Cracking The SAT Premium

(Marvins-Underground-K-12) #1

CHAPTER DRILL ANSWERS AND EXPLANATIONS


Algebra Strategies Drill 1: No Calculator Section

5. C Plug   in  the answers.    If  you start   with    (B),    the length  is  8,  and the width   is  half    that,   or  4.  Area    is
length × width. The area of this rectangle is 8 × 4, which is nowhere near 128. Eliminate (A)
and (B), as both are too small. Try (C): If the length is 16, the width is 8. So, does 128 = 16 ×
8? You could write it all out, since you can’t use your calculator, but you can also estimate. 16
× 10 = 160, so 16 × 8 would be about 130. The number in (D) is too large and will give a
weird fraction, so (C) is correct. Alternatively, write an equation. The equation is area = w ×
2 w. So, 128 = 2w^2 . Divide by 2 to get 64 = w^2 . Take the square root of both sides to find w = 8.
The length is twice this width, so length = 2 × 8 = 16, and the answer is (C).

10. B A question    with    unknown variables   indicates   a   good    place   to  plug    in. You need    numbers for x
and y that will give you a negative product. Try x = 1 and y = –2. If you plug these into the
statements in the Roman numerals, you find that (I) is false, but (II) and (III) are true. You can
eliminate any answer choice that contains (I). This leaves (B) and (D). Now try different
numbers to see if you can eliminate another choice. If you try x = –1 and y = 2, you find that (II)
is false and (III) is still true. This leaves you with (B), which is correct.

13. D Plug  in  the answers,    starting    with    (B).    If  x   =   16, the left    side    of  the equation    is      =       =   2.

Does    that    equal   2   ?   No—it’s too small.  Choice  (C) is  ugly    to  work    with,   so  try (D) next.   If  it

is  too big,    (C) is  your    answer. For (D),    x   =   32, and the left    side    of  the equation    becomes 

. It’s a match, so (D) is correct. You could also solve this


algebraically.  Multiply    both    sides   by  2   to  get  .  Square  both    sides   to  get x   =   16  ×   2   =


  1. Choice (D) is correct.


15. C We’ve got variables   in  the answer  choices,    which   means   this    is  a   perfect Plug    In  problem.
Since we can’t use calculators, let’s make up an easy value for x, such as 2. 9x + 3x+1 then
becomes 9^2 + 32+1 = 81 + 27 = 108. We plugged in x = 2, so let’s use that to find y: y = 3x, so y
= 3^2 = 9. Now plug in y = 9 to each answer choice to see which one gives you 108. Choice (C)
is y(y + 3), which is 9(12) = 108, which is the correct answer.

Algebra Strategies Drill 2: Calculator-Permitted Section

8. C Two    variables   tells   you this    is  a   great   place   to  plug    in. Let’s   pick    numbers that    make    the math
easy. You can try x = 30 and y = 2. So in 2 hours there are 4 periods of 30 minutes each: 12 × 4
= 48. Alex can fold 48 napkins in 2 hours. 48 is your target. Plugging into your answer choices
gives you (C).
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