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If x and y are positive, is 81^3 = 27^(y − x) ?
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11 May 2018, 11:58
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Re: If x and y are positive, is 81^3 = 27^(y − x) ?
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Updated on: 11 May 2018, 12:43
The question in the premise can be further reduced to is \(3^{12}\) \(=\) \(3^{3(y  x)}\) 12=3(yx) yx = 4? Statement 1 :  When y = 5, x=1 Then YES When y = 1 , x = 5 Then NO INSUFFICIENTStatement 2 : You can simplify it further to \(2^y\) \(=\) \(2^{3x}\) y=3x Now, yx = 3xx = 2x It depends on the value of x to get yx = 4 INSUFFICIENTCombining both together,we get  yx  = 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 yx =4 SUFFICIENTAnswer 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) ?
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11 May 2018, 12:30
The statement can be reduced to yx=4 ?
1) This gives us y=x+4 OR y=x4. Therefore, yx=4 OR yx=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 & yx=4 OR 4. Not Sufficient. Hence, E.



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Re: If x and y are positive, is 81^3 = 27^(y − x) ?
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11 May 2018, 12:43
urvashis09 wrote: The statement can be reduced to yx=4 ?
1) This gives us y=x+4 OR y=x4. Therefore, yx=4 OR yx=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 & yx=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) ?
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11 May 2018, 12:52
pikolo2510 wrote: urvashis09 wrote: The statement can be reduced to yx=4 ?
1) This gives us y=x+4 OR y=x4. Therefore, yx=4 OR yx=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 & yx=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) ?
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11 May 2018, 15:33
I) yx=4 But yx 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 yx=2x since x is positive 2x will always be positive So we can remove mod 2x=4 x=2
Y=3x=6 So yx=4 i.e. 3^12=27^yx Hence sufficient
So C is answer
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If x and y are positive, is 81^3 = 27^(y − x) ?
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11 May 2018, 20:13
Given \(x, y > 0\) & asked is \(81^{3} = 27^{(yx)}\)?
Reduce the equation in question to a simpler form, by reducing it to a common base.
\(81^{3} =( 3^{4})^{3} = 3^{12}\)
&
\(27^{(yx)} = 3^{3(yx)}\)
So we have, \(3^{12} = 3^{3(yx)}\)
hence, \(12 = 3(yx)\)
simplified to \(4 = yx\), hence our rephrased question is to find out whether \((yx) = 4\) ?
Stmt 1: \(/ yx / = 4\)
that means \((yx) = 4\) or \((yx) = 4\)
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, \((yx) = 3x  x = 2x\)
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,
\((yx) = 4\) or \(4\)
&
\((yx) =2x\)
Therefore,
\(2x = 4\) or \(4\)
\(x = 2\) or \(2\)
however \(x>0\), hence \(x=2\), therefore \((yx)=4\)
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) ?
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11 May 2018, 22:29
Option C is correct.
From the question we can deduce that 3^12 = 3^3(yx) so yx = 4
1. From statement 1 . yx = 4 means yx = 4 or 4 Since we have two choices we cannot infer if yx=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 yx = 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 yx 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) ?
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13 May 2018, 02:25
The question in the premise can be further reduced to is \(3^{12}\) \(=\) \(3^{3(y  x)}\) Equating the powers 12=3(yx) 1 yx = 4. Statement 1 :  yx=4.
==> y = 5, x=1. Statement 1 is correct. Let's check if equation 1 satisfies or not.
12 = 3(51) 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(15) 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(yx) 12 = 3(3xx) 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
 yx  = 4 & y = 3x.
3xx = 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) ?
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14 May 2018, 10:00
Let's do this stepwise: 1. Break the stem down: \(81^3\) can be written as \(3^{4(3)}\) = \(3^{12}\) Likewise, 27^{yx} can be written as \(3^{3(yx)}\) Put together, we now have \(3^{12}\) = \(3^{3(yx)}\) Because the bases are the same, this is the same as 12 = 3(yx) or 12 or yx = 4. Statement 1) yx = 4. Absolute value can have 2 answers. The first and (positive) form is yx = 4 which satisfies above so sufficient; however, we need to consider (yx) = 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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