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2^(x+y)/2^(x-y)=?

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MathRevolution
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2^(x+y)/2^(x-y)=? [#permalink] New post  12 Jul 2016, 19:05
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2^(x+y)/2^(x-y)=?
1) x=1
2) y=2

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msk0657
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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  12 Jul 2016, 22:41
MathRevolution wrote:
2^(x+y)/2^(x-y)=?
1) x=1
2) y=2

*An answer will be posted in 2 days


Given 2^(x+y)/2^(x-y).

We can write it as 2^(x+y)-(x-y)

=> 2^2y

So Stat 2 : clearly this is Sufficient as y = 2.

Option B is correct.
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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  13 Jul 2016, 01:29
Lets solve the stem first

2(x+y)-(x-y) = 2^2y

Hence Option B satisfies it
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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  13 Jul 2016, 02:24
Solve the stem - you will get 2^2y
Hence B
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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  13 Jul 2016, 04:09
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if GMAT will ask such questions then I'll score 800.


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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  13 Jul 2016, 06:53
MathRevolution wrote:
2^(x+y)/2^(x-y)=?
1) x=1
2) y=2

*An answer will be posted in 2 days


2^(x+y)/2^(x-y)= 2^(x+y)- (x-y)= 2^2y

So, it depends on the value of y.

B is the answer
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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  13 Jul 2016, 13:45
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MathRevolution wrote:
2^(x+y)/2^(x-y)=?

1) x=1
2) y=2


Target question: What is the value of 2^(x+y)/2^(x-y)?
This is a great candidate for rephrasing the target question.

Let's apply the division law, (x^a)/(x^b) = x^(a-b)
So, 2^(x+y)/2^(x-y) = 2^[(x+y) - (x-y)]
= 2^(2y)

We can now REPHRASE the target question...
REPHRASED target question: What is the value of 2^(2y)?

Statement 1: x = 1
This information has no bearing on the REPHRASED target question
So, statement 1 is NOT SUFFICIENT

Statement 2: y = 2
PERFECT.
This means that 2^(2y) = 2^4 = 16
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer =
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B


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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  14 Jul 2016, 21:23
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If we modify the original condition and the question, we get 2^(x+y)/2(x-y)=? --> 2^((x+y)-(x-y))=2^2y. We only have to know y and the correct answer is B.


- Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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2^x+y/2^x-y=? [#permalink] New post  31 Jan 2017, 02:57
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\(2^x^+^y/2^x^-^y=?\)

1) x=1
2) y=2
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Re: 2^x+y/2^x-y=? [#permalink] New post  31 Jan 2017, 04:23
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MathRevolution wrote:
\(2^x^+^y/2^x^-^y=?\)

1) x=1
2) y=2




\(2^x^+^y/2^x^-^y=\frac{2^x*2^y}{2^x}*2^y=2^{2y}\)

So we require to know only y..

Statement II gives us value of y..
Sufficient

B
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2^x+y/2^x-y=? [#permalink] New post  01 Feb 2017, 07:36
MathRevolution wrote:
\(2^x^+^y/2^x^-^y=?\)

1) x=1
2) y=2



2^(x+y-(x-y))----{ as x^m/x^n = X^(m-n)}
=2^(x+y-x+y)
=2^2y=??

(1) x=1
no info on Y
insuff

(2) y=2
thus 2^2y = 2^4 =16
suff

Ans B
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Re: 2^x+y/2^x-y=? [#permalink] New post  02 Feb 2017, 02:34
MathRevolution wrote:
\(2^x^+^y/2^x^-^y=?\)

1) x=1
2) y=2



2^(x+y-x+y)=2^2y
A-not suff
B-suff , as we know value of Y.

So B.
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Re: 2^x+y/2^x-y=? [#permalink] New post  02 Feb 2017, 02:51
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==> If you modify the original condition and the question, from \(2^x^+^y/2^x^-^y=2^x^+^y^-^(^x^-^y^)=2^x^+^y^-^x^+^y=2^2^y\), you only need to know y. From con 2) y=2, it is sufficient.

Therefore, the answer is B.
Answer: B
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Re: 2^x+y/2^x-y=? [#permalink] New post  04 Feb 2017, 07:02
MathRevolution wrote:
\(2^x^+^y/2^x^-^y=?\)

1) x=1
2) y=2


\(2^x^+^y/2^x^-^y=?\) can be re-written as:

\(2^x * 2^y / 2^x * 2^{-y}\) =?

\(2^y / 2^{-y}\) = ?

\(2^{2y}\) =?

We now only need to know the value of one variable i.e. \(y\). Hence statement-2 is sufficient. Answer is B.
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Re: 2^x+y/2^x-y=? [#permalink] New post  31 Jul 2017, 08:01
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Bunuel Can you please merge the above topic. :-)

https://gmatclub.com/forum/2-x-y-2-x-y-221942.html
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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  31 Jul 2017, 09:34
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NandishSS wrote:
Bunuel Can you please merge the above topic. :-)

https://gmatclub.com/forum/2-x-y-2-x-y-221942.html

_______________
Done. Thank you.
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Re: 2^(x+y)/2^(x-y)=? [#permalink] New post  31 Jul 2020, 21:37
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