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Re: If a and b are integers, is a + b = 12? [#permalink]
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stne wrote:
chetan2u wrote:
If a and b are integers, is a + b = 12?

(1) \(a-b=7\)
(2) \(\frac{a}{b}=8\)

New question!!...


(1) \(a-b=7\)
Let \(a + b = 12\) and \(a-b = 7\) on solving we see that \(a\) and \(b\) are not integers, So it is not possible that \(a-b =7\) and \(a+b= 12\) both be true and \(a\)and \(b\) be integers. SUFF

(2) \(\frac{a}{b}=8\)
So question becomes is \(8b +b = 12\) or is \(9b= 12\) ? we know that \(b\) is an integer hence \(9b\) can never be equal to \(12\), SUFF

Ans - D
chetan2u, please can you show how statement 2 is NOT SUFF. IMO ans should be D.


Yes, it is D.
Edited. Kudos for the solution.
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Re: If a and b are integers, is a + b = 12? [#permalink]
Expert Reply
chetan2u wrote:
If a and b are integers, is a + b = 12?

(1) \(a-b=7\)
(2) \(\frac{a}{b}=8\)

New question!!...



Now a and b are integers, so a+b will be EVEN only when either BOTH rare even or BOTH are odd..

(1) \(a-b=7\)
a-b will be ODD only when one is odd and the other is even, so a+b will also be ODD and hence \(a+b\neq{12}\)

(2) \(\frac{a}{b}=8\)
a=8b... so a+b=8b+b=9b, and \(9b\neq{12}\), as b is an integer

D
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Re: If a and b are integers, is a + b = 12? [#permalink]
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