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01 Sep 2021, 04:30
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A
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Difficulty:
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56% (01:44) correct
44%
(02:13)
wrong
based on 162
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What is the value of \(\frac{(2a + 6a − 3b)}{(2a − b)}\)?
(1) \(\frac{−2a}{b − 2a}\)=8
(2) \(6a – 3b = 3\)
(1) \(\frac{−2a}{b − 2a}\)=8
(2) \(6a – 3b = 3\)
bhavika01
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01 Sep 2021, 04:35
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C - both equations have two variables, equating them will help us find value for one and eventually another.
Please explain if I'm wrong?
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Please explain if I'm wrong?
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01 Sep 2021, 04:56
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Bunuel
What is the value of \(\frac{(2a + 6a − 3b)}{(2a − b)}\)?
(1) \(\frac{−2a}{b − 2a}\)=8
(2) \(6a – 3b = 3\)
(1) \(\frac{−2a}{b − 2a}\)=8
(2) \(6a – 3b = 3\)
The very last DS question in OG 2018 (and probably other recent editions, that's just the one I have at hand) is very similar to this question. You could rewrite S1, and then substitute into the expression in the question, or we can rewrite the expression in the question:
\(\\
\frac{2a + 6a - 3b}{2a - b} = \frac{2a}{2a - b} + \frac{6a - 3b}{2a - b} = \frac{2a}{2a - b} + \frac{(3)(2a - b)}{2a - b} = \frac{2a}{2a - b} + 3\\
\)
Now if we just cancel -1 in the numerator and denominator on the left side of Statement 1, we learn that 2a/(2a - b) = 8, so Statement 1 is sufficient. Statement 2 lets us find the value of 2a - b, but not of 2a alone, so we can't use it to evaluate 2a/(2a - b) + 3. So the answer is A.
This is one of hundreds of questions that illustrate why you shouldn't count equations and count unknowns in GMAT DS. If we needed to find a or to find b here, we'd need both Statements, but the question doesn't ask us to find a or b; it asks for some combination of a and b that we can find without knowing what a or b are individually.
