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integers values

by ketkoag » Tue Jul 21, 2009 10:04 am
If x and y are positive integers, and x<y, which of the following could be the value of (x/x+y)*10+(y/x+y)*20?
(a) 10
(b) 14
(c) 16
(d) 21
(e) 30

OA : 10
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by ASS1991 » Tue Jul 21, 2009 5:30 pm
Sry, I get answer (c), hope you find my error.

(x/x+y)*10+(y/x+y)*20
=10*((x+y)/(x+y) + (y/x+y))
=10 + (y/x+y)*10

=> 15<10 + (y/x+y)*10<20

=> Answer (c) 16.

Thanks,
Antoni
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by Ian Stewart » Wed Jul 22, 2009 4:04 am
It's definitely C. The equation is just that of a weighted average (group x has an average of 10, group y has an average of 20, and we're combining them). The combined average must therefore be between 10 and 20, and if y is greater than x, it must be between 15 and 20. Of course you can also work that out algebraically.

I'm curious about the source of the question. It is identical in every respect to Q148 in the PS section of OG12, except that some of the answer choices have been changed. Is it a prep company question?
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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by ketkoag » Sat Jul 25, 2009 11:37 am
Thanks Ian for ur explanation..
yes, it is a prep company question..
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