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2xy-x-2x^2
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16 Nov 2021, 01:20
16 Nov 2021, 01:20
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Question Stats:
25% (01:14) correct
75% (01:37) wrong based on 12 sessions
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Quantity A |
Quantity B |
\(\frac{2xy-x-2x^2}{2x-4x} \) |
\(x-y \) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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2xy-x-2x^2
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16 Nov 2021, 08:19
16 Nov 2021, 08:19
1
\(\frac{2xy-x-2x^2}{2x-4x} = \frac{2x^2 - 2xy + x}{4x - 2x} = \frac{x(2x - 2y + 1)}{x(2)} = \frac{2x - 2y + 1}{2} = \frac{2(x-y)}{2} + \frac{1}{2} = \textbf{x-y + 0.5} > \textbf{x-y}\)
Hence, Answer is A
Hence, Answer is A
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2xy-x-2x^2
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16 Nov 2021, 09:45
16 Nov 2021, 09:45
1
Carcass wrote:
Quantity A |
Quantity B |
\(\frac{2xy-x-2x^2}{2x-4x}\) |
\(x-y \) |
I don't believe the official GRE test would ever have a question of this nature, unless it included some kind of proviso that says x can't equal 0.
Here's why:
Since there are no restrictions on the value of x, it could be the case that x = 0 and y = -1
In this case, we get:
QUANTITY A: \(\frac{2xy-x-2x^2}{2x-4x}= \frac{2(0)y-0-2(0^2)}{2(0)-4(0)}= \frac{0}{0}\)
QUANTITY B: \(x-y =0-1=-1\)
So which quantity is greater?
Unfortunately 0/0 isn't an actual number. So we might as well be comparing -1 with a banana (another non-number).
