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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]
s is non negative

which means, s>=0. This is they key statement here.

Option E
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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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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.
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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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