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Is x^2-y^2 < x-y?

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Math Revolution GMAT Instructor
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Is x^2-y^2 < x-y?  [#permalink]

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New post Updated on: 08 Dec 2017, 01:03
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

46% (01:39) correct 54% (00:52) wrong based on 52 sessions

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[GMAT math practice question]

Is \(x^2-y^2 < x-y?\)

1) \(x - y < 0\)
2) \(x + y > 0\)

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Originally posted by MathRevolution on 06 Dec 2017, 01:08.
Last edited by MathRevolution on 08 Dec 2017, 01:03, edited 1 time in total.
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Re: Is x^2-y^2 < x-y?  [#permalink]

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New post 06 Dec 2017, 01:15
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Imo E

please correct me if I am wrong in my approach.

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Re: Is x^2-y^2 < x-y?  [#permalink]

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New post 06 Dec 2017, 05:00
I think its E.
but :) we can not cancel out (x-y) on both sides of the equation till we know the signs of x and y or x-y not equals to 0.

the expression x2-y2< x-y => (x-y)(x+y-1)<0 { transferring x-y to the LHS }

Stat #1
x-y < 0. as we do not know the sign of (x+y-1) we can not conclude that (x-y)(x+y-1)<0
so insuff
Stat #2
x+y >0 tells that (x+y-1) will be +ve if x+y>1 and -ve if 0< (x+y) < 1
so insuff

Even both the equations do not convey the sign of (x+y-1) to conclude that x2-y2< x-y or (x-y)(x+y-1)<0

thanks
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Is x^2-y^2 < x-y?  [#permalink]

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New post Updated on: 10 Dec 2017, 23:21
=>

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 method is to modify the original condition and the question, and then recheck the question.

Modifying the question yields
\(x^2-y^2 < x-y?\)
\(⇔ x^2 - y^2 – ( x - y ) < 0?\)
\(⇔ ( x + y )( x – y ) – ( x - y ) < 0?\)
\(⇔ ( x + y – 1 )( x – y ) < 0?\)

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer, and so we should consider conditions 1) & 2) together first.

Conditions 1) & 2)
Since \(x – y < 0\), asking if \(( x + y – 1 )( x – y ) < 0\) is equivalent to asking if \(x + y – 1 > 0\).
For \(x = 1, y = 1, x + y – 1 > 0\) and the answer is ‘yes’.
For \(x = \frac{1}{4}\), \(y = \frac{1}{4}\), \(x + y – 1 < 0\) and the answer is ‘no’.

Thus, both conditions together are not sufficient.

Therefore, the answer is E.

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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Originally posted by MathRevolution on 08 Dec 2017, 01:04.
Last edited by MathRevolution on 10 Dec 2017, 23:21, edited 1 time in total.
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Re: Is x^2-y^2 < x-y?  [#permalink]

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New post 08 Dec 2017, 10:37
MathRevolution wrote:
=>

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 method is to modify the original condition and the question, and then recheck the question.

Modifying the question yields
\(x^2-y^2 < x-y?\)
\(⇔ x^2 - y^2 – ( x - y ) < 0?\)
\(⇔ ( x + y )( x – y ) – ( x - y ) < 0?\)
\(⇔ ( x + y – 1 )( x – y ) < 0?\)

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer, and so we should consider conditions 1) & 2) together first.

Conditions 1) & 2)
Since \(x – y < 0\), asking if \(( x + y – 1 )( x – y ) < 0\) is equivalent to asking if \(x + y – 1 > 0\).
For \(x = 1, y = 1, x + y – 1 > 0\) and the answer is ‘yes’.
For \(x = \frac{1}{4}\), \(y = \frac{1}{4}\), \(x + y – 1 < 0\) and the answer is ‘no’.

Thus, both conditions together are not sufficient.

Therefore, the answer is E.

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.

Answer: C


Hi

In your post, your answer says E (before the last paragraph of your post) as well as C (in the end). And in the question, OA is C. Can you please rectify this.
Re: Is x^2-y^2 < x-y? &nbs [#permalink] 08 Dec 2017, 10:37
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