OG13 Question #41 on page 278

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OG13 Question #41 on page 278

by tobiasaebi » Fri Oct 19, 2012 12:22 pm
Hi all,

I was solving question #41 of the OG13 and don't understand the solutions...

The question asks:
Is 4^(x+y)=8^10?

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

The answer is C

My solution was:

1. Restate the question into: 2^2(x+y)=2^30 which equals x+y=15
2. Looked at statement 1 which i restated into: x=9+y
3. I substituted step 2 into step 1 and got x=12 and y=3, therefore statement 1 should be sufficient?
4. I checked whether statment 2 is sufficient. Statement 2 was transformed to x=4y, which I substituted into the question and got 4y+y=15 equals y=3 and x=12. Therefore I thought that statement 2 is also sufficient. Thus I went for D.

Why is it not D?

Thank you for your answers! :)

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by neelgandham » Fri Oct 19, 2012 12:47 pm
Let me rephrase the question.
Is 4^(x+y)=8^10
Is 2^2(x+y)=2^30
Is (x+y) = 15.
(1) x-y=9
If x = y + 9,then x + y need not necessarily be 15.
If y = 1, then x = 1+9 = 10. x+y = 11.
If y = 3, then x = 3+9 = 12. x+y = 15.
(2) y/x=1/4
x = 4y,then x + y need not necessarily be 15.
If y = 1, then x = 4. x+y = 5.
If y = 3, then x = 12. x+y = 15.
From 1 + 2
x = 4y and
x = y + 9
So, 3y = 9 and y = 3. x = 12. x+y = 15

Answer C

Do you now know where you went wrong ? If No, please let me know.
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by tobiasaebi » Fri Oct 19, 2012 1:02 pm
Thank you for your answer, Neel.

I understand that if I'm looking at each statement, without considering x+y=15 from the question stem, then there are more than one possible solution. However, I included x+y=15 from the question stem and hence had two equations with two unknowns.
For instance:
(I)x+y=15
(II)x-y=9
If I substitute (II) into (I) I think I get the unique solution of y=3 and x=12?

Is it not allowed to combine an additional information from the question stem (here x+y=15) with each of the separate statements or where am I wrong?

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by GMATGuruNY » Fri Oct 19, 2012 6:26 pm
tobiasaebi wrote:Thank you for your answer, Neel.

I understand that if I'm looking at each statement, without considering x+y=15 from the question stem, then there are more than one possible solution. However, I included x+y=15 from the question stem and hence had two equations with two unknowns.
For instance:
(I)x+y=15
(II)x-y=9
If I substitute (II) into (I) I think I get the unique solution of y=3 and x=12?

Is it not allowed to combine an additional information from the question stem (here x+y=15) with each of the separate statements or where am I wrong?
Is 4^(x+y)=8^10?

(2²)^(x+y) = (2³)¹�
2^(2x+2y) = 2³�
Since the bases are equal, so must be the exponents:
2x+2y = 30
x+y = 15.

Question rephrased: Is x+y = 15?

This is not INFORMATION but a QUESTION: IS x+y=15?
It is not KNOWN that x+y=15.
Quite the opposite: whether x+y=15 is what we are being ASKED to determine.
Since it is not KNOWN that x+y=15, we can't combine x+y=15 with the statements.
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by TheAnuja55 » Fri Oct 19, 2012 8:03 pm
tobiasaebi wrote:Hi all,

I was solving question #41 of the OG13 and don't understand the solutions...

The question asks:
Is 4^(x+y)=8^10?

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

The answer is C

My solution was:

1. Restate the question into: 2^2(x+y)=2^30 which equals x+y=15
2. Looked at statement 1 which i restated into: x=9+y
3. I substituted step 2 into step 1 and got x=12 and y=3, therefore statement 1 should be sufficient?
4. I checked whether statment 2 is sufficient. Statement 2 was transformed to x=4y, which I substituted into the question and got 4y+y=15 equals y=3 and x=12. Therefore I thought that statement 2 is also sufficient. Thus I went for D.

Why is it not D?

Thank you for your answers! :)
tobiasaebi,

Rephrasing the question:

Is 4^(x+y)=8^10 ?? or is 2^2(x+y)=2^3*10 ?? or is 2(x+y)=30 ??

(1) x-y=9 Insufficient
since, if x=12 y=3, then YES it satisfies x+y=15.
if x=11 y=2, then No it doesn't satisfies x+y=15.

(2) y/x=1/4. or x=4y Also insufficient
Since, if x=12 y=3, then YES it satisfies x+y=15.
if x=4 y=1, then No it doesn't satisfies x+y=15.

And if we combine (1)&(2)
we two equations with 2 unknowns (x-y=9 and y/x=1/4) hence we can solve x and y.
hence we get x=12 y=3, which satisfies x+y=15.