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$5$ liters of wine are drawn from a case  containing $50$ liters of wine. It is replaced by a water. The process is repeated in $3$ times.  What is the final quantity of wine left in the case?

  1. $29.16$ liters
  2. $36.45$ liters
  3. $16.29$ liters
  4. $45.36$ liters
in Quantitative Aptitude recategorized by
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1 Answer

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$\begin{array}{ll} \text{Final amount of ingredient that is not replaced} & = \text{Initial amount} \times \bigg( 1 - \frac{\text{replace}}{\text{total}} \bigg) ^n \\ & = 50 \times \bigg( 1 - \frac{5}{50} \bigg) ^3 \\ & = 50 \times \frac{9}{10} \frac{9}{10} \frac{9}{10} \\ & = 36.45 \text{ltrs} \end{array}$
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