## If p, q, r, and s are consecutive integers, with

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### If p, q, r, and s are consecutive integers, with

by VJesus12 » Thu Mar 15, 2018 4:19 am
If p, q, r, and s are consecutive integers, with p<q<r<s, is pr<qs?

(1) pq<rs

(2) ps<qr

The OA is A.

I got confused with the inequalities. Experts, can you show me how you'd solve it? Please.

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by [email protected] » Thu Mar 15, 2018 10:37 am
VJesus12 wrote:
If p, q, r, and s are consecutive integers, with p < q < r < s, is pr < qs?

(1) pq < rs

(2) ps < qr

The OA is A.

I got confused with the inequalities. Experts, can you show me how you'd solve it? Please.
Interesting question. We could do a little algebra here. If the numbers are consecutive, and p is the smallest, we can designate the other values as p + 1, p + 2, and p + 3.

Our rephrased question: Is p(p +2) < (p+1)(p+3)?
Simplify: p^2 + 2p < p^2 +4p + 3?
2p < 4p + 3?
-2p < 3?
Is p > -3/2?

Statement 1 rephrased: p(p+1) < (p+2)(p+3)
p^2 + p < p^2 +5p + 6
p < 5p + 6
-4p < 6
p> -6/4
p > -3/2 --> that's exactly what we were trying to determine! Statement 1 alone is sufficient to answer the question.

Statement 2 rephrased: p(p+3) < (p+1)(p+2)
p^2 + 3p < p^2 +3p + 2
0 < 2.
We already know that. Not sufficient.