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If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]
17 Jul 2010, 11:57

00:00

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D

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Difficulty:

5% (low)

Question Stats:

60% (02:45) correct
40% (02:48) wrong based on 10 sessions

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???

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.

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.

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)

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.