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I get why 1 is NSF, no info on x. But with stmt 2, isnt it also NSF b/c we cant confirm if x is neg or positive or not?

If \(x^2y^3=200\), what is \(xy\)?

(1) \(y\) is an integer --> \(y\) can be any positive integer and there will be two values of \(x\) for each \(y\) to satisfy \(x^2y^3=200\). For example \(y=10\) --> \(x^2y^3=x^2*1000=200\) --> \(x=\frac{1}{\sqrt{5}}\) or \(x=-\frac{1}{\sqrt{5}}\). So there are infinite values of \(xy\). Not sufficient.

(2) \(\frac{x}{y}=2.5\) --> \(x=2.5y\). We have two equations and two variables --> \(x^2y^3=6.25y^5=200\) --> \(y^5=32\) --> \(y=2\), \(x=5\) --> \(xy=10\). Sufficient.

First of all \(x^2y^3=2^3*5^2\) has infinitely many solutions for \(x\) and \(y\). For ANY (nonzero) value of \(x\) there exist some \(y\) to satisfy \(x^2y^3=200\) and vise-versa. For example \(x=1\) and \(y=\sqrt[3]{200}\), or \(x=10\) and \(y=\sqrt[3]{2}\), ... As you can see it's not necessary for \(x\) and \(y\) to be integers (3 and 2) to satisfy the given equation.

As for (1): \(y\) is an integer. \(y\) can be any positive integer and there will be two values of \(x\) for each \(y\) to satisfy \(x^2y^3=200\). For example \(y=10\) --> \(x^2y^3=x^2*1000=200\) --> \(x=\frac{1}{\sqrt{5}}\) or \(x=-\frac{1}{\sqrt{5}}\). So there are infinite values of \(xy\). Not sufficient.

Re: If x^2y^3=200 , what is xy ? 1. y is an integer 2. x/y=2.5 B [#permalink]

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23 May 2012, 03:52

ashish8 wrote:

I got this problem wrong. Bunuel, can you explain if I approached it the wrong way.

I factored out 200, which = \(2^3 * 5^2\).

After this its obvious \(xy = 10\)

What am i missing here?

Be careful when 1.the number type is not stated-that could be any type,integer,real etc 2.you see factorization type question-they are not always integer or prime number,unless explicitly stated.

common mistakes for me..even after months of studying

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No, that's totally wrong. You cannot solve x/y = 2.5 for x and y. Why do you say x = 5 and y = 2? Why not x = 2.5 and y = 1? Or x = 25 and y = 10? x/y = 2.5 has infinitely many solutions for x and y - we have one equation and two unknowns!
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