Problem Solving

How many liters of oil must be added to x liters of an oil-water solution that is y percent oil to produce a solution that is z percent oil?

xz - xy
________ (A
100

xz - xy
________ (B
z - 100

xy - xz
________ (C
z - 100

100y - xz
_________ (D
z - 100

xz - 100y
_________ (E
z - 100

OA is C

Abdulla wrote:How many liters of oil must be added to x liters of an oil-water solution that is y percent oil to produce a solution that is z percent oil?

xz - xy
________ (A
100

xz - xy
________ (B
z - 100

xy - xz
________ (C
z - 100

100y - xz
_________ (D
z - 100

xz - 100y
_________ (E
z - 100

Here's a question that we definitely want to attack by picking numbers.

Let's let:

x = 100
y = 25
z = 50

In other words, we're starting with a solution of 100 litres that's 25% oil and we want a solution that's 50% oil.

Since we started with 100l total, 25 of those were oil. If we want 50% oil, we solve:

oil/total = 1/2

(25 + new oil)/(100 + new oil) = 1/2

let's call "new oil" n.

(25 + n)/(100 + n) = 1/2

cross multiplying:

2(25 + n) = 100 + n
50 + 2n = 100 + n
n = 50

So, we need to add 50l of oil. Now we plug in our x, y and z values to the choices to find one that = 50.

xz - xy
________ = ((100*50) - (100*25))/100 = 50 - 25 = 25... wrong
100

xz - xy
________ = ((100*50) - (100*25))/(50-100) = negative... wrong
z - 100

xy - xz
________ = ((100*25) - (100*50))/(50-100) = -2(25-50) = 50.. yay!
z - 100

100y - xz
_________ = ((100*25) - (100*50))/(50-100) = -2(25-50) = 50.. yay?
z - 100

xz - 100y
_________ = same as above but +/- = -.... wrong.
z - 100

Both (C) and (D) worked out, so let's look for the difference. In (D) we have "100y" and in (C) we have "xy". (C) should be our clear winner, since we're definitely going to get a different result by changing the value of x. Changing the value of x will give us a different result in (C), but not so much in (D). Choose (C)!

* * *

If you're an algebra savant you may have been able to set up the equation, but it certainly won't be self-evident to most test takers.

Let's start with the basic percent formula:

% = (part/whole) * 100%

Applying the formula to this question:

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

Let's convert to fractions:

z/100 = ((xy/100) +n)/(x+n)

(x+n)z/100 = xy/100 + n
(x+n)z = xy + 100n
xz + nz - xy = 100n
xz - xy = 100n - nz
xz - xy = n(100 - z)

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

and to change this to match one of the answers, we need to multiply the left side by -1/-1 (which is purely a cosmetic change) to get:

(xy - xz)/(z - 100) = n.

Uuuuuuuuugly!