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GMAT Reivew 12th Edition, Prob76 Pg279

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GMAT Reivew 12th Edition, Prob76 Pg279

by WannabeGeek » Sun Feb 28, 2010 9:48 am
I just started some practice questions in the Official Guide and I am stiuck on the below problem. I would think the answer is C since (2) t-m=16 tells me that t and m are both even or both odd. In which case going back to (1) if I rewrite it to be 2p=t+m, p is even based on the conclusion from (2). Which means using both (1) and (2) the initial question is satisfied.

But the Answer Key has provided E as the answer.

If m, p, and t are positive integers and m < p < t, is
the product mpt an even integer?
(1) t-p=p-m
(2) t- m = 16
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by Ian Stewart » Sun Feb 28, 2010 10:41 am
WannabeGeek wrote:I just started some practice questions in the Official Guide and I am stiuck on the below problem. I would think the answer is C since (2) t-m=16 tells me that t and m are both even or both odd. In which case going back to (1) if I rewrite it to be 2p=t+m, p is even based on the conclusion from (2). Which means using both (1) and (2) the initial question is satisfied.

But the Answer Key has provided E as the answer.

If m, p, and t are positive integers and m < p < t, is
the product mpt an even integer?
(1) t-p=p-m
(2) t- m = 16
From S2, you have correctly concluded that t and m are either both even or both odd. If you rewrite Statement 1 as you've done, 2p = t + m, since t and m are both even or both odd, the right side of this equation must be even. That doesn't tell us much, since the left side clearly must be even regardless of whether p is even or odd; 2p will always be even.

The two Statements together are not sufficient here; S1 tells you the numbers are equally spaced, and S2 tells you the distance between the smallest and largest numbers is 16. Our numbers could easily be 2, 10 and 18 here, or they could be 1, 9 and 17. So mpt can be even or can be odd, and the answer is E.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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