A Container contains 192 litres of Milk. A seller draws out x% of Milk and replaced it with the same quantity of water. He repeated the same process for 3 times. And thus Milk content in the mixture is only 81 litres. Then how much percent he withdraw every time?

10%

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15%

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18%

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20%

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25%

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Solution

The correct option is E 25%
Using the formula, 192 ${(1 - (\frac{x}{100}))}^{3} = 81 \Rightarrow {(1 - (\frac{x}{100}))}^{3} = \frac{81}{192} {(1 - \frac{x}{100})}^{3} = \frac{3^{4}}{4^{3} \times 3} = {(\frac{3}{4})}^{3} \Rightarrow \frac{x}{100} = \frac{1}{4} X = 25$

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A Container contains 192 litres of Milk. A seller draws out x% of Milk and replaced it with the same quantity of water. He repeated the same process for 3 times. And thus Milk content in the mixture is only 81 litres. Then how much percent he withdraw every time?

The correct option is E 25%Using the formula, 192 (1−(x100))3=81⇒(1−(x100))3=81192(1−x100)3=3443×3=(34)3⇒x100=14X=25

The correct option is E 25%
Using the formula, 192 ${(1 - (\frac{x}{100}))}^{3} = 81 \Rightarrow {(1 - (\frac{x}{100}))}^{3} = \frac{81}{192} {(1 - \frac{x}{100})}^{3} = \frac{3^{4}}{4^{3} \times 3} = {(\frac{3}{4})}^{3} \Rightarrow \frac{x}{100} = \frac{1}{4} X = 25$