A vessel contains a mixture of acid and water in the ratio 13 : 4. 15.5 litres of the mixture is taken out, and then 1.5 litres of pure water and 3 litres of acid are added to the vessel. If the resulting mixture contains 25% water, what was the initial quantity (in litres) of the mixture in the vessel?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 41
Mixture taken out has the same ratio 13 : 4 as the vessel, so water forms \(\frac{4}{17}\) of whatever is removed. The water left plus the water added must then be 25% of the final volume.
Let the initial mixture be \(17k\) litres: water \(=4k\)
Water removed \(\displaystyle =\frac{4}{17}\times15.5=\frac{62}{17}\)
Final water \(\displaystyle =4k-\frac{62}{17}+1.5\)
Final volume \(=17k-15.5+1.5+3=17k-11\)
\(\displaystyle 4k-\frac{62}{17}+\frac{3}{2}=\frac{17k-11}{4}\)
Multiplying by 68: \(272k-248+102=289k-187\), so \(17k=41\)
Check: final water \(=\frac{164}{17}-\frac{62}{17}+1.5=7.5\) litres out of \(41-11=30\) litres, which is 25%. The vessel initially held 41 litres.