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Re: y > 0 [#permalink]
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After simplification, equation becomes 8/y^3 and 9/y^2.

Since y>0 so multiply both sides by y^2. You will get 8/y and 9. If y is an integer then quantity A will always be less than B. If y is a proper fraction,say 0.8, quantity A will be 10. Here comes the contradiction of first case so relationship cannot be established.
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Re: y > 0 [#permalink]
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y as a fraction vs ya as a whole number can change the answer options hence the answer
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Re: y > 0 [#permalink]
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mohan514 wrote:
y as a fraction vs ya as a whole number can change the answer options hence the answer


Correct,however it is a bit simple. While shortcuts are great time saver but use them wisely they are not very reliable unless you have a decent command of the topic.

Preferably try to solve the question specially with exponential and percentages they can be tricky.
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Re: y > 0 [#permalink]
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Explanation

Plug in values for y. If y = 1, then Quantity A is 8 and Quantity B is 9.

In this case, Quantity B is larger, so eliminate choices (A) and (C). If \(\frac{1}{2}\), then Quantity A is 64 and Quantity B is 36; eliminate choice (B).

You are left with choice (D).
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Re: y > 0 [#permalink]
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