M12-32

Official Solution:


If \(x^3*y^4 = 2000\), what is the value of \(y\)?

Let's begin by factorizing the expression: \(x^3*y^4 = 2^4*5^3\). Although it may seem that \(y=2\) and \(x=5\) would be the solution, it's important to note that we are not given any information about whether \(x\) and \(y\) are integers. Therefore, for any positive value of \(x\), there will exist two corresponding values of \(y\) that satisfy the equation. For instance:

When \(x = \frac{1}{2}\), we get \(y = \sqrt[4]{16000}\) or \(y = -\sqrt[4]{16000}\).

When \(x = 1\), we get \(y = \sqrt[4]{2000}\) or \(y = -\sqrt[4]{2000}\).

When \(x = 2\), we get \(y = \sqrt[4]{250}\) or \(y = -\sqrt[4]{250}\).

...

Similarly, for any nonzero value of \(y\), there will exist corresponding values of \(x\) that satisfy the equation. For instance:

When \(y = \frac{1}{2}\), we get \(x = \sqrt[3]{32000}\).

When \(y = -1\), we get\(x = \sqrt[3]{2000}\).

When \(y = \sqrt{10}\), we get \(x = \sqrt[3]{20}\).

...

Moreover, even if we were given that \(x\) and \(y\) are integers, we cannot deduce that \(y=2\) and \(x=5\) because of the even power of \(y\). It could also be that \(y=-2\).

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

As explained earlier, knowing that \(x\) is an integer does not provide us with sufficient information to determine a unique value for \(y\). Not sufficient.

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

As explained earlier, knowing that \(y\) is an integer does not provide us with sufficient information to determine a unique value for \(y\). Not sufficient.

(1)+(2) We know that both \(x\) and \(y\) are integers. However, as we discussed earlier, this still does not provide us with sufficient information to determine a unique value for \(y\). For instance, it could be that \(y=2\) and \(x=5\), or it could be that \(y=-2\) and \(x=5\). Not sufficient.


Answer: E
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I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation. good question

great question, so tricky

Great Question!!

Great question

Superb question!
Easy to forget that 2 and -2 both have to be considered.

got it wrong, but question is great .

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).

I think this is a high-quality question and I agree with explanation.

I think this is a high-quality question and I agree with explanation.

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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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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