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# 3^(-(x + y))/3^(-(x - y)) =

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Intern
Joined: 17 Jan 2010
Posts: 21
3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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Updated on: 27 Jul 2015, 14:47
00:00

Difficulty:

45% (medium)

Question Stats:

56% (01:07) correct 44% (01:10) wrong based on 87 sessions

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

(1) $$x = 2$$
(2) $$y = 3$$

Originally posted by rahulms on 22 Jan 2010, 01:29.
Last edited by Bunuel on 27 Jul 2015, 14:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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Updated on: 22 Jan 2010, 12:03
The given expression is same as $$3^{-2x}$$. So we do not need value of y, only value of x. Therefore Statement 2 is useless and while Statement 1 is Sufficient.

(A)

Originally posted by shalva on 22 Jan 2010, 04:32.
Last edited by shalva on 22 Jan 2010, 12:03, edited 1 time in total.
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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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22 Jan 2010, 11:35
use

a^x/a^y = a^x-y

thus this becomes 3^-2y

this is dependent upon y only... thus B
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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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22 Jan 2010, 11:47
1
bhumika wrote:
:shock:

got it!

it's quite simple bhumika. here's how:

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

$$=3^{-(x+y)-(x-y)}$$

$$=3^{-x-y-x+y}$$

$$=3^{-2x}$$

so, we need the value of $$x$$, which is exactly what statement 1 gives us!

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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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22 Jan 2010, 11:50
1
3^-(x+y)/3^-(x-y)
= 3^-(x+y)-(x-y) ....((eg. t^-4 = 1/t^4))
=3^-x-y-x+y ....((opening the brackets))
=3^-2x....((-y+y gets cancelled))
Therefore ,
Statement I itself is sufficient since we only require the value of x

IMO A

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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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22 Jan 2010, 12:01
it should be 3^-2y instead of 3^-2x

3^-(x+y)/ 3^-(x-y)

= 3^ -(x+y) - (-(x-y))

=3^ -(x+y) +(x-y)

=3^-2y

Thus this depends upon y only...not x
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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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22 Jan 2010, 13:27
3^ -(x+y) /3^ -(x-y)
{1/3^(x+y)} / {1/3^(x-y)}
3^(x-y) / 3^(x+y)
3^(x-y)-(x+y)
3^x-y-x-y
3^-2y

OA B

rahulms kindly post the OA answers and also make the correction in this case.

Regards,
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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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22 Jan 2010, 13:38
1
rahulms wrote:
$$\frac{3^{-(x+y)}}{3^{-(x-y)}}=$$

1) $$x = 2$$
2) $$y = 3$$

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

So we need to find $$Y=?$$

1. Not Sufficient
2. Sufficient

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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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24 Jan 2010, 19:11
Consider doing

3^2/3^-2

same way it will be 3^-(x+y)+(x-y) = 3^(-x-y+x-y)=3^-2y

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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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02 Jan 2018, 03:34
An easy question, can be solved using laws of exponents
3^-(x+y) / 3^-(x-y)
=>3^(-x-y+x-y)
=> 3^-2y
(1) Insufficient : x is not required
(2) Sufficient : Value of y is given

Correct choice: B
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Re: 3^(-(x + y))/3^(-(x - y)) =  [#permalink]

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07 Jan 2018, 14:27
rahulms wrote:
$$\frac{3^{-(x+y)}}{3^{-(x-y)}}=$$

(1) $$x = 2$$
(2) $$y = 3$$

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

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

Since the condition 2) only has a clue about $$y$$, the answer is B.
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Re: 3^(-(x + y))/3^(-(x - y)) =   [#permalink] 07 Jan 2018, 14:27
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