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]
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A vessel ( capacity 10 ltrs) is filled entirely with petrol
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 shivanshchauhan94
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$$Initial\ volume\ of\ petrol=\frac{3}{5}\cdot10=\frac{30}{5}=6$$
Let the volume of petrol after 3 replacement = x
$$x=6\left(1\frac{2}{10}\right)^{^3}$$
$$x=6\left(\frac{102}{10}\right)^{^{^3}}$$
$$x=6\left(\frac{8}{10}\right)^{^{^3}}$$
$$x=6\left(\frac{4}{5}\right)^{^{^3}}$$
$$x=6\cdot\frac{64}{125}=6\cdot0.512=3.072$$
$$\%\ of\ petrol\ in\ the\ final\ solution=\frac{3.072}{10}\cdot100\ =30.72\%$$
Hence, option B is the correct answer.
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Let the volume of petrol after 3 replacement = x
$$x=6\left(1\frac{2}{10}\right)^{^3}$$
$$x=6\left(\frac{102}{10}\right)^{^{^3}}$$
$$x=6\left(\frac{8}{10}\right)^{^{^3}}$$
$$x=6\left(\frac{4}{5}\right)^{^{^3}}$$
$$x=6\cdot\frac{64}{125}=6\cdot0.512=3.072$$
$$\%\ of\ petrol\ in\ the\ final\ solution=\frac{3.072}{10}\cdot100\ =30.72\%$$
Hence, option B is the correct answer.
Thanks
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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.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]
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
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Since the question stem asks for a PERCENTAGE, the original amount of petrol is irrelevant.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%
In the original solution, petrol:kerosene = 3:2, implying that petrol constitutes 3 of every 5 liters.
Thus, petrol = 3/5.
Each time 2 liters are removed from the 10liter vessel, 8 liters remain.
Since 8/10 = 4/5, left in the vessel after each removal process will be 4/5 of the amount of petrol.
Since the 4/5 process is performed THREE TIMES upon the original 3/5 portion of petrol, we get:
3/5 * 4/5 * 4/5 * 4/5 = 192/625 â‰ˆ 200/600 = 1/3 â‰ˆ 30%
The correct answer is B.
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As a tutor, I don't simply teach you how I would approach problems.
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