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Explaination of an answer

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Explaination of an answer

by Heuters » Sun Nov 08, 2009 4:35 pm
Here's the DS problem:

What is the value of (2^x + 2^x)/2^y?
(1) x-y=8
(2) x/y=-3

The correct answer is (A) Statement 1 alone...

When I tried to solve this problem, I didn't simplify the equation, I took statement (1) & (2) and solved for one variable... then solved for the other so I could technically have two integers to plug into the original equation. Therefore I chose (C) Both statements together....

The book says that you should simplify the original equation to 2^x-y+1... therefore you only need to find the answer for x-y to solve. Statement (1) answers this so that is how they solved it.

Why is the approach I took wrong? Should I have simplified the original equation first?

Thanks!
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Source: — Data Sufficiency |
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by palvarez » Sun Nov 08, 2009 7:42 pm
Just one is sufficient entails that both are sufficient. (one can derive p or q from p being true)


But look at the wording of what A, B, C, D and E are.


C says that both are sufficient and that either alone is not sufficient.


Yes, you should simplify enough to answer "each alone being sufficient".

There is a grid approach, AD/BCE or BD/ACE.
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by bigmikefr » Tue Nov 24, 2009 11:56 am
The way I approach this is (2^x + 2^x)/2^y I try to work out the stimulus to make my life easier when I get to the statements.

(2^x + 2^x)/2^y yields (2^x)*(2)/(2^y)= (2^x-y)*2. This is why x-y is all you need to solve this problem.

(1) It gives you exactly what we worked out above, so A or D

(2) it tells you that x/y= -3 or that x= -3y. If you plug that into the original equation, you will get (2^-3y-y)*(2) or that (2^-4y)*2 and you cannot simplify further

So the answer is (A). Hopefully this helps
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by mp2437 » Tue Nov 24, 2009 12:46 pm
(1) tells you that x - y = 8,

value of (2^x + 2^x) / 2^y = 2^(x+1) / 2 ^ y = 2 ^ (x + 1 - y)...Rearranging, you get 2 ^ (1 + x -y) = 2 ^ (1 + 8) = 2^9. Sufficient.

Statement 2 gives a ratio, which others have mentioned above is insufficient to solve.
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