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If x^2y^3 = 200, what is xy ? [#permalink]
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09 Apr 2010, 20:14
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If \(x^2y^3 = 200\), what is xy ? (1) y is an integer (2) x/y = 2.5 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?
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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09 Apr 2010, 23:14
x^2 * y^3 = 200 since x^2 is +ve, then y must be positive to satisfy this. now if x/y = 2.5 , if y is +_ve then x is also +ve. I hope this helps
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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10 Apr 2010, 10:09
Thanks gurpreet. This is another good example of a question that isnt that hard, but when ive been studying for too long I'll get confused. Thanks again.
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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10 Apr 2010, 10:48
solve question only when your mind is fresh....if u feel sluggish, take a small nap. Good luck.
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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13 Apr 2010, 12:34
Stmt 1. y is an integer to satisfy x^2 y^3 = 200 ............ y should be positive integer but x can be negative or positive Insufficient stmt. 2 x/y =2.5 ....... now to satify the given condition y should be positive then x has to be positive Sufficient IMO B
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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14 Apr 2010, 02:53
TheSituation wrote: If x^2y^3=200 , what is xy ? 1. y is an integer 2. x/y=2.5 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. Answer: B.
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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08 May 2012, 08:24
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?



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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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09 May 2012, 02:06
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? Check my post above: ifx2y3200whatisxy1yisaninteger2xy25b92486.html#p713396First 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 viseversa. 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. Hope it's clear.
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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23 May 2012, 04:52
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 statedthat could be any type,integer,real etc 2.you see factorization type questionthey are not always integer or prime number,unless explicitly stated. common mistakes for me..even after months of studying



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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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19 Oct 2014, 01:34
Hi Bunuel,
With regards to the problem above, for st 2, is it fine to approach it the following way 
(2) x/y = 2.5 x/y = 25/10 => 5/2 therefore, x = 5 & y = 2. giving xy = 10 Suff.
please do let me know! Thanks in advance!



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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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20 Oct 2014, 03:27



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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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20 Oct 2014, 09:28
Thank you so much for clearing that doubt!



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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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27 Mar 2018, 06:56
TheSituation wrote: If x^2y^3 = 200, what is xy ? (1) y is an integer (2) x/y = 2.5 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? The question needs to be worded better. X^2 * Y^3 I interpreted the question as (X^2y)^3
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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27 Mar 2018, 07:29



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If x^2y^3 = 200, what is xy ? [#permalink]
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27 Mar 2018, 10:59
Bunuel wrote: MT1988 wrote: TheSituation wrote: If x^2y^3 = 200, what is xy ? (1) y is an integer (2) x/y = 2.5 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? The question needs to be worded better. X^2 * Y^3 I interpreted the question as (X^2y)^3 Edited, thank you. Though mathematically x^2y^3 can only mean \(x^2y^3\). If it were \((x^2y)^3\) it would be written as (x^2y)^3 and if it were \((x^{2y})^3\) it would be written as (x^(2y))^3 PEMDAS.... Was wondering what if my interpretation was correct? Are there questions like that? Took for ever to solve but if my interpretation was right would B be the answer? Am I unnecessarily complicating the question?
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Re: If x^2y^3 = 200, what is xy ? [#permalink]
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23 Jun 2018, 09:07
Bunuel wrote: TheSituation wrote: If x^2y^3=200 , what is xy ? 1. y is an integer 2. x/y=2.5 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. Answer: B. For Statement 2, I think we can stop at this step: \(6.25y^5=200\) y is raised to an odd power, so there will be one solution. we can find y... we can find x.




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