A chemist has two acid solutions, one with 10% acid and the other with 30% acid. He wants to mix them to get 50 litres of a solution with 22% acid. The quantity of the 10% solution he must use is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 20 litres
Let x litres of the 10% solution and y litres of the 30% solution be used. The total volume and the total quantity of acid give
\(x + y = 50\)
\(0.10x + 0.30y = 0.22 \times 50 = 11\)
Multiplying the second equation by 10:
\(x + 3y = 110\)
In matrix form, with \(A = \begin{bmatrix} 1 & 1 \\ 1 & 3 \end{bmatrix}\) and \(|A| = 3 - 1 = 2\):
\( \displaystyle \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 3 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 50 \\ 110 \end{bmatrix} \)
\( \displaystyle = \frac{1}{2}\begin{bmatrix} 150 - 110 \\ -50 + 110 \end{bmatrix} = \begin{bmatrix} 20 \\ 30 \end{bmatrix} \)
So 20 litres of the 10% solution (and 30 litres of the 30% solution) are needed.
Check: \(2 + 9 = 11\) litres of acid in 50 litres, which is 22%.