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# If x^2y^3=200, what is xy?

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Manager
Joined: 06 Jul 2010
Posts: 71
If x^2y^3=200, what is xy?  [#permalink]

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Updated on: 12 Jun 2013, 04:09
3
00:00

Difficulty:

65% (hard)

Question Stats:

58% (02:02) correct 42% (01:47) wrong based on 112 sessions

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

(1) y is an integer
(2) x/y= 2.5

Originally posted by heyholetsgo on 21 Aug 2011, 10:23.
Last edited by Bunuel on 12 Jun 2013, 04:09, edited 2 times in total.
Renamed the topic and edited the question.
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21 Aug 2011, 10:47
Stmt1: y is an integer. y=0. xy=0
y=1 x^2*1=200
x^2=200, x=10sqrt(2)
xy= 10sqrt(2)*1=10sqrt(2).

Stmt2: x/y=2.5
x=2.5y
x^2*y^3=200
(2.5y)^2*y^3=200
y^5=200/6.25
y^5=32
y^5=2^5
y=2
x=2.5y=2.5*2=5
xy=5*2=10
Sufficient.

OA B
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21 Aug 2011, 10:50
heyholetsgo wrote:
If x^2y^3=200, what is xy?
1.) y is an integer
2.) x/y= 2.5
why is it not e? Don't we get 2 solutions for x once we get the value for y?

From the stem, we know y=+ve. thus x is +ve, because x/y=x/+ve=2.5; numerator +ve as well.

$$x=2.5y$$

$$(2.5y)^2*y^3=200$$

$$6.25y^5=200$$

$$y^5=\frac{200}{6.5}$$

$$y=\sqrt[5]{\frac{200}{6.5}}$$

$$x=2.5*\sqrt[5]{\frac{200}{6.5}}$$

$$xy=2.5*{(\frac{200}{6.5})}^{\frac{2}{5}}$$

Sufficient.

Ans: "B"
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21 Aug 2011, 18:58
1
x^2 * y^3 = 200

1. not sufficient

as different y will yield different x. hence different xy.

2. sufficient

x/y = 2.5

(2.5y)^2 * y ^3 = 200

25/4 y ^5 = 200 = >y =2

x =5

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21 Aug 2011, 21:13
(xy)^2*y=200

1. Insufficient

2.

(2.5y^2)^2*y=200
6.25y^5=200
y^5=200/(2.5*2.5)
y^5=80/2.5
y^5=32
y=2
xy=2.5*y^2
xy=10
B is sufficient
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21 Aug 2011, 21:44
1
heyholetsgo wrote:
If x^2y^3=200, what is xy?
1.) y is an integer
2.) x/y= 2.5
why is it not e? Don't we get 2 solutions for x once we get the value for y?

Even though, the solution has already been provided above, I want to point out something. The silver lining of DS questions is that you don't need to solve. If you know you will get a unique value, you are done.

Question stem: If $$x^2y^3=200$$, what is xy?

Statement 1: y is an integer but x needn't be.
If y = 1, $$x = +-\sqrt{200}$$
If y = 2, x = +-5 etc
We do not know what xy is.

Statement 2:$$\frac{x}{y} = \frac{5}{2}$$
This means $$x = (\frac{5}{2})y$$
This value of x you can put above to get a unique value of y. (Since you will get y^5 = ... An odd power will give you only the positive solution here. )
Once you have the value of y, you get a unique value for x and hence you get a value for xy.
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05 Sep 2011, 12:18
Quote:
If x^2y^3=200, what is xy?
1.) y is an integer
2.) x/y= 2.5

This is my approach...

from question:
prime factors of 200: 2, 2, 2, 5, 5
---> x=5; y = 2

From Statement 1

From Statement 2
x/y=5/2=2.5
Sufficient.

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20 Oct 2011, 06:25
1
gmatopoeia wrote:
Quote:
If x^2y^3=200, what is xy?
1.) y is an integer
2.) x/y= 2.5

This is my approach...

from question:
prime factors of 200: 2, 2, 2, 5, 5
---> x=5; y = 2

From Statement 1

From Statement 2
x/y=5/2=2.5
Sufficient.

Good explanation
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25 Sep 2012, 05:42
After i get Using B the value of y=2

when I substitute in the main equation ,

x^2* y^3=200
i get
x=+/- 5

so is it not wrong?
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Joined: 02 Sep 2009
Posts: 52431

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25 Sep 2012, 05:47
shankar245 wrote:
After i get Using B the value of y=2

when I substitute in the main equation ,

x^2* y^3=200
i get
x=+/- 5

so is it not wrong?

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.

Hope it helps.
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Re: If x^2y^3=200, what is xy?  [#permalink]

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01 Apr 2018, 12:18
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Re: If x^2y^3=200, what is xy? &nbs [#permalink] 01 Apr 2018, 12:18
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