• 7 CATs FREE!
    If you earn 100 Forum Points

    Engage in the Beat The GMAT forums to earn
    100 points for $49 worth of Veritas practice GMATs FREE

    Veritas Prep
    VERITAS PRACTICE GMAT EXAMS
    Earn 10 Points Per Post
    Earn 10 Points Per Thanks
    Earn 10 Points Per Upvote
    REDEEM NOW

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

This topic has expert replies
Legendary Member
Posts: 563
Joined: 14 Oct 2017
Followed by:3 members
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.

User avatar
Legendary Member
Posts: 2666
Joined: 14 Jan 2015
Location: Boston, MA
Thanked: 1153 times
Followed by:125 members
GMAT Score:770

by DavidG@VeritasPrep » 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.

The answer is A
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course