bmwhype2 wrote:
Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
Another weighted average approach, expressed a bit differently.
Track on milk. We know the desired concentration of milk in the resultant mixture.
Milk is a fraction (or percentage or concentration) of all three mixtures of milk
and water. The weighted average formula accounts for water by way of volume.
This formula is easy (concentration can be a percentage or a fraction):
\((Concentration_{A})(Vol_{A}) + (Concentration_{B})(Vol_{B})\)
\(=(Concentration_{A+B})(Vol_{A+B})\)Let
A = # of gallons of A (volume)
1) Use ratios and desired percentage to find the concentration of milk in A, B, and end mixture
(With ratios, remember to find \(\frac{part}{whole}\))
In mixture A, \(\frac{Milk}{Water}=\frac{2}{5}\)
2 parts milk, 5 parts water, total parts = 7
So in
A, milk is \(\frac{2parts}{7parts}=\frac{2}{7}\)
In the second mixture, B, milk is what fraction?
\(\frac{M}{W}=\frac{5}{4}\)
B, concentration of milk \(=\frac{5}{4+5}=\frac{5}{9}\)
Resultant mixture, desired concentration =
40% milk \(=\frac{40}{100}=\frac{2}{5}\)
The volume of B is 90 gallons.
The volume of the resultant mixture is (A + B).
What is the volume of A?
2) Use weighted average to find the unknown volume of A (steps can be combined)
\(\frac{2}{7}A + \frac{5}{9}(90)=\frac{2}{5}(A+90)\)
\(\frac{2}{7}A + 50=\frac{2}{5}A +
\frac{2}{5}(90)\)
\(\frac{2}{7}A + 50=\frac{2}{5}A + 36\)
\(14=\frac{2}{5}A-\frac{2}{7}A\)
\(14 = \frac{4}{35}A\)
\(A = (14*\frac{35}{4})=122.5\) gallons of A
Answer B
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