In what ratio must a mixture of $40\%$ acid be mixed with that of $60\%$ acid to get a mixture of $49\%$ acid?

Using the method of alligation:Concentration of first mixture ($C_1$) = $40\%$ Concentration of second mixture ($C_2$) = $60\%$ Mean concentration required ($M$) = $49\%$ $
$\text{Ratio}=\frac{C_2-M}{M-C_1}=\frac{60-49}{49-40}=\frac{11}{9}$$Thus,therequiredratiois$ 11:9$.