Manhattan GMAT CAT - DS : root(xy) = xy

[jtg0101]

This is a Manhattan GMAT CAT - DS question. Please help and reconfirm the answer

If root (xy) = xy, what is the value of x + y?

(1) x = -1/2
(2) y is not equal to zero

I solved it with option 'A' only and arrived at the value of X+Y that is given in the answer explanation. Official answer is 'C'. Could you please look into it?

Here is my solution

Take square of both sides:
(root(xy) )^2 = (xy)^2
xy = (xy)^2
With Option (1): x = -1/2
-1/2y = 1/4 * y^2
y = 4 * (-1/2)
y = -2
Hence, X+Y = -1/2 - 2 = -5/2.

Please reconfirm the answer and let me also now if I am committing any mistake.

Thanks in advance

[shaji]

This is a Manhattan GMAT CAT - DS question. Please help and reconfirm the answer

If root (xy) = xy, what is the value of x + y?

(1) x = -1/2
(2) y is not equal to zero

I solved it with option 'A' only and arrived at the value of X+Y that is given in the answer explanation. Official answer is 'C'. Could you please look into it?

Here is my solution

Take square of both sides:
(root(xy) )^2 = (xy)^2
xy = (xy)^2
With Option (1): x = -1/2
-1/2y = 1/4 * y^2
y = 4 * (-1/2)
y = -2
Hence, X+Y = -1/2 - 2 = -5/2.

Please reconfirm the answer and let me also now if I am committing any mistake.

Thanks in advance

The answer is definitely C

The flaw in the above reasoning is if x=-1/2, the y could be 0 leading to x+y=-1/2 ...INSUFF

[Guest]

Shaji,

Y could be 0, would lead to some other value agreed. But that would have been the case, if I had assumed any value of y. In this case, I am deriving the value of Y by replacing the value of X.

Hence, no question of y being any other value, as long as x remains -1/2.

Through option (1), we can deduce the value of Y so no need to consider Option (2). Option (2) anyway doesn't give any result. Hence, answer (A).

Let me know, if I am assuming anything wrong here!!

[shaji]

Shaji,

Y could be 0, would lead to some other value agreed. But that would have been the case, if I had assumed any value of y. In this case, I am deriving the value of Y by replacing the value of X.

Hence, no question of y being any other value, as long as x remains -1/2.

Through option (1), we can deduce the value of Y so no need to consider Option (2). Option (2) anyway doesn't give any result. Hence, answer (A).

Let me know, if I am assuming anything wrong here!!

The very fact that y=0 is also a plausible solution makes Statement 1 INSUFFICIENT.
Yes! U have derived the value of y by substitutig the value of x, whch gives you the other possible value of y. As I explained earlier since there are 2 possible values to x+y, Statement 1 becomes INSUFFICIENT
Please notice by combining both Statements, you are doing away with the first possibility of y=0; and thus driving U to definitive value of x+y justifying Answer C.

[StaceyKoprince]

Here's your error:

-1/2y = 1/4 * y^2
y = 4 * (-1/2)

You divided out a y. You can't do that when you're looking for factors because you divided out one of your possible factors! (zero)

the correct math is:
-1/2y = 1/4 * y^2 quadratic, so set equal to zero
0 = 1/4 * y^2 +1/2y multiply by 4 b/c fractions are annoying
0 = y^2 + 2y
0 = y(y+2)
y = 0, -2

Remember that you should NEVER divide out a variable when you are trying to find factors. If you do, you will LOSE one of your factors.

[Guest]

Thank you.

cheers
JTG

[StaceyKoprince]

You're welcome! :)

[viksnme]

Here's your error:

-1/2y = 1/4 * y^2
y = 4 * (-1/2)

You divided out a y. You can't do that when you're looking for factors because you divided out one of your possible factors! (zero)

the correct math is:
-1/2y = 1/4 * y^2 quadratic, so set equal to zero
0 = 1/4 * y^2 +1/2y multiply by 4 b/c fractions are annoying
0 = y^2 + 2y
0 = y(y+2)
y = 0, -2

Remember that you should NEVER divide out a variable when you are trying to find factors. If you do, you will LOSE one of your factors.

Hi Stacey,
I just figured out another way to solve this problem, which also avoids the b/c form. Hope you agree.

xy=(xy)^2
=> (xy)(xy-1) = 0
=> either xy = 0 or xy = 1
when xy = 0, y=0 since x=-1/2 ; but ruled out due to statement 2
when xy=1, y=-2 ; the correct answer hence C.

[RonPurewal]

Hi Stacey,
I just figured out another way to solve this problem, which also avoids the b/c form. Hope you agree.

heh heh, that's funny... never saw that coming

'b/c' in that text is an abbreviation for 'because', not a fraction.

your solution is a lot like stacey's: both of you are solving quadratics. her solution is more 'standard' in that she's reducing the problem to a single variable and then solving, whereas you're taking the more unconventional approach of solving a quadratic with the grouped variable xy. either approach is sound.

[viksnme]

hmm !!! Thanks Ron for highlighting the b/c fiasco. However, that sentence still talks about fractions if not about b/c form :)

Maths is an interesting subject as there can be 'n' number of ways to solve a problem.

[StaceyKoprince]

Yeah, I guess I should spell out "because" from now on (at least, when typing about math problems!).

Math is both great and infuriating because there are always multiple ways to do a problem. I encourage my students to find at least two ways to do any math problem, plus a way to make an educated guess, and then decide which way is the best way, why it's the best way, and how to recognize on a future (different) problem of the same type that the student should use that same "best" method, whatever it is.

[shyamprasadrao]

How can we consider X or Y to be zero? root(zero) is not defined.
As the question states root(xy), it means netheir X nor Y nor XY is zero.
Am I wrong?

[Ben Ku]

Hi shyamprasadrao,

sqrt(0) is not undefined: sqrt(0) = 0, because 0 * 0 = 0.

There are rules about numbers that you do need to pay attention to:
sqrt (x) is undefined when x < 0
1 / x is undefined when x = 0

Hope that helps.