Discussion
A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
Answer: B
The target is 12 L with water : milk \( = 3 : 5 \), so water is \( \tfrac{3}{8} \times 12 = 4.5 \) L.
Let \( x \) litres come from the first can (25% water) and \( 12 - x \) from the second (50% water):
\[ 0.25x + 0.5(12 - x) = 4.5 \]
\[ 0.25x + 6 - 0.5x = 4.5 \quad\Longrightarrow\quad -0.25x = -1.5 \]
\[ x = 6 \quad\Longrightarrow\quad 6 \text{ litres from each can} \]
The target of 37.5% water is exactly midway between 25% and 50%, which is why the two cans contribute equally.
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