for now. Choice (H) becomes = −2 , which is
not equal to 0; eliminate (H). Choice (J) becomes = −2, which is not greater than 0;
eliminate (J). Choice (K) becomes = 1, which is not less than 0; eliminate (K) and
choose (G).
49 . C This question is best approached by Plugging In the Answers. Start with (C): if n = 2, then the
equation becomes 272(2) = 812 + 1 or 27^4 = 81^3 . Put these numbers into your calculator to
determine whether they are, in fact, equal. Both 27^4 and 81^3 are equal to 531,441, so (C) is
correct.
55 . C The problem asks for an irrational solution, so while (B) would work, it is rational, so cross
it off. Then Plug In the Answers. When you Plug In the value in (C), the equation reads |
− 30| − 6 = 0. When you square , you get 24, so it’s |24 − 30| − 6 = 0. The
value inside the absolute value symbol becomes −6, the absolute value of which is 6. The
final equation is now much simpler: 6 − 6 = 0.
59 . D First, figure out all the pairs of positive integers which add up to 6 and Plug In. 1^5 and 5^1
aren’t equal, so try 2 and 4: 2^4 = 4^2 . Be sure to keep track of what the problem asks: How
many values of a satisfy the equation? Consider that a could occupy the place of either the 2
or the 4, so those are 2 possible values for a. The last pair of numbers is 3 and 3. Since 3^3 =
33 , 3 is the third value for a.
