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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8003
GMAT 1: 760 Q51 V42 GPA: 3.82

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Difficulty:   35% (medium)

Question Stats: 57% (01:30) correct 43% (00:58) wrong based on 7 sessions

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Is $$xy < -(x/y)$$?

1) $$xy < 0$$

2) $$y < 0$$

_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8003
GMAT 1: 760 Q51 V42 GPA: 3.82

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Official Solution:

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 (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

Modifying the question:

$$xy < -(x/y)$$

⇔ $$xy^3 < -xy$$ (multiplying both sides by $$y^2$$)

⇔ $$xy^3 + xy$$ < 0

⇔ $$xy(y^2+1)$$ < 0

⇔ $$xy < 0$$ since $$y^2+1 > 0$$

Condition 1): $$xy < 0$$

Condition 1) is same as the question.

This condition is sufficient.

Condition 2):

Since this condition tells us nothing about $$x$$, it is not sufficient.

Therefore, the answer is A.

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.

_________________
Manager  G
Joined: 23 Jan 2018
Posts: 221
Location: India
WE: Information Technology (Computer Software)

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MathRevolution wrote:
Official Solution:

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 (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

Modifying the question:

$$xy < -(x/y)$$

⇔ $$xy^3 < -xy$$ (multiplying both sides by $$y^2$$)

⇔ $$xy^3 + xy$$ < 0

⇔ $$xy(y^2+1)$$ < 0

⇔ $$xy < 0$$ since $$y^2+1 > 0$$

Condition 1): $$xy < 0$$

Condition 1) is same as the question.

This condition is sufficient.

Condition 2):

Since this condition tells us nothing about $$x$$, it is not sufficient.

Therefore, the answer is A.

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.

chetan2u, Bunuel

Dear experts :
While modifying the question, instead of multiplying by y^2, if I multiply by y only, or say we cross multiply and make the equation as x(y2+1)=0.
Am I making some mistake? Please advice.

regards,
Arup
Manager  G
Joined: 23 Jan 2018
Posts: 221
Location: India
WE: Information Technology (Computer Software)

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ArupRS wrote:
MathRevolution wrote:
Official Solution:

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 (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

Modifying the question:

$$xy < -(x/y)$$

⇔ $$xy^3 < -xy$$ (multiplying both sides by $$y^2$$)

⇔ $$xy^3 + xy$$ < 0

⇔ $$xy(y^2+1)$$ < 0

⇔ $$xy < 0$$ since $$y^2+1 > 0$$

Condition 1): $$xy < 0$$

Condition 1) is same as the question.

This condition is sufficient.

Condition 2):

Since this condition tells us nothing about $$x$$, it is not sufficient.

Therefore, the answer is A.

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.

chetan2u, Bunuel

Dear experts :
While modifying the question, instead of multiplying by y^2, if I multiply by y only, or say we cross multiply and make the equation as x(y2+1)=0.
Am I making some mistake? Please advice.

regards,
Arup

MathRevolution Bunuel Bunuel

Could you plase help?

Regards,
Arup
Math Expert V
Joined: 02 Sep 2009
Posts: 58334

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1
ArupRS wrote:
MathRevolution wrote:
Official Solution:

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 (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

Modifying the question:

$$xy < -(x/y)$$

⇔ $$xy^3 < -xy$$ (multiplying both sides by $$y^2$$)

⇔ $$xy^3 + xy$$ < 0

⇔ $$xy(y^2+1)$$ < 0

⇔ $$xy < 0$$ since $$y^2+1 > 0$$

Condition 1): $$xy < 0$$

Condition 1) is same as the question.

This condition is sufficient.

Condition 2):

Since this condition tells us nothing about $$x$$, it is not sufficient.

Therefore, the answer is A.

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.

chetan2u, Bunuel

Dear experts :
While modifying the question, instead of multiplying by y^2, if I multiply by y only, or say we cross multiply and make the equation as x(y2+1)=0.
Am I making some mistake? Please advice.

regards,
Arup

We cannot multiply xy < -x/y by y and write xy^2 < -x because we don't know the sign of y. If y is positive, then yes, from xy < -x/y, we can get xy^2 < -x (keep the sign when multiplying by positive value) but if y is negative, then when we multiply by negative value, we should flip the sign and we'll get xy^2 > -x (recall that we should flip the sign of an inequality if we multiply/divide it by negative value)..

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.
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