problem with sum and products

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problem with sum and products

by Taniuca » Tue Aug 17, 2010 5:49 pm
Someone help me out to find out, the easiest and best way to approach the following problem, to get the right answer.

Is the product of x and y greater than the sum of x and y?

(1) xy < 0

(2) x > -y
Source: — Data Sufficiency |

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by Rahul@gurome » Tue Aug 17, 2010 6:32 pm
Taniuca wrote:Someone help me out to find out, the easiest and best way to approach the following problem, to get the right answer.

Is the product of x and y greater than the sum of x and y?

(1) xy < 0

(2) x > -y
The question states: Is xy > x + y?

(1) xy < 0 implies that either x should be positive and y negative or x should be negative and y positive.
If x = -2 and y = 5, then xy = -10 and x + y = 3. Here xy < x + y
If x = -4 and y = 2 then xy = -8 and x + y = -2. Here also xy < x + y
If x = 1/2 and y = -1 then xy = -1/2 and x + y = -1/2. Here xy = x + y

So, we don't get a unique answer. Hence (1) is NOT SUFFICIENT to answer the question.

(2) x > -y implies x + y > 0.
If x = 8, y = -2, xy = -16 and x + y = 6. so, xy < x + y
If x = 3 and y = 2 then xy = 6 and x + y = 5. Here, xy > x + y
No unique answer. So, (2) is NOT SUFFICIENT.

Combining (1) and (2), we know that one of x and y is negative such that x + y > 0. So, the product xy is negative and x + y is positive implies that xy < x + y

The correct answer is (C).
Hope this helps?
Last edited by Rahul@gurome on Tue Aug 17, 2010 8:22 pm, edited 2 times in total.
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by Taniuca » Tue Aug 17, 2010 7:30 pm
According to Kaplan the answer is (c) both statement together.
So, now I'm confussed would Kaplan have the wrong choice?
HELP!!!

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by Rahul@gurome » Tue Aug 17, 2010 8:13 pm
Taniuca wrote:According to Kaplan the answer is (c) both statement together.
So, now I'm confussed would Kaplan have the wrong choice?
HELP!!!
Let me check again.

Is xy > x + y?

(1) If x = -1 and y = 1/2 then xy = -1/2 and x + y = -1/2
Here, xy = x + y, which implies we don't get a unique answer.

Let me edit the original post, so that you are not confused. Thanks!
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