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

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

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15 Oct 2009, 15:08
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If x, y, and k are positive numbers such that $$(\frac{x}{x+y})(10)+(\frac{y}{x+y})(20) = k$$ 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
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Re: What is the value of k [#permalink]

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15 Oct 2009, 15:34
answer is 10 . ( A)

10 ( x+2 y) = k (x+y)

10 / k = x+2y / x+y
...... might be wrong from this step ------
10 = x+ y and k = x+ y + y
k = 10+ y

not a , not e , not d,

either b or c...

so x = 5 and y = 5 .... then x+ 2y 15...
x = 8 and y = 2 then x+ 2y 12...

..some one can take from here... hope so ..
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Re: What is the value of k [#permalink]

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15 Oct 2009, 17:36
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Good question. (Edited to be easier for reading)

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

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

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

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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Last edited by Bunuel on 26 Oct 2009, 18:07, edited 1 time in total.
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Re: What is the value of k [#permalink]

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15 Oct 2009, 22:57
eresh wrote:
If x, y, and k are positive numbers such that $$(\frac{x}{x+y})(10)+(\frac{y}{x+y})(20) = k$$ and if x < y, which of the following could be the value of k?

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

Couldn't find a way to solve it.

Ans D. the above expression can be written as [10(x+y)/(x+y)] + 10y/(x+y) = 10 + 10y/(x+y)
now we know y>x then to simplify we can take values for x and y such that they sum up to 10 and we are left with y. If x=2 and y=8 we get 18. Other options will not satisfy.
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Re: What is the value of k [#permalink]

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16 Oct 2009, 15:33
x boxes that weight 10lbs
y boxes that weight 20lbs

That expression is average weight of a box. If there are more 20lbs boxes taht 10lbs boxes, obviously average box > 15 lbs.
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Re: What is the value of k [#permalink]

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16 Oct 2009, 17:58
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We can pick nos. for this and solve it.
Let x=1 & y=4, solve the equation answer is 18, Option D
Let x=2 & y=3,solve the equation, answer is 16, no option available.
Let x=1 & y=9, solve the equation, answer is 19, no option available.

Correct option D........
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Re: What is the value of k [#permalink]

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16 Oct 2009, 18:33

however I like Bunuel's explanation..
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Re: What is the value of k [#permalink]

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17 Oct 2009, 15:29
Thanks a lot all. The answer is 18.
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Re: What is the value of k [#permalink]

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18 Oct 2009, 14:47
eresh wrote:
If x, y, and k are positive numbers such that $$(\frac{x}{x+y})(10)+(\frac{y}{x+y})(20) = k$$ and if x < y, which of the following could be the value of k?

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

Given that: $$x < y$$
$$(\frac{x}{x+y})< (\frac{y}{x+y})< 1$$

Also given that: $$(\frac{x}{x+y})(10)+(\frac{y}{x+y})(20) = k$$
$$(10)(\frac{x+2y}{x+y}) = k$$
$$(10)(\frac{x+y+y}{x+y}) = k$$
$$(10)(1+\frac{y}{x+y}) = k$$

The value of the above expression range from 10 to 20 but with the given fact, it cannot be 10, 12 and 30 because all are out of scope. Now its between 15 and 18 however 15 is also out for being the average of 10 and 20. That would be true only if x and y were equal. That leaves out 18 as the only choice as answer..
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Re: What is the value of k [#permalink]

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18 Oct 2009, 15:43
D;
Prefer Hunt's style; makes abstraction quicker.
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Re: What is the value of k [#permalink]

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18 Oct 2009, 18:20
Yup....Hunts approach seems faster.....
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Re: What is the value of k [#permalink]

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18 Oct 2009, 23:04
Thanks Hunt...Indeed a less than 2 min solution..best to adopt for the exams..
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Re: What is the value of k [#permalink]

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19 Oct 2009, 00:19
PussInBoots's Solution is also good enough. If you this x boxes that weight 10lbs
y boxes that weight 20lbs, and x<y, then the number is obviously greater than the average of 10 and 20 (if there were same number of boxes). Since 20lb boxes are more in number, the average weight must be between 15-20 lbs. Less than 1 min solution
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Re: What is the value of k [#permalink]

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20 Oct 2009, 21:27
After rearranging the equation

k=10+10*(y/(x+y))
hence k>10

y/(x+y) < y/2y as x<y
hence y/(x+y)<1/2

hence 10< k < 15

the only option matches is k=12

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Re: What is the value of k [#permalink]

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20 Oct 2009, 21:39
After rearranging the equation

k=10+10*(y/(x+y))
hence k>10

y/(x+y) < y/2y as x<y
hence y/(x+y)<1/2

hence 10< k < 15

the only option matches is k=12

Part in red is not correct:
y/(x+y) > y/2y as x<y -->15< k < 20
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Re: What is the value of k [#permalink]

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20 Oct 2009, 21:46
Thanks ---- I was wrong 18 is the answer
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Re: What is the value of k [#permalink]

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20 Oct 2009, 22:38
After rearranging the equation

k=10+10*(y/(x+y))
hence k>10

Could not figure out how you arrived at the rearrangement!
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Re: What is the value of k [#permalink]

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20 Oct 2009, 23:39
k=10*x/(x+Y) + 20* y/(x+y)
= (10*x+20*y)/(x+y)
= (10(x+Y) + 10*y)/(x+y)
= 10+ 10*y/(x+y)
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Re: What is the value of k [#permalink]

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21 Oct 2009, 00:07
k=10*x/(x+Y) + 20* y/(x+y)
= (10*x+20*y)/(x+y)
= (10(x+Y) + 10*y)/(x+y)
= 10+ 10*y/(x+y)

Ooh! I was to lazy to figure that out
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Re: What is the value of k [#permalink]

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26 Oct 2009, 12:35
I got that the expression reduces to:
k =10(1+ y/x+y)
Can somebody please elaborate after that?How are we getting values?
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Re: What is the value of k   [#permalink] 26 Oct 2009, 12:35

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