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In a chemistry lab, there is some quantity of $50\%$ acid solution and some quantity of $25\%$ acid solution. How much of each should be mixed to make $10$ litres of $40\%$ acid solution ?
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Let quantity of $50\%$ and of $25\%$ acid solution be '$x$'L and '$y$'L respectively.
Therefore, $x + y = 10$ ----- (i)
and $\frac{50}{100} \times x + \frac{25}{100} \times y = \frac{40}{100} \times 10$ or $2x + y = 16$ ----- (ii)
Solving (i) and (ii) to get $x = 6, y = 4$
Hence, $6$L of $50\%$ and $4$L of $25\%$ acid solution are mixed.
Therefore, $x + y = 10$ ----- (i)
and $\frac{50}{100} \times x + \frac{25}{100} \times y = \frac{40}{100} \times 10$ or $2x + y = 16$ ----- (ii)
Solving (i) and (ii) to get $x = 6, y = 4$
Hence, $6$L of $50\%$ and $4$L of $25\%$ acid solution are mixed.