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# Is uv > 0? (1) u^2v^3 < 0 (2) uv^2 < 0

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Re: Is uv > 0? (1) u^2v^3 < 0 (2) uv^2 < 0 [#permalink]
Is uv>0?

(1) u2v3<0
(2) uv2<0

Modify the question asked - are u and v in the same sign?

(1) u^2 will always be positive.. so if the product of u^2 and v^3 is negative --> v is negative - we know nothing about u - Insufficient
(2) Same as (1) - v^2 will always be positive.. so if the product of u and v^2 is negative --> u is negative - we know nothing about v - Insufficient

Combined - (1) & (2) u & v are in the same sign - Sufficient
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Re: Is uv > 0? (1) u^2v^3 < 0 (2) uv^2 < 0 [#permalink]
Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 2 variables: Let the original condition in a DS question contain 2 variables. Now, 2 variables would generally require 2 more equations for us to be able to solve for the value of the variable.

We know that each condition would usually give us an equation, and Since we need 2 more equations to match the numbers of variables and equations in the original condition, the logical answer is C.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find whether uv > 0 ?.

Second and the third step of Variable Approach: From the original condition, we have 2 variables (u, and v).To match the number of variables with the number of equations, we need 2 more equations. Since conditions (1) and (2) will provide 2 equations, C would most likely be the answer.

Let’s take look at both condition together.

Condition(1) tells us that $$u^2v^3$$ < 0.

Condition(2) tells us that $$uv^2$$ < 0.

=> Since, $$u^2$$ will always be positive, for the product $$u^2v^3$$ < 0, 'v' has to be v < 0.

=> Since, $$v^2$$ will always be positive, for the product $$uv^2$$ < 0, 'u' has to be u < 0.

=> negative * negative = positive: uv > 0 - is uv > 0 - YES

Since the answer is unique YES , both conditions combined together are sufficient by CMT 1.

Both conditions combined together are sufficient.

So, C is the correct answer.