cfc198 wrote:
When 20% of a solution of milk and water is removed and replaced with water, the ratio of milk and water becomes 2 : 3. Find the ratio of milk and water in the original solution.
(1) 1 : 1
(2) 1 : 2
(3) 2 : 1
(4) 1 : 3
(5) 3 : 2
This is a global ratio problem, which we can solve by using the weighted arithmetic average formula.
First, focus on the ratio of the volume of milk to the total volume, \(M:Total\).
Let \(x\) be the partial ratio for the original solution in the above context. The partial ratio for the water added is obviously 0.
The weights used in the formula will be the volumes of the remaining original solution and the water added, respectively. In this case, we don't know the weights in natural units, but fortunately only the relative size of the weights matter in weighted averages. If 20% of the original solution is replaced with water, then 80% of the new solution will be from the original one.
\(\frac{80x+20\cdot 0}{100}=\frac{2}{2+3}\)
\(\frac{4}{5}x=\frac{2}{5}\)
\(x=\frac{1}{2} \implies M:Total=1:2 \implies M:W:Total=1:1:2\)
Answer: A
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