Jasonammex wrote:

if (y+3)(y-1)-(y-2)(y-1)=r(y-1), what is the value of y?

(1) r^2=25

(2) r=5

I chose A...

In the equation above we see that (y-1) is the common factor in all terms. So let's take (y-1) common out of all the terms.

(y+3)(y-1) - (y-2)(y-1) - r(y-1) = 0

(y-1)[(y+3) -(y-2) - r] = 0

(y-1)(5 - r) = 0

Now, the product of these two is 0. This means that at least one of them has to be 0.

Either (y-1) = 0 or (5 - r) = 0 or both are 0.

So, either y = 1 or r = 5 or both.

Only if we know that r is not 5, then we can say that y must be 1. If r is 5, y may be 1 or may not be 1.

Stmnt 1: r^2 = 25

So r = +- 5

This statement tells us that r can be 5.

If r = 5, y may or may not be 1.

If r is not 5, y will be 1.

Since we do not know whether r is 5 or not, we cannot say what the value of y is. Not sufficient.

Stmnt 2: r = 5

If r = 5, y may or may not be 1.

Not sufficient.

Both together, r = 5. Again, not sufficient. Answer E

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Karishma

Veritas Prep GMAT Instructor

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