340 MCGRAW-HILL’S SAT Concept Review 2 1. It should look like this: 2. When two of the three values are given in a prob- lem, write them in the pyramid and perform the operation between them. The result is the other value in the pyramid. 3. A median is the “middle” number when the num- bers are listed in order. If there are an even num- ber of numbers in the set, it is the average of the two middle numbers. 4. The number that appears the most frequently in a set. 5. When the numbers are evenly spaced, the mean is always equal to the median. This is true more generally if the numbers are distributed “symmetrically” about the mean, as in [−10, −7, 0, 7, 10]. 6. If the average of four numbers is 15, then their sum must be (4)(15) = 60. If one of the numbers is 18, then the sum of the other three is 60 − 18 = 42. So the average of the other three is 42/3 = 14. 7. You need to read this problem super-carefully. If the average of the five numbers is 25, then their sum is (5)(25) = 125. If none of the numbers is less than 10, and since they are all different inte- gers, the least that four of them can be is 10, 11, 12, and 13. Therefore, if x is the largest possible number in the set, x + 10 + 11 + 12 + 13 = 125 Simplify: x + 46 = 125 Subtract 46: x = 79 8. If the 20 students in Ms. Appel’s class averaged 90%, then they must have scored a total of (20)(90) = 1,800 points. Similarly, Mr. Bandera’s class scored a total of (30)(80) = 2,400. The combined average is just the sum of all the scores divided by the number of scores: (1,800 + 2,400)/50 = 84. Notice, too, that you can get a good estimate by just noticing that if there were an equal number of students in each class, the overall average would simply be the average of 80 and 90, which is 85. But since there are more students in Mr. Bandera’s class, the average must be weighted more heavily toward 80. Answer Key 2: Mean/Median/Mode Problems SAT Practice 2 1. D The average of 2x, 2x, y, and 3y is (2x + 2x + y + 3y)/4 = (4x + 4y)/4 = x + y. Substitut- ing 2x + 1 for y gives x + (2x + 1) = 3x + 1. 2. C If the average of seven integers is 11, their sum is (7)(11) = 77. If each of these integers is less than 20, then the greatest any can be is 19. The question doesn’t say that the integers must be dif- ferent, so if x is the least possible of these integers, x + 19 + 19 + 19 + 19 + 19 + 19 = 77. Simplify: x + 114 = 77 Subtract 114: x =−37 3. 4 The median is the average of the two middle numbers. A little trial and error shows that 1, 4, 6, and 8 have a median of 5, so k must be 4. 4. C Call the number you are looking for y. The av- erage of x and y is z, so set up the equation and solve: (x + y)/2 = z Multiply by 2: x + y = 2z Subtract x: y = 2z − x 5. A Just fill in the pyramid: n = 12m/3k = 4m/k. 6. B The average is 0, so (5 + 8 + 2 + k) = 0. Solving for k gives us k =−15. So we put the numbers in order: −15, 2, 5, 8. Since there are an even number of numbers, the median is the average of the two middle numbers: (2+5)/2 = 3.5. 7. B The most frequent number is 1, so c = 1. This means that statement III is untrue, and you can eliminate choices (D) and (E). To find the median, you need to find the average of the 10th and 11th numbers, when you arrange them in order. Since average how many sum × ÷ ÷ both of these are 3, b = 3. Therefore, statement II is true, and you can eliminate choice (A). To find the average, just divide the sum by 20: ((1)(5) + (2)(3) + (3)(3) + (4)(3) + (5)(3) + (6)(3))/20 = 3.25, so a = 3.25. Therefore, statement I is not true, so the answer is (B). 8. B When a 30% solution and a 50% solution are combined, the concentration must be anywhere between 30% and 50%, depending on how much of each you added. It can’t be 50%, though, because the 30% solution dilutes it. 9. A If Set A were {0, 0, 10, 10, 10}, then its median, m, would be 10. Set B would be {2, 2, 12, 12, 12}. Inspection of Set B shows that it is a counterexam- ple to statements II and III, leaving (A) as a possible answer. CHAPTER 9 / SPECIAL MATH PROBLEMS 341 Arithmetic Reasoning 342 MCGRAW-HILL’S SAT Lesson 3: Numerical Reasoning Problems The first statement, m < n < p < r, tells you that the al- phabetical order is also the numerical order of the numbers. The second statement, mnpr = 0, tells you that one of the numbers must be 0. (This is the zero product property!) The third statement, m + n + p + r = 0, tells you that you must have at least one positive and one negative, and all the numbers must “cancel out.” This means that m can’t be 0 because then none of the numbers would be negative, and r can’t be 0, because then none of the numbers would be positive. Thus, either n or p is 0. This means that both I and II are necessarily true, so you can eliminate choices (A), (B), and (D). The example m =−3, n = 0, p = 1, r = 2 shows that statement III is not necessarily true, so the answer is (C). Digit Problems Some of the most common problems on the SAT are numerical reasoning problems, which ask you to think about what happens to numbers when you perform basic operations on them. You just need to know the common numerical and arithmetic rules and think logically. Example: If a + b is negative, which of the following CANNOT be negative? (A) ab (B) ab 2 (C) a 2 b (D) a 2 b 2 (E) a − b Start by thinking about what might be true about a and b and what must be true about a and b. First think of possible values for a and b. −2 and 1 work, because a + b =−2 + 1 =−1. Notice that this proves that (A), (B), and (E) are incorrect, because they can be negative: (A) ab = (−2)(1) =−2, (B) ab 2 = (−2)(1) 2 =−2, and (E) a − b = (−2) − (1) =−3. But (C) a 2 b = (−2) 2 (1) = 4 is positive, so does that mean the answer is (C)? Not so fast! Your job is not to find which one can be positive, but rather which cannot be negative. Notice that (C) can be negative if a and b are, say, 1 and −2 (notice that a + b is still negative, so those values work): (C) a 2 b = (1) 2 (−2) =−2. Therefore, by process of elimination, the answer is (D). This question is much easier if you remember a sim- ple fact: If x is a real number, then x 2 is never nega- tive. If you don’t know this already, play around with possible values of x until you see why this is true. Then look at choice (D) a 2 b 2 . a 2 can’t be negative, and neither can b 2 , so a 2 b 2 can’t be negative. Example: If m < n < p < r, mnpr = 0, and m + n + p + r = 0, then which of the following must be true? I. If m and n are negative, then p = 0. II. np = 0 III. m + r = 0 (A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III You may see a question on the SAT like the one below, where letters represent digits. Remember that digits can only take the values 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Also remember that you may have to consider “carried” digits when looking at a sum or product. Lastly, you may find it best to work from left to right rather than right to left. Example: 1BA + 8B 211 If A and B represent distinct digits in this addi- tion problem, what is the value of A − B? (A) Ϫ9 (B) Ϫ7 (C) 2 (D) 7 (E) 9 Look at the left (hundreds) column first. Since the sum has a 2 in the hundreds place, there must be a carry of 1 from the tens place. Therefore, B + 8 + (carry from ones column, if any) = 11. This means B = 2 or 3. Trying each one shows that only B = 2 and A = 9 works, giving 129 + 82 = 211. There- fore A − B = 9 − 2 = 7, so the answer is (D). CHAPTER 9 / SPECIAL MATH PROBLEMS 343 1. If neither a nor b is 0, what is the relationship between a ÷ b and b ÷ a? 2. What is the relationship between a − b and b − a? Complete the following “parity rules.” 3. Odd × even = _____ 4. Even × even = _____ 5. Odd × odd = _____ 6. Even + even = _____ 7. Odd + even = _____ 8. Odd + odd = _____ Complete the following “sign rules.” 9. If n is odd, (−1) n = _____. 10. If n is even, (−1) n = _____. 11. If x + y = 0 and x ≠ 0, then x/y = _____. 12. Dividing by x is the same as multiplying by __________. 13. Subtracting (x + 1) is the same as adding __________. 14. When a number is multiplied by its reciprocal, the result is __________. 15. When a number and its opposite are added, the result is __________. 16. When a number (other than 0) is divided by its opposite, the result is __________. 17. If a positive number is multiplied by a number greater than 1, what happens to it? _______________ 18. If a positive number is multiplied by a number between 0 and 1, what happens to it? _______________ 19. If a negative number is multiplied by a number greater than 1, what happens to it? _______________ 20. Is x always bigger than −x? Explain. 21. Is x 2 always bigger than x? Explain. 22. Is x 3 always bigger than x 2 ? Explain. 23. If x is between 0 and 1, then 1/x is _________________________. 24. If a > b > 0, then is ____________________. 25. If b > a > 0, then is ____________________. a b a b Concept Review 3: Numerical Reasoning Problems 344 MCGRAW-HILL’S SAT SAT Practice 3: Numerical Reasoning Problems 1. If m and n are both odd integers, which of the following must be true? I. m 2 + n 2 is even II. m 2 + n 2 is divisible by 4 III. (m + n) 2 is divisible by 4 (A) none (B) I only (C) I and II only (D) I and III only (E) I, II, and III 2. 6AA × 8 50B4 If A and B represent distinct digits in this cor- rectly worked multiplication problem, what is the value of B? (A) 2 (B) 3 (C) 5 (D) 6 (E) 8 3. If j is the number of integers between 1 and 500 that are divisible by 9 and k is the number of in- tegers between 1 and 500 that are divisible by 7, what is j + k? (A) 126 (B) 127 (C) 128 (D) 129 (E) 130 4. If 60 is written as the product of four integers, each greater than 1, then what is the sum of those integers? 5. If n is an integer and 2 n is a factor of 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9, what is the greatest possible value of n? (A) 5 (B) 6 (C) 7 (D) 8 (E) 9 6. If p + pq is 4 times p − pq, and pq ≠ 0, which of the following has exactly one possible value? (A) p (B) q (C) pq (D) p + pq (E) p − pq 7. If a, b, c, d, and e are whole numbers and a(b(c + d) + e) is odd, then which of the follow- ing CANNOT be even? (A) a (B) b (C) c (D) d (E) e a + b + c = 7 c + d + e = 9 8. If each letter in the sums above represents a different positive integer, then c = (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 ABB +9B7 AA7C 9. If A, B, and C are distinct digits in the correctly worked addition problem above, what is the value of A + B + C? (A) 4 (B) 9 (C) 14 (D) 16 (E) 17 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 SAT Practice 3 1. D Start with the simplest odd values for m and n: m = n = 1. (There’s no reason why m and n can’t equal the same number!) Notice that m 2 + n 2 = 1 2 + 1 2 = 2, which isn’t divisible by 4, so statement II is not necessarily true, and you can eliminate choices (C) and (E). Next, notice that m 2 and n 2 must both be odd, so m 2 +n 2 must be even, so statement I is necessarily true, and you can eliminate choice (A). (m + n) 2 must be a multiple of 4 because m + n must be even (odd + odd = even), so it is a multiple of 2. When it is squared, it becomes a multi- ple of 4. So III is true, and the answer is (D). 2. D Trial and error should show that A =3. If A is less than 3, the product is too small. If A is greater than 3, the product is too large. Since 633 × 8 = 5,064, B = 6. 3. A 500 ÷ 9 =55.55, so there are 55 multiples of 9 be- tween 1 and 500. 500 ÷ 7 = 71.43, so there are 71 mul- tiples of 7 between 1 and 500. So j + k = 55 + 71 = 126. 4. 12 Trial and error shows that the only way to write 60 as the product of four integers, each greater than 1, is 2 × 2 × 3 × 5. Their sum is 2 + 2 + 3 + 5 = 12. 5. C Do the prime factorization: 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 = 1 × 2 × 3 × (2 × 2) × 5 × (2 × 3) × 7 × (2 × 2 × 2) × (3 × 3) Since there are seven factors of 2, the greatest power of 2 that is a factor is 2 7 . 6. B p + pq = 4(p − pq) Distribute: p + pq = 4p − 4pq Divide by p: 1 + q = 4 − 4q (This is okay as long as p is anything but 0.) Add 4q: 1 + 5q = 4 Subtract 1: 5q = 3 Divide by 5: q = 3/5 Because p can have many possible values but q can only equal 3/5, (B) q is the only expression that has only one possible value. 7. A a cannot be even, because an even number times any other integer yields an even number, but a(b(c + d) + e) is odd. 8. A The only three different positive integers that have a sum of 7 are 1, 2, and 4. The only three dif- ferent positive integers that have a sum of 9 are 1, 3, and 5 or 1, 2, and 6. But 1 + 2 + 6 doesn’t work, since that would have two numbers in common with the first set, but it may only have one (C). Since (C) is the only number they may have in common, it must be 1. 9. C The only solution is 188 + 987 = 1,175, so A + B + C = 1 + 8 + 5 = 14. CHAPTER 9 / SPECIAL MATH PROBLEMS 345 Concept Review 3 1. They are reciprocals, so their product is 1. 2. They are opposites, so their sum is 0. 3. Odd × even = even 4. Even × even = even 5. Odd × odd = odd 6. Even + even = even 7. Odd + even = odd 8. Odd + odd = even 9. If n is odd, (−1) n =−1. 10. If n is even, (−1) n = 1. 11. If x + y = 0 and x ≠ 0, then x/y =−1. 12. Dividing by x is the same as multiplying by 1/x. 13. Subtracting (x + 1) is the same as adding −x − 1. 14. When a number is multiplied by its reciprocal, the result is 1. 15. When a number and its opposite are added, the result is 0. 16. When a number (other than 0) is divided by its opposite, the result is −1. 17. It gets bigger. 18. It gets smaller. 19. It gets smaller (more negative). 20. No. If x is 0, then −x is equal to x, and if x is neg- ative, then −x is greater than x. 21. No. If x is between 0 and 1, then x 2 is smaller than x. And if x is 0 or 1, then they are the same. If x is negative, then x 2 is positive, and therefore greater than x. 22. No. If x is between 0 and 1, then x 3 is smaller than x 2 . And if x is 0 or 1, then they are the same. If x is negative, then x 2 is positive, and therefore greater than x 3 . 23. greater than 1. 24. greater than 1. 25. between 0 and 1. Answer Key 3: Numerical Reasoning Problems 346 MCGRAW-HILL’S SAT What Are Rates? The word rate comes from the same Latin root as the word ratio. All rates are ratios. The most common type of rate is speed, which is a ratio with respect to time, as in miles per hour or words per minute, but some rates don’t involve time at all, as in miles per gallon. Rate units always have per in their names: miles per gallon, meters per second, etc. Per, remember, means divided by, and is like the colon (:) or fraction bar in a ratio. The Rate Pyramid Watch Your Units Whenever you work with formulas, you can check your work by paying attention to units. For instance, the problem above asks how long, so the calculation has to produce a time unit. Check the units in the calculation: miles miles hours miles hours miles hours=× = Lesson 4: Rate Problems The name of any rate is equivalent to its formula. For instance, speed is miles per hour can be translated as or Speed = distance time Speed = number of miles number of hours Since this formula is similar to the “average” formula, you can make a rate pyramid. This can be a great tool for solving rate problems. If a problem gives you two of the quantities, just put them in their places in the pyramid, and do the operation be- tween them to find the missing quantity. Example: How long will it take a car to travel 20 miles at 60 miles per hour? Simply fill the quantities into the pyramid: 20 miles goes in the distance spot, and 60 miles an hour goes in the speed spot. Now what? Just do the division the way the diagram says: 20 miles ÷ 60 miles per hour = 1/3 hour. Two-Part Rate Problems Rate problems are tougher when they involve two parts. When a problem involves, say, two people working together at different rates and times, or a two-part trip, you have to analyze the problem more carefully. Example: Toni bicycles to work at a rate of 20 miles per hour, then takes the bus home along the same route at a rate of 40 miles per hour. What is her average speed for the entire trip? At first glance, it might seem that you can just aver- age the two rates: (20 + 40)/2 = 30 miles per hour, since she is traveling the same distance at each of the two speeds. But this won’t work, because she isn’t spending the same time at each speed, and that is what’s important. But if that’s true, you might notice that she spends twice as much time going 20 miles per hour as 40 miles per hour (since it’s half as fast), so instead of taking the average of 20 and 40, you can take the average of two 20s and a 40: (20 + 20 + 40)/3 = 26.67 miles per hour. Simple! But if that doesn’t make sense to you, think of it this way: Imagine, for simplicity’s sake, that her trip to work is 40 miles. (It doesn’t matter what number you pick, and 40 is an easy number to work with here.) Now the average speed is simply the total distance divided by the total time (as the pyramid says). The total distance, there and back, is 80 miles. The total time is in two parts. Getting to work takes her 40 miles ÷ 20 miles per hour = 2 hours. Getting home takes her 40 miles ÷ 40 miles per hour = 1 hour. So the total time of the trip is 3 hours. The average speed, then, must be 80 miles ÷ 3 hours = 26.67 miles per hour! speed time distance × ÷ ÷ CHAPTER 9 / SPECIAL MATH PROBLEMS 347 For each of the following rates, write the formula of the rate and the corresponding “rate pyramid.” 1. Speed is miles per hour. 2. Efficiency is miles per gallon of fuel. 3. Typing speed is pages per minute. Find the missing quantity, including the units, in each of these rate situations. 4. A train travels for 375 miles at 75 mph. 5. A car that gets 28 miles per gallon uses 4.5 gallons of fuel. 6. Harold can type 600 words in 5 minutes. 7. A landscaper who cuts 1.6 acres of grass per hour cuts an 8-acre lot. 8. A train leaves New York at 1:00 pm, going 50 mph, bound for Philadelphia, which is 90 miles away. If it makes no stops, at what time should it be expected to arrive? 9. Anne can paint a room in 2 hours, and Barbara can paint a room in 3 hours. When they work together, their work rate is the sum of their rates working separately. How long should it take them to paint a room if they work together? Concept Review 4: Rate Problems 1. Janice and Edward are editors at a newspaper. Janice can edit 700 words per minute and Edward can edit 500 words per minute. If each page of text contains 800 words, how many pages can they edit, working together, in 20 minutes? 2. Two cars leave the same point simultaneously, going in the same direc- tion along a straight, flat road, one at 35 mph and one at 50 mph. After how many minutes will the cars be 5 miles apart? 3. What is the average speed, in miles per hour, of a sprinter who runs 1 ⁄4 mile in 45 seconds? (1 hour = 60 minutes) (A) 11.25 mph (B) 13.5 mph (C) 20 mph (D) 22 mph (E) 25 mph 4. A car travels d miles in t hours and arrives at its destination 3 hours late. At what average speed, in miles per hour, should the car have gone in order to have arrived on time? (A) t – 3 (B) (C) (D) (E) t d − 3 d t − 3 d t − 3 t d − 3 348 MCGRAW-HILL’S SAT 5. If x > 1, how many hours does it take a train traveling at x − 1 miles per hour to travel x 2 − 1 miles? (A) (B) (C) x (D) x – 1 (E) x + 1 6. In three separate 1-mile races, Ellen finishes with times of x minutes, y minutes, and z min- utes. What was her average speed, in miles per hour, for all three races? (1 hour = 60 minutes) (A) (B) (C) (D) (E) 7. A hare runs at a constant rate of a mph, a tor- toise runs at a constant rate of b mph, and 0 < b < a. If they race each other for d miles, how many more hours, in terms of a, b, and d, will it take the tortoise to finish than the hare? (A) (B) (C) (D) ad – bd (E) a – b 8. Sylvia drives 315 miles and arrives at her desti- nation in 9 hours. If she had driven at an average rate that was 10 mph faster than her actual rate, how many hours sooner would she have arrived? (A) 1.75 (B) 2.00 (C) 2.25 (D) 2.50 (E) 2.75 b d a d − d b d a − ab d + 2 xyz++ 20 180 xyz++ xyz++ 180 3 xyz++ xyz++ 3 1 1x + 1 1x − SAT Practice 4: Rate Problems 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 1 0 2 3 4 5 7 8 9 6 CHAPTER 9 / SPECIAL MATH PROBLEMS 349 Concept Review 4 1. Speed =#miles ÷#hours 2. Efficiency =#miles ÷#gallons 3. Typing speed =#pages ÷#minutes 4. 375 miles ÷ 75 mph = 5 hours for the trip. 5. 28 miles per gallon × 4.5 gallons = 126 miles the car can go before it runs out of fuel. 6. 600 words ÷ 5 minutes = 120 words per minute is Harold’s typing speed. 7. 8 acres ÷ 1.6 acres per hour = 5 hours for the job. 8. 90 miles ÷ 50 mph = 1.8 hours, or 1 hour 48 min- utes for the entire trip. At 1 hour and 48 minutes after 1:00 pm, it is 2:48 pm. 9. Anne can paint one room in 2 hours, so her rate is 1 ⁄2 room per hour. Barbara can paint one room in 3 hours, so her rate is 1 ⁄3 room per hour. When they work together, their rate is 1 ⁄2 + 1 ⁄3 = 5 ⁄6 room per hour. So to paint one room would take one room ÷ 5 ⁄6 room per hour = 6 ⁄5 hours, or 1.2 hours, or 1 hour 12 minutes. Answer Key 4: Rate Problems SAT Practice 4 1. 30 Working together, they edit 700 + 500 = 1,200 words per minute. Since each page is 800 words, that’s 1,200 words per minute ÷ 800 words per page = 1.5 pages per minute. In 20 minutes, then, they can edit 1.5 × 20 = 30 pages. 2. 20 Since the two cars are traveling in the same di- rection, their relative speed (that is, the speed at which they are moving away from each other) is 50 −35 =15 mph. In other words, they will be 15 miles farther apart each hour. Therefore, the time it takes them to get 5 miles apart is 5 miles ÷15 miles per hour = 1/3 hour, which is equivalent to 20 minutes. 3. C Since there are (60)(60) = 3,600 seconds in an hour, 45 seconds = 45/3,600 hour. Speed = dis- tance ÷ time = 1/4 mile ÷ 45/3,600 hour = 3,600/180 = 20 miles per hour. 4. C To arrive on time, the car must take t − 3 hours for the whole trip. To travel d miles in t − 3 hours, the car must go d/(t − 3) miles per hour. 5. E According to the rate pyramid, time = distance ÷ speed = (x 2 − 1) miles ÷ (x − 1) miles per hour = hours. Or you can pick a simple value for x, like 2, and solve numerically. 6. D Speed = miles ÷ hours. Her total time for the three races is x + y + z minutes, which we must convert to hours by multiplying by the conver- sion factor (1 hour/60 minutes), which gives us (x + y + z)/ 60 hours. Since her total distance is 3 miles, her overall speed is 3 miles ÷ (x + y + z)/60 hours = 180/(x + y + z) miles per hour. 7. B If the hare’s rate is a mph, then he covers d miles in d/a hours. Similarly, the tortoise covers d miles in d/b hours. The difference in their finish- ing times, then, is d/b − d/a. 8. B Sylvia’s speed is 315 miles ÷ 9 hours = 35 mph. If she were to go 10 mph faster, then her speed would be 45 mph, so her time would be 315 miles ÷ 45 mph = 7 hours, which is 2 hours sooner. x x xx x x 2 1 1 11 1 1 − − = − ( ) + ( ) − =+ speed time distance × ÷ ÷ speed minutes pages × ÷ ÷ efficiency gallons miles × ÷ ÷ . hour. Two -Part Rate Problems Rate problems are tougher when they involve two parts. When a problem involves, say, two people working together at different rates and times, or a two -part trip,. that it is a counterexam- ple to statements II and III, leaving (A) as a possible answer. CHAPTER 9 / SPECIAL MATH PROBLEMS 341 Arithmetic Reasoning 342 MCGRAW-HILL’S SAT Lesson 3: Numerical. then is ____________________. a b a b Concept Review 3: Numerical Reasoning Problems 344 MCGRAW-HILL’S SAT SAT Practice 3: Numerical Reasoning Problems 1. If m and n are both odd integers, which