If x and y are integers such that x < y < 0, what is x

[gmattesttaker2]

Hello,

Can you please assist with this? This is from MGMAT.

If x and y are integers such that x < y < 0, what is x - y?

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

(2) xy = 6

OA: A


I was going through the Official explanation and tried to approach as follows:

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

Both (x + y) and (x - y) are integers

So, ( 1 )( 5 ) = 5 or ( 5 )( 1 ) = 5
or ( -1 )( -5 ) = 5 or ( -5 )( -1 ) = 5


I was stuck after this point. Can you please assist?


Thanks,
Sri

[GMATGuruNY]

Hello,

Can you please assist with this? This is from MGMAT.

If x and y are integers such that x < y < 0, what is x - y?

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

(2) xy = 6

Statement 1: (x + y)(x - y) = 5
x² - y² = 5.
Look for two PERFECT SQUARES that have a difference of 5:
1, 4, 9, 16, 25...
Only the values in red have a difference of 5.
Thus, x²=9 and y²=4.
Since x<y<0, x=-3 and y=-2, implying that x-y = -3-(-2) = -1.
SUFFICIENT.

Statement 2: xy=6
If x=-3 and y=-2, then x-y = -3-(-2) = -1.
If x=-6 and y=-1, then x-y = -6-(-1) = -5.
Since x-y can be different values, INSUFFICIENT.

The correct answer is A.

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Hi gmattesttaker2,

Mitch offers a good math explanation (showcasing the importance of knowing your Classic Quadratics), so I won't rehash that here. Instead, I'll offer a Number Property rule that you could have used to solve this problem, even if you didn't notice the "difference of squares."

Since x and y are negative integers (the prompt says so),

(x + y) MUST be an integer AND
(x - y) MUST be an integer

So, if (x + y)(x - y) = 5 and both parentheses are integers, then what could they be???

There's just 1 option: they'd have to be -1 and -5 in some combination. With that insight, you can quickly deduce what the x and y are.

GMAT assassins aren't born, they're made,
Rich

[Brent@GMATPrepNow]

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