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# M60-18

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

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11 Jun 2018, 06:39
00:00

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) $$y < 0$$

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8003 GMAT 1: 760 Q51 V42 GPA: 3.82 Re M60-18 [#permalink] ### Show Tags 11 Jun 2018, 06:39 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. Answer: A Answer: A _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
Manager
Joined: 23 Jan 2018
Posts: 221
Location: India
WE: Information Technology (Computer Software)

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23 Feb 2019, 10:20
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.

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.

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

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15 Apr 2019, 13:31
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.

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.

regards,
Arup

MathRevolution Bunuel Bunuel

Could you plase help?

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

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15 Apr 2019, 21:17
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.

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.

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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Re: M60-18   [#permalink] 15 Apr 2019, 21:17
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# M60-18

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