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3^-(x+y)/3^-(x-y)=? 1) x=3 2) y=2

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3^-(x+y)/3^-(x-y)=? 1) x=3 2) y=2  [#permalink]

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New post 29 Jun 2017, 02:58
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  25% (medium)

Question Stats:

71% (00:52) correct 29% (01:16) wrong based on 55 sessions

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\(3^{-(x+y)}/3^{-(x-y)}=?\)

1) x=3
2) y=2

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3^-(x+y)/3^-(x-y)=? 1) x=3 2) y=2  [#permalink]

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New post 29 Jun 2017, 03:11
1
MathRevolution wrote:
\(3^{-(x+y)}/3^{-(x-y)}=?\)

1) x=3
2) y=2


Simplifying the fraction, we get;

\(\frac{3^{-(x+y)}}{3^{-(x-y)}} = 3^{-(x+y)}*3^{(x-y)}\)

\(3^{(-x - y + x - y)}\) \(= 3^{ (-2 y) }\)

So we just need the value of \(y\) to find the value of the fraction \(\frac{3^{-(x+y)}}{3^{-(x-y)}}\).

1) \(x=3\)

Value of \(x\) is not needed. Hence I is Not Sufficient.

2) \(y=2\)

\(3^{( -2 y )}\) = \(3^{( -2 *2 )}\) = \(3^{ -4}\) = \(\frac{1}{3^{ 4}}\) \(= \frac{1}{81}\) .

Hence II is Sufficient.

Answer (B)...
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Re: 3^-(x+y)/3^-(x-y)=? 1) x=3 2) y=2  [#permalink]

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New post 29 Jun 2017, 03:32
\(\frac{3^{-(x+y)}}{3^{-(x-y)}}=?\)

Let's simplify the above equation:

= \(\frac{3^{(x-y)}}{3^{(x+y)}}\)

= \(\frac{3^x * 3^-y}{3^x * 3^y}\)

Cancel out \(3^x\)

\(= \frac{3^-y}{3^y}\)

\(= 3^-2y = ?\)

\(= \frac{1}{3^2y} = ?\)

As per above if we know the value of y we should be able to find the answer.

\(1) x=3\)

Clearly not sufficient as we are not aware of the value of x

Hence, (1) =====> is NOT SUFFICIENT

\(2) y=2\)

As we know the value of y now, we can plug this value in the above equation

\(= \frac{1}{3^2y}\)

\(= \frac{1}{3^2*2} = \frac{1}{3^4} = \frac{1}{81}\)

As we are able to determine the value, B is sufficient

Hence, (2) =====> is SUFFICIENT

Hence, Answer is B
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Re: 3^-(x+y)/3^-(x-y)=? 1) x=3 2) y=2  [#permalink]

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New post 02 Jul 2017, 18:08
==> If you modify the original condition and the question, you get \(\frac{3^{-(x+y)}}{3^{-(x-y)}}=3^{x-y+x-y}=3^{-2y}\), so you only need to know y.

The answer is B.
Answer: B
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Re: 3^-(x+y)/3^-(x-y)=? 1) x=3 2) y=2   [#permalink] 02 Jul 2017, 18:08
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