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If x, y, and k are positive numbers such that [X/(X+Y)][10]

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If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink] New post 17 Jul 2010, 11:57
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64% (02:32) correct 36% (02:48) wrong based on 11 sessions
If x, y, and k are positive numbers such that [X/(X+Y)][10] + [Y/(X+Y)][20] = k and if x < y, which of the following could be the value of k?

A. 10
B. 12
C. 15
D. 18
E. 30

I posted this question earlier but somehow it disappeared from the forum... I dont know why... Anyhow, I am posting again. Can someone come up with some solution???
[Reveal] Spoiler: OA
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Re: Can someone come up with some solution of this question??? [#permalink] New post 17 Jul 2010, 12:04
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It didn't disappear, it was merged with similar topic: must-be-an-easier-way-88620.html#p668499

Below is my solution from there:

10*\frac{x}{x+y}+20*\frac{y}{x+y}=k

10*\frac{x+2y}{x+y}=k

10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k

Finally we get: 10*(1+\frac{y}{x+y})=k


We know that x<y

Hence \frac{y}{x+y} is more than 0.5 and less than 1

0.5<\frac{y}{x+y}<1

So, 15<10*(1+\frac{y}{x+y})<20


Only answer between 15 and 20 is 18.

Answer: D (18)

There can be another approach:

We have: \frac{10x+20y}{x+y}=k, if you look at this equation you'll notice that it's a weighted average.

There are x red boxes and y blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?

k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.

Answer: D (18)
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Simplification [#permalink] New post 30 Aug 2010, 14:18
If x, y, and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and if x < y, which of the following could be the value of k ?
(A) 10
(B) 12
(C) 15
(D) 18
(E) 30


The explantion seems to be time consuming. Help for easy process.
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Re: Can someone come up with some solution of this question??? [#permalink] New post 30 Aug 2010, 15:34
Bunuel wrote:
It didn't disappear, it was merged with similar topic: must-be-an-easier-way-88620.html#p668499

Below is my solution from there:

10*\frac{x}{x+y}+20*\frac{y}{x+y}=k

10*\frac{x+2y}{x+y}=k

10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k

Finally we get: 10*(1+\frac{y}{x+y})=k


We know that x<y

Hence \frac{y}{x+y} is more than 0.5 and less than 1

0.5<\frac{y}{x+y}<1

So, 15<10*(1+\frac{y}{x+y})<20


Only answer between 15 and 20 is 18.

Answer: D (18)

There can be another approach:

We have: \frac{10x+20y}{x+y}=k, if you look at this equation you'll notice that it's a weighted average.

There are x red boxes and y blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?

k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.

Answer: D (18)


actually, you don't need to write all them down to find the answer is D.
especially in your test you don't have much time, so just pick any number you want. In this case, just pick x as 1 and y as 2, so k would be around 18.333, which is near to 18. If you little bit hesitate then try different numbers, such x as 1 and y as 3.
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Re: Can someone come up with some solution of this question??? [#permalink] New post 31 Aug 2010, 10:55
[X/(X+Y)][10] + [Y/(X+Y)][20] = k
So, 10 (X+2Y) = K (X+Y)

Put the options in place of K, and solve till the end. Make sure X<Y.

For eg – K=10,

Then 10X+20Y = 10X+10Y
Y=0; makes no sense

K=12;
10X+20Y=12X+12Y
8Y=2X
4Y=X
But X<Y Therefore Not Possible

Similarly, K=18.
10X+20Y=18X+18Y
5X+10Y=9X+9Y
Y=4X
According to question, X<Y. Therefore, K=18. :wink:

In the case of K=30,
Finally you will reach to -2X=Y.
Since X and Y are +ve, this case is inapplicable… :-D
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Re: Can someone come up with some solution of this question??? [#permalink] New post 04 Sep 2010, 06:25
Good question. Bunuel has a great solution - per the usual.
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Re: Can someone come up with some solution of this question??? [#permalink] New post 04 Sep 2010, 11:42
i've seen a lot of approaches where you guys use the weighted average strategy, nicely done!
Re: Can someone come up with some solution of this question???   [#permalink] 04 Sep 2010, 11:42
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