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

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If x, y, and k are positive numbers such that (x/x+y )(10) + [#permalink]  26 Jul 2008, 05:51
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: unknown [#permalink]  26 Jul 2008, 08:20
I kind of cheated on this question because I used excel, but I was stuck.

First we have the following:

\frac{x}{x+y}(10) + \frac{y}{x+y}(20) = k

\frac{10x}{x+y} + \frac{20y}{x+y} = k

\frac{10x + 20y}{x+y} = k

10x + 20y = k(x+y)

Now, becuase we know that x<y and it's easier to deal with 2 variables, I made y = x + 2.
In the spreadsheet attached, I made 2 a value that can be changed, as well as the starting value for x.

I came up with D as the answer, but I have no clue how I would have done this on the actual GMAT with the marker and dry erase board. Probably would have guessed C and moved on getting it wrong.

I did notice a relationship. When y = 3x, k = 18. I'm not sure how to go about finding that realationship without excel.

Any suggestions?
Attachments

NihitProblem.xls [16 KiB]

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Re: unknown [#permalink]  30 Jul 2008, 08:36
The answer is correct but is there a better way of doing this?
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Re: unknown [#permalink]  30 Jul 2008, 09:00
Here we go..

10x/(x+y) + 20y/(x+y) = 10 + 10y/(x+y) = k ....... (1)

For k to be integer, 10y/(x+y) has to be integer. So x+y = 2 or 5 or 10 or some higher multiple of 10.

you can eliminate 2 because y>x and all are +ve.

for x+y=5, you can have (x,y) = (2,3) or (1,4)
plug in the values in 1 and you get k=16 or k=18

At this point you have your answer as 18. Since the question asks which could be the value of k, and 18 is one of the options, you won't be wrong.
Besides they would give options like only 1and3 , 1,2,and3 etc etc had there been multiple correct answers.

For trials, i checked with x+y=10
here you have (x,y)= (1,9), (2,8), (3,7), (4,6). again because x<y
plug into (1) and you get
k= 19, 18, 17, or 16.

So i guess that you will have 18 in all the valid x+y combinations.
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Re: unknown [#permalink]  30 Jul 2008, 09:09
2
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Nihit 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

(x/x+y )(10) + (y/x+y )(20) = k
(10)[(x + y)/(x + y) + y/(x+y)] = k
10 + 10y/(x + y) = k ---- (a)

As x < y,
y/(x + y) > y/(y + y) = 1/2 = 0.5

Go back to (a):
k = 10 + 10y/(x + y) > 10 + 10(0.5) = 15

So the ans is either D or E

As y/(x + y) must be smaller than 2,

the ans is D
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Re: unknown [#permalink]  30 Jul 2008, 09:13
Nihit 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

(x/x+y )(10) + (y/x+y )(20) = k

--> (10x+20y)/(x+y) =k
--> (10x+10y+10y)/(x+y) =k
--> 10 + 10 (y/x+y)=k

From the above equations and x<y conditions K should be >10 an d< 20
A,E are out.

All other three choice aren integers.. so x+y must be multiple of 5
also y>2 ..
(X,Y)=(2,3) --> leads to 10 +10 (2/5) =16 .. answer not present
(X,Y)=(1,4) --> leads to 10 +10 (4/5) =18 .. correct answer.
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Re: unknown [#permalink]  30 Jul 2008, 09:16
I don't follow this part. Could you please elaborate ?
Thanks

judokan wrote:
As y/(x + y) must be smaller than 2,

the ans is D
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Re: unknown [#permalink]  30 Jul 2008, 09:19
judokan wrote:
Nihit 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

(x/x+y )(10) + (y/x+y )(20) = k
(10)[(x + y)/(x + y) + y/(x+y)] = k
10 + 10y/(x + y) = k ---- (a)

As x < y,
y/(x + y) > y/(y + y) = 1/2 = 0.5

Go back to (a):
k = 10 + 10y/(x + y) > 10 + 10(0.5) = 15

So the ans is either D or E

As y/(x + y) must be smaller than 2,

the ans is D

you mean As y/(x + y) must be smaller than 1,

Good logic ...
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Re: unknown [#permalink]  30 Jul 2008, 09:20
x2suresh wrote:
......
All other three choice aren integers.. so x+y must be multiple of 5
......

Not quite. 5 is the only case. other than than x+y needs to be a multiple of 10.

eg. if you take x=7 y=8 then you dont get an integer value for k.
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Re: unknown [#permalink]  30 Jul 2008, 09:20
bhushangiri wrote:
I don't follow this part. Could you please elaborate ?
Thanks

judokan wrote:
As y/(x + y) must be smaller than 2,

the ans is D

10 + 10 (y/x+y)=k
y/(x + y) <1 --> so k<20.
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Re: unknown [#permalink]  30 Jul 2008, 09:25
bhushangiri wrote:
x2suresh wrote:
......
All other three choice aren integers.. so x+y must be multiple of 5
......

Not quite. 5 is the only case. other than than x+y needs to be a multiple of 10.

eg. if you take x=7 y=8 then you dont get an integer value for k.

you caught me..
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Re: unknown [#permalink]  30 Jul 2008, 09:26
x2suresh wrote:

10 + 10 (y/x+y)=k
y/(x + y) <1 --> so k<20.

Ah.. >0.5 but less than 1. Hence >15 and less than 20. Elegant. +1 to judo boy

Thanks.
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Re: unknown [#permalink]  30 Jul 2008, 09:42
bhushangiri wrote:
x2suresh wrote:

10 + 10 (y/x+y)=k
y/(x + y) <1 --> so k<20.

Ah.. >0.5 but less than 1. Hence >15 and less than 20. Elegant. +1 to judo boy

Thanks.

Judo,
You deserve +1
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Re: unknown [#permalink]  30 Jul 2008, 09:51
1
KUDOS
Nihit 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

k is weighted average of 10 and 20.... the weight of 10 and 20 is given by x and y ....
since its an average it has to between 10 and 20

wt average will be 10 when y = 0
wt average will be 20 when x = 0
wt average will be 15 when x = y

we know y>x, so wt average will be closer to 20 than to 10. only option D fits the condition
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Re: unknown [#permalink]  30 Jul 2008, 10:00
durgesh79 wrote:
Nihit 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

k is weighted average of 10 and 20.... the weight of 10 and 20 is given by x and y ....
since its an average it has to between 10 and 20

wt average will be 10 when y = 0
wt average will be 20 when x = 0
wt average will be 15 when x = y

we know y>x, so wt average will be closer to 20 than to 10. only option D fits the condition

another "work of art" solution.. +1 to u.
Re: unknown   [#permalink] 30 Jul 2008, 10:00
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