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Is x^2-y^2>x+y?

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by Max@Math Revolution » Mon Apr 09, 2018 1:22 am
[GMAT math practice question]

$$Is\ x^2-y^2>x+y?$$

$$1)\ x-y<0$$
$$2)x+y<0$$
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Source: — Data Sufficiency |
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by GMATGuruNY » Mon Apr 09, 2018 3:57 am
Max@Math Revolution wrote:[GMAT math practice question]

$$Is\ x^2-y^2>x+y?$$

$$1)\ x-y<0$$
$$2)x+y<0$$
Question stem, rephrased:
Is (x+y)(x-y) > x+y?

Statement 1: x - y < 0
If x-y = -1 and x+y = 1, then (x+y)(x-y) = -1 and x+y = 1, so the answer to the question stem is NO.
If x-y = -1 and x+y = -1, then (x+y)(x-y) = 1 and x+y = -1, so the answer to the question stem is YES.
INSUFFICIENT.

Statement 2: x+y < 0
If x+y = -1 and x-y = -1, then (x+y)(x-y) = 1 and x+y = -1, so the answer to the question stem is YES.
If x+y = -1 and x-y = 1, then (x+y)(x-y) = -1 and x+y = -1, so the answer to the question stem is NO.
INSUFFICIENT.

Statements combined:
The question stem can be rephrased as follows:
Is (negative)(negative) > negative ?
Is positive > negative?
The answer to the rephrased question stem is YES.
SUFFICIENT.

The correct answer is C.
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by Mo2men » Mon Apr 09, 2018 10:04 am
GMATGuruNY wrote:
Max@Math Revolution wrote:[GMAT math practice question]

$$Is\ x^2-y^2>x+y?$$

$$1)\ x-y<0$$
$$2)x+y<0$$
Question stem, rephrased:
Is (x+y)(x-y) > x+y?

Statement 1: x - y < 0
If x-y = -1 and x+y = 1, then (x+y)(x-y) = -1 and x+y = 1, so the answer to the question stem is NO.
If x-y = -1 and x+y = -1, then (x+y)(x-y) = 1 and x+y = -1, so the answer to the question stem is YES.
INSUFFICIENT.

Statement 2: x+y < 0
If x+y = -1 and x-y = -1, then (x+y)(x-y) = 1 and x+y = -1, so the answer to the question stem is YES.
If x+y = -1 and x-y = 1, then (x+y)(x-y) = -1 and x+y = -1, so the answer to the question stem is NO.
INSUFFICIENT.

Statements combined:
The question stem can be rephrased as follows:
Is (negative)(negative) > negative ?
Is positive > negative?
The answer to the rephrased question stem is YES.
SUFFICIENT.

The correct answer is C.
Dear GMATGuru,
Can the analysis be different when combining both statements?

From S2: We know that (x +y) is Negative and does not equal to zero, so the question could rephrased:

Is (x+y)(x-y) > (x+y)? ....divide both by (x+y) and change sign
Is (x-y) < 1??
From S1 we know that (x - y) < 0 the it is less than 1.

Answer would be Yes and choice is C

Is my reasoning above valid?
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by Max@Math Revolution » Wed Apr 11, 2018 1:16 am
=>

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.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be 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) together:
x^2 - y^2 = (x+y)(x-y) > 0 since x+y < 0 and x-y < 0.
Therefore,
x^2 - y^2 > 0 > x + y
Thus, both conditions together are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1):
If x = -2 and y = -1, then the answer is "yes".
If x = 1 and y = 2, then the answer is "no".

Thus, condition 1) is not sufficient.


Condition 2):
If x = -3 and y = -1, then the answer is "yes".
If x = -1 and y = -3, then the answer is "no".
Thus condition 2) is not sufficient.

Therefore, C is the answer.

Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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same posting

by Max@Math Revolution » Wed Apr 11, 2018 1:16 am
same posting
Last edited by Max@Math Revolution on Wed Apr 11, 2018 1:19 am, edited 1 time in total.
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by Max@Math Revolution » Wed Apr 11, 2018 1:17 am
=>

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.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be 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) together:
x^2 - y^2 = (x+y)(x-y) > 0 since x+y < 0 and x-y < 0.
Therefore,
x^2 - y^2 > 0 > x + y
Thus, both conditions together are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1):
If x = -2 and y = -1, then the answer is "yes".
If x = 1 and y = 2, then the answer is "no".

Thus, condition 1) is not sufficient.


Condition 2):
If x = -3 and y = -1, then the answer is "yes".
If x = -1 and y = -3, then the answer is "no".
Thus condition 2) is not sufficient.

Therefore, C is the answer.

Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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by Max@Math Revolution » Wed Apr 11, 2018 1:17 am
=>

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.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be 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) together:
x^2 - y^2 = (x+y)(x-y) > 0 since x+y < 0 and x-y < 0.
Therefore,
x^2 - y^2 > 0 > x + y
Thus, both conditions together are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1):
If x = -2 and y = -1, then the answer is "yes".
If x = 1 and y = 2, then the answer is "no".

Thus, condition 1) is not sufficient.

Condition 2):
If x = -3 and y = -1, then the answer is "yes".
If x = -1 and y = -3, then the answer is "no".
Thus condition 2) is not sufficient.

Therefore, C is the answer.
Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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by GMATGuruNY » Wed Apr 11, 2018 1:59 am
Mo2men wrote:Dear GMATGuru,
Can the analysis be different when combining both statements?

From S2: We know that (x +y) is Negative and does not equal to zero, so the question could rephrased:

Is (x+y)(x-y) > (x+y)? ....divide both by (x+y) and change sign
Is (x-y) < 1??
From S1 we know that (x - y) < 0 the it is less than 1.

Answer would be Yes and choice is C

Is my reasoning above valid?
Perfect!
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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