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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

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

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

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

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.

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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14 Apr 2012, 01:19

Bunuel, I didn't understand how you got the 15 and 20 at the extreme ends in the equation. Could you please explain? thanks! 15<10*(1+(y/x+y)<20 _________________

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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14 Apr 2012, 02:09

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mymbadreamz wrote:

Bunuel, I didn't understand how you got the 15 and 20 at the extreme ends in the equation. Could you please explain? thanks! 15<10*(1+(y/x+y)<20

Since \(x<y\) then \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\): \(0.5<\frac{y}{x+y}<1\).

In this case for \(1+\frac{y}{x+y}\) is more than 1.5 and less than 2, so \(10*(1+\frac{y}{x+y})\) is more than 15 and less than 20: \(15<10*(1+\frac{y}{x+y})<20\).

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

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

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

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

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)

Very well solved by Bunuel, this is question no. 148 in OG 12 P.S, which has been solved in a very complicated manner in O.G.

(A) \(\frac{-10}{0}\) Out! x and y are non-zeroes. (B) \(\frac{x}{y}=\frac{-8}{-2}=\frac{4}{1}\) Out! x should be less than y. (C) \(\frac{-5}{-5}\) Out! x should not be equal to y. (D) \(\frac{-2}{-8}=\frac{1}{4}\) Bingo!

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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11 Dec 2012, 04:10

I use this approach: X=1 Y=2 1/3*10 + 2/3*20 10/3+40/3= 50/3--> 16...... Te only outcome greater than 15 is 18 (D). 30 is too much. am I correct? thanks

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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15 Dec 2012, 00:05

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fabrizio1983 wrote:

I use this approach: X=1 Y=2 1/3*10 + 2/3*20 10/3+40/3= 50/3--> 16...... Te only outcome greater than 15 is 18 (D). 30 is too much. am I correct? thanks

The expression is all about recognizing that it represents "weighted averages" and that k could have a range of values. In this case, the range of values is 15 < k < 20. Perhaps this link would be helpful: Weighted Averages Pattern on the GMAT _________________

Re: Question to OG 12th Edition PS 148. [#permalink]

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29 Jan 2013, 04:05

0haxx wrote:

Hey guys, this is my first post here. Please pardon me if I am doing anything wrong I am currently preparing for the test and I stumbled upon the task in the title. It is like \(\frac{x}{x+y} * 10 + \frac{y}{x+y} * 20 = k ; y>x\) What is a possible value of k? The OG has a very complicated and time consuming explanation for this and I hope that it can be solved more quickly. I also don't know if it is really that difficult.. Can anyone help?

Thank you in advance!

The above equation can be written as \(\frac{10x}{x+y} + \frac{20y}{x+y} = k\)

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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06 Sep 2013, 12:37

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Another very easy way to do this question is to understand that (x+y) is the total sum in a sample space, and treat this as a question of "Averages".. Now, applying the weighted average concept, 'k', i.e. the average should ideally be closer to 20 (between 10 and 20) - since we have just once choice that is between 15 and 20, the answer can easily be deduced to option (D), i.e. 18. _________________

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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12 Jan 2014, 09:24

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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? Solved by trail and error The denominator, x+y which is common to both the terms must be a factor of 10 and 20 to get a integer value. First minimum value of x+y which can be used is 2, and using 2 with the possible values of x and y ==> (0,2) or (1,1) is not possible as 0 is not a positive integer in the first possible set and x is equal to y in the second set. The next minimum value of x+y possible is 5. The possible values of x and y with sum of 5 are (2,3) and (1,4) By substituting (2,3) as (x,y) in the equation, we get k=16. This is not given in the answer options By substituting (1,4) as (x,y) in the equation, we get k=18. This is given in the answer option and hence the possible value of k with all conditions satisfied. So, option D

gmatclubot

Re: If x, y, and k are positive numbers such that (x/(x+y))(10)
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12 Jan 2014, 09:24

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