Container I contains a mixture of 2 liters of oil and 3 liters of wate

Container I contains a mixture of 2 liters of oil and 3 liters of water. Container II contains a mixture of 6 liters of oil and 7 liters of water. 50% of the mixture of container I and 25% of the mixture of container II are transferred to container III. What is the proportion of oil and water in container III ?

4 : 5
5 : 4
13 : 10
10 : 13
This question was previously asked in
UPSC CISF-AC-EXE – 2020
The correct proportion of oil and water in container III is 10 : 13.
– Container I has 2 L oil and 3 L water (total 5 L). 50% of this mixture is transferred, which is 2.5 L. This 2.5 L contains 50% of the oil from container I (0.5 * 2 L = 1 L oil) and 50% of the water from container I (0.5 * 3 L = 1.5 L water).
– Container II has 6 L oil and 7 L water (total 13 L). 25% of this mixture is transferred, which is 13/4 L = 3.25 L. This 3.25 L contains 25% of the oil from container II (0.25 * 6 L = 1.5 L oil) and 25% of the water from container II (0.25 * 7 L = 1.75 L water).
– Container III receives the transferred amounts. Total oil in container III = 1 L (from I) + 1.5 L (from II) = 2.5 L. Total water in container III = 1.5 L (from I) + 1.75 L (from II) = 3.25 L.
– The proportion of oil to water in container III is 2.5 L : 3.25 L.
– To simplify the ratio, we can multiply both parts by 100 to remove decimals: 250 : 325.
– Dividing both by their greatest common divisor, 25: 250/25 = 10 and 325/25 = 13.
– The ratio is 10 : 13.
When a portion of a mixture is taken out, the ratio of components in the taken portion is the same as the ratio in the original mixture. The amount of each component transferred is calculated based on this ratio and the total amount transferred.