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If x, y, and k are positive numbers........

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by selango » Wed Oct 20, 2010 12:31 am
10x/x+y+20y/x+y=k

10(x+2y)/(x+y)=k

there are 2 scenarios possible.

1.k is multiple of 10.

k=10 or 30

k=10-->x+2y/x+y=1

y=0,not possible as y is postive.

k=30-->x+2y/x+y=3

2x=-y,not possible as x and y are positive.

2.x+y=10 so that 10 x+y cancels out

k=12,15,18

x=y,y=5,k=15,not possible as x<y

x=2,y=8,k=18,possible and x<y

Pick D
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by Rahul@gurome » Wed Oct 20, 2010 12:33 am
We have 10x/(x+y) + 10y/(x+y) + 10y/(x+y) = k.
Or 10 + 10y/(x+y) = k.
Now it is given that x < y.
Or x + y < 2y.
Or y/(x+y) > ½.
Or 10y/(x+y) > 5.
Or 10 + 10y/(x+y) > 15.
Also since both x and y are positive, y/(x+y) < 1.
Or 10 + 10y/(x+y) < 20.
Therefore 15<k<20.
The only possible value of k is 18 and the correct answer is D.
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by GMATGuruNY » Wed Oct 20, 2010 3:47 am
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
This is a tricky problem. For many test-takers, the easiest and most 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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by davedecibel » Wed Dec 29, 2010 5:14 pm
when i first saw the problem my first approach was to plug in the answer choices, it kind of takes time, but it should work. i got a little unsure though, after i plugged answer choice E, k=30, which results in x = - (1/2) y.

this might come as an obvious question but i'd like to be sure... in this case, can i exclude this answer choice because x and y can't be both positive, and the problem says "x, y and k are positive numbers [...]", right?
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by GMATGuruNY » Wed Dec 29, 2010 5:43 pm
davedecibel wrote:when i first saw the problem my first approach was to plug in the answer choices, it kind of takes time, but it should work. i got a little unsure though, after i plugged answer choice E, k=30, which results in x = - (1/2) y.

this might come as an obvious question but i'd like to be sure... in this case, can i exclude this answer choice because x and y can't be both positive, and the problem says "x, y and k are positive numbers [...]", right?
Exactly right. If k=30, then x and y can't both be positive. Since the problem states that x and y are both positive integers, answer choice E can be eliminated.
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by Scott@TargetTestPrep » Fri Dec 15, 2017 10:10 am
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
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.

Answer: D

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