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# If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)

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Joined: 02 Sep 2009
Posts: 62298
Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)  [#permalink]

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20 Jun 2017, 21:01
uvee wrote:
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 $$1$$

$$\frac{y}{x+y}$$

Since both x and y are positive, then the denominator is greater than numerator. So, $$\frac{y}{x+y}<1$$.

Next, if x were equal to y, then we'd have y/(2y) = 1/2 but x < y, so denominator is less than 2y, which makes the fraction greater than 1/2.

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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19 Jan 2018, 07:31
Could you guys please explain if the answer holds true when x<0 and y>0 ? 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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19 Jan 2018, 19:26
Hi Phlaryu,

The prompt tells us that X, Y and K are all POSITIVE, so you shouldn't be wasting time on the outcome if X < 0.

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

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03 Dec 2018, 23:58
1
1
This one's all about pattern recognition.

Notice that the equation is giving us a weighted average. Because y is bigger than x, the average must be weighted towards y. Visually:

x---------avg---y
10--------avg---20

Thus, the answer must be between 15 and 20, exclusive.

Among the answer choices, only 18 fits the bill.

Choose D.
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Re: If x, y, and k are positive numbers such that 10*x/(x+y)+ 20  [#permalink]

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03 Dec 2018, 23:59
2
pzazz12 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 suspect that many test-takers would not recognize that this is a weighted average problem. For those test-takers, a straightforward and efficient approach would be to plug in the answer choices, which represent the value of k.

(10x + 20y)/(x+y) = 15
10x + 20y = 15x + 15y
5y = 5x
y = x
Doesn't work because the problem states that x<y.

We need y to be larger, so let's try answer choice D: k=18
10x + 20y = 18x + 18y
2y = 8x
y/x = 8/2
Success! x<y.

The process would be very quick if you recognized right away that since k = one of the answer choices, you could simply plug them in for 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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01 Sep 2019, 08:08
BANON 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:
1. x, y, and k are positive numbers such that $$\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k$$
2. x < y

Asked: Which of the following could be the value of k?

$$\frac{x}{x + y}*10 + \frac{y}{x + y}*20 = k$$
$$\frac{x+y}{x + y}*10 + \frac{y}{x + y}*10 = k$$
$$10 + \frac{y}{x + y}*10 = k$$
SInce y>x
$$y> \frac{(x+y)}{2}$$
$$\frac{y}{x+y}>\frac{1}{2}$$
$$k > 10 + \frac{1}{2}*10$$
$$k > 15$$
$$10 + \frac{y}{x + y}*10 = k$$
Since $$\frac{y}{x+y}< 1$$
$$k< 10 + 1*10$$
k<20
15<k<20
Only option D 18 satisfies these conditions

IMO D
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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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01 Sep 2019, 20:58
10x/ (X+Y) + 20Y/(X+Y) = K

10*{ (X+2Y)/ (X+Y) } = K

(X+2Y) / (X+Y) = K/10 and we know that X<Y, so here we plug in values.

(X+2Y) / (X+Y) = 10/10......Can't be true as Y will have to be 0 for this.
(X+2Y)/ (X+Y) = 12/10.... Gives us 2 equations. X+ 2Y = 12 and X+Y = 10...solving this we get X= 8 and Y =2, so incorrect.
(X+2Y) / (X+Y) = 15/10...similar to above, we will get two equations and by solving them, we get X= 5 and Y= 5, so incorrect.
(X+2Y) / (X+Y) = 18/10.......X+ 2Y= 18 and X+Y = 10........solving them we get, X= 2 and Y = 8...This is the correct solution.
(X+2Y) / (X+Y) = 30/10......we get -ve value of X. Hence the answer is (D)
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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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15 Nov 2019, 23:24
I encountered this in a GMATFocus test.

Here's the GMATFocus solution.
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Re: If x, y, and k are positive numbers such that x/(x + y)*10 + y/(x + y)   [#permalink] 15 Nov 2019, 23:24

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