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MathRevolution
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

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
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MathRevolution
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

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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If y <0,

then by changing the sign of the inequality and solve the equation as follows:

xy < -(x/y)

xy^2 > -x (since y is -ve sign changed)

xy^2 + x > 0 (adding x on both sides)

x (y^2 + 1) > 0

since y ^ 2 + 1 > 0, x >0

Now, if y<0 and x> 0 and we get the answer as "no'

Experts please help if the thought is correct or no
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chetan2u yashikaaggarwal

Need your help on this
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Is \(xy < -(x/y)\)?

1) \(xy < 0\)

2) \(y < 0\)

main point to remember......xy and x/y will have same property when it comes to being positive or negative. Either both are positive or both are negative.
Thus xy and -(x/y) will have OPPOSITE signs.

1) xy<0
So -(x/y)>0......xy<0<-(x/y).......xy<-(x/y)
Suff

2) y<0
If x<0, then xy>0 but -(x/y)<0. So xy>-(x/y)
If x>0, then same as in statement I...xy<-(x/y)
Insufficient

A

Also you can solve the initial inequality
xy<-(x/y)......xy+x/y<0....(xy^2+x)/y<0 or x(y^2+1)/y<0
Now y^2+1 is always positive, so the question becomes
Is x/y<0, that is ‘Do x and y have opposite sign?’

1) xy<0
Yes, x and y have opposite signs

2) y<0
Nothing about x

A
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