leaving 70 − 14 = 56 pages left to type. He plans to type of the remaining pages in the afternoon. Since of 56 is 8, John
plans to type 8 pages in the afternoon, leaving 56 − 8 = 48 pages for him to type in the evening, choice (E).
G
Plug in the number 3 for each y in this expression. The numerator of the fraction becomes 8(3 + 5), and the denominator
becomes 3[(3 × 3) + 7]. Now we follow the order of operations (PEMDAS). In the numerator, 3 + 5 is 8, and 8 × 8 is 64, so
our numerator is 64. In the denominator, 3 × 3 is 9, 9 + 7 is 16, and 3 × 16 is 48, making our denominator 48. So the fraction
is which can be reduced to choice (G).
86.
E
Since all of the five answer choices are prime numbers, the easiest way to approach this problem is to try each answer
choice and see which one does not divide evenly into 210. Because 210 is even, 2 must be a factor; 210 is divisible by 3,
because 210 is 10 × 21, and 21 is divisible by 3; 210 ends with a zero, so it is divisible by 5; 210 is divisible by 7 (210 ÷ 7
= 30). So the only choice left is 11, choice (E).
87.
G
To solve this problem, we will pick even and odd numbers to plug in for n and try them in each answer choice. First we’ll try
choice (F). If n = 2, an even number, then n + 1 = 2 + 1, or 3, an odd number. If n = 3, an odd number, then n + 1 = 3 + 1, or 4,
an even number. As you can see, n + 1 is not always odd, so (F) is not the correct answer. Now we’ll try choice (G). If n = 2,
then 2 n + 1 = 2(2) + 1, or 4 + 1, or 5, an odd number. If n = 3, then 2 n + 1 = 2(3) + 1 = 7, an odd number. This time we got an
odd answer whether we started with an odd number or an even number, so 2 n + 1 must be odd, and choice (G) is the correct
answer.
88.
C
This is a translation problem. “A number x is multiplied by 5,” can be represented as 5x. “The result is” is represented by an
equal sign. And “8 less than the result of multiplying a number y by 3” means 3 y − 8. So the whole sentence can be written as
5 x = 3y − 8. This does not match any of the answer choices, but if we add 8 to both sides, we get 5 x + 8 = 3y, which is (C).
89.
J
Since Tony bought z stamps, 5 times the number of stamps Tony bought is 5z stamps. The amount that John bought is 3 more
than this— 5 z + 3. The total that both bought is 5 z + 3 + z = 6z + 3, choice (J).
90.
D
To find all possible values of y, plug in each possible value of x and then find the corresponding value of y. If x is 3, then our
equation reads 2(3) = 3y. After multiplying 2 and 3, we have 6 = 3y. Now we can divide both sides by 3 and find that y = 2.
This rules out choices (A) and (E), which do not include 2. If x = 6, then our equation becomes 2(6) = 3y, or 12 = 3y.
Dividing both sides by 3, we find that y = 4. This rules out choice (B). If x = 12, then 2(12) = 3y, or 24 = 3y and y = 8. So our
three values of y are 2, 4, and 8, choice (D).
91.
