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# If x, y and k are positive numbers such that

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Senior Manager
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If x, y and k are positive numbers such that [#permalink]

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01 Nov 2009, 21:01
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Question Stats:

57% (01:59) correct 43% (01:17) wrong based on 121 sessions

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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

Open discussion of this question is here: if-x-y-and-k-are-positive-numbers-such-that-x-x-y-128231.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Apr 2012, 22:30, edited 1 time in total.
Topic is locked
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Re: which could be the value of K? [#permalink]

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01 Nov 2009, 21:11
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Expert's post
$$10*\frac{x+2y}{x+y}=k$$

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

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)

This question was discussed before, please refer to the other approaches at: what-is-the-value-of-k-85382.html#p640124
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Senior Manager
Joined: 18 Aug 2009
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Kudos [?]: 232 [0], given: 9

Re: which could be the value of K? [#permalink]

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01 Nov 2009, 21:29
Bunuel wrote:
$$10*\frac{x+2y}{x+y}=k$$

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

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)

This question was discussed before, please refer to the other approaches at: what-is-the-value-of-k-85382.html#p640124

Thanks for the solution and the pointer to the earlier thread.
I'd run a search but didn't hit that thread... maybe due to the mathematical formula.

hit a roadblock right in the first step of simplifying the equation... $$10(x+2y)/(x+y)$$
But now understood the approach
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Re: which could be the value of K? [#permalink]

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09 May 2011, 05:13
Let us try x = 1, y = 4

=> 1/5 * 10 + 4/5 * 20 = k

=> 2 + 16 = k

=> k = 18

Answer - D
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Re: which could be the value of K? [#permalink]

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27 Apr 2012, 19:46
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Hi Bunuel,

Is this approach ok?
since it is of the weighted averages form

10x + 20 y/x+y= k

and 20 pulls the averagge more twds itself we can eliminate other options and oly 18 wud be possible?
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Re: which could be the value of K? [#permalink]

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27 Apr 2012, 22:28
shankar245 wrote:
Hi Bunuel,

Is this approach ok?
since it is of the weighted averages form

10x + 20 y/x+y= k

and 20 pulls the averagge more twds itself we can eliminate other options and oly 18 wud be possible?

Yes, you can use weighted average approach for this question:

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.

In case of any question please post it here: if-x-y-and-k-are-positive-numbers-such-that-x-x-y-128231.html
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Re: which could be the value of K?   [#permalink] 27 Apr 2012, 22:28
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# If x, y and k are positive numbers such that

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