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Intern  Joined: 20 Feb 2012
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If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 62374
If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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

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

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

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

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 + y/(x + y)  [#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!

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

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1
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)
Manager  Joined: 15 Apr 2011
Posts: 58
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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

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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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Joined: 02 Sep 2009
Posts: 62374
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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 + y/(x + y)  [#permalink]

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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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Intern  Joined: 12 Nov 2012
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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 + y/(x + y)  [#permalink]

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1
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: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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 + y/(x + y)  [#permalink]

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3
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 + y/(x + y)  [#permalink]

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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? Math Expert V
Joined: 02 Sep 2009
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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kanusha wrote:
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? $$\frac{x+2y}{x+y}=\frac{x+y}{x+y}+\frac{y}{x+y}=1+\frac{y}{x+y}$$.

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

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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.

GMAT assassins aren't born, they're made,
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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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.

what will happen when x>y in this question? since we dont know the value of x and y they can be 1:100 or something like that. what will be the ans then?
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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Hi anik19890,

You ask a couple of different questions.

First, the prompt tells us that X < Y, so we CAN'T have a situation in which X > Y (or even X = Y). THAT specific inequality helps to define the correct answer.

Second, the prompt asks for which of the following COULD be the value of K, meaning that there is more than one possible answer (but only one of the 5 answer choices is correct). Based on the inequality we're given, we COULD have X = 1, Y = 100 (as you described), which would give us...

(1/101)(10) + (100/101)(20) = K

10/101 + 2000/101 = K
2010/101 = K

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

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

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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

We are given:

(x/(x+y))(10) + (y/(x+y))(20) = k

[10x/(x+y)] + [20y/(x+y)] = k

We can combine the two fractions on the left side of the equation because they have the same denominator, (x + y).

[(10x + 20y)/(x+y)] = k

We see that we have a weighted average equation in which x items have an average of 10, and another y items have an average of 20 and a weighted average of k. In this case, the value of k must be between 10 and 20. However, since x is less than y, the weighted average (or k) must be closer to 20 than to 10. Thus k must be 18.

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

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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.

Hi, Bunuel! I have a silly question. Can you please explain how you arrived at the following?
We know that $$x<y$$

Hence $$\frac{y}{x+y}$$ is more than $$0.5$$ and less than $$1$$ Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)   [#permalink] 20 Jun 2017, 17:19

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