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Is 4^(x+y)=8^(10) ?

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Re: Is 4^(x+y)=8^(10) ? [#permalink]

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New post 25 Aug 2016, 16:07
Bunuel wrote:
Is \(4^{x+y}=8^{10}\) ?

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


We need to determine whether 4^(x+y) = 8^10. We start by breaking down our two bases into prime factors.

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

8^10 = (2^3)^10 = 2^30

We can now rephrase the question as:

Is 2^(2x+2y) = 2^30 ?

Because the bases are the same, we can drop them and set the exponents equal to each other. The question becomes:

Is 2x+2y = 30 ?

Is x + y = 15 ?

After simplifying the equation, we see that we need to determine whether the sum of x and y is equal to 15.

Statement One Alone:

x – y = 9

Knowing the difference of x and y is not the same as knowing the sum of x and y; thus, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

y/x = ¼

When we cross multiply obtain:

4y = x

4y = x is not enough information to determine the value of x + y. Statement two alone is not sufficient. We can eliminate answer choice B.

Statements One and Two Together:

Using statements one and two we know the following:

x – y = 9 and 4y = x

Since 4y = x, we can substitute 4y for x into the equation x – y = 9 and we have:

4y – y = 9

3y = 9

y = 3

Since y = 3, x = 4(3) = 12.

Thus, x + y = 12 + 3 = 15. We can answer yes to the question. Both statements together are sufficient.

The answer is C.
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Re: Is 4^(x+y)=8^(10) ? [#permalink]

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New post 17 Jul 2017, 07:24
Is \(4^{x+y}=8^{10}\) ?


\(4^{x+y}=8^{10}\)

\(2^{2(x+y)}=2^{30}\)

\(2(x+y) = 30\)

\(x+y = 15\)

So, we will find if \(x+y = 15\)?

(1) \(x - y = 9\)

Clearly not sufficient as we can have multiple values for x and y

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

(2) \frac{y}{x = 1/4}

\(x = 4y\)

Again this is not sufficient as we can get multiple values of x and y

Hence, (2) ===== is NOT SUFFICIENT

Combining (1) & (2)

When we substitute \(x = 4y\) in equation \(x - y = 9\)

we get,

\(4y - y = 9\)

\(3y = 9\)

\(y = 3\)

\(x = 4y\)

\(x = 4 * 3\)

\(x = 12\)

\(x + y = 12 + 3 = 15\)

Hence, (1) & (2) ===== is SUFFICIENT

Hence, Answer is C

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Is 4^(x+y)=8^(10) ? [#permalink]

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New post Updated on: 15 Jan 2018, 09:00
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Bunuel wrote:
Is \(4^{x+y}=8^{10}\) ?

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



Target question: Is 4^(x + y) = 8^10?
This is a good candidate for rephrasing the target question.

Given equation: 4^(x + y) = 8^10
Rewrite 4 and 8 as powers of 2 to get: (2²)^(x + y) = (2³)^10
Apply power of a power law to get: 2^(2x +2y) = 2^30
This means that: 2x + 2y = 30
Divide both sides by 2 to get: x + y = 15
In other words, asking whether 4^(x + y) = 8^10 is the SAME as asking whether x + y = 15
REPHRASED target question: Is x + y = 15?

Statement 1: x - y = 9
Is this enough information to answer the REPHRASED target question? No.
Consider these two CONFLICTING cases:
Case a: x = 12 and y = 3. In this case, x + y = 12 + 3 = 15. So, x + y DOES equal 15
Case b: x = 10 and y = 1. In this case, x + y = 10 + 1 = 11. So, x + y does NOT equal 15
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y/x = 1/4
Is this enough information to answer the REPHRASED target question? No.
Consider these two CONFLICTING cases:
Case a: x = 12 and y = 3. In this case, x + y = 12 + 3 = 15. So, x + y DOES equal 15
Case b: x = 8 and y = 2. In this case, x + y = 8 + 2 = 10. So, x + y does NOT equal 15
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that x - y = 9
Statement 2 tells us that y/x = 1/4
Since we have 2 different linear equations with 2 variables, we COULD solve the system for the individual values of x and y, which means we COULD answer the REPHRASED target question with certainty. Of course, we wouldn't waste precious time performing such calculations, since our sole goal is to determine the sufficiency of the combined statements.
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

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Originally posted by GMATPrepNow on 03 Aug 2017, 14:58.
Last edited by GMATPrepNow on 15 Jan 2018, 09:00, edited 1 time in total.
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Re: Is 4^(x+y)=8^(10) ? [#permalink]

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New post 04 Aug 2017, 12:02
Bunuel wrote:
Is \(4^{x+y}=8^{10}\) ?
(1) x - y = 9
(2) y/x = 1/4


Q Stem ->\(2^2(x+y)\) = \(2^30\)
=> x+y = 15 ?

1) x - y = 9
x = 10, y = 1 => x + y = 11 => NO
x = 12, y = 3 => x + y = 15 => YES
Insufficient.

2) x = 4y
x = 4, y = 1 => x + y = 5 => NO
x = 12, y = 3 => x + y = 15 => YES
Insufficient.

1+2)
4y - y = 9
=> y = 3 => x = 12
Sufficient.

C is the answer.
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Re: Is 4^(x+y)=8^(10) ? [#permalink]

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New post 07 Apr 2018, 16:14
I understand that statement one alone is not sufficient, however when we plug in for statement 2 we end up with 4y=x. So wouldn’t it be as follows,
4y+y= 15
5y= 15
Y= 3 and then we can solve for x
Isn’t the answer B instead?
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Re: Is 4^(x+y)=8^(10) ? [#permalink]

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New post 08 Apr 2018, 04:00
Ekjotb wrote:
I understand that statement one alone is not sufficient, however when we plug in for statement 2 we end up with 4y=x. So wouldn’t it be as follows,
4y+y= 15
5y= 15
Y= 3 and then we can solve for x
Isn’t the answer B instead?


Hello

It is NOT given that 4^(x+y) = 8^10. We have to determine whether this is actually true or not.

You are assuming that 4^(x+y) equals 8^10.
Re: Is 4^(x+y)=8^(10) ?   [#permalink] 08 Apr 2018, 04:00

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