Is a b ? (1) ax bx (2) x*a^2 + x*b^2

Question

Is  a > b  ?

(1)  ax > bx

(2)  x*a^2 + x*b^2 < 0

M36-43

Bunuel · Accepted Answer

Official Solution: Is a > b ? (1) ax > bx If x > 0 , the after reducing by x we'd get a > b (keep the sign when reducing by a positive value). If x < 0 , the after reducing by x we'd get a < b (flip the sign when reducing by a negative value). Not sufficient. (2) x*a^2 + x*b^2 < 0 Factor out x to get x(a^2 + b^2) < 0 a^2 + b^2 is positive, thus x must be negative for the product to be negative. That's all we can get from this statement. Not sufficient. (1)+(2) Since from (2) x < 0 , the from (1) a < b . Sufficient. Answer: C

parijit · Answer

From statement two it is clear that x's value is negative, so b is greater than ,if compare x's value with statement 1.
C is the answer.

amanvermagmat · Answer

(1) ax > bx or ax - bx > 0 or x*(a-b) > 0. So either both x > 0 and a > b OR both x < 0 and a < b. So we cant be sure if a > b only since we dont know about x. Not sufficient . (2) x*(a^2 + b^2) < 0 Here a^2 + b^2 cannot be negative, whether a/b are positive or negative doesnt matter. So the only possibility is that x < 0. So we know that x is negative but we cannot say anything about a & b. Not sufficient . Combining both the statements, x is negative so a < b. Then only ax can be greater than bx. So this is sufficient to give NO as an answer to the question. Hence C answer .

Bunuel · Answer

Par of  GMAT CLUB'S New Year's Quantitative Challenge Set

GMATin · Answer

Statement 1: ax>bx

Since it is an inequality and we don't the value of x, we cannot cancel x just like that. Instead, let's take all the variables on one side.
Thus, ax-bx>0
This implies that x(a-b)>0

Again, since its an inequality, we can either have x>0 and a>b (i.e. both elements are positive) or x<0 and a<b (i.e. both elements are negative). Since, we cannot conclusively say whether a<b or a>b, this statement is insufficient.

Statement 2: \(x*a^2 + x*b^2 < 0\)

This implies [m]x*(a^2 + b^2) < 0

Now, [m](a^2 + b^2) will be either 0 (if a=b=0) or positive. Sum of squares of two numbers can never be negative.

Thus, for statement 2 to be true, we will necessarily have x<0. However, this statement doesn't give us any knowledge of whether a<b or a>b. Thus, insufficient

Combining 1& 2, we definitely know that x<0. Thus, according to statement 1, a<b. This means that the question will be false and 1& 2 together are sufficient to answer the question.

Hence, answer is C

Kinshook · Answer

Asked: Is a > b ? (1) ax > bx If x< 0; a < b But if x>0; a>b NOT SUFFICIENT (2) x*a^2 + x*b^2 < 0 x < 0 a < b SUFFICIENT IMO B

Is a > b ? (1) ax > bx (2) xa^2 + xb^2 < 0

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Is a > b ? (1) ax > bx (2) x*a^2 + x*b^2 < 0