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Source: — Data Sufficiency |
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by Jim@Grockit » Sun Sep 26, 2010 7:17 pm
ruthm_24 wrote:If x and y are integers greater than 1, is x a multiple of y?

1. 3y^2 + 7y = x

2. x^2 - x is a multiple of y.

Answer is A.
Statement 1: y^2 is y times y, therefore a multiple of y. Three times that (3y^2) will still be a multiple of y. Adding 7 more y will still make it a multiple of y, and that's what x equals . . . SUFFICIENT.

Statement 2: x (x-1) is a multiple of y, therefore x (x-1) = [a positive integer]*y. IF (and only if) you can divide both sides by (x-1) and be left with a positive integer coefficient of y (being left with x = [a positive integer]*y) is x a multiple of y. We don't know whether that's the case, so INSUFFICIENT.
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by hi.itz.mani » Mon Sep 27, 2010 1:37 am
for x to be amultiple of Y , we should be able to write x = YK where K is an integer

statement 1 . 3y^2 + 7y = x => y ( 3y + 7 ) = x . From this we can deduce that if 3y + 7 is an integer then we can write yK = x. Now whether 3y + 7 is an integer? We know Y is an integer hence 3 multiply by an integer will be an integer hence 3y is an integer. an integr + integer = integer. Which gives us 3y + 7 also an integer.

x = Y K

hence Statement 1 is sufficient.

Statement 2: x ( x - 1 ) = YK , here we may need to prove x = y ( k / x-1) .. hence k / (x-1) is an integer, which may or may not be true.

Hence Answer is A
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