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

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New post 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 \(18\).

Answer: D (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.

Answer: D (18)


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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New post 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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New post 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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New post 03 Dec 2018, 23:58
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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New post 03 Dec 2018, 23:59
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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.

Answer choice C: k= 15
(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 correct answer is D.

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 10*x/(x+y)+ 20 &nbs [#permalink] 03 Dec 2018, 23:59

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