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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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17 Jul 2010, 12:57
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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]

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17 Jul 2010, 13:04
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Expert's post
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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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30 Aug 2010, 15:18
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]

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30 Aug 2010, 16:34
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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]

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31 Aug 2010, 11:55
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[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]

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04 Sep 2010, 07:25
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]

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04 Sep 2010, 12:42
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]

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16 Jan 2015, 00:16
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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]

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16 Jan 2015, 18:02
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]

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09 Mar 2015, 05:56
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]

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09 Mar 2015, 05:57
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.
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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 Apr 2016, 03:04
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If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]

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22 Apr 2016, 05:56
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20 second solution. This is a weighted average question:

K is the weighted average of 10 and 20.

The answer must be between the two (10 and 20): A and D are out.

With Y > X, they are not equally weighted so 15 cannot be the answer

Also since Y> X, the answer sould be closer to 20 than to 10. So the only possible answer 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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01 Apr 2017, 09:58
amitjash 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

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???

10 and 20 multiplied by fractions so sum has to be >10 and < 30 - A, E out.
15 possible ONLY when both the ratios are 1/2 -> not possible as x != y - C Out
Greater number y is being multiplied by 20 so > 15 makes sense - B Out

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

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01 Apr 2017, 10:44
amitjash 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

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???

Let $$f(x) = \frac{10x}{x + y} + \frac{20y}{x+y}$$ be a continuous function on $$[0,y]$$.
Then $$f(0) = 20$$ and $$f(y) = \frac{10y}{2y} + \frac{20y}{2y} = 15$$.
For any number between $$15$$ and $$20$$, there must be a number $$k$$ in $$(0,20)$$ by Intermediate Value Theorem.

Only $$18$$ from the choices is between $$15$$ and $$20$$.
Therefore $$18$$ is the correct answer.

The following is Intermediate Value Theorem.
It simply says that for an intermediate value between $$y$$ values at end points, there is another point between $$x$$ values at end points that has the intermediate value as a function value.

Let $$f$$ be a continuous function $$[a,b]$$ and $$f(a) < f(b)$$, where $$a < b$$.
For any $$k$$ in $$(f(a),f(b))$$, there must be $$c$$ in $$(a,b)$$ such that $$f(c) = k$$.
In the case that $$f(a) > f(b)$$, it is also valid.
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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]

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02 Apr 2017, 04:56
amitjash 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

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???

hi
i am happy to share my thoughts on this problem
as kaplan says when we see which of the following in PS problems , we should go with D and E options means we should always check them first., well here is my shot
see if we solve the above equation we come with this solution
E.10x+20y=30x+30y ( taken k =30) now we simplify this we get -2x=y , as it is mentioned in stem that x,y or k are +ve numbers. so here x is -ve so not an answer.
D. taken k=18 and after simplifying as above we get 4x=y here we can say x would be +ve as well as would be less than y ,hold on
C. take k=15 we get x=y so wrong (as x must be less than Y)
B. take k=12 now we get x=4y (here also x is not less than y) wrong
A. take k=10 in this we get y=0 so wrong as like B and C reason .

hence option D is the right answer

please kudos if you liked my post

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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10]   [#permalink] 02 Apr 2017, 04:56
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