Question : A vessel contains 20 L of acid. 4 L of acid is taken out of the vessel and replaced by the same quantity of water. Next, 4 L of the mixture is withdrawn and again the vessel is filled with the same quantity of water. The ratio of the quantity of acid left in the vessel with the quantity of acid initially in the vessel is:
Option 1: 4 : 5
Option 2: 4 : 25
Option 3: 16 : 25
Option 4: 1 : 5
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Correct Answer: 16 : 25
Solution :
Here,
Remaining acid = Initial quantity × $(1–\frac{\text{Quantity taken out}}{\text{Original quantity}})^{\text{number of times removed}}$
⇒ Remaining acid $=20×(1-\frac{4}{20})^2=20×\frac{4}{5}×\frac{4}{5}= 12.8$ L
So, the required ratio
= (12.8) : (20)
= 128 : 200
= 16 : 25
Hence, the correct answer is 16 : 25.
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