next, create two more columns and label them as “L + 10” and “C + 10” for their respective ages in 10 years. Now, Plug In starting with (C): if Lori is 20 years now, Carol must be 5 because Lori is 15 years older than Carol. This means that in 10 years, Lori will be 30 (20 + 10 = 30), and Carol will be 15 (5 + 10 = 15). The question states that in 10 years, Lori will be twice as old as Carol. 30 = 2 × 15, so this satisfies the statement. The correct answer is (C).

10. ALet the choices help here. The choices represent how many people are currently in the group. Eliminate (D) immediately since 30 cannot be divided 12 ways (hopefully there aren’t partial people in the car!). Try (B), which is the middle answer of the remaining choices: if there are 6 people renting the car now, the cost would be $5 each since Image = 5. If a seventh person joined, the cost per person would be Image, which is not an integer. Since the problem stated that adding 1 person to the group would result in each person owing $1 less, this does not satisfy the question, so eliminate (B) and try another choice. Try (A): if there are 5 people renting the car, they will each pay $6 since Image = 6. If a sixth person joins, they will each pay $5 because Image = 5. $5 is exactly one dollar less than $6, so this satisfies the question stem. Choice (A) is correct.

Extra Practice

1. BBallparking is one way to work through this problem. The shaded region looks to be about half the square, and half of 144 is 72, (B). To be more precise, the side of the square must be 12 since A = s2 and 122 = 144, so the height of each shaded triangle is 12. The base for one triangle ends at P and the base for the other triangle ends at Q, and PQ = 12 since it’s a side of the square. So make the base for each triangle 6 since the two bases must add up to 12. Plug those values into the formula: A = Imagebh = Image(6)(12) = 36. Since both areas equal 36, the total area of the shaded region is 36 + 36 = 72. The correct answer is (B).

2. BTake this question in bite-sized pieces. The question asks for the shaded region, so you want the part inside the square but outside the circle. In other words, if you find the area of the square and the area of the circle, you can find the shaded region by removing what you do not need (the area of the circle). First, find the area of the square. The side of the square is equal to 4, so A = s2 = 42 = 16. Eliminate (A), (C), and (D), since these do not contain 16. For added security, find the area of the circle. The radius is 2, so A = πr2 = π(2)2 = 4π. Remember to subtract that from the area of the square, so the full answer is 16 — 4π. The correct answer is (B).

Practice Drill 8—Functions

1. BDon’t be scared off by these types of questions! Simply follow the directions and plug numbers into the equation where specified. In this case, replace n with the given number (7). Thus, the equation should read $7 = 10(7) — 10. Simplify the equation to $7 = 70 — 10 = 60. The correct answer is (B).

2. CThe question asks which of the choices will yield a result of 120. Therefore, use PITA to solve, starting with (C). If n = 13, then replace 13 for n in the given equation: $13 = 10(13) — 10. Simplify to find that $13 = 130 — 10 = 120. This works, so the correct answer is (C).

3. DIn this function, simply plug in the number to the left of the weird symbol for d and the number to the right of the weird symbol for y exactly as the example directs. The function should read d ¿ y = 10 ¿ 2 = (10 × 2) — (10 + 2), which simplifies to (20) — (12) = 8. The correct answer is (D).

4. DThis question asks to first find the result of the function, and then to find the unknown K. Take this question in bite-sized pieces. Start with the parentheses first. Solve for 4 ¿ 3: (4 × 3) — (4 + 3) = (12) — (7) = 5. Next, plug 5 into the equation to find K: K(5) = 30. Divide by 5 on both sides to find that K equals 6. The correct answer is (D).

5. AYou will need to set the equation up based on the function defined above, and then use PEMDAS to simplify and solve for the end result. Take this question in bite-sized pieces. Start with the first set of parentheses: (2 ¿ 4) = (2 × 4) — (2 + 4) = (8) — (6) = 2. Next, work with the second set of parentheses: (3 ¿ 6) = (3 × 6) — (3 + 6) = (18) — (9) = 9. Put these values back in the original equation: (2 ¿ 4) × (3 ¿ 6) = (2) × (9) = 18. Now, test the choices to see which expression yields 18 as well. Try (A): (9 × 3) — (9 + 3) + 3 = (27) — (12) + 3 = 18. Since this matches, stop here. The correct answer is (A). Remember, if you find a question too time consuming, skip it and move on! You can come back to it later if you have time.

Practice Drill 9—Charts and Graphs

1. DFirst, find what District A spent in 1990: $400,000 (pay attention to the note below the table: the numbers are in thousands of dollars). Look for double this amount. $800,000 is listed in the table for the value in 1991 for District E. The correct answer is (D).

2. DAdd across to find which district spent the most, keeping in mind that these are all in the thousands (though this doesn’t really matter to find the largest sum). District E has the largest sum: $600,000 + $800,000 = $1,400,000. The correct answer is (D).

3. DRemember that these numbers are in the thousands. Add down to find the sum of the values in 1990: $1,800,000. Do the same with the values in 1991 to find a sum of $2,600,000. Find the difference of these values: $2,600,000 — $1,800,000 = $800,000. The correct answer is (D).

4. DCheck the graph. Carl owns 5 CDs, so the other two people together must own a total of 5 CDs. Eliminate (A) because Abe has 2 and Ben has 4, totaling 6. Eliminate (B) as well because Ben has 4 and Dave has 3, which is 7. Choice (C) is incorrect since Abe and Ed both have 2, so this amounts only to 4. Choice (D) works because Abe has 2 and Dave has 3, amounting to 5. The correct answer is (D).

5. BTo find which student owns one-fourth of all the CDs, first add all the CDs to find a total. Your work from the previous question will help! Abe = 2, Ben = 4, Carl = 5, Dave = 3, and Ed = 2, which yields a total of 16 CDs. Image of 16 is Image × 16 = Image = 4, so Ben is the student who has 4 CDs. The correct answer is (B).

6. DTo find Matt’s earnings for the week, first add up all his hours and then multiply by his hourly salary ($6/hour). He works 3.5 + 4 + 3.5 + 3 = 14 hours over the week, so 14 × 6 = 84. The correct answer is (D).

7. BRemember, if you see the word average, you can use an Average Pie Image. The previous question helped you find the total number of hours Matt worked: 14. Put that number in the total place. He worked 4 days—note the question says on the days he worked not the number of days in a week. Put 4 in the # of items place. Divide these two numbers to find the average: Image = 3.5. If you’re pressed for time, instead of doing the long division, let the answer choices help! Since 14 is not divisible by 4, eliminate all the integers. The correct answer is (B).

8. COne way to solve this problem is to translate the words into math: The hours he worked on Monday is 3.5, accounted for is equals, what percent is Image, and the total hours he worked is 14. The equation is 3.5 = Image × 14. Simplify the right side: Image × 14 = Image = Image. Multiply both sides by 100 to get 350 = 14x. Divide both sides by 14, and x = 25. The correct answer is (C). You can also find a percent by dividing the desired amount by the total amount: Matt worked 3.5 hours on Monday and a total of 14 hours, so Image = Image = Image, or 25%.

Practice Drill 10—Quantitative Comparison—Middle and Upper Levels

1. CLook at Column B first. 17 × 2 + 17 is the same as 17 + 17 + 17, or 17 × 3. Thus, the two columns are equal. The correct answer is (C).

2. ADraw a picture. For instance, draw a pie and divide it into eight parts since eight is a common denominator for Image and Image. Since Image = Image, four of the eight parts of the pie would be colored in. Only three out of eight would be filled in for Image. Therefore, Column A is greatest, and the correct answer is (A).

3. BThere are variables in the columns, so Plug In twice. For instance, try b = 10. 10 + 80 = 90, and 10 + 82 = 92. In this case, Column B is greater, so eliminate (A) and (C). Try another number, perhaps a negative number: —10. Perform the necessary calculations: —10 + 80 = 70 and —10 + 82 = 72. Column B is still greater, so the correct answer is (B).

4. BPlug In a value here. Say that Matt is 60 inches tall, making Rob 58 inches tall since Rob is two inches shorter than Matt. The question stem also states that Joel is four inches taller than Matt, so Joel must be 64 inches tall. This makes Joel taller than Rob, so Column B is greater. The correct answer is (B).

5. CWhen dealing with exponents, write it out! Column A can be rewritten as 16 × 16 × 16. Column B can be rewritten as 4 × 4 × 4 × 4 × 4 × 4. Notice that 16 is the same as 4 × 4. Therefore, Column A can also be written as (4 × 4) × (4 × 4) × (4 × 4). Since each column contains 6 fours, the two columns are equal. The correct answer is (C).

6. DThe information given does not indicate the direction or orientation of either girl’s house. However, the information does state that Kimberly lives two miles from school, so Column A is 2. However, Jennifer could live another two miles past Kimberly’s house, four miles in the other direction from the school (making the two houses 6 miles apart), or she could even live 4 miles north or south of the school (making the distance between their houses yet another value). Since there is no way to determine the distance between their houses without more information, the solution cannot be determined. The correct answer is (D).

Practice Drill 11—Quant Comp

1. BSince there are variables in the columns, Plug In a number. Pay attention to the restriction given: Plug In a number greater than 1 for x. Let x = 4. Column A is equal to 4, and column B is equal to 42, or 16. Since column B is greater, eliminate choices (A) and (C). Try a different number to see if Column A could be greater or if the quantities could be equal. Since x > 1, x cannot be negative, zero, or one. Try a very large number. 1,0002 is much larger than 1,000, so Column B is still greater. You could also try a decimal, like 2.5. In this case, Column B is still greater since 2.52 = 6.25, which is greater than 2.5. Therefore, since Column B is always greater, the correct answer is (B).

2. CRead the question carefully: b is an integer and —1 < b < 1. There is only one integer between —1 and 1. Therefore, b must be 0. Plug 0 in for b into each of the columns. Column A is Image. Column B is Image. The quantities are equal, so (C) is the correct answer.

3. DSince there are variables in the columns, Plug In values for p and m. For instance, let p = 16 and m = 3. Since it takes 4 quarts to make one gallon, Column B is less than 1 gallon while Column A is 16 gallons. This makes Column A greater. However, the question does not state anything about requirements for these numbers, and the values could easily be reversed, that p = 3 and m = 16. The 16 quarts in Column B is equal to 4 gallons, which is greater than the 3 gallons in Column A. Since this could be true as well, it cannot be determined which quantity is larger. The correct answer is (D).

4. ASince there are variables in the columns, Plug In a number. Pay attention to the restriction given: if x must be a positive integer, Plug In a positive integer for x. For example, let x = 3. Column A is Image while Column B is Image. If you’re not sure which value is greater, draw a picture. You can also use Bowtie to compare fractions. Column A becomes Image, and Column B becomes Image. Thus, Column A is greater. Eliminate (B) and (C). Try Plugging In another value for x to see if another outcome is possible. Remember the restriction given, so x cannot be negative or zero, so try a large integer. Make x = 100. Column A is Image = 25, and Column B is Image = 20. Column A is still greater. You could also try x = 1, but you will get the same result. Column A will be greater since Image = 0.25 is greater than Image = 0.20. The correct answer is (A).

5. DSince there are variables in the columns, Plug In values for p and w, according to the information given: w is an integer less than 4, so let w = 3. You are also given that p is an integer greater than 10, so let p = 11. Therefore, Column A is (3)(11) = 33 while Column B is equal to 3. In this case, Column A is greater. Eliminate (B) and (C). Now, try Plugging In different numbers to see if another outcome is possible. Let w = 0 and p = 12. In Column A, (12)(0) = 0. This is equal to Column B since w = 0. Since Column A isn’t always greater nor are the two columns always equal, the correct answer is (D).

6. DSince there are variables in the columns, Plug In a value for c. Let c = 2. In Column A, 4(2) + 6 = 8 + 6 = 14. Do the same for Column B: 3(2) + 12 = 6 + 12 = 18. In this case, Column B is greater, so eliminate (A) and (C). Now, try a different number, perhaps a negative number. Let c = —10. Now, Column A will read 4(—10) + 6 = —40 + 6 = —34. Do the same to Column B: 3(—10) + 12 = —30 + 12 = —18. In this case, —18 > —34, so Column A is now greater. Since neither column is always greater, the correct answer is (D).

Practice Drill 12—Quantitative Comparison—Middle and Upper Levels

1. CFind the total cost in each of the columns. Column A contains the statement The total cost of 3 plants that cost $4 each, so 3($4) = $12. Column B contains the statement The total cost of 4 plants that cost $3 each, so 4($3) = $12. The columns are equal, so the correct answer is (C).

2. DFirst, simplify the expression in Column A. Distribute the 30 in the expression 30(1 — 2n) to get 30 — 60n. Now, Plug In a value for n to solve each column. Let n = 2. In Column A, 30 — 60(2) = —90. In Column B, 30 — 2(2) = 26. Column B is greater, so eliminate (A) and (C). Now, Plug In a second time, trying a different number (remember, try 1, 0, fractions/decimals, negatives, or large or small numbers to find different outcomes). Try a negative number here. Let n = —3. Column A will now read 30 — 60(—3) = 210, and Column B will read 30 — 2(—3) = 36. Column A is greater in this case. Since neither column is always greater, the correct answer is (D).

3. DYou are given the statement The product of 3 integers is 48. There are many ways to reach a product of 48. For instance, 2 × 3 × 8 = 48. Of the three integers, the smallest is 2, so Column A is 2. Compared to Column B, Column A is greater. Eliminate (B) and (C). However, this is not the only way to multiply integers to get a product of 48. For example, 1 × 2 × 24 = 48. In this case, the smallest of the three integers is 1, which means the two columns are equal. Since Column A is not always greater nor are the two columns always equal, the correct answer is (D).

4. CFirst, simplify Column A: (x + y)(xy) can be FOILed out to be x2 + xyxyy2. The two middle values cancel each other out, so the expression reads x2y2, which is the same expression in Column B. Since the two columns are equal, the correct answer is (C). Note that you can also Plug In values for x and y and solve the problem this way. You should try more than one set of numbers to check for other possible outcomes.

5. AUse correct PEMDAS to evaluate the expression in Column A. First, work within the parentheses: (7 − 4) × 3 − 3 = (3) × 3 − 3. There are no exponents, so the next step is to do the multiplication and division from left to right: (3) × 3 − 3 = 9 − 3. Finally, add and subtract from left to right: 9 — 3 = 6. Since 6 is greater than 0, Column A is greater. The correct answer is (A).

6. ASince line m is equal to y = x + 4, the slope of line m is equal to 1. Remember, the slope is the coefficient of x in linear equations. Therefore, Column A is 1. Perpendicular lines will have slopes that are negative reciprocals of one another. Therefore, line l, which is perpendicular to line m, will have a slope of —1 since the negative reciprocal of Image is −Image. Thus, Column B is —1. Since 1 is greater than —1, Column A is greater. The correct answer is (A).

7. BWork through the information provided to find the value of Column A. If the shoes are $100 and the price is increased by 20%, find 20% of $100 and add that result to the total. Image(100) reduces to Image(100) = Image = 20. The price increased $20, so $100 + $20 = $120. The shoes are now $120. However, the price was reduced by 20%. Find 20% of 120 and subtract that result from the total. Image(120) reduces to Image(120) = Image = 24. The price decreased $24, so $120 — $24 = $96. The final price of the shoes is $96, so Column A is $96. Since Column B is $100, it is greater. The correct answer is (B).

8. BRemember, a negative sign will stay negative with an odd exponent. Both columns contain negative numbers and odd exponents, so both columns will remain negative. With negative numbers, the value that is closest to zero will be the greater value (e.g., −1 > −4). When working with fractions, remember that as the denominator gets larger, the fraction will get smaller (e.g., Image > Image). However, with negative fractions, the one that is “less negative” will be greater (e.g., −Image > −Image). In Column B, the exponent is greater, so the denominator in Column B will be larger and thus the value of the fraction will be smaller. Since the fraction in Column B will be less negative than the fraction in Column A, the value in Column B is greater. The correct answer is (B).

9. ABoth fractions are positive and both are being raised to a positive, even power. However, be careful when working with fractions less than 1. When those numbers are raised to a positive power, they become smaller since the denominator increases. Remember, as the denominator gets larger, the fraction will get smaller (e.g., Image > Image). Thus, the denominator in Column A (64) will be smaller than the denominator in Column B (66), which means the value in Column A will be greater. The correct answer is (A).

10. AThe negative sign in both columns will become positive since both columns are being raised to an even power. With fractions less than 1, the larger the denominator, the smaller the fraction. Since the denominator in Column A (62) will be smaller than the denominator in Column B (64), the value in Column A will be greater. The correct answer is (A).

11. AWith fractions less than 1, the larger the denominator, the smaller the fraction. Since the denominator in Column A (63) will be smaller than the denominator in Column B (65), the value in Column A will be greater. The correct answer is (A).

12. DThere are no instructions as to the values of x and y, so Plug In. For instance, x and y could be equal. Let both x and y equal 4. Column A would be C = 2(4)π, or 8π, and Column B would be A = π(4)2 = 16π. In this case, Column B is greater, so eliminate (A) and (C). Since x and y do not have to be equal, plug in a second time to see if a different outcome is possible. Say that x = 2 and y = 1. Then Column A will be C = 2(2)π = 4π, and Column B will be A = π(1)2 = π. Now, Column A is greater. Since neither column is always greater, the correct answer is (D).

13. CList the prime numbers on the 6-sided die: 2, 3, and 5 (Note! 1 is not a prime number). Since there are 6 sides on the die, the probability of rolling a prime number is Image, so Column A is Image. This is the same value listed in Column B. Since the columns are equal, the correct answer is (C).

14. DSince there are variables in the columns, Plug In values for a and b. If a and b are integers and a + b = 5, choose numbers to make both statements true. Let a = 3 and b = 2. In this case, Column A is greater, so eliminate (B) and (C). However, these numbers could easily be reversed because there are no restrictions listed. If a = 2 and b = 3, then Column B would be greater. Since neither column is always greater, the correct answer is (D).

15. AEvaluate the expressions, using order of operations. Column A is Image = Image = 4, and Column B is ImageImage = 5 − 3 = 2. Therefore, Column A is greater. The correct answer is (A).

16. CStart with Set B, since it is a finite set. Set B consists of 5, 10, 15, 20, 25, 30, 35, 40, and 45. Set A contains all prime numbers. The only prime number contained in Set B is 5, so the intersection of these two sets (i.e., Set C) will contain only the number 5. There is only one number in Set C, so the columns are equal. The correct answer is (C).

17. BWhen multiplying fractions, multiply the numerators across and the denominators across. Therefore, the value of Column A is Image × Image = Image = Image. When adding fractions with a common denominator, add the numerators. Thus, the value of Column B is Image + Image = Image = Image = 1Image. Column B is greater. The correct answer is (B).

18. CSince there are variables in the columns, Plug In values for a and b, paying attention to restrictions given. Since a > 0, let a = 2. Since b < 0, let b = —3. Now, plug in these numbers to the expressions. Column A is −(2 × −3) = −(−6) = 6. Column B is −2 × −3 = 6 as well. The two columns are equal. You can try testing different values for a and b, but the columns will always be equal because the negative signs will always cancel out and the same numbers are being multiplied in each column, so the result will not vary. The correct answer is (C).

19. BThis question is testing math vocabulary. The median of a set is the middle number when the numbers are listed in order from least to greatest. In Set A, the numbers are already in order, so find the middle number: 8. Thus, Column A is 8. In Set B, the numbers are also already in order. However, there is an even number of items in this set, so the median will be found by taking the average of the two middle numbers. The two middle numbers are 8 and 9, so the median is Image = Image = 8.5. The value in Column B is 8.5. Since Column B is greater, the correct answer is (B).

20. CInclusive means to include the outer limits—in this case, 1 and 10. In Column A, add up all the integers, including 1 and 10: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55. Simplify Column B to get 5 × 11 = 55. Since the columns are equal, the correct answer is (C).

21. BRemember, with exponents, write it out! Thus, Column A can be rewritten as (2 × 2 × 2) + (2 × 2 × 2) + (2 × 2 × 2) = 8 + 8 + 8 = 24. Note that in Column A, the quantities are being added, not multiplied. Column B can be rewritten as 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2, which will equal a value larger than 24. Therefore, Column B is greater. Another way to solve this problem is to use exponent rules. When exponents with the same base and same exponent are being added, simply treat them like any variable. For example, x + x + x = 3x. Thus, 23 + 23 + 23 = 3(23) = 3(8) = 24. This is not the same as 29, which is a much larger number. Since Column B is greater, the correct answer is (B).

22. DSince there are variables in the columns, Plug In. First, simplify Column A: 7(x — 3) = 7x — 21. Now, Plug In a value for x. For instance, let x = 3. Plug In 3 to Column A to find that 7(3) — 21 = 21 — 21 = 0. Now, do the same for Column B: 21 — 7(3) = 21 — 21 = 0. The columns are equal, so eliminate (A) and (B). Remember, Plug In a second time to see if a different outcome is possible. This time let x = 2. In Column A, the expression will read 7(2) — 21, which simplifies to 14 — 21 = —7. In Column B, the expression will read 21 — 7(2), which simplifies to 21 — 14 = 7. Now Column B is greater. Since the columns are not always equal nor is Column B always greater, the correct answer is (D).

23. BFirst, take the instructions in Column A into bite-sized pieces. The smallest positive factor of 25 is 1, and the biggest positive factor of 16 is 16. Thus, 1 × 16 = 16. Column A is 16. Since Column B is greater, the correct answer is (B).

24. AThe probability of getting heads on any single flip is Image, since there are two sides of a coin and only one of those is heads. Column A is the probability of getting heads on 2 consecutive flips, so you need to multiply the probability of the first flip by the probability of the second flip (i.e., Event 1 × Event 2): Image × Image = Image = Image. To find Column B, include the probability of a getting heads on third flip (i.e., Event 1 × Event 2 × Event 3): Image × Image × Image = Image = Image. Since Column A is greater, the correct answer is (A). Note: If you know that the probability of getting the same result (in this case, heads) on consecutive flips of a coin diminishes with each subsequent flip, you can find the correct answer quickly!

25. CUse the formula given to find the areas of each figure. In Column A, the height of Cylinder A is 10 cm and the radius is 6 cm. Plug these values into the volume formula to find that V = π(6)2(10) = π(36)(10) = 360π. Column A is 360π. For Column B, you can find the volumes of Cylinder B and Cylinder C separately and then add the results together. Alternatively, since the two figures have the same dimensions, you can multiply the volume formula by 2 to find the total volume. Thus, Column B is V = 2π(6)2(5) = 2π(36)(5) = 2π(180) = 360π. Since the columns are equal, the correct answer is (C).

26. CSince all the angles are labeled as x, they are all equal. The angles in a triangle must add up to 180º; therefore, each angle is equal to 60º since Image = 60. Thus, the value of Column A is 60, which makes the columns equal. The correct answer is (C).

27. ATo find percent change, use this formula: % change = Image × 100. In Column A, the difference from 1 to 2 is 1. Since you are looking for percent increase, the smaller number will be the original number. Thus, the original number is 1. Column A is Image × 100, which simplifies to Image × 100 = 100. For Column B, the difference from 2 to 1 is still 1. However, since you are looking for percent decrease, the larger number will be the original number. Thus, the original value is now 2. Column B is Image × 100, which simplifies to Image × 100 = 50. The value in Column A is greater, so the correct answer is (A).

28. CRemember, if you see the word average, you can use an Average Pie Image. First, find the sum: 4 + 6 + 8 + 10 = 28. That number goes in the total place. There are 4 numbers, so put 4 in the # of items spot. Divide to get Image = 7. Column A is 7. Remember that the median of a set is the middle number when the numbers are listed in order from least to greatest. Since there is an even number of items in this set, the median will be found by taking the average of the two middle numbers, 6 and 8. The average of 6 and 8 is Image = Image = 7, so the value in Column B is 7. Since the two columns are equal, the correct answer is (C).

29. DSince there are variables in the columns, Plug In values for the variables and evaluate the expressions. For instance, let x = 3 and y = 5 since x and y must be positive numbers. Column A is Image = Image = 7.5 whereas Column B is Image = Image, which is a little bit less than 4 since Image = 4. Since Column A is greater, eliminate (B) and (C). Now, Plug In a second time to see if a different outcome is possible. Let x = 1 and y = 1. Column A is Image = Image, and Column B is Image = Image = 1. This time, Column B is greater. Since neither column is always greater, the correct answer is (D).

30. CThis is a common trick on the ISEE. Notice that all the numbers are exactly the same and in the exact same order. All that is different is the placement of the decimals. 567.83 and 5.6783 are off by two decimals, or a factor of 100. Similarly, 0.40 and 40.0 are related by a factor of 100, or two decimal places as well. This cancels itself out, and the product will be the same for both columns. Check if you want to see for yourself, but if you know this trick it will save you some time on the test! The correct answer is (C).

31. CFirst, find the probability of not picking red shoes, which is the same as picking purple or white shoes, and then compare the result to Column B’s value. There are 10 total pairs to pick from, and there are 8 pairs of shoes that are purple or white (i.e., not red), so the probability is Image = Image. Column A is Image which is the value of Column B. Since the columns are equal, the correct answer is (C).

32. BFind the total of each column. Column A is 10 × 8 = 80, and Column B is 20 × 4.5 = 90. Column B is greater, so the correct answer is (B). Remember, you can estimate Column B’s value if you’re running short on time. 20 × 4 = 80; however, since 20 is really being multiplied by 4.5, not 4, you know the product has to be more than 80.

33. CRemember, if you see exponents, you can always write it out! You could also Plug In since there are variables in the columns. However, another way to find the values of the two columns is to use the exponent rules (MADSPM). For Column A, Image = Image = Image = x7−4 = x3. Column B is also x3. Since the two columns are equal, the correct answer is (C).

34. DThe information given is x2 = 36. Remember that 6 is not the only solution to this equation. 62 is equal to 36, but (—6)2 is also equal to 36 since two negatives multiplied together will equal a positive number. Since Column A can be greater than Column B but could also be equal to Column B, the correct answer is (D).

35. CRemember that factors are numbers you can multiply together to equal another number. The largest positive factor of 16 is 16, since 1 × 16 = 16. Thus, Column A is 16. Multiples are the result of multiplying two numbers together. The smallest positive multiple of 16 is also 16, since 16 × 1 = 16. Therefore, Column B is also 16. Since the columns are equal, the correct answer is (C).

36. BSince there are variables in the figure, Plug In a value for x. For instance, let x = 2. The perimeter of square ABCD is equal to 2 + 2 + 2 + 2 = 8. The perimeter of MNOP is equal to 2 + 2(2) + 2 + 2(2), or 2 + 4 + 2 + 4 = 12. In this case, Column B is greater, so eliminate (A) and (C). Try Plugging In a second time to see if a different outcome is possible. Try x = Image. The perimeter of square ABCD is equal to Image + Image + Image + Image, or 1 + 1 = 2. The perimeter of MNOP is equal to Image + 2Image + Image + 2Image, or 1 + 1 + 1 = 3. Column B is still greater. Since there are figures, it is not possible to Plug In a negative number or zero; the only possible numbers will be positive. All positive numbers will yield the same outcome since the rectangle has two sides that are double the side length of the square, meaning that the rectangle will always have a perimeter a little bit (or a lot) bigger than the square’s. Therefore, Column B will always be greater, and the correct answer is (B).

37. CDraw the triangle on the figure provided and write down the formula for the area of a triangle: A = Image bh. Next find the values for the base and height. To find the base, find the value of DE. AD and BC are congruent since the shape is a rectangle. From there, the base must be 4, since E is the midpoint of AD, which splits 8 into two equal parts. The height of the triangle must meet the base at a right angle, so the height of the triangle is the same as the measure of side AB, which has a length of 6. Plug In these values to the formula to find that A = Image(4)(6) = (2)(6) = 12. Column A is 12, which is equivalent to the value in Column B. Since the columns are equal, the correct answer is (C).

38. CRemember, with exponents, you can always write it out. However, this is a common trick on the ISEE. Another way to evaluate these expressions is to rewrite them with the same base. In Column B, 64 is the same as 43, so the whole expression can be rewritten as (43)4. Using the exponent rules (MADSPM), you can further simplify: (43)4 = 43×4 = 412. Since the value in Column B is the same as the value of Column A, the correct answer is (C).

39. AWhen the question asks about ratios, use a Ratio Box. The ratio of blue to red tickets is 3:5, so fill in that information into the top row of the Ratio Box. Remember to add the two numbers together to get the total for the ratio row: 3 + 5 = 8. No other information is provided. To make things simple, let’s say that there are a total of 8 actual tickets. That makes the multiplier 1. Column B asks for the fractional part of all the tickets that are blue. Use the numbers from the Ratio Box. There are 3 blue tickets and 8 total tickets: Image = Image. Thus, Column B is Image. Remember that when the numerators are the same, the fraction with the bigger denominator is actually the smaller number. Therefore, Column A is greater, and the correct answer is (A).

Image

40. CRemember, if you see the word average, you can use an Average Pie Image. Luke made two trips, so the # of items is 2. Column A represents the average speed for the entire trip, so to find the total speed, add the speeds from both parts of the trip: 50 + 60 = 110. Put that value in the total spot. Finally, divide to find the average: Image = Image = 55. Thus, Column A is 55, which is the same as Column B. Since the two columns are equal, the correct answer is (C).

41. AFirst, rearrange the equation in Column A to y = mx + b form. Subtract 12x from both sides to get — 4y = —12x + 16. Divide by —4 on both sides to get y = 3x — 4. The slope is the coefficient of x, so the slope is 3. Column A is 3. Now, find the slope from the two points given in Column B using the formula m = Image. Thus, Image = Image = Image = 1, and Column B is 1. The value of Column A is greater, so the correct answer is (A).

42. BDon’t be intimidated by the decimals under the square root! You probably know that 92 = 81, right? To go from 81 to 0.81, you need to move the decimal to the left two places, so move the decimal one place to the left for each 9 and multiply: 0.9 × 0.9 = 0.81. Thus, Image = 0.9. Column A is 0.9. For Column B, the decimal needs to be moved only one place. That means the two numbers that will be multiplied together will have to be bigger than 0.9. Therefore, Column B will be greater. The correct answer is (B). Note, you can also ballpark. Image = 2 and Image = 3, so the Image will be less than 3 but greater than 2, which is still larger than 0.9.

43. DThere are infinitely different ways in which the length and width could multiply together to equal 36. For instance, the length of y could be 4 and the width of z could be 9, or vice versa. The values of y and z could also each be 6, which would make the two columns equal. Since it is impossible to know if y or z is the greater value, or if they are equal, without more information, the correct answer must be (D).

44. AFirst, list the nonnegative even integers less than 10. Don’t forget about zero! The list should contain 0, 2, 4, 6, and 8. Therefore, Column A is 5. Since this is greater than the value of Column B, the correct answer is (A).

45. DDon’t make assumptions! As for all geometry questions, use the rules of the shape you are given to help you solve and ignore anything unstated about the illustration. Isosceles triangles have 2 equal sides. There are therefore three possibilities for the two equal sides: 1) both AB and BC equal 2, 2) both BC and AC equal 2, or 3) BC equals 2 and both AB and AC equal some other number. Because the shape is not drawn to scale, it’s possible that AB and AC could be equal to, say, 100, which would make the area of the triangle much greater than 4. In other words, there are so many possibilities, many of which have an area greater than 4, that the correct answer must be (D).

Math Review

1.Yes

2.It is neither positive nor negative.

3.Addition

4.Multiplication

5.The quotient

6.Yes 3 + 1 + 2 = 6, which is divisible by 3;

No 3 + 1 + 2 = 6, which is not divisible by 9.

7.Exponents

8.Yes 3 goes into 12 evenly 4 times;

No 12 cannot go into 3.

9.No No integer times 12 is equal to 3;

Yes 3 × 4 = 12

10.0  The tens digit is two places to the left of the decimal.

11.2  The tenths digit is one place to the right of the decimal.

12.8  Write it out! 2 × 2 × 2 = 8

13.Over 100 or Image

14.Multiplication

15.Total column

16.Average pie

17.Plug In a number

18.Add; all four

19.Multiply; two (or square one side, since the sides of a square are the same)

20.180

21.3; 180

22.360

23.2; equilateral

24.Hypotenuse; right angle

25.A = Image(base)(height)

REVIEW—THE VERBAL PLAN

Pacing and Verbal Strategy

I will do the verbal questions in this order.

1.Synonyms with words I know

2.Synonyms with words I sort of know

3.Sentence completions

I should spend less than ten minutes on synonyms.

I will always eliminate wrong (or “worse”) answers.

If possible, I will eliminate choices before I guess, but even if I can’t eliminate any, I will still guess productively.

No, I cannot eliminate choices that contain words I do not know.

SYNONYMS

Practice Drill 1—Write Your Own Definition

Possible D efinitions

1.weird

2.introduction

3.giving

4.doing the right thing

5.change

6.circle around

7.optimistic

8.stick around

9.help

10.build

11.bend down

12.honest

13.tease

14.rough

15.self-centered

16.c alm

17.use

18.full of life

19.stretch out

20.help

Practice Drill 2—Write Another Definition

Look up these seven words in a dictionary to see how many different meanings they can have.

Practice Drill 3—Basic Synonym Techniques

1.C

2.A

3.C

4.D

5.D

6.D

7.B

8.C

9.D

10.A

11.C

12.B

13.D

14.B

15.A

16.C

17.D

18.C

19.B

20.C

Practice Drill 4—Making Your Own Context

Possible Contexts (Answers Will Vary)

1.Common cold; common man

2.Competent to stand trial

3.Abridged dictionary

4.Untimely demise; untimely remark

5.Homogenized milk

6.Juvenile delinquent; delinquent payments

7.Inalienable rights

8.Paltry sum

9.Auspicious beginning; auspicious occasion

10.Prodigal son

Practice Drill 5—Using Your Own Context

1.C

2.D

3.D

4.C

5.A

6.B

7.C

8.A

9.C

10.A

Practice Drill 6—All Synonyms Techniques

1.C

2.A

3.C

4.A

5.D

6.A

7.D

8.B

9.C

10.B

11.B

12.C

13.D

14.C

15.D

16.B

17.C

18.B

19.D

20.B

21.C

22.B

23.B

24.D

25.D

26.D

27.C

28.C

29.A

30.D

31.B

32.B

33.B

34.D

35.D

36.C

37.C

38.D

39.B

40.D

SENTENCE COMPLETIONS

Practice Drill 7 —Coming Up with Your Own Word

These words are just to give you an idea of what you could use. Any words that accurately fill the blank, based on the clue and the direction word, will do.

1.good

2.rare

3.remarkable

4.awake

5.lucky

6.thoughtful

7.alike

8.waste time

9.frugal

10.movement

11.simple

12.produce

13.changed

14.strong

15.not necessary

16.repetitive

17.risky

18.outgoing

19.balanced

20.generous

21.steadfast

22.intimidated; shy

23.annoyed

24.on time

25.skill

26.variety

27.creative

28.sharing

29.flexible

30.inborn

31.affable; talkative

32.awestruck

Practice Drill 8 —Eliminating Answers Based on Your Word

Below are the correct answers to the problems. You should have eliminated the other choices.

1.D

2.B

3.C

4.D

5.A

6.A

7.D

8.C

9.B

10.A

11.D

12.D

13.B

14.C

15.D

16.C

17.B

18.A

19.C

20.C

21.B

22.D

23.A

24.B

25.A

26.C

27.B

28.B

29.C

30.B

31.A

32.C

Practice Drill 9 —Using Positive/Negative

1.+

2.+

3.—

4.+

5.—

6.+

7.—

8.—

9.—

10.+

Practice Drill 10 —Eliminating Based on Positive/Negative

1.A

2.C

3.B

4.D

5.C

6.A

7.C

8.D

9.A

10.C

Practice Drill 11 —Two-Blank Sentence Completions

1.B

2.C

3.A

4.A

5.D

6.A

7.C

8.C

9.A

10.C

Review—The Sentence Completions Plan

For each and every sentence completion, the first thing I do is cover the answers.

I look for the clue, and I mark it by underlining it.

I look for any direction words, and I circle them.

Then I come up with my own word for the blank. If I have trouble coming up with a word for the blank, I decide if the blank is positive or negative (or neither).

Then I eliminate choices, and I guess from the remaining choices.

For each and every sentence completion, the first thing I do is cover the answers.

I look for the clue, and I mark it by underlining it.

I look for any direction words, and I circle them.

If the sentence completion has two blanks, I do them one at a time.

I do the blank that is easier first—the one that has the better clue.

I come up with a word for one of the blanks, and when I uncover the choices, I uncover only the words for the blank that I am working on, and I eliminate based on those.

Then, I go back to the sentence and come up with a word for the other blank, uncover the choices that are left, and eliminate.

No, I cannot eliminate choices that contain words I do not know.

If I can eliminate only one or two choices, then I guess from the remaining choices.

If the sentence or vocabulary looks so difficult that I can’t come up with a word or decide if the blank is positive or negative, then I fill in my “letter-of-the-day.”

I spend my last minute filling in the “letter-of-the-day” for any questions I have not gotten around to answering.

I should never leave a question unanswered because there is no penalty for guessing.

Practice Drill 12 —All Sentence Completion Techniques

1.A

2.C

3.B

4.A

5.D

6.B

7.B

8.C

9.C

10.B

11.D

12.A

13.B

14.C

15.A

16.B

17.C

18.D

19.A

20.D

READING COMPREHENSION

For detailed explanations, go to your Student Tools.

Practice Drill 1 —Getting Through the Passage

You should have brief labels like the following:

1st Label:

Norway → Iceland

2nd Label:

Iceland → Greenland

3rd Label:

Lost

4th Label:

Saw America; landed Greenland

What?

A Viking

So What?

Found America early

Passage type?

History of an event—social studies

Practice Drill 2—Answering a General Question

1.D

2.D

Practice Drill 3 —Answering a Specific Question

1.C

2.A

3.B

4.D

5.C

Review—The Reading Plan

After I read each paragraph, I label it.

After I read an entire passage, I ask myself: What? and So what?

The five main types of general questions, and the questions I can ask myself to answer them, are:

· Main idea: What was the “What? So what?” for this passage?

· Tone/attitude: How did the author feel about the subject?

· General interpretation: Which answer stays closest to what the author said and how he said it?

· General purpose: Why did the author write this?

· Prediction: How was the passage arranged? What will come next?

To find the answer to a specific question, I can use three clues.

· Paragraph labels

· Line or paragraph reference

· Lead words

If the question says “In line 22,” then I begin reading at approximately line 17. On a general question, I eliminate answers that are:

· Too small

· Not mentioned in the passage

· In contradiction to the passage

· Too big

· Too extreme

· Against common sense

On a specific question, I eliminate answers that are:

· Too extreme

· Contradicting passage details

· Not mentioned in the passage

· Against common sense

When I’ve got it down to two possible answers, I:

· Reread the question

· Look at what makes the two answers different

· Go back to the passage

· Eliminate the answer that is worse

Practice Drill 4 —All Reading Techniques—Lower Level

What? Tides

So what? Are caused by the moon

1.A

2.D

3.D

4.C

Practice Drill 5 —All Reading Techniques—Lower Level

What? Brooklyn Bridge

So what? There were problems building it.

1.D

2.B

3.C

4.A

5.C

Practice Drill 6 —All Reading Techniques—All Levels

What? William Levitt

So what? Built homes efficiently

1.C

2.D

3.C

4.D

5.A

6.C

Practice Drill 7 —All Reading Techniques—All Levels

What? Etymology

So what? Has many words to explore

1.B

2.C

3.A

4.C

5.B

Practice Drill 8 —All Reading Techniques—Upper Level

What? Bob Dylan

So what? Was destined to be a musician

1.D

2.B

3.D

4.A

5.C

Practice Drill 9 —All Reading Techniques—Middle and Upper Levels

What? Science

So what? Doesn’t have all the answers

1.A

2.D

3.D

4.A

5.C