Exercise Answers
Discrete Quantitative Questions
- D Since a and b are negative, their product and quotient must be positive.
Thus eliminate A and C. The sum of two negatives is a negative. Thus eliminate
B. For D and E, choose values. Let a = –4 and b = –3. Evaluate the expression
in D: –3 – (–4) = –3 + 4 = 1. 1 > 0, so D is true. Evaluate the expression in E:
–4 – (–3) = –4 + 3 = –1. –1 is not greater than zero. Eliminate E. - B If xy is positive, then x and y must have the same sign. If yz is negative, then
y and z must have different signs. If x and y have the same sign and y and z
have different signs, then x and z must have different signs, meaning xz < 0.
Use this information in each choice. For A, you know that xz < 0, but you do
not know the sign of y, so you do not know whether the product is negative.
In B, you know that y^2 > 0 (because of the even exponent) and that xz < 0.
(–)(+) = (–), so Choice B is negative. In C, you know that x^2 y^2 must be positive
because of the even exponents, but you do not know the sign of z. In D, all the
variables are raised to even exponents, so the result must be positive. In E, you
know that xy > 0, but you do not know the sign of the denominator. The only
choice that must be negative is B. - C, D, and E In the first inequality, you know that a must be positive since b^2
is positive. In the second inequality, you know that a and c must have different
signs. If a > 0, and a and c have different signs, then c < 0. - A, B, and D Since this is a “must be true” question, it is helpful to plug in
values for x and y. First, plug in integers, and then fractions.
Case 1: x = –0.5 and y = –0.75. Plug these values into the choices:
A: –0.5^2 – (– 0.75)^2 < 0 → Tr ue
B: –0.75 – (–0.5) < 0 → Tr ue
C: (^) –0.5^12 < 1 → False
D: –0.5 + –0.75–0.5 > 0 → Tr ue
E: –0.75^2 – (– 0. 5)^2 < 0 → False
Now plug in integers for the choices that yielded true in Case 1. Let x = –2
and y = –3.
A: –2^2 – (–3)^2 < 0 → Tr ue
B: –3 – (–2) < 0 → Tr ue
D: –2 + –3–2 > 0 → Tr ue
Since 1, 2, and 4 remain true for both conditions, those are the answers.
- C For a fraction to be positive, the numerator and denominator must have the
same sign. Since z is raised to an even exponent, you know that z^2 must equal
at least zero. Therefore, z^2 + 1 must be positive. If the denominator of this
fraction is positive, then the numerator is also positive. Thus x – a > 0.
Add a to both sides: x > a.
196 PART 4 ■ MATH REVIEW03-GRE-Test-2018_173-312.indd 196 12/05/17 11:51 am