MathRevolution wrote:
[GMAT math practice question]
(Algebra) \(x, y,\) and \(z\) are real numbers. What is the value of \(\frac{y}{x}\)?
1) \(x\) and \(y\) are positive numbers.
2) \(\frac{x}{y}=\frac{2y}{x-z}=\frac{2x+y}{z}\)
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Since we have \(3\) variables (\(x, y,\) and \(z\)) and \(0\) equations, E is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
We have \(x^2 - zx = 2y^2\) when we cross multiply \(\frac{x}{y}=\frac{2y}{x-z}.\)
We have \(2xy + y^2 = zx\) when we cross multiply \(\frac{x}{y}=\frac{2x+y}{z}\).
When we subtract the second equation from the first equation, we have
\(x^2 – zx – (2xy + y^2) = 2y^2 – zx\)
\(x^2 – zx – 2xy – y^2 = 2y^2 – zx\) (multiplying \(-1\) through the bracket)
\(x^2 – zx – 2xy – y^2 – 2y^2 + zx = 0\) (moving all terms to the left side of the equation)
\(x^2 - 2xy - 3y^2 = 0\) (gathering like terms)
\((x - 3y)(x + y) = 0 \)(factoring)
Since \(x > 0\) and \(y > 0\) from condition 1), we have \(x + y ≠ 0\) and \(x = 3y.\)
Thus \(\frac{y}{x} = \frac{y}{3y} = \frac{1}{3}.\)
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2) when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.
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