A sample of x litres from a container having a 90 litre mixture of milk and water containing milk and water in the ratio of 4 : 5 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

12 L

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15 L

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18 L

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9 L

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Solution

The correct option is D 9 L
Amount of milk and water initially 40:50 After removing x litres of solution $40 - (\frac{4}{9} \times x) : 50 - (\frac{5}{9} \times x)$ After adding x litres of milk back to it $40 + (\frac{5}{9} \times x) : 50 - (\frac{5}{9} \times x) 40 + (\frac{5}{9} \times x) = 50 - (\frac{5}{9} \times x) \Rightarrow x = 9 L$

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A sample of x litres from a container having a 90 litre mixture of milk and water containing milk and water in the ratio of 4 : 5 is replaced with pure milk so that the container will have milk and water in equal proportions. What is the value of x?

The correct option is D 9 LAmount of milk and water initially 40:50 After removing x litres of solution 40−(49×x):50−(59×x) After adding x litres of milk back to it 40+(59×x):50−(59×x)40+(59×x)=50−(59×x)⇒x=9L