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Percent Problem

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Percent Problem

by Fractal » Fri Sep 24, 2010 6:08 am
If x percent of y is equal to y percent less than z, what is y in terms of x and z?

(A) 100(x+z)/x
(B) 100x/(x+z)
(C) 100(x+z)/(x-z)
(D) 100z/(x-z)
(E) 100z/(x+z)

I had no problems with building the equation:

xy/100 = z - yz/100

But if I solve this equation for y, I get the following: y = z(100-y)/x, which is not one of the solution.


Steps:

xy/100 = z - yz/100

1. xy = 100z - yz

2. y = (100z - yz)/x or z(100-y)/x

Could anybody help please?


thx
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by kmittal82 » Fri Sep 24, 2010 6:41 am
>x percent of y

xy/100

>y percent less than z
z - yz/100

>If x percent of y is equal to y percent less than z,

xy/100 = z - yz/100

=> xy/100 + yz/100 = z

=> y(x+z)/100 = z

=> y = 100z/(x+z)

(E)

>what is y in terms of x and z
Your method also has y on the right hand side, you need to find y purely in terms of x and z
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by Fractal » Fri Sep 24, 2010 7:53 am
kmittal82 wrote:>x percent of y

xy/100

>y percent less than z
z - yz/100

>If x percent of y is equal to y percent less than z,

xy/100 = z - yz/100

=> xy/100 + yz/100 = z

=> y(x+z)/100 = z

=> y = 100z/(x+z)

(E)

>what is y in terms of x and z
Your method also has y on the right hand side, you need to find y purely in terms of x and z
ah I got it! thanks a lot! therefore one should always bring the relevant variable to one side of the equations!

greets
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