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

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Math Expert
Joined: 02 Sep 2009
Posts: 59587
Is a > b ? (1) ax > bx (2) x*a^2 + x*b^2 < 0  [#permalink]

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22 Nov 2018, 03:21
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Difficulty:

45% (medium)

Question Stats:

58% (01:21) correct 42% (01:40) wrong based on 149 sessions

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Is $$a > b$$ ?

(1) $$ax > bx$$

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

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Joined: 02 Sep 2018
Posts: 117
Re: Is a > b ? (1) ax > bx (2) x*a^2 + x*b^2 < 0  [#permalink]

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22 Nov 2018, 06:16
1
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.
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Re: Is a > b ? (1) ax > bx (2) x*a^2 + x*b^2 < 0  [#permalink]

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22 Nov 2018, 11:27
1
Bunuel wrote:
Is $$a > b$$ ?

(1) $$ax > bx$$

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

(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.

Math Expert
Joined: 02 Sep 2009
Posts: 59587
Re: Is a > b ? (1) ax > bx (2) x*a^2 + x*b^2 < 0  [#permalink]

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24 Dec 2018, 01:47
Bunuel wrote:
Is $$a > b$$ ?

(1) $$ax > bx$$

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

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Re: Is a > b ? (1) ax > bx (2) x*a^2 + x*b^2 < 0  [#permalink]

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03 Jan 2019, 03:10
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.

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Re: Is a > b ? (1) ax > bx (2) x*a^2 + x*b^2 < 0  [#permalink]

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08 Oct 2019, 05:35
Bunuel wrote:
Is $$a > b$$ ?

(1) $$ax > bx$$

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

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
Re: Is a > b ? (1) ax > bx (2) x*a^2 + x*b^2 < 0   [#permalink] 08 Oct 2019, 05:35
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