32 . K When you have one function inside another, start with the one on the inside and work your
way out. And of course, you should Plug In! So plug in 2, and start with f(2) = 2^2 + 3. That
means f(2) = 7. That’s now the value you can plug in to the other function: g(7) = 3(7) − 1, so
your target answer is 20. Plug 2 into the answer choices, and (K) will read 3(2)^2 + 8, which
is 20.
40 . F Plug In values for the number of teams in the league and the number of teams that qualify for
the playoffs. Say there are 15 teams in the league (represented by x) and 8 of them qualify
(represented by y). That leaves 7 for the number of teams that do not qualify, which means
that your target answer is . Choice (F) becomes , or .
41 . C Plugging In is the easiest way to solve this problem, but we need to choose numbers that
work with the given equations. Start with the first equation, 3x = 4y . The easy thing to do
here is choose x = 4 and y = 3 because (3)(4) = (4)(3). Now we can solve the second
equation: (3) = z, so 2 = z and z = 6. Now it’s a simple matter to put them in order: 3 <
4 < 6, so y < x < z.
45 . C To make things a little easier, you can cube both sides to get rid of the radical. The right side
of the equation is now 27. Now, Plug In the Answers and see which one works. In (C), (−3)^2
− 6(−3) = 9 − (−18) = 27, and (9)^2 − 6(9) = 81 − 54 = 27.
46 . J Plug In anything you want as long as x > y, such as x = 3 and y = 2. In the original problem
−|2 − 3| = −|− 1|, so the target answer is −1. Try this out in the answer choices. Choice (J)
works: −(3−2) = −1
48 . G This question can be approached by thinking about abstract number properties, but it is easier
to plug in a simple number for x and check whether the answer choices are true. Try an easy
number, such as x = . Choice (F) becomes = 0, which is not greater than 0;
eliminate (F). Choice (G) becomes = = 2, which is greater than 1, so keep (G)
