shivanshchauhan94 wrote:A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2. Two ltrs of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.
A) 20%
B) 30.72%
C) 40.23%
D) 52.15%
E) 63.2%
[spoiler](B)[/spoiler]
We can represent the initial amount of petrol and kerosene in the mixture by 3x and 2x, for some positive number x. Since 3x + 2x = 10, we find x = 2 and thus, there are 3 * 2 = 6 liters of petrol and 2 * 2 = 4 liters of kerosene in the mixture initially.
Since 2 liters of mixture is replaced by 2 liters of kerosene, the total mixture is always 10 liters.
Notice that the number of liters of kerosene decreases by the amount of kerosene in the 2 liters of mixture and increases by 2 liters after each iteration, whereas the amount of petrol decreases by the amount of petrol in 2 liters of mixture after each iteration and never increases; therefore, it will be easier to calculate the amount of petrol in the mixture after each iteration.
After the first iteration, 2 * (6/10) = 1.2 liters of petrol is removed and thus, there are 6 - 1.2 = 4.8 liters of petrol remaining in the mixture.
After the second iteration, 2 * (4.8/10) = 0.96 liters of petrol is removed and thus, there are 4.8 - 0.96 = 3.84 liters of petrol remaining in the mixture.
After the third iteration, 2 * (3.84/10) = 0.768 liters of petrol is removed and thus, there are 3.84 - 0.769 = 3.072 liters of petrol remaining in the mixture.
Thus, after the third iteration, the mixture contains (3.072/10) * 100 = 30.72% petrol.
Answer: B
