Algebra: Manipulation
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1. Jason collects x dollars by renting out his family's house while they are on vacation. He also receives a 2-dollar tip from the renters. Jason's family has five members including Jason, and all of the money Jason receives from the renters including the tip is divided evenly amongst the five family members. Jason spends six dollars of his share leaving him with y dollars from the rental venture. Which of the following represents the amount, x, that Jason charged for the house rental?

(A) 5 · (y + 6) / 2
(B) ((y + 6) / 5) – 2
(C) 5y + 28
(D) 5y + 8
(E) 2 · (y + 6) / 5

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Start with x, then add 2, then divide by 5, then subtract 6. You get y = (x+2)/5 — 6. Rearrange to get x = 5y + 28 or (C).

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2. If a=2·b2 and b=4·c3 and c=8·d4 then:

(A) a = 223 · d36
(B) a = 223 · d24
(C) a = 211 · d36
(D) a = 211 · d24
(E) a = 211 · d12

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Note that the numbers in the answer choices are all powers of 2. First write c-cubed as 29 · d12. So b is 211 · d12. This means b2 = 222 · d24. Since a = 2b2, the answer is (B).

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3. If x=2b and a=3 then 2(a–b)/2x =

(A) 8–b
(B) 8(b–1)
(C) 8(1–b)
(D) 8b
(E) 8

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Note that there is no x in the answer choices so replace x with 2b and put all the b's in the denominator to get 2a/23b. Since a=3 and 23 = 8 the expression can be written as 8/8b which is the same as choice (C). Note that SATan loves the fact that 23=8.

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4. If r + s + t = r – u = s – t then which of the following statements are true?

I. r = –2t
II. s + t + u = 0
III. r + 2s = 2 · (s – t)

(A) I, only
(B) II, only
(C) III, only
(D) I and III, only
(E) I, II, and III

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The problem gives you an equation in the form A = B = C. Use A = C to solve for r in terms of t (the s cancels). This shows that statement I is true. Use A = B to get s+t+u = 0. This shows that II is true. It's hard to get statement III directly, but if you realize that the 2s terms cancel then you see that SATan has made III the same as I. So III is true also and the answer is (E).

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5. If x = (ab + c) / a then xb =

(A) xa – c
(B) (xa – c) / a
(C) x2 – cx
(D) x2a – cx
(E) (x2a – cx) / a

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Note that the answer choices don't have b. This means you have to substitute for b. If you rearrange the given equation, you get b = (ax-c)/a. If you multiply this by x, you get the answer (E).

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6. If x(8/3) = y(6/7) (y > 0) then y(3/7) =

(A) x(7/2)
(B) x(4/3)
(C) x(5/3)
(D) x(5/2)
(E) x(7/3)

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Note that y3/7 is the square root of y6/7. Therefore the answer is the square root of x8/3 which is x4/3 or (B).

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7. If a = 2b = 3c = 6d then the average of a, b, c, and d is:

(A) 3a
(B) 2a
(C) a
(D) a / 2
(E) a / 6

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The answers are in terms of a so you have to average a, a/2, a/3, and a/6. Since 1 + 1/2 + 1/3 + 1/6 = 2, the average is 2a/4 = a/2 or (D).

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8. If 4 · √(x–1) = 9 then which of the following is a true statement about the value of x?

(A) 6 < x < 7
(B) 7 < x < 8
(C) 8 < x < 9
(D) 9 < x < 10
(E) 10 < x < 11

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Divide both sides by 4 and square both sides to get x - 1 = 81/16. Since 81/16 is a little more than 5, that means x is a little more than 6 and the answer is (A).

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9. If (2x + y) / (x – y) = 3 then x/y =

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

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You can't get x or y individually but you can get x/y. The first step is 2x+y = 3x-3y. Therefore, 4y = x and x/y = 4 (C).

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10. If (x + y) / (x – y) = a and a, x, and y are positive numbers greater than 1 and y≠x then y =

(A) ax / (a – 1)
(B) a / (ax + x)
(C) (ax – x) / (a + 1)
(D) (a + 1) / (ax – x)
(E) ax / (a + 1)

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As in problem 9, start with x+y = ax-ay. Then get y(a+1) = x(a-1). This leads to y = x(a-1)/(a+1) which is the same as (C) if you distribute the x.

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