SSC CPO SI and ASI Model Paper 9

(4) Let the quantity of the mixturebe 16 litres of liquid.
Let x litre of this liquid be replacedwith water.
Quantity of water in new mixture$=\left(5\mathit{–}\frac{5\mathit{X}}{16}+\mathit{x}\right)$ litres.
Quantity of milk in new mixture$=\left(11\mathit{–}\frac{11\mathit{x}}{16}\right)$ litres.
$\therefore \left(5\mathit{–}\frac{5\mathit{x}}{16}+\mathit{x}\right)=\left(11\mathit{–}\frac{11\mathit{x}}{16}\right)$
$\Rightarrow 80+11\mathit{x}=176\mathit{–}11\mathit{x}$
$\Rightarrow 22\mathit{x}=96$
⇒ $\mathit{x}=\frac{48}{11}$
∴Part of mixture that must bereplaced $=\frac{48}{11}÷16=\frac{3}{11}$