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If x and y are positive, is 81^3 = 27^(y − x) ?

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If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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If \(x\) and \(y\) are positive, is \(81^3\) \(=\) \(27^{y - x}\)?


(1) |\(y\) \(-\) \(x\)| \(=\) \(4\)

(2) \(2^y\) \(=\) \(8^x\)

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Re: If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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The question in the premise can be further reduced to is
\(3^{12}\) \(=\) \(3^{3(y - x)}\)

12=3(y-x)

y-x = 4?


Statement 1 : -

When y = 5, x=1
Then YES

When y = 1 , x = 5
Then NO

INSUFFICIENT

Statement 2 : -

You can simplify it further to
\(2^y\) \(=\) \(2^{3x}\)
y=3x

Now, y-x = 3x-x
= 2x

It depends on the value of x to get y-x = 4

INSUFFICIENT

Combining both together,we get

| y-x | = 4
| 2x | = 4
Now, we can only get this if x=2 (remember x cannot be -2 as x and y are both positive)
Hence x has to be 2. If x is 2, y = 6

Hence y-x =4

SUFFICIENT

Answer is C
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Originally posted by pikolo2510 on 11 May 2018, 12:11.
Last edited by pikolo2510 on 11 May 2018, 12:43, edited 2 times in total.
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Re: If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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The statement can be reduced to y-x=4 ?

1) This gives us y=x+4 OR y=x-4. Therefore, y-x=4 OR y-x=-4. Not Sufficient.

2) This gives us y=3x. Not sufficient as a number of values are possible.

Combining 1 and 2, we get |2x|=4. Therefore, 2x=4 OR 2x=-4. x=2 OR x=-2. Hence, y=-6 OR 6 & y-x=4 OR -4. Not Sufficient. Hence, E.
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Re: If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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urvashis09 wrote:
The statement can be reduced to y-x=4 ?

1) This gives us y=x+4 OR y=x-4. Therefore, y-x=4 OR y-x=-4. Not Sufficient.

2) This gives us y=3x. Not sufficient as a number of values are possible.

Combining 1 and 2, we get |2x|=4. Therefore, 2x=4 OR 2x=-4. x=2 OR x=-2. Hence, y=-6 OR 6 & y-x=4 OR -4. Not Sufficient. Hence, E.



x and y are both positive. hence x=-2 case is invalid :-)
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Re: If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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New post 11 May 2018, 12:52
pikolo2510 wrote:
urvashis09 wrote:
The statement can be reduced to y-x=4 ?

1) This gives us y=x+4 OR y=x-4. Therefore, y-x=4 OR y-x=-4. Not Sufficient.

2) This gives us y=3x. Not sufficient as a number of values are possible.

Combining 1 and 2, we get |2x|=4. Therefore, 2x=4 OR 2x=-4. x=2 OR x=-2. Hence, y=-6 OR 6 & y-x=4 OR -4. Not Sufficient. Hence, E.



x and y are both positive. hence x=-2 case is invalid :-)



Oh missed that part, thanks! :) It should be C then.
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If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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New post 11 May 2018, 15:33
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I) |y-x|=4
But y-x can be positive or negative as we don't know whether x>y or y>x
If y >x then it it is sufficient
If x>y then it's insufficient
We have multiple cases here so this is insufficient

II)it gives y=3x
We can have different values of y for different values of x
We don't know value of x so insufficient

Combining both statements
Y=3x
So y-x=2x since x is positive 2x will always be positive
So we can remove mod
2x=4
x=2

Y=3x=6
So y-x=4
i.e. 3^12=27^y-x
Hence sufficient

So C is answer

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If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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New post 11 May 2018, 20:13
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Given \(x, y > 0\) & asked is \(81^{3} = 27^{(y-x)}\)?

Reduce the equation in question to a simpler form, by reducing it to a common base.

\(81^{3} =( 3^{4})^{3} = 3^{12}\)

&

\(27^{(y-x)} = 3^{3(y-x)}\)


So we have, \(3^{12} = 3^{3(y-x)}\)


hence, \(12 = 3(y-x)\)

simplified to \(4 = y-x\), hence our rephrased question is to find out whether \((y-x) = 4\) ?

Stmt 1: \(/ y-x / = 4\)

that means \((y-x) = 4\) or \((y-x) = -4\)

hence Stmt 1 alone is Insufficient. Therefore Choice A & D are not possible.

Stmt 2: \(2^{y} = 8^{x}\)

reduced to \(2^{y} = 2^{3x}\)

hence \(y = 3x\)

so, \((y-x) = 3x - x = 2x\)

given \(x>0\) , hence x can take any positive value.

hence Stmt 2 alone is Insufficient. Therefore Choice B is not possible.


Combining Stmt 1 & 2, we get,

\((y-x) = 4\) or \(-4\)

&

\((y-x) =2x\)


Therefore,

\(2x = 4\) or \(-4\)

\(x = 2\) or \(-2\)

however \(x>0\), hence \(x=2\), therefore \((y-x)=4\)

So combining Stmt 1 & 2 is sufficient, hence C is the answer choice.

Thanks,
Gym
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Re: If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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Option C is correct.

From the question we can deduce that 3^12 = 3^3(y-x)
so y-x = 4

1. From statement 1 .
|y-x| = 4 means y-x = 4 or -4
Since we have two choices we cannot infer if y-x=4. -- Insufficient

2. From statement 2 .
2^y = 8^x -> 2^y = 2^3x so y = 3x
But since y & x are both unknowns quantities they can take any value. So we cannot infer y-x = 4. -- Insufficient

3. ST1 & ST2 together.
Since y & x are positive & Y is greater than X ( Arrived from ST2 y = 3x which makes y the larger quantity ) we can say in ST1 y-x cannot be -4.
So Y - X = 4.
Both ST1 & ST2 together sufficient.
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If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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New post 13 May 2018, 02:25
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The question in the premise can be further reduced to is
\(3^{12}\) \(=\) \(3^{3(y - x)}\)

Equating the powers
12=3(y-x) --------1

y-x = 4.


Statement 1 : -

|y-x|=4.

==> y = 5, x=1. Statement 1 is correct. Let's check if equation 1 satisfies or not.

12 = 3(5-1)
12 =12.
For y = 5, x=1. It is YES.

==> y = 1, x=5. Statement 1 is correct. Let's check if equation 1 satisfies or not.
12 = 3(1-5)
12 = 3 * (-4)
12 = -12. not true.

Statement 1 does not give definite answer. Not sufficient.


Statement 2 : -

You can simplify it further to
\(2^y\) \(=\) \(2^{3x}\)
y=3x

Using this in Eq 1.
12=3(y-x)
12 = 3(3x-x)
12 = 6x.

x value is unknown. Therefore it is insufficient.

NOTE - We don't need to find the values of x. We have to prove that y=3x satisfies the equation 1.

Combining both together,we get

| y-x | = 4 & y = 3x.

|3x-x| = 4.
2x =4
x=2
Note - x cannot be -2 as x and y are both positive
Hence x has to be 2. If x is 2, y = 6

If you substitute it in - \(3^{12}\) \(=\) \(3^{3(y - x)}\). It will satisfy the condition.

Answer is C
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If x and y are positive, is 81^3 = 27^(y − x) ? [#permalink]

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New post 14 May 2018, 10:00
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Let's do this step-wise:

1. Break the stem down:

\(81^3\) can be written as \(3^{4(3)}\) = \(3^{12}\)
Likewise, 27^{y-x} can be written as \(3^{3(y-x)}\)

Put together, we now have \(3^{12}\) = \(3^{3(y-x)}\)

Because the bases are the same, this is the same as 12 = 3(y-x) or 12 or y-x = 4.

Statement 1) |y-x| = 4. Absolute value can have 2 answers. The first and (positive) form is y-x = 4 which satisfies above so sufficient; however, we need to consider -(y-x) = 4 which makes this statement insufficient. Thus, this answer is insufficient as there is not a single clear answer.
Statement 2) Using the same math in the stem above, \(2^y\) = \(2^{3x}\) thus y = 3x. Plugging this in the equation above (y - x) = 4, we notice that x is unknown; thus, insufficient.

Combine both statements: |3x - x| = 4 --> |2x| = 4. Same as above: x can take a negative and positive value; however, the stem says that both x and y are positive; hence, only 1 answer and therefore C.
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If x and y are positive, is 81^3 = 27^(y − x) ?   [#permalink] 14 May 2018, 10:00
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