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.

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

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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.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.

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.

The answer is A