Question

A mixture containing 43% of acid is mixed with another mixture containing 82% acid in the ratio 10 : 9. Find the percentage of acid in the mixture formed.

• A. \(60.45 \% \)
• B. \(61.25 \% \)
• C. \(64.25 \% \)
• D. \(61.47 \% \)

Hint:

You can also use Alligation: The ratio of weights is inversely proportional to the deviation of their concentrations from the mean.

Answer 1

The Correct Option is D

Concept: The weighted average formula is used for mixtures: \[ \text{Resultant Percentage} = \frac{n_1p_1 + n_2p_2}{n_1 + n_2} \] Plug the ratios and percentages into the formula.
\(n_1 : n_2 = 10 : 9\)
\(p_1 = 43\%\), \(p_2 = 82\%\) \[ \text{Acid \%} = \frac{(10 \times 43) + (9 \times 82)}{10 + 9} \] \[ \text{Acid \%} = \frac{430 + 738}{19} = \frac{1168}{19} \approx 61.47\% \]