From a container having a 70 litres mixture of milk and water containing milk and water in the ratio of 3:4, a sample of x litres is replaced with pure milk so that the container will have milk and water in the ratio 5:2. What is the value of x?

16 litres

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35 litres

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30 litres

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None of these

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Solution

The correct option is B 35 litres
Quantity of milk in the solution initially $= \frac{3}{7} \times 70 = 30$ litres
Quantity of milk after adding pure milk $= \frac{5}{7} \times 70 = 50$ litres
Quantity of milk removed $= \frac{3}{7} \times x$
Quantity of milk added = x
Therefore, $30 - \frac{3}{7} x + x = 50 \Rightarrow \frac{3}{7} \times x = 20 \Rightarrow x = 35$ litres

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From a container having a 70 litres mixture of milk and water containing milk and water in the ratio of 3:4, a sample of x litres is replaced with pure milk so that the container will have milk and water in the ratio 5:2. What is the value of x?

The correct option is B 35 litresQuantity of milk in the solution initially =37×70=30litres Quantity of milk after adding pure milk =57×70=50 litres Quantity of milk removed =37×x Quantity of milk added = x Therefore, 30−37x+x=50⇒37×x=20⇒x=35 litres