If x, y, and k are positive numbers such that (x/(x+y))(10) : GMAT Problem Solving (PS)
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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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27 Feb 2012, 07:42
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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
[Reveal] Spoiler: OA
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Re: If x, y, and k are positive [#permalink]

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27 Feb 2012, 07:50
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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 $$18$$.

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.

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

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27 Feb 2012, 07:56
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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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14 Apr 2012, 00:19
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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14 Apr 2012, 00:48
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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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14 Apr 2012, 01:09
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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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23 Apr 2012, 09:24
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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

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

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23 Apr 2012, 16:00
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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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14 May 2012, 09:35
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 $$18$$.

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.

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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Re: Value of k, given x,y and k are positive integers [#permalink]

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30 Sep 2012, 13:22
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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 $$15<k<20$$.

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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!

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

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11 Dec 2012, 03:10
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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14 Dec 2012, 23:05
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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
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Re: Question to OG 12th Edition PS 148. [#permalink]

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29 Jan 2013, 03:05
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?

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.

Please give a kudo if you like my explanation.
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Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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03 Jul 2013, 00:24
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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06 Sep 2013, 11:37
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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.
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If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]

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09 Sep 2013, 04:57
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

This is from a thread posted last year.

Could someone please explain how to deduce since x<y;

Hence
[y][/x+y] is more than 0.5 and less than 1

0.5<{y}/{x+y}<1

Only answer between 15 and 20 is 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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09 Sep 2013, 05:04
smartyman 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

This is from a thread posted last year.

Could someone please explain how to deduce since x<y;

Hence
[y][/x+y] is more than 0.5 and less than 1

0.5<{y}/{x+y}<1

Only answer between 15 and 20 is 18.

Merging similar topics.

As for your question. It's given in the stem: "... if 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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12 Jan 2014, 06:53
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!!

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

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12 Jan 2014, 08:24
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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
Re: If x, y, and k are positive numbers such that (x/(x+y))(10)   [#permalink] 12 Jan 2014, 08:24

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