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kripalkavi
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45*2 is < 100 and 45*6 is > 200 so has to be 45*4 so B ?

36*3 and 36*5 are >100 and <200 so cant be D
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Right, even I got B. But the OA said C....go figure :roll:
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kripalkavi
Right, even I got B. But the OA said C....go figure :roll:


A mistake ;)... What is the source of the question? :)
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kripalkavi
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Fig
kripalkavi
Right, even I got B. But the OA said C....go figure :roll:

A mistake ;)... What is the source of the question? :)


The source is a set of 'tough' quant questions that I got from a friend. He got it from the Kaplan classes that he attended. Hmm...doesn't speak to well about the quality of their tutoring, does it?! :wink:
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I guess most of those independent tutoring services give a question with a mistake answer once in a while...

And yes, it´s B.
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If n is an integer and 100 < n <200, what is the value of n?
(1) n/36 is an odd integer.
(2) n/45 is an even integer.

From Statement 1:

n/36 to be an odd integer first multiple 36 * 5 (approx) and you will get 180 which means that 36*6 will give you a number beyond 200. The previous odd integer is 3 (36*3 = 108) which is in between 100 and 200.

2 values of n = 3 and 5. Cannot determine a single value for n (INSUFF)

Therefore BCE

From Statement 2. To get an even integer multiply 45 by 4 and 2

45*4 = 180 which seems to be the highest number divisible by 45 in that range of 100 and 200.

45 * 2 = 90 is a number out of the range of 100 and 200.

Bingo... One value of 4 for n between 100 and 200. B is the BEST CHOICE

:)



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