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Question
$$68 \text{ kg}$$ of a mixture contains milk and water in the ratio $$27 : 7$$. How much more water is to be added to this mixture to get a new mixture containing milk and water in the ratio $$3 : 1$$?
Sum
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Solution
Total quantity of mixture = $$68\text{ kg}$$.
Sum of ratio terms = $$27 + 7 = 34$$.
Quantity of milk = $$\frac{27}{34} \times 68 = 54\text{ kg}$$.
Quantity of water = $$\frac{7}{34} \times 68 = 14\text{ kg}$$.
Let $$x\text{ kg}$$ of water be added to the mixture.
Then, the new ratio of milk to water is: $$\frac{54}{14 + x} = \frac{3}{1}$$
Cross-multiply: $$3(14 + x) = 54$$
$$42 + 3x = 54$$
$$3x = 12$$
$$\implies x = 4$$
Therefore, $$4\text{ kg}$$ more water must be added.
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