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by niketdoshi123 » Thu Jul 05, 2012 1:30 am
Bhupisuhag wrote:If x and y are integers such that x < y < 0, what is x - y?

(1) (x + y)(x - y) = 5

(2) xy = 6
As x<y<0 , we know that both are negative integers and the |x|>|y|.

Statement 1: Sufficient

we know that (a+b)(a-b) = a^2-b^2

=>x^2-y^2 = 5 (

check the squares the numbers
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16

we can clearly see that the difference between 3^2 & 2^2 is 5 and the difference increases as we go for bigger numbers.
So x=-3 & y=-2 since x<y<0.
and (x-y) = -1

Statement 2: INSUFFICIENT

xy=6
(-3)*(-2) = 6
(-6)*(-1) = 6
There are two possible pairs of xy
so a unique solution for (x-y) is not possible
Last edited by niketdoshi123 on Fri Jul 06, 2012 10:27 am, edited 1 time in total.
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by Brent@GMATPrepNow » Thu Jul 05, 2012 6:39 am
Bhupisuhag wrote:If x and y are integers such that x < y < 0, what is x - y?
(1) (x + y)(x - y) = 5
(2) xy = 6
Target question: What is x - y?

Statement 1: (x + y)(x - y) = 5
First, if x and y are integers, then x+y and x-y are both integers.
Second, if x < y < 0, then x+y is negative and x-y is negative.
Third, (x + y)(x - y) = 5, then there are only 2 cases:
case 1: (x+y) = -1 and (x-y) = -5
case 2: (x+y) = -5 and (x-y) = -1

Case 1 is impossible. If x and y are 2 different negative integers, then (x+y) cannot equal -1.
So, if case 1 is not possible, then case 2 must be true. This means that it must be the case that x+y= -5 and x-y= -1
So, statement 1 is sufficient.

Statement 2: xy = 6
If xy=6 (and x<y<0), there are two cases to consider
case 1: x = -3 & y = -2, in which case x-y = -1
case 2: x = -6 & y = -1, in which case x-y = -5
Since we get two conflicting answers to the target question, statement 2 is not sufficient.

Answer = A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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