Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4 : GMAT Data Sufficiency (DS)
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# Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4

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Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4 [#permalink]

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31 May 2012, 02:49
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Is 4^(x+y) = 8^10?

(1) x-y=9
(2) y/x=1/4

[Reveal] Spoiler:
Acc. to the question stem, we can figure out x+y=9

St#1. x-y=9, acc. to the OG explanation, it says that it is possible that x+y can have more than one value. But afterall, acc. to the question and st.# 1, if we can solve x=12, and y=3 which is sufficient. But OG says X can have many different possible value for X. Can anyone please clearify it. Thank you in advance.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-4-x-y-139120.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 24 Mar 2013, 01:26, edited 2 times in total.
Edited the question
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Re: Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4 [#permalink]

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31 May 2012, 03:06
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Is 4^(x+y) = 8^10?

Is $$4^{x+y}=8^{10}$$? --> is $$2^{2*(x+y)}=2^{3*10}$$? --> is $$2(x+y)=3*10$$ --> is $$x+y=15$$?

(1) x-y=9. Clearly insufficient to answer the question whether $$x+y=15$$. Consider $$x=9$$ and $$y=0$$ for a NO answer and $$x=12$$ and $$y=3$$ for an YES answer.

(2) y/x=1/4. Also insufficient. Consider $$x=4$$ and $$y=1$$ for a NO answer and $$x=12$$ and $$y=3$$ for an YES answer.

(1)+(2) We have two distinct linear equations with two unknowns (x-y=9 and y/x=1/4), so we can solve for $$x$$ and $$y$$ and answer the question. Sufficient.

Hope it's clear.
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Re: Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4 [#permalink]

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01 Jun 2012, 19:24
Thanks Bunuel, you are Awesome!!
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Re: Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4 [#permalink]

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01 Jan 2013, 23:30
Bunuel wrote:
Is 4^(x+y) = 8^10?

Is $$4^{x+y}=8^{10}$$? --> is $$2^{2*(x+y)}=2^{3*10}$$? --> is $$2(x+y)=3*10$$ --> is $$x+y=15$$?

(1) x-y=9. Clearly insufficient to answer the question whether $$x+y=15$$. Consider $$x=9$$ and $$y=0$$ for a NO answer and $$x=12$$ and $$y=3$$ for an YES answer.

(2) y/x=1/4. Also insufficient. Consider $$x=4$$ and $$y=1$$ for a NO answer and $$x=12$$ and $$y=3$$ for an YES answer.

(1)+(2) We have two distinct linear equations with two unknowns (x-y=9 and y/x=1/4), so we can solve for $$x$$ and $$y$$ and answer the question. Sufficient.

Hope it's clear.

by componendo-dividendo rule,
y/x=1/4

=>

(y+x)/x=5/4

so we get y+x..
is it not sufficient to answer the prompt..

Guess it is insufficient because we don't get the exact value of y+x..
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Re: Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4 [#permalink]

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23 Mar 2013, 17:39
Don't understand why the answer is "C"

We can simplify 4 ^ x + 4 = 8 ^1- to --> x + y = 15

With item (1), we now have 2 equations with 2 variables. (1) is sufficient (x + y = 15 and x-y = 9 - > 2 x = 24; x = 12 . plug in and get y = 3)

With item (2), we have 2 equations with 2 variables. (2) is sufficient (x +y = 15 and 4y=x -> 5y=15; y = 3. plug in and get x =12 )

Answer should be "D"
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Re: Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4 [#permalink]

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24 Mar 2013, 01:26
jordonswain wrote:
Don't understand why the answer is "C"

We can simplify 4 ^ x + 4 = 8 ^1- to --> x + y = 15

With item (1), we now have 2 equations with 2 variables. (1) is sufficient (x + y = 15 and x-y = 9 - > 2 x = 24; x = 12 . plug in and get y = 3)

With item (2), we have 2 equations with 2 variables. (2) is sufficient (x +y = 15 and 4y=x -> 5y=15; y = 3. plug in and get x =12 )

Answer should be "D"

Check here: is-4-x-y-139120.html#p1191376

Hope it helps.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-4-x-y-139120.html
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Re: Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4   [#permalink] 24 Mar 2013, 01:26
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# Is 4^(x+y) = 8^10? (1) x-y=9 (2) y/x=1/4

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