330 MCGRAW-HILL’S SAT 7. When the Apex Pet Store first opened, the ratio of cats to dogs was 4 to 5. Since then, the number of cats has doubled, while the number of dogs has increased by 12. If the ratio of dogs to cats is now 1 to 1, how many cats did the store have when it opened? 8. Hillside High School has 504 students. One-quarter of the students are sophomores, and 3/7 of the sophomores are boys. If one-third of the sopho- more girls take French, how many sophomore girls do not take French? (A) 24 (B) 36 (C) 48 (D) 72 (E) 126 9. A jar contains only red, green, and blue marbles. If it is three times as likely that you randomly pick a red marble as a green marble, and five times as likely that you pick a green one as a blue one, which of the following could be the number of marbles in the jar? (A) 38 (B) 39 (C) 40 (D) 41 (E) 42 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 8 / ESSENTIAL ALGEBRA I SKILLS 331 Concept Review 7 1. m = Mike’s current age, d = Dave’s current age; m = 2(d − 5) 2. a = the population of town A, b = the population of town B; a = 1.4b 3. n = number of marbles in the jar; n − (2/3)n = 5 + (1/6)n 4. b = number of blue marbles, r = number of red marbles; b = 4 + 2r 5. c = cost of one candy bar, l = cost of one lollipop; 3c + 2l = 2.20, and 4c + 2l = 2.80. Subtract: 4c + 2l = 2.80 – (3c + 2l = 2.20) c = .60 Plug in to find l: 3(.60) + 2l = 2.20 Simplify: 1.80 + 2l = 2.20 Subtract 1.80: 2l = .40 Divide by 2: l = .20 6. n = number of seats in the stadium; (2/3)n − 1,000 = (3/7)n Subtract (2/3)n: −1,000 =−(5/21)n Multiply by −(21/5): 4,200 = n 7. Simplify: Multiply by 4: 2m + 2n = s + t Subtract t: 2m + 2n − t = s 8. b = value of blue chip, r = value of red chip, g = value of green chip; b = 2 + r, r = 2 + g, and 5g = m, so Cost of 10 blue and 5 red chips: 10b + 5r Substitute b = 2 + r: 10(2 + r) + 5r Simplify: 20 + 15r Substitute r = 2 + g: 20 + 15(2 + g) Simplify: 50 + 15g Substitute g = m/5: 50 + 3m mn st+ = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 24 mn st+ = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 1 22 Answer Key 7: Word Problems SAT Practice 7 1. C You could test the choices here, or do the algebra: Multiply by x: 24 − x = 3x Add x: 24 = 4x Divide by 4: 6 = x 2. n = Nora’s current age, m = Mary’s current age. Interpret first sentence: n − 3 = (1/2)m Interpret second sentence: m = n + 4 Subtract 4: m − 4 = n Substitute n = m − 4: m − 4 − 3 = (1/2)m Simplify: m − 7 = (1/2)m Subtract m: −7 =−(1/2)m Multiply by −2: 14 = m 3. E p/q = 9/7, q/r = 14/3. Multiply: 4. 40 J = number of books Joan had originally. E = number of books Emily had originally. J = 2 E. After the exchange, Emily has E + 5 and Joan has J − 5 books, so J − 5 = 10 + (E + 5). Simplify: J − 5 = E + 15 Subtract 15: J − 20 = E Substitute into J = 2(J − 20) first equation: Solve for J: J = 40 (Reread and check) 5. C Let xbe the cost of living in 1960. In 1970, the cost of living was 1.2x, and in 1980 it was 1.5x. Use the per- cent change formula: (1.5x − 1.2x)/1.2x = .25 = 25%. p q q r p r ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ == ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 9 7 14 3 6 1 24 3 − = x x 6. 8 Let w = the number of games won and l = the number of games lost. w/l = 7/5 and w + l = 48. Multiply by l: w = (7/5)l Substitute into 2nd eq.: (7/5)l + l = 48 Simplify: (12/5)l = 48 Multiply by 5/12: l = 20 Plug in to find w: w + 20 = 48 Subtract 20: w = 28 How many more games won than lost? w − l = 28 − 20 = 8 7. 16 Let c = number of cats originally, d = number of dogs originally. c/d = 4/5. Now the number of cats is 2c and the number of dogs is d + 12. If the ratio of dogs to cats is now 1 to 1, 2c = d + 12. Cross-multiply: 5c = 4d Divide by 4: (5/4)c = d Substitute: 2c = (5/4)c + 12 Subtract (5/4)c: (3/4) c = 12 Multiply by 4/3: c = 16 (Reread and check) 8. C Number of sophomores = (1/4)(504) = 126. If 3/7 of the sophomores are boys, 4/7 are girls: (4/7)(126) = 72. If 1/3 of the sophomore girls take French, 2/3 do not: (2/3)(72) = 48. 9. E r, g, and b are the numbers of red, green, and blue marbles. r = 3g and g = 5b. Total marbles = r + g + b. Substitute r = 3g:3g + g + b = 4g + b Substitute g = 5b: 4(5b) + b = 21b So the total must be a multiple of 21, and 42 = 2(21). SPECIAL MATH PROBLEMS CHAPTER 9 332 1. New Symbol or Term Problems 2. Mean/Median/Mode Problems 3. Numerical Reasoning Problems 4. Rate Problems 5. Counting Problems 6. Probability Problems ✓ Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use. CHAPTER 9 / SPECIAL MATH PROBLEMS 333 New Symbol or Term Problems Example: For all real numbers a and b, let the expression a ¿ b be defined by the equation a ¿ b = 10a + b. Question 3: What is 5 ¿ 10? Just substitute 5 for a and 10 for b in the given equa- tion: 5 ¿ 10 = 10(5) + 10 = 60. Question 4: If 2.5 ¿ x = 50, what is the value of x? Just translate the left side of the equation: 2.5 ¿ x = 10(2.5) + x = 50 Then solve for x: 25 + x = 50 x = 25 Question 5: What is 1.5 ¿ (1.5 ¿ 1.5)? According to the order of operations, evaluate what is in parentheses first: 1.5 ¿ (1.5 ¿ 1.5) Substitute: 1.5 ¿ (10(1.5) + 1.5) Simplify: 1.5 ¿ (16.5) Substitute again: 10(1.5) + 16.5 Simplify: 15 + 16.5 = 31.5 Lesson 1: New Symbol or Term Problems Don’t be intimidated by SAT questions with strange symbols, like Δ, φ, or ¥, or new terms that you haven’t seen before. These crazy sym- bols or terms are just made up on the spot, and the problems will always explain what they mean. Just read the definition of the new sym- bol or term carefully and use it to “translate” the expressions with the new symbol or term. Example: Let the “kernel” of a number be defined as the square of its greatest prime factor. For instance, the kernel of 18 is 9, because the greatest prime factor of 18 is 3 (prime factorization: 18 = 2 × 3 × 3), and 3 2 equals 9. Question 1: What is the kernel of 39? Don’t worry about the fact that you haven’t heard of a “kernel” before. Just read the definition carefully. By the definition, the kernel of 39 is the square of its greatest prime factor. So just find the greatest prime factor and square it. First, factor 39 into 3 × 13, so its greatest prime factor is 13, and 13 2 = 169. Question 2: What is the greatest integer less than 20 that has a kernel of 4? This requires a bit more thinking. If a number has a kernel of 4, then 4 must be the square of its greatest prime factor, so its greatest prime factor must be 2. The only numbers that have a greatest prime factor of 2 are the powers of 2. The greatest power of 2 that is less than 20 is 2 4 = 16. There’s a lot of detail to learn and understand to do well on the SAT. For more tools and resources that will help, visit our Online Practice Plus at www.MHPracticePlus.com/SATmath. 334 MCGRAW-HILL’S SAT Concept Review 1: New Symbol or Term Problems For questions 1–6, translate each expression into its simplest terms, using the definition of the new symbol. The following definition pertains to questions 1–3: For any real number x, let § x be defined as the greatest integer less than or equal to x. 1. §−4.5 = __________ 2. §−1.5 + §1.5 = __________ 3. = __________ The following definition pertains to questions 4–6: If q is any positive real number and n is an integer, let q @ n be defined by the equation . 4. 8 @ 3 = __________ 5. 9 @ (k − 1) = __________ 6. x 2 @ 0 = __________ 7. If q is any positive real number and n is an integer, let q @ n be defined by the equation . If y @ 2 = 64, what is the value of y? 8. For any integer n and real number x, let x ^ n be defined by the equation x ^ n = nx n−1 . If y ^ 4 =−32, what is the value of y? 9. For any integer n, let Ωn be defined as the sum of the distinct prime factors of n. For instance, Ω36 = 5, because 2 and 3 are the only prime factors of 36 and 2 + 3 = 5. What is the smallest value of w for which Ωw = 12? qn q n @ = +1 qn q n @ = +1 §§15 17+ CHAPTER 9 / SPECIAL MATH PROBLEMS 335 1. For all real numbers d, e, and f, let d * e * f = de + ef + df. If 2 * 3 * x = 12, then x = (A) (B) (C) (D) 2 (E) 6 2. If b ≠ 0, let . If x # y = 1, then which of the following statements must be true? (A) x = y (B) x =|y| (C) x =−y (D) x 2 − y 2 = 0 (E) x and y are both positive 3. On a digital clock, a time like 6:06 is called a “double” time because the number representing the hour is the same as the number represent- ing the minute. Other such “doubles” are 8:08 and 9:09. What is the smallest time period between any two such doubles? (A) 11 mins. (B) 49 mins. (C) 60 mins. (D) 61 mins. (E) 101 mins. 4. Two numbers are “complementary” if their reciprocals have a sum of 1. For instance, 5 and are complementary because . If x and y are complementary, and if , what is y? (A) −2 (B) (C) (D) (E) 3 1 3 − 1 3 − 1 2 x = 2 3 1 5 4 5 1+= 5 4 ab a b # = 2 2 8 5 6 5 5 6 5. For x ≠ 0, let . What is the value of $$5? 6. For all nonnegative real numbers x, let ◊x be defined by the equation . For what value of x does ◊x = 1.5? (A) 0.3 (B) 6 (C) 12 (D) 14 (E) 36 7. For any integer n, let [n] be defined as the sum of the digits of n. For instance, [341] = 3 + 4 + 1 = 8. If a is an integer greater than 0 but less than 1,000, which of the following must be true? I. [10a] < [a]+1 II. [[a]] < 20 III. If a is even, then [a] is even (A) none (B) II only (C) I and II only (D) II and III only (E) I, II, and III 8. For all integers, n, let What is the value of 13&&? (A) 10 (B) 13 (C) 20 (D) 23 (E) 26 n nn nn & = − ⎧ ⎨ ⎪ ⎩ ⎪ 2 3 if is even if is odd ◊=x x 4 $x x = 1 SAT Practice 1: New Symbol or Term 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 6 336 MCGRAW-HILL’S SAT Answer Key 1: New Symbol or Term Problems 8. y^4 =−32 Translate: 4y 4–1 =−32 Simplify and divide by 4: y 3 =−8 Take the cube root: y =−2 9. If Ωw = 12, then w must be a number whose dis- tinct prime factors add up to 12. The prime num- bers less than 12 are 2, 3, 5, 7, and 11. Which of these have a sum of 12? (Remember you can’t re- peat any, because it says the numbers have to be distinct.) A little trial and error shows that the only possibilities are 5 and 7, or 2, 3, and 7. The smallest numbers with these factors are 5 × 7 = 35 and 2 × 3 × 7 = 42. Since the question asks for the least such number, the answer is 35. SAT Practice 1 1. B 2 * 3 * x = 12 Translate: (2)(3) + (3)(x) + (2)(x) = 12 Simplify: 6 + 5x = 12 Subtract 6: 5x = 6 Divide by 5: x = 6/5 2. D If x # y = 1, then (x 2 /y 2 ) = 1, which means x 2 = y 2 . Notice that x =−1 and y = 1 is one possible solu- tion, which means that (A) x = y (B) x =⏐y⏐ (E) x and y are both positive is not necessarily true. Another simple solution is x = 1 and y = 1, which means that (C) x =−y is not necessarily true, leaving only (D) as an answer. 3. B All of the consecutive “double times” are 1 hour and 1 minute apart except for 12:12 and 1:01, which are only 49 minutes apart. 4. A If 2 ⁄3 and y are complementary, then the sum of their reciprocals is 1: 3 ⁄2 + 1/y = 1 Subtract 3 ⁄2:1/y =−1/2 Take the reciprocal of both sides: y =−2 5. 5 The “double” symbol means you simply per- form the operation twice. Start with 5, then $5 = 1/5. Therefore, $$5 = $(1/5) = 1/(1/5) = 5. 6. E Multiply by 4: Square both sides: Plug in x = 36 to the original and see that it works. 7. C If a is 12, which is even, then [12] = 1 + 2 = 3 is odd, which means that statement III is not necessarily true. (Notice that this eliminates choices (D) and (E).) Statement I is true because [10a] will always equal [a] because 10a has the same digits as a, but with an extra 0 at the end, which con- tributes nothing to the sum of digits. Therefore, [10a] < [a] + 1 is always true. Notice that this leaves only choice (C) as a possibility. To check statement II, though (just to be sure!), notice that the biggest sum of digits that you can get if a is less than 1,000 is from 999. [999] = 9 + 9 + 9 = 27; therefore, [[999]] = [27] = 2 + 7 = 9. It’s possible to get a slightly bigger value for [[a]] if a is, say, 991: [[991]] = [19] = 10, but you can see that [[a]] will never ap- proach 20. 8. C Since 13 is odd, 13& = 13 − 3 = 10. Therefore, 13&& = 10&. Since 10 is even, 10& = 2(10) = 20. ◊= = = = x x x x 4 15 6 36 . Concept Review 1 1. §−4.5 =−5 2. §−1.5 + §1.5 =−2 + 1 =−1 3. 4. 5. 6. 7. Simplify: Take the cube root: Square: yy y y y @2 64 64 4 16 21 3 = () = () = = = + xxx 22 01 0@ = () = + 919 3 11 @ k k k − ( ) = () = −+ 83 8 64 4 @ = () = §15 17 3 4 7+=+= CHAPTER 9 / SPECIAL MATH PROBLEMS 337 Average (Arithmetic Mean) Problems You probably know the procedure for finding an average of a set of numbers: add them up and divide by how many numbers you have. For instance, the av- erage of 3, 7, and 8 is (3 + 7 + 8)/3 = 6. You can de- scribe this procedure with the “average formula”: Since this is an algebraic equation, you can manipu- late it just like any other equation, and get two more formulas: Sum = average × how many numbers This is a great tool for setting up tough problems. To find any one of the three quantities, you simply need to find the other two, and then perform the operation between them. For instance, if the problem says, “The average (arithmetic mean) of five numbers is 30,” just write 30 in the “average” place and 5 in the “how many” place. No- tice that there is a multiplication sign between them, so multiply 30 ×5 =150 to find the third quantity: their sum. Medians How many numbers = sum average Average = sum how many numbers Lesson 2: Mean/Median/Mode Problems Just about every SAT will include at least one question about averages, otherwise known as arithmetic means. These won’t be simplistic questions like “What is the average of this set of numbers?” You will have to really under- stand the concept of averages beyond the basic formula. Occasionally the SAT may ask you about the mode of a set of numbers. A mode is the num- ber that appears the most frequently in a set. (Just remember: MOde = MOst.) It’s easy to see that not every set of numbers has a mode. For instance, the mode of [−3, 4, 4, 1, 12] is 4, but [4, 9, 14, 19, 24] doesn’t have a mode. The average (arithmetic mean) and the me- dian are not always equal, but they are equal whenever the numbers are spaced symmetri- cally around a single number. it splits the highway exactly in half. The median of a set of numbers, then, is the middle number when they are listed in increasing order. For in- stance, the median of {−3, 7, 65} is 7, because the set has just as many numbers bigger than 7 as less than 7. If you have an even number of numbers, like {2, 4, 7, 9}, then the set doesn’t have one “middle” number, so the median is the average of the two middle numbers. (So the median of {2, 4, 7, 9} is (4+7)/2 = 5.5.) All three of these formulas can be summarized in one handy little “pyramid”: When you take standardized tests like the SAT, your score report often gives your score as a percentile, which shows the percentage of students whose scores were lower than yours. If your percentile score is 50%, this means that you scored at the median of all the scores: just as many (50%) of the students scored below your score as above your score. Example: Consider any set of numbers that is evenly spaced, like 4, 9, 14, 19, and 24: Notice that these numbers are spaced symmetrically about the number 14. This implies that the mean and the median both equal 14. This can be helpful to know, because finding the median of a set is often much easier than calculating the mean. Modes 41492419 A median is something that splits a set into two equal parts. Just think of the median of a highway: average how many sum × ÷ ÷ 1. Draw the “average pyramid.” 2. Explain how to use the average pyramid to solve a problem involving averages. 3. Define a median. 4. Define a mode. 5. In what situations is the mean of a set of numbers the same as its median? 6. The average (arithmetic mean) of four numbers is 15. If one of the numbers is 18, what is the average of the remaining three numbers? 7. The average (arithmetic mean) of five different positive integers is 25. If none of the numbers is less than 10, then what is the greatest possible value of one of these numbers? 8. Ms. Appel’s class, which has twenty students, scored an average of 90% on a test. Mr. Bandera’s class, which has 30 students, scored an average of 80% on the same test. What was the combined average score for the two classes? 338 MCGRAW-HILL’S SAT Concept Review 2: Mean/Median/Mode Problems 1. If y = 2x + 1, what is the average (arithmetic mean) of 2x, 2x, y, and 3y, in terms of x? (A) 2x (B) 2x + 1 (C) 3x (D) 3x + 1 (E) 3x + 2 2. The average (arithmetic mean) of seven inte- gers is 11. If each of these integers is less than 20, then what is the least possible value of any one of these integers? (A) −113 (B) −77 (C) −37 (D) −22 (E) 0 3. The median of 8, 6, 1, and k is 5. What is k? 4. The average (arithmetic mean) of two numbers is z. If one of the two numbers is x, what is the value of the other number in terms of x and z? (A) z − x (B) x − z (C) 2z − x (D) x − 2z (E) 5. A set of n numbers has an average (arithmetic mean) of 3k and a sum of 12m, where k and m are positive. What is the value of n in terms of k and m? (A) (B) (C) (D) (E) 36km 6. The average (arithmetic mean) of 5, 8, 2, and k is 0. What is the median of this set? (A) 0 (B) 3.5 (C) 3.75 (D) 5 (E) 5.5 m k4 k m4 4k m 4m k xz + 2 7. A die is rolled 20 times, and the outcomes are as tabulated above. If the average (arithmetic mean) of all the rolls is a, the median of all the rolls is b, and the mode of all the rolls is c, then which of the following must be true? I. a = b II. b > c III. c = 5 (A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III 8. If a 30% salt solution is added to a 50% salt so- lution, which of the following could be the con- centration of the resulting mixture? I. 40% II. 45% III. 50% (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III 9. Set A consists of five numbers with a median of m. If Set B consists of the five numbers that are two greater than each of the numbers in Set A, which of the following must be true? I. The median of Set B is greater than m. II. The average (arithmetic mean) of Set B is greater than m. III. The greatest possible difference between two numbers in Set B is greater than the greatest possible difference between two numbers in Set A. (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III SAT Practice 2: Mean/Median/Mode Problems CHAPTER 9 / SPECIAL MATH PROBLEMS 339 Roll Frequency 15 23 33 43 53 63 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 . understand to do well on the SAT. For more tools and resources that will help, visit our Online Practice Plus at www.MHPracticePlus.com/SATmath. 334 MCGRAW-HILL’S SAT Concept Review 1: New Symbol. is odd ◊=x x 4 $x x = 1 SAT Practice 1: New Symbol or Term 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 6 336 MCGRAW-HILL’S SAT Answer Key 1: New Symbol. 330 MCGRAW-HILL’S SAT 7. When the Apex Pet Store first opened, the ratio of cats to dogs was 4 to 5. Since then,