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If the operation % is defined by x % y = px^2 + qy^2 (p and q are cons

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Math Revolution GMAT Instructor
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If the operation % is defined by x % y = px^2 + qy^2 (p and q are cons  [#permalink]

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New post 05 Jul 2018, 00:47
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  45% (medium)

Question Stats:

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[GMAT math practice question]

If the operation \(%\) is defined by \(x % y = px^2 + qy^2\) (\(p\) and \(q\) are constants), what is the value of \(2 % 4?\)

(1) 1 % 2 = 5
(2) 1 % 1 = 2

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If the operation % is defined by x % y = px^2 + qy^2 (p and q are cons  [#permalink]

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New post 05 Jul 2018, 01:28
MathRevolution wrote:
[GMAT math practice question]

If the operation \(%\) is defined by \(x % y = px^2 + qy^2\) (\(p\) and \(q\) are constants), what is the value of \(2 % 4?\)

(1) 1 % 2 = 5
(2) 1 % 1 = 2


Given, x % y = \(px^2 + qy^2\)
Question stem:-2 % 4=4p+16q=?
Statement-1:-
1 % 2 = 5
Or, p+4q=5
or, 4(p+4q)=20
Or. 4p+16q=20
Sufficient.

Statement-2:-
1 % 1 = 2
Or, p+q=2
2 variables and one equation, we can't find p & q.

hence insufficient.

Ans. A
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Re: If the operation % is defined by x % y = px^2 + qy^2 (p and q are cons  [#permalink]

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New post 08 Jul 2018, 04:15
=>

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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
2 % 4 = \(4p + 16q = 4 ( p + 4q ) = 4( p* 1^2 + q*2^2 )\) = 4(1%2)
Thus, condition 1) is sufficient.

Condition 2)
Now, 1%1 = \(p + q = 2.\)
If p = 1 and q = 1, then 2 % 4 = \(1*2^2 + 1*4^2 = 20.\)
If p = 3 and q = -1, then 2 % 4 =\(3*2^2 + (-1)*4^2 = 12 – 16 = -4.\)
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A
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Re: If the operation % is defined by x % y = px^2 + qy^2 (p and q are cons &nbs [#permalink] 08 Jul 2018, 04:15
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