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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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11 May 2018, 11:58
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45% (01:45) correct 55% (01:47) wrong based on 169 sessions

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GST Week 5 Day 4 Manhattan Prep Question 4

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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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Updated on: 11 May 2018, 12:43
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1
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

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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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11 May 2018, 12:30
1
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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11 May 2018, 12:43
1
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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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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11 May 2018, 15:33
1
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

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

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11 May 2018, 20:13
1
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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11 May 2018, 22:29
1
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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13 May 2018, 02:25
1
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.

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

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14 May 2018, 10:00
1
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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