| Several litres of acid were drawn off a 54 L vessel full of acid and an equal amount of water added. |
| Again the same volume of the mixture was drawn off and replaced by water. As a result, the vessel contained 24 L of pure acid. |
| How much of the acid was drawn off initially? |
A) 12 L
B) 16 L
C) 18 L
D) 24 L
Correct Answer: C
Solution :
Let x L of several litres of acid drawn off initially. \[\therefore \]Remaining acid in the vessel \[=(54-x)L\] and quantity of water in the vessel \[=x\,L\] Now, x L of mixture is drawn off. \[\therefore \]Quantity of acid drawn off \[=\left( \frac{54-x}{54}\times x \right)L\] and quantity of water drawn off\[=\frac{{{x}^{2}}}{54}L\] Now, the quantity of acid \[=\left[ 54-x-\left( \frac{54-x}{54} \right)x \right]L\] \[54-x-\frac{(54-x)}{54}x=24\] \[\therefore \] \[{{x}^{2}}-108+1620=0\] \[\therefore \] \[x=90,\,\,18\] Since, 90 > 54 there x = 90 is ruled out. Hence, x = 18 L
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