M60-11

Question

If  xy>0 , is  x^3 y^4>0?

1)  x>0

2)  y>0

MathRevolution · Answer

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:

x^3 y^4>0?

⇔  x>0

since  y^4>0  is always true. We can ignore even exponents in inequalities that have 0 on one side.

So, the question becomes, ' is x>0? '.

Condition 1) is certainly sufficient.

Condition 2):

x > 0  since  y > 0  and  xy > 0 . Thus, condition 2) is also sufficient.

Therefore, D is the answer.

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).

Answer: D

Pranithare · Answer

can you please check for the condition y <0 and x <0 in the question. In that case only b is not sufficient.
could you please explain in detail for option B

Praveenapk · Answer

Same query? do we answer for both being negative and we could stil end up xy>0

pranavK77 · Answer

Option B explanation:

given, xy>0, X and Y should either be both negative or positive. Since, y>0, ie, Y is positive, it implies X has to be positive (x>0), to meet this condition xy>0.

Hope this answers your query

CrackverbalGMAT · Answer

Hello Pranithare, Although the question data is the most important data in a DS question, when you use it in conjunction with the data given in the statements, you need to consider the implications of both, when you are trying to answer the question using a particular statement. From the question data, xy>o. This can happen in two ways: If x>0, y>0 and If x<0, y<0. From statement II alone, y>0. This is like giving the first case to you on a platter, right? By telling that y>0, the statement is automatically eliminating the second case for you, isn’t it? If y>0, x>0. Therefore, y^4 >0 and x^3 >0 and hence x^3*y^4 >0. Hope that clears up things for you! Thanks

ProfChaos · Answer

This is tricky
I marked A initially and thought that I solved this one correctly.

Anyway,
Given: xy>0
Thus the product of x and y is positive --> Two scenarios such that product xy is positive
Scenario 1: x +ve and y+ve
Scenario 2: x -ve and y -ve

We are asked if x^3y^4>0

Statement 1: x>0 --> x +ve 
Since we know that xy>0 and now we know that x>0 thus we can infer that y>0 
Thus x^3y^4>0
Sufficient

Statement 2: y>0 --> y +ve
Since we know that xy>0 and now we know that y>0 thus we can infer that x>0
Thus x^3y^4>0
Sufficient

Option D is correct

MISTAKE I MADE:
I noticed x^3y^4>0 and instantly thought that since y^4 it will always be positive and we just need to know about x to answer this and marked A because A told me exactly what I was looking for
Didnt think a lot about Statement 2

RPer · Answer

Isn't this a question where actually no statement would be needed to prove the sufficiency?
Because xy greater 0 can only be achieved if neither x nor y is negative... therefore I don't even need Statement 1 to know to know that x is greater than 0. This is already inherently given with xy greater 0

Bunuel · Answer

xy > 0 means that x and y have the same sign - either both are negative (for example, -2 and -3) or both are positive (for example 4 and 1.2). So, x^3y^4 could be negative, if x and y are negative, and positive, if x and y are positive. Each statement individually proves that x and y are both positive, which means that x^3y^4 > 0.

Hope it's clear.

bumpbot · Answer

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