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Bunuel
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albertmercu19
What is the value of \(\frac{x^2}{y}\)?

(1) \(x^2y = 4\)

(2) \(y^2 = 4 \)


Is the answer should be E?

Diving statement 1 with statement 2, we will get:
) x^2 y = 4 >square these figures > x^4 y^2 = 16 > x^4 4 = 16 > x^4 = 4 > x^2 = +-2
Statement 2: y^2 = 4 > y = +-2

Insert to the x^2 / y = +-2 / +-2
There are two options
(1) +2 / -2 or -2 / +2 = the answer will be -1
(2) +2 / +2 or -2 / -2 = the answer will be 1

Hence, it gives multiple answer, the answer is E

The point is that dividing \(x^2y = 4\) by \(y^2 = 4 \) directly gives:

    \(\frac{x^2y}{y^2} = \frac{4}{4}\)

    \(\frac{x^2}{y}=1\)

Also, I'm not sure I follow everything you've written above, but remember that x^2, being the square of a number, cannot be negative. Keep in mind that by default, all numbers in GMAT questions are considered to be real numbers.

Hope it helps.
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Bunuel
albertmercu19
What is the value of \(\frac{x^2}{y}\)?

(1) \(x^2y = 4\)

(2) \(y^2 = 4 \)


Is the answer should be E?

Diving statement 1 with statement 2, we will get:
) x^2 y = 4 >square these figures > x^4 y^2 = 16 > x^4 4 = 16 > x^4 = 4 > x^2 = +-2
Statement 2: y^2 = 4 > y = +-2

Insert to the x^2 / y = +-2 / +-2
There are two options
(1) +2 / -2 or -2 / +2 = the answer will be -1
(2) +2 / +2 or -2 / -2 = the answer will be 1

Hence, it gives multiple answer, the answer is E

The point is that dividing \(x^2y = 4\) by \(y^2 = 4 \) directly gives:

    \(\frac{x^2y}{y^2} = \frac{4}{4}\)

    \(\frac{x^2}{y}=1\)

Also, I'm not sure I follow everything you've written above, but remember that x^2, being the square of a number, cannot be negative. Keep in mind that by default, all numbers in GMAT questions are considered to be real numbers.

Hope it helps.

Great answer and just realized I had a miscalculations. Thanks a lot Bunuel!
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