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Re: M12-32 [#permalink]
I think this is a high-quality question and I agree with explanation. good question
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Re: M12-32 [#permalink]
great question, so tricky
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Re: M12-32 [#permalink]
Great Question!!
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Re: M12-32 [#permalink]
Great question
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Re: M12-32 [#permalink]
Superb question!
Easy to forget that 2 and -2 both have to be considered.
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Re: M12-32 [#permalink]
got it wrong, but question is great .
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Re: M12-32 [#permalink]
Great question. Took me 1 min; after I had realized it's y^4 there, I knew I can't get the answer with either (1), (2) or (1) + (2) as there will always be more than 1 possibility for y (provided that y is not equal to 0, which in this case isn't the y).
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Re: M12-32 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M12-32 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M12-32 [#permalink]
Bunuel wrote:
If \(x^3*y^4 = 2000\), what is \(y\)?


(1) \(x\) is an integer.

(2) \(y\) is an integer.


Hi Bunuel VeritasKarishma. I have one query, had the stem mentioned x and y are positive, Answer would have been B right (that is statement 2 would have been sufficient)
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Re: M12-32 [#permalink]
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RohitSaluja wrote:
Bunuel wrote:
If \(x^3*y^4 = 2000\), what is \(y\)?


(1) \(x\) is an integer.

(2) \(y\) is an integer.


Hi Bunuel VeritasKarishma. I have one query, had the stem mentioned x and y are positive, Answer would have been B right (that is statement 2 would have been sufficient)


No, the answer would become C in this case. (2) would not sufficient because for each value of y, there would exist such x that would satisfy \(x^3*y^4 = 2000\). For example:

If y = 1, then \(x = \sqrt[3]{2000}\)

If y = 2, then \(x = \sqrt[3]{\frac{2000}{16}}\)

If y = 3, then \(x = \sqrt[3]{\frac{2000}{81}}\)
...
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Re: M12-32 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M12-32 [#permalink]
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