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A vessel contains 60 litre of milk. 12 litres of milk taken out from it and replaced with water. Then again from mixture, 12 litres is taken out and replaced with water. The ratio of milk and water in the resultant mixture is ?
Answer & Solution
Correct Answer:
Option
A
Total milk = 60 litres
Drawn off = 12 litres
$$\frac{{{\text{Final quantity}}}}{{{\text{Initial quantity}}}}$$ $$ = {\left( {1 - \frac{x}{c}} \right)^t}$$
X = Replaced quantity
C = Capacity
T = Number of process
$$\eqalign{ & \frac{{{\text{Final quantity}}}}{{{\text{Initial quantity}}}} = {\left( {1 - \frac{{12}}{{60}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{4}{5}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{{25}} \cr} $$
Ratio of milk and water in the resultant mixture :
= 16 : 9
Drawn off = 12 litres
$$\frac{{{\text{Final quantity}}}}{{{\text{Initial quantity}}}}$$ $$ = {\left( {1 - \frac{x}{c}} \right)^t}$$
X = Replaced quantity
C = Capacity
T = Number of process
$$\eqalign{ & \frac{{{\text{Final quantity}}}}{{{\text{Initial quantity}}}} = {\left( {1 - \frac{{12}}{{60}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{4}{5}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{{25}} \cr} $$
Ratio of milk and water in the resultant mixture :
= 16 : 9
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LoginVolume of milk in the final stage=(4/5)^2=16/25
Volume of water in the final stage=9/25
Now, milk:water=16/25:9/25
=16:9