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Re: If p and q are integers, such that p < 0 < q, and s is a [#permalink]
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Hi All,

While this is an older question, it's still based around some essential Number Property rules that still appear on the GMAT. This question can be solved by TESTing VALUES.

We're told that P < 0 < Q and that S is a NON-NEGATIVE INTEGER. We're asked which of the following MUST be true.

Let's TEST:
P = -2
Q = 1
S = 0

Answer A: P^2 < Q^2.... 4 < 1 NOT TRUE
Answer B: P+Q=0... -2+1 = -1 NOT TRUE
Answer C: SP < SQ... 0 < 0 NOT TRUE
Answer D: SP ≠ SQ... 0 = 0 NOT TRUE

There's only one answer left...

Answer E: P/Q < S... -2/1 < 0 This IS TRUE

Final Answer:

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Re: If p and q are integers, such that p < 0 < q, and s is a [#permalink]
Thank you for adding me as a member.

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Re: If p and q are integers, such that p < 0 < q, and s is a [#permalink]
The best approach for this question and all similar questions is to go with broader set of conditions rather than plugging in values. Here p is negative and q is positive, for all values p/q is negative and will be less than any non negative number.
Thus option E.
Thanks
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Re: If p and q are integers, such that p < 0 < q, and s is a [#permalink]
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iwillcrackgmat wrote:
If p and q are integers, such that p < 0 < q, and s is a non-negative integer, which of the following must be true?

A. p² < q²
B. p + q = 0
C. sp < sq
D. sp ≠ sq
E. p/q < s


Approach #1
If p < 0 < q, then p is NEGATIVE and q is POSITIVE

So, p/q = NEGATIVE/POSITIVE, which means p/q is NEGATIVE

If s is a non-negative integer, then we can be certain that p/q < s

Answer: E

Approach #2
The question asks, "Which of the following must be true?"
So, if we can find a counterexample that shows an answer choice can be false, then we can eliminate that answer choice.

A. p² < q²
If p = -2 and q = 1, we get: (-2)² < 1²
Simplify to get: 4 < 1, which is not true.
Eliminate A

B. p + q = 0
If p = -2 and q = 1, we get: (-2) + 1 = 0, which is not true.
Eliminate B

C. sp < sq
p = -2, q = 1, and s = 0, we get: (0)(-2) < (0)(1)
Simplify to get: 0 < 0, which is not true.
Eliminate C

D. sp ≠ sq
p = -2, q = 1, and s = 0, we get: (0)(-2) ≠ (0)(1)
Simplify to get: 0 ≠ 0, which is not true.
Eliminate D

By the process of elimination, the correct answer must be E

Cheers,
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Re: If p and q are integers, such that p < 0 < q, and s is a [#permalink]
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iwillcrackgmat wrote:
If p and q are integers, such that p < 0 < q, and s is a nonnegative integer, which of the following must be true?

A. p^2 < q^2
B. p + q = 0
C. sp < sq
D. sp ≠ sq
E. p/q < s


We know that p is negative and q is positive, so p/q is always negative. Since s is a nonnegative integer, then p/q is always less than s.

Answer: E
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Re: If p and q are integers, such that p < 0 < q, and s is a [#permalink]
Given

P and Q are integers.
So they can be 0, +ve or -ve.
P<0
Q>0
S is a non-negative integer. So S>=0
This deduction is very important to this question.

Options:

A. p^2 < q^2

While p is negative and square of a negative number is always positive, this means p^2 and q^2 could be any number of possible numbers, hence P^2 cannot always be less than Q^2
Incorrect

B. p + q = 0

Again, could be any number. P could be -5 and Q could be 2. p+q not equal to 0; whereas when P is -3 and Q is +3, their sum is 0.
Incorrect

C. sp < sq

s could be 0, any number divided by 0 is always 0. Hence both sides are 0. 0 is not less than 0. Hence, incorrect

D. sp ≠ sq

as mentioned above, 0 on either sides is possible. Hence, sp ≠ sq being "must be true" is incorrect

E. p/q < s

Since this is the only option remaining, you could save time and choose E but here goes:

negative/positive is always negative.
S is either 0 or more than 0
Hence, p/q < s always holds true!
E is the right answer
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Re: Q1: A parking garage has places for a c ertain number of [#permalink]
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Re: Q1: A parking garage has places for a c ertain number of [#permalink]
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