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If p,q and r are positive numbers (Simpler solution)

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by melguy » Fri Aug 09, 2013 9:14 am
Please suggest a simpler solution to this problem. Thanks
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by GMATGuruNY » Fri Aug 09, 2013 9:59 am
If p, q, and r are positive numbers such that 40 * (r+r²)/(q+r) -17 = p, and 1 < q < r, which of the following could be the value of p?

A 10
B 15
C 17
D 23
E 24
Since r > q > 1, r² > q.

Thus:
(r+r²) / (q+r) = (r + greater value) / (r + smaller value) = (greater value)/(smaller value) = more than 1.

Thus:
p = 40 * (r+r²)/(q+r) - 17 = [40 * (more than 1)] - 17 = (more than 40) - 17 = more than 23.
Eliminate A, B, C and D.

The correct answer is E.
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by Matt@VeritasPrep » Fri Aug 09, 2013 10:28 am
Melguy, where did you find this problem? This is pretty complex.

Here's my reasoning:

Simplify the equation first.

(r+r²)/(q+r) = (p+17)/40

Since r > q > 1, (r+r²) > (q+r). Hence (p+17) > 40, so p > 23, so the answer is E. This also gives you a lower bound on p, which is cool.

The values you're going to get for q and r are not pleasant: one solution I found and verified for p = 24 is q = 61/60 and r = (1 + √((20,011)/3))/80.
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by melguy » Tue Aug 20, 2013 11:10 pm
Hi Matt

This problem is from Knewton CAT. Certainly one of the difficult ones. I read all the explanations but took me a while to understand the solution

Thanks everyone for your input :-)
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