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. Formula:

\((Concentration_{A})(Vol_{A}) + (Concentration_{B})(Vol_{B})\)

\(=(Concentration_{A+B})(Vol_{A+B})\)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 A, \(\frac{M}{W}=\frac{2}{5}\)

2 parts milk, 5 parts water, total parts = 7

So milk is \(\frac{2parts}{7parts}=\frac{2}{7}\)

B: \(\frac{M}{W}=\frac{5}{4}.\) Milk \(=\frac{5}{4+5}=\frac{5}{9}\)

Resultant mixture: 40% milk \(=\frac{2}{5}\)

We have 90 gallons for the volume of B. How much A?

2) Weighted average to find 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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