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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]

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

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equation can be simplified to
10y = (x+y)(K-10)

Hence Ans = D as only if k=18, we have a positive ratio that is less than 1 for x/y, {where x<y)

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

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

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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


main funda is

y/(x+y) is equal to 1/2 if x=y but when x<y then x+y<2y i.e. y/x+y >1/2 (x,y>0)

also the maximum value is x tends to 0 i.e. as x>0 say x be 0.0000...(infinite times )1 then the expression x+y tends to y i.e. y/ (x+y) <1

hence 0.5 < y/(x+y) < 1
=> 1+0.5 < 1+y/(x+y) < 1+1
=> 10*1.5 < 10*(1+(y/x+y) <2 *10
=> 15<10*(1+(y/x+y)<20

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

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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: \(15<10*(1+\frac{y}{x+y})<20\).

Hope it's clear.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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(x/(x+y))(10) + (y/(x+y))(20)=k this equation is equal to

{(x/(x+y))(10) + (y/(x+y))(10)} + (y/(x+y))(10)=k

{ 10 }+ (y/(x+y))(10)=k


10+(y/(x+y))(10)=k

if x < y but still y< x+y so (y/(x+y))(10) < 10 ==> 10+(y/(x+y))(10) <20, k<20

if x=y again y< x+y but it is certain that (y/(x+y))(10) = 5 ==> 10 +(y/(x+y))(10) =15 k=15

x and y are not equal, therefore k must be bigger than 15 , less than 20

15< k < 20

k= 18 valid answer

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

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I like the weighted box answer approach!

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Re: If x, y, and k are positive [#permalink]

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New post 14 May 2012, 10:35
Bunuel wrote:
BANON wrote:
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.

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ausadj18 wrote:
Can someone show me the shortcut to solving number 148 from the OG:

If x,y and k are positive numbers such that (x/x+y)(10)+(y/x+y)(20)=k and x <y, which of following could be value for k?

a) 10
b) 12
c)15
d)18
e) 30


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

\(\frac{10y}{x+y}>\frac{10y}{2y}=5\) and \(\,\,\frac{10y}{x+y}<\frac{10y}{y}=10\), therefore \(10+5<k<10+10\) or \(15<k<20\).

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

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\(\frac{10x}{x+y}+ \frac{20y}{x+y}=k\)
\(10x + 20y = kx + ky\)
\((10-k)x=(k-20)y\)
\(\frac{x}{y}=\frac{k-20}{10-k}\)

(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!

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

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New post 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

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

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

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

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

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

Now since x<y try to come up with a relation between x and y so that we can eliminate the fraction \(\frac{10y}{x+y} = k\)

In this case 4x = y or x = y/4 then the fraction will become 10y/(y/4 + y) = 10y/5y/4 = 8.

So the final answer would be 10 + 8 = 18.

We are trying to eliminate the fraction because we need a integer value for k.

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

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

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New post 12 Jan 2014, 07:53
Hi,

I did it this way:

Y>X

Then start with X=1 and Y=2 and continue until you found an integer...
- X=1 and Y=3 then K is not an integer
-X=1 and Y=4 then K=18!!

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

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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

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

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New post 12 Jan 2014, 11:13
I got this wrong

I chose to solve it this way but I guess it doesn't work and I dont know how why.

x=1
y=2

\(\frac{1}{3} x \frac{10}{1} + \frac{2}{3} x \frac{10}{1}\)

\(\frac{10}{3} + \frac{20}{3}=\frac{30}{3}=10\)

but I guess that's flawed

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

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New post 06 May 2014, 22:18
Hi Bunnel ,
Can you pls Explain how these Steps Came in your Solution

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

I Didn't Understood how The Steps 3 and 4 Came? :(
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10)   [#permalink] 06 May 2014, 22:18

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