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If x and y are non-zero numbers and x ≠ ± y, then ( x^2 +

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by Max@Math Revolution » Thu Jan 24, 2019 5:07 am

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[Math Revolution GMAT math practice question]

If x and y are non-zero numbers and x ≠ ± y, then ( x^2 + y^2 ) / ( x^2 - y^2 )=?

1) |x/y| = 1/3
2) y = -3x
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by fskilnik@GMATH » Thu Jan 24, 2019 5:46 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If x and y are non-zero numbers and x ≠ ± y, then ( x^2 + y^2 ) / ( x^2 - y^2 )=?

1) |x/y| = 1/3
2) y = -3x
\[x,y\,\, \ne 0\]
\[\left( {x + y} \right)\left( {x - y} \right)\,\, \ne \,\,0\]
\[?\,\, = \,\,\frac{{{x^2} + {y^2}}}{{\left( {x + y} \right)\left( {x - y} \right)}}\]
\[\left( {1 + 2} \right)\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {1, - 3} \right)\,\,\,\, \Rightarrow \,\,\,? = \frac{{10}}{{\left( { - 2} \right)\left( 4 \right)}} = - \frac{5}{4} \hfill \\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( { - 2,6} \right)\,\,\,\, \Rightarrow \,\,\,? = \frac{{10}}{{\left( 4 \right)\left( { - 8} \right)}} \ne - \frac{5}{4} \hfill \\
\end{gathered} \right.\]


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Max@Math Revolution » Sun Jan 27, 2019 5:05 pm
=>

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.
The question asks for the value of ( x^2 + y^2 ) / ( x^2 - y^2 )= ( (x/y)^2 + 1 ) / (x/y)^2 - 1 ).

When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient.

Condition 1)

Since |x/y|=1/3, x/y = ± (1/3), and ( x^2 + y^2 ) / ( x^2 - y^2 )= ( (x/y)^2 + 1 ) / ( (x/y)^2 - 1 ) = ( (1/3)^2 + 1 ) / ( (1/3)^2 - 1) = (1/9 + 1)/(1/9-1) = (10/9)/(-8/9) = -10/8 = -5/4.
Condition 1) is sufficient since it gives a unique solution.


Condition 2)
Since y = -3x, x/y = -1/3, and ( x^2 + y^2 ) / ( x^2 - y^2 )= ( (x/y)^2 + 1 ) / ( (x/y)^2 - 1 ) = ( (-1/3)^2 + 1 ) / ( (-1/3)^2 - 1) = (1/9 + 1)/(1/9-1) = (10/9)/(-8/9) = -10/8 = -5/4.
Condition 2) is sufficient since it gives a unique solution.

Therefore, the answer is D.
Answer: D

FYI: Tip 1) of the VA method states that D is most likely to be the answer if conditions 1) and 2) provide the same information.
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