value of x*y

[vishal.pathak]

Given that A = 3y + 8x, B = 3y - 8x, C = 4y + 6x, and D = 4y - 6x, what is the value of x*y?

1. AB + CD = -275
2. AD - BC = 420

[rijul007]

Statement 1
AB+CD = (9y^2 - 64x^2) + (16y^2 - 36x^2) = 25y^2 - 100x^2
y^2 - 4x^2 = -11

Insifficient

Statement 2
AD -BC = 420
(12y^2 - 18xy + 32xy - 48x^2) - (12y^2 + 18xy -32xy - 48x^2) = 420

28xy = 420
xy = 420/28 = 15

Sufficient

IMO:Option B

[tpr-becky]

This question is based on your knowledge of quadratics and FOIL, the only way to find the value of xy is to cancel out teh x^2 and y^2 values-

statement 1) will never work because the two foiled equations each have positive values and you are adding the two equations thus you will never be able to cancel them out.


Statement 2) might work becuase you are now subtracting those two positive values of x^2 and y^2 so let's foil them out

AD foils to 12y^2 + 14xy - 48x^2 and BC foils out to 12y^2 + 6xy - 48x^2 now we see that we have equal version of y^2 and x^2 and if you subract them then you will only be left for versions of xy and thus you have 20xy = 420 so you can solve for xy and B is sufficient.

[vishal.pathak]

This question is based on your knowledge of quadratics and FOIL, the only way to find the value of xy is to cancel out teh x^2 and y^2 values-

statement 1) will never work because the two foiled equations each have positive values and you are adding the two equations thus you will never be able to cancel them out.


Statement 2) might work becuase you are now subtracting those two positive values of x^2 and y^2 so let's foil them out

AD foils to 12y^2 + 14xy - 48x^2 and BC foils out to 12y^2 + 6xy - 48x^2 now we see that we have equal version of y^2 and x^2 and if you subract them then you will only be left for versions of xy and thus you have 20xy = 420 so you can solve for xy and B is sufficient.

Yes, I got tricked. I looked at statement 2 and found that, when expanded, AD - BC will result in 3 types foils. Type 1 (with xy in it), type 2 (with x^2 in it) and type 3(with y^2 in it). I didnt think of the possibility that y^2 and x^2 terms could cancel each other. Yes, I am stupid