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by vaibhav101 » Wed Aug 01, 2018 7:23 am
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how many liters of oil must be added to x liters of an oil water solution that is y percent oil to produlution that is z percent oil?

A xz-xy/100

B xz-xy/z-100

C xy-xz/z-100

D xz-100y/z-100
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by Jay@ManhattanReview » Fri Aug 03, 2018 1:50 am
vaibhav101 wrote:how many liters of oil must be added to x liters of an oil water solution that is y percent oil to produlution that is z percent oil?

A xz-xy/100

B xz-xy/z-100

C xy-xz/z-100

D xz-100y/z-100
Currently, the quantity of oil in the solution = xy/100 liters and the quantity of water in the solution = x – xy/100 liters

Say p liters of oils is added; now, the total quantity of oil = xy/100 + p liters and the total quantity of solution = x + p liters

Thus, the percentage of oil now = [(xy/100 + p) / (x + p)]*100% = z

Upon solving, p = xy – xz/(z – 100) liters

The correct answer: C

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Sun Aug 26, 2018 6:38 pm
vaibhav101 wrote:how many liters of oil must be added to x liters of an oil water solution that is y percent oil to produlution that is z percent oil?

A xz-xy/100

B xz-xy/z-100

C xy-xz/z-100

D xz-100y/z-100

Let n = the amount, in liters, of oil needed to be added to x liters of an oil water solution. So we have:

(y/100 * x + n)/(x + n) = z/100

y/100 * x + n = z/100 * (x + n)

Multiplying both sides of the equation by 100, we have:

xy + 100n = z(x + n)

xy + 100n = zx + zn

100n - zn = zx - xy

n(100 - z) = zx - xy

n = (zx - xy)/(100 - z)

If we multiply the numerator and denominator each by -1, we have (xy - zx)/(z - 100).

Answer: C

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