akshayk
A factory cleans its machinery with a solution that is 87% water and the remainder vinegar. If the factory manager starts with 260 Litres of a 40% vinegar solution, how much water must the manager add to the existing solution to attain the proper vinegar concentration for cleaning the machinery?
A) 104 Litres
B) 120 Litres
C) 470 Litres
D) 540 Litres
E) 800 Litres
Weighted AverageIn mixture problems that involve percent water and percent "something else," we can use a straightforward weighted average.
Track on only one concentration, percent water or percent vinegar. Either works.
(The formula below works for any two liquids whose ratio we must find. I mention "water" because in a mixture problem, if pure water is added or subtracted, its concentration of the other thing such as vinegar is 0%. If we track on water, water is 100% water.)
We have a solution that is 40% vinegar.
The original is thus 60% water
We need a resultant solution that is 87% water
Let A = original solution with 60% water
Let w = amount of water to be added
w = 100% water, in decimal form = 1
A + w = Total volume of resultant mixture
% = concentration / percent water
If A and B are the two liquids to be mixed, one weighted average formula is
(% A)(Vol A) + (% B)(Vol B) = (Desired % of A+B)(Vol A+B)Original 60% water solution = A
Water to be added = B = w
\(.60(A) + (1)w = .87(A + w)\)
\(.60(260) + 1w = .87(260 + w)\)
\(156 + 1w = (.87)(260) + .87w\)
\(1w - 0.87w = 226.2 - 156\)
\(.13w = 70.2\)
\(w=\frac{70.2}{.13}=540\)
\(w\) = amt of water to be added = \(540\)
Answer D