A container has a 100-litre mixture of acid and water in the ratio 4:1. 20 litres is removed and replaced with water. Then 20 litres is again removed and replaced with pure acid. What is the final percentage of acid?
71.2%
Start: the 100 L mixture is acid : water \(= 4:1\), so acid \(= 80\) L and water \(= 20\) L.
Step 1 — remove 20 L of mixture. Since 20 L is \(\dfrac{20}{100} = \dfrac{1}{5}\) of the tank, every component drops to \(\dfrac{4}{5}\) of its value. Acid becomes \(80 \times \dfrac{4}{5} = 64\) L.
Adding 20 L of water does not change the acid amount, so acid \(= 64\) L (volume back to 100 L).
Step 2 — remove 20 L of the new mixture. Again each component scales by \(\dfrac{4}{5}\), so acid becomes \(64 \times \dfrac{4}{5} = 51.2\) L.
Now add 20 L of pure acid: acid \(= 51.2 + 20 = 71.2\) L, total volume \(= 100\) L.
Final percentage of acid \(= \dfrac{71.2}{100} \times 100 = 71.2\%\).
Hence, the final percentage of acid is 71.2%.
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Important Questions from Mixture and Alligation
My father is presently 25 years older than me. The sum of our ages 5 years ago was 39 years. Find my present age.
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If a + b + c = 0, then the value of (a² + b² + 2ab) is equal to:
