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$$68 \text{ kg}$$ 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

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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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Chapter 7: Ratio and Proportion - EXERCISE 7A [Page 94]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 7 Ratio and Proportion
EXERCISE 7A | Q 31. | Page 94
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