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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]  17 Jul 2010, 11:57
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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

I posted this question earlier but somehow it disappeared from the forum... I dont know why... Anyhow, I am posting again. Can someone come up with some solution???
[Reveal] Spoiler: OA
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Re: Can someone come up with some solution of this question??? [#permalink]  17 Jul 2010, 12:04
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Expert's post
3
This post was
BOOKMARKED
It didn't disappear, it was merged with similar topic: must-be-an-easier-way-88620.html#p668499

Below is my solution from there:

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

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

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

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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Simplification [#permalink]  30 Aug 2010, 14:18
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

The explantion seems to be time consuming. Help for easy process.
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Re: Can someone come up with some solution of this question??? [#permalink]  30 Aug 2010, 15:34
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Bunuel wrote:
It didn't disappear, it was merged with similar topic: must-be-an-easier-way-88620.html#p668499

Below is my solution from there:

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

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

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

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.

actually, you don't need to write all them down to find the answer is D.
especially in your test you don't have much time, so just pick any number you want. In this case, just pick x as 1 and y as 2, so k would be around 18.333, which is near to 18. If you little bit hesitate then try different numbers, such x as 1 and y as 3.
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Re: Can someone come up with some solution of this question??? [#permalink]  31 Aug 2010, 10:55
[X/(X+Y)][10] + [Y/(X+Y)][20] = k
So, 10 (X+2Y) = K (X+Y)

Put the options in place of K, and solve till the end. Make sure X<Y.

For eg – K=10,

Then 10X+20Y = 10X+10Y
Y=0; makes no sense

K=12;
10X+20Y=12X+12Y
8Y=2X
4Y=X
But X<Y Therefore Not Possible

Similarly, K=18.
10X+20Y=18X+18Y
5X+10Y=9X+9Y
Y=4X
According to question, X<Y. Therefore, K=18.

In the case of K=30,
Finally you will reach to -2X=Y.
Since X and Y are +ve, this case is inapplicable…
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Re: Can someone come up with some solution of this question??? [#permalink]  04 Sep 2010, 06:25
Good question. Bunuel has a great solution - per the usual.
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Re: Can someone come up with some solution of this question??? [#permalink]  04 Sep 2010, 11:42
i've seen a lot of approaches where you guys use the weighted average strategy, nicely done!
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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]  15 Jan 2015, 23:16
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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]  16 Jan 2015, 17:02
Expert's post
Hi All,

We're told that X, Y and K are POSITIVE and that X < Y.

We're also told that 10X/(X+Y) + 20Y/(X+Y) = K. We're asked what COULD be the value of K, which means that there's more than one possible answer. Since the answers are NUMBERS, one of the them MUST be a possible answer.

We can manipulate the given equation into:

10X + 20Y = K(X+Y)

Since all of the variables are POSITIVE, K MUST be between 10 and 20. Here's the proof:

IF.....K=10, then the equation becomes...
10X + 20Y = 10X + 10Y
20Y = 10Y
Since Y is positive, 20Y = 10Y is NOT possible.

In that same way, K can't be 20 (or greater) because the end equation would be an impossibility.

With the remaining 3 answers, we can TEST the possibilities...

IF...K = 12, then the equation becomes...
10X + 20Y = 12X + 12Y
8Y = 2X
4Y = X
In this scenario, X > Y which is the OPPOSITE of what we were told. This is NOT the answer.
Eliminate B.

IF....K=15, then the equation becomes...
10X + 20Y = 15X + 15Y
5Y = 5X
Y = X
In this scenario, X = Y, which is NOT a match for what we were told. This is NOT the answer.
Eliminate C.

IF....K=18, then the equation becomes....
10X + 20Y = 18X + 18Y
2Y = 8X
Y = 4X
Here, X < Y, which IS a match for what we were told. This IS a POSSIBLE answer.

[Reveal] Spoiler:
D

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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]  09 Mar 2015, 04:56
Th equation can be solved into-

X/Y=[ 20-K / K-10]---------------------------------------(1)

Since X ,Y and K are positive so 10 < k < 20 -----------(2)

X< Y and so X/Y <1 and also X/Y >0 because both X and Y are positive.

0 < [ 20-K / K-10] <1--------------------------(3)

which solves into

15 < k < 20------------------- Combining (1) ,(2) and (3)

and only 18 satisfies the given inequality.
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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]  09 Mar 2015, 04:57
Bunuel wrote:
It didn't disappear, it was merged with similar topic: must-be-an-easier-way-88620.html#p668499

Below is my solution from there:

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

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

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

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.

Th equation can be solved into-

X/Y=[ 20-K / K-10]---------------------------------------(1)

Since X ,Y and K are positive so 10 < k < 20 -----------(2)

X< Y and so X/Y <1 and also X/Y >0 because both X and Y are positive.

0 < [ 20-K / K-10] <1--------------------------(3)

which solves into

15 < k < 20------------------- Combining (1) ,(2) and (3)

and only 18 satisfies the given inequality.
_________________

Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time.

I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.-Mohammad Ali

Re: If x, y, and k are positive numbers such that [X/(X+Y)][10]   [#permalink] 09 Mar 2015, 04:57
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