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If x, y, and k are positive numbers such that 10 + 20 = k

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If x, y, and k are positive numbers such that 10 + 20 = k [#permalink]

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New post 05 Sep 2008, 23:46
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A
B
C
D
E

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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

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Re: Zumit PS 004 [#permalink]

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New post 06 Sep 2008, 00:13
D.
Simplify the equation : 10 +10y/(x+y)=k. Since x, y,k are positive #s. and X<y.

Only D satisfies all constraints. Check it.
If k=18
then 10y/(x+y)=8
10y =8x +8y
2y=8x
y=4x, so y(or 4x) > x

Other choices doesn't satisfy X<Y condition.

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Re: Zumit PS 004 [#permalink]

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New post 06 Sep 2008, 09:37
dancinggeometry 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


Given eq can be rewritten as (10/x+y) [x+2y] =k x <y

For the equation to result in a+ve integer k (x+y) must divide 10 or x+2y without remainder.

Lets choose x=1, y=9 could result in x+2y being 19
x=2 y=8 results in K equal to 18

Hence D

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Re: Zumit PS 004 [#permalink]

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New post 06 Sep 2008, 10:01
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This is a weighted average of 10 and 20. Since y > x, more weight is placed on the 20 so the answer must be 18, since this is the only answer between 10 and 20 that is closer to 20

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Re: Zumit PS 004 [#permalink]

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New post 06 Sep 2008, 22:54
OA is D. Bin 2 problem.

Good alternative explanation lsmv479!

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Re: Zumit PS 004 [#permalink]

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New post 08 Sep 2008, 09:25
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lsmv479 wrote:
This is a weighted average of 10 and 20. Since y > x, more weight is placed on the 20 so the answer must be 18, since this is the only answer between 10 and 20 that is closer to 20


Agree with this solution. this good approach.
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Re: Zumit PS 004 [#permalink]

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New post 19 Oct 2011, 03:42
dancinggeometry 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


Ans - D

This question is very well explained on MGMAT forum. Here is the explaination:

1. Solving the equation a little will yield following that represent a weighter Avg. equation
[10X + 20Y]/(X + Y)

2. Now, 10 < K < 20. This will eleminate option A & E
3. Since, Y > X and 20 > 10 therefore K should be quite close to 20. This eleminates B & C and D is the best guess
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Re: If x, y, and k are positive numbers such that 10 + 20 = k [#permalink]

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

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New post 23 Oct 2017, 06:50
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x, y, and k are positive numbers such that 10 + 20 = k   [#permalink] 23 Oct 2017, 06:50
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