Average problem 5 by Dhaval Furia

Let $x$ litre be the volume of liquid $A$ in the mixture. Then the volume of liquid $B$ in the mixture is $0.5 - x$ litre. And we have

$\begin{aligned} 1000x + 800(0.5-x) & = 480 \\ 200x + 400 & = 480 \\ 200x & = 80 \\ \implies x & = \frac {80}{200} = 0.4 \end{aligned}$

Therefore the percentage of liquid $A$ in the mixture is $\dfrac {0.4}{0.5} \times 100 \% = \boxed{80} \%$ .

Let $V_1$ liters of first liquid be mixed with $V_2$ liters of the second. Then $\dfrac {V_1+0.8V_2}{V_1+V_2}=0.96$

$\implies V_1=4V_2\implies \dfrac {V_1}{V_1+V_2}=\dfrac {4}{5}=\boxed {80\%}$ .