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possible value of K

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by missrochelle » Sun Aug 22, 2010 2:49 pm
I got stuck after manipulating the equation...Was going to try to plug in values but nothing made sense --

I simplified it to be -
10X / (X+Y) + 20Y (X+Y) = K
10X + 20Y
(X+Y)

Get rid of the denominator you have
(10x + 20y) = K(x+y)
10x+20y = Kx +Ky.

Which doesn't make sense since K can't be both 10 and 20.

Any help?
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by Gurpinder » Sun Aug 22, 2010 3:27 pm
missrochelle wrote:I got stuck after manipulating the equation...Was going to try to plug in values but nothing made sense --

I simplified it to be -
10X / (X+Y) + 20Y (X+Y) = K
10X + 20Y
(X+Y)

Get rid of the denominator you have
(10x + 20y) = K(x+y)
10x+20y = Kx +Ky.

Which doesn't make sense since K can't be both 10 and 20.

Any help?
I agree. I am suspicious about the source of this question.

if you simplify the equation, you get:
10x + 20y
x+Y

which is suppose to equal k. but lets say x=1,y=2 you get k = 10.
or lets say x=2,y=3 then k = 16

so not whether i am doing something wrong or, this question is flawed.
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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by limestone » Sun Aug 22, 2010 5:18 pm
After simplified, the equal is (10x+20y)/(x+y)=k, or ((10x+10y) +10y)/(x+y) =k
that equal to 10 + 10y/(x+y)=k

Since x<y then x+y <2y, then 10y/(x+y) > 10y/2y
It means 10y/(x+y)>5, then 10 + 10y/(x+y)>15
Eliminate A,B,C

Since x is positive, x+y >y, then 10y/(x+y) < 10y/y
We get 10y/(x+y)<10, then 10+10y/(x+y)<20
Eliminate E; D remains.
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by missrochelle » Sun Aug 22, 2010 5:21 pm
How are you going from (10x+10y) +10y)/(x+y) =k
to 10 + 10y/(x+y)=k ? Not sure I follow that manipulation.
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by limestone » Sun Aug 22, 2010 6:33 pm
(10x+20y)/(x+y) = (10x+10y+10y)/(x+y) = ((10x+10y)+10y)/(x+y)
= (10x+10y)/(x+y) + 10y/(x+y)
As x and y are positive, then x+y is not zero, we can simplify (10x+10y)/(x+y) = 10*(x+y)/(x+y) =10
So (10x+20y)/(x+y) = (10x+10y)/(x+y) +10y/(x+y) = 10 + 10y/(x+y)
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by srajan » Sun Aug 22, 2010 9:59 pm
This is little confusing question. But the question asks possible value of K?
You guys have already solved the equation as:
10*((x+2y)/(x+y)) = k
Now let's further simplify it:

y/(x+y) = (k/10)-1
or
y/(x+y)= (k-10)/10 ---- (1)

Since x, y, K are positive numbers, the left hand side of the above equation is positive.
i.e. K-10>0
i.e. k>10

Also, since y/(x+y) < 1 (all are positive nos)
(K-10)/10 < 1
i.e. K<20

so far we have eliminated 2 choices: A and E

Now let's use the last help given in the question x<y

let's subtract from 1 both sides, i.e.

1-(y/(x+y))= 1-((k-10)/10)

i.e. x/(x+y) = (20-k)/10

since x<y, x/(x+y) < 1/2
we can prove the above too.

x<y
x+x < x+y
2x < (x+y)
x/(x+y) < 1/2 (we can cross multiply since all are positive nos.)

therefore, (20-K)/10 < 1/2
i.e. 20-k < 5
i.e. 15 < k
combining all the above results for K we can write

15< k < 20

There is only one answer, which satisfies this inequality: D

I hope I was clear in my explanation.
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