Let p, q, and r be non-zero integers satisfying the following three conditions: I. p^3 q r… — GRE General Test

Let $p$ , $q$ , and $r$ be non-zero integers satisfying the following three conditions:

I. $p^3 q r^2 < 0$
II. $(-p)^q < 0$
III. $(-1)^{p + r} = 1$

Which of the following expressions MUST be negative?

$q \cdot p^r$ Answer
B
$p \cdot (-q)^r$
C
$(-p)^r \cdot q$
D
$(p - q)^r$
E
$q \cdot (r - p)^q$

Answer

The expression

q \cdot p^r

MUST be negative.

Condition II establishes that

p > 0

and

q

is odd. Condition I establishes that

q < 0

. Since

p > 0

, raising

p

to any integer exponent

r

results in a strictly positive value (

p^r > 0

). Multiplying this positive value by negative

q

guarantees a negative outcome regardless of the value or sign of

r

Step-by-Step Solution

Analyze Condition II:

(-p)^q < 0

Base

-p < 0 \implies p > 0

, and exponent

q

must be an odd integer.

A negative number raised to an integer power is negative if and only if the exponent is odd.

Analyze Condition I:

p^3 q r^2 < 0

q < 0

(negative odd integer).

Since

p > 0

p^3 > 0

. Also

r \neq 0 \implies r^2 > 0

. For the overall product

p^3 q r^2

to be negative,

q

must be negative.

Analyze Condition III:

(-1)^{p + r} = 1

p + r

is an even integer, so

p

and

r

have the same parity.

(-1)^k = 1

requires

k

to be an even integer.

Evaluate the sign of

q \cdot p^r

q \cdot p^r < 0

for all valid values.

Since

p > 0

, any integer power

p^r > 0

. Multiplying positive

p^r

by negative

q

yields a negative result.

Key Concept

Even-odd exponent rules and negative base sign determination

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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