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555-605 Level|   Algebra|   Exponents|      
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If x and y are integers, what is the value of 2x^(6y) - 4? 11 Nov 2017, 08:01
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67% (01:22) correct 33% (02:06) wrong based on 122 sessions

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If x and y are integers, what is the value of \(2x^{(6y)} - 4\)?


(1) \(x^{(2y)} = 16\)

(2) \(xy = 4\)
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A
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If x and y are integers, what is the value of 2x^(6y) - 4? 11 Nov 2017, 08:08
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Bunuel
If x and y are integers, what is the value of \(2x^{(6y)} - 4\)?


(1) \(x^{(2y)} = 16\)

(2) \(xy = 4\)
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Statement 1: \(x^{(2y)} = 16\), cube both sides to get

\(x^{(6y)} = 16^3\). we got the value of the variable. hence we can calculate the value of the equation. Sufficient

Statement 2: Two variables and one equation. Cannot be solved. Hence Insufficient

Option A
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If x and y are integers, what is the value of 2x^(6y) - 4? 14 Nov 2017, 21:13
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Statement 1: x^(2y)=16, cube both sides to get

x^(6y)=16^3. Hence 2*x^(6y)-4 = 2*16^3-4. Sufficient

Statement 2: xy = 4; possible values are (2,2), (1,4), (4,1). Each gives different answers. So insufficient.

Answer A
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If x and y are integers, what is the value of 2x^(6y) - 4? 16 Nov 2017, 11:03
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Bunuel
If x and y are integers, what is the value of \(2x^{(6y)} - 4\)?


(1) \(x^{(2y)} = 16\)

(2) \(xy = 4\)
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given expression an be deduced to 2*x^(2y)^3 -4

From statement 1 we can find the value of x^(2y)

So answer is A
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If x and y are integers, what is the value of 2x^(6y) - 4? 18 Nov 2017, 16:33
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Bunuel
If x and y are integers, what is the value of \(2x^{(6y)} - 4\)?


(1) \(x^{(2y)} = 16\)

(2) \(xy = 4\)
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 2 variables and 0 equations, C is most likely to be the answer and so we should consider both conditions 1) & 2) together first.

By CMT(Common Mistake Type) 4(A), we need to consider A or B as an answer.

Condition 1)
Since \(x^2y = 16\), \(x^6y = (x^2y)^3 = 16^3\) and so \(2x^6y - 4 = 2\cdot16^3-4\).
This is sufficient.

Condition 2)
\(xy = 4\).
From \(xy = 4\), we have \((2,2)\), \((-2,-2)\), \((1,4)\), \((-1,-4)\), \((4,1)\) and \((-4,-1)\) as pairs of \((x,y)\).
\(x^{6y} = 2^{12} = 2048\) for \(x=2\), \(y=2\)
\(x^{6y} = 1^{24} = 1\) for \(x=1\),\(y=4\).
Since we don't have unique solutions, this is not sufficient.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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