Alligation - Aptitude MCQ Questions and Solutions with Explanations

Total milk = 60 litres
Drawn off = 12 litres
$$\frac{{{\text{Final quantity}}}}{{{\text{Initial quantity}}}}$$   $$ = {\left( {1 - \frac{x}{c}} \right)^t}$$
X = Replaced quantity
C = Capacity
T = Number of process
$$\eqalign{ & \frac{{{\text{Final quantity}}}}{{{\text{Initial quantity}}}} = {\left( {1 - \frac{{12}}{{60}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( {\frac{4}{5}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{16}}{{25}} \cr} $$
Ratio of milk and water in the resultant mixture :
= 16 : 9

Water content in 40 litres of mixture.
$$\eqalign{ & = 40 \times \frac{{10}}{{100}} \cr & = 4{\text{ litres}} \cr} $$
∴ Milk in the mixture.
$$\eqalign{ & {\text{ = 40}} - 4 \cr & = 36{\text{ litres}} \cr} $$
Let x litres of water is mixed
$$\eqalign{ & \Rightarrow \frac{{4 + x}}{{40 + x}} = \frac{{20}}{{100}} \cr & \Rightarrow x = 5 {\text{ litres}} \cr} $$

According to the question,
Mixture of copper and aluminium = 2000 gms
30% is copper means
= $$\frac{30}{100}$$ × 2000
= 600 gms copper
$$\eqalign{ & \Rightarrow \frac{{600}}{{1400 + x}} = \frac{{20\% }}{{80\% }} \cr & \Rightarrow 1400 + x = 2400 \cr} $$
⇒ x = 1000 gms

Percentage of sugar in 4 litres mixture.
$$\eqalign{ & = \frac{9}{{5 \times 4}} \times 100 \cr & = 45\% \cr} $$

According to the question,
Selling price of the mixture = Rs. 320, Gain = 20%
∴ Cost price of the mixture
$$\eqalign{ & = 320 \times \frac{{100}}{{120}} \cr & = {\text{Rs}}{\text{. }}\frac{{800}}{3} \cr} $$
Now using alligation method

$$\eqalign{ & {\text{Ratio of quantity }} \cr & \to 40:260 \cr & \,\,\,\,\,\,\,\,\,\,\,2:13 \cr} $$

Let the quantity of liquid drawn out = x
$$\eqalign{ & \Rightarrow \frac{{3 - \frac{3}{4}x}}{{1 - \frac{1}{4}x + x}} = \frac{1}{1} \cr & \Rightarrow 12 - 3x = 4 - x + 4x \cr & \Rightarrow 8 = 6x \cr & \Rightarrow x = \frac{4}{3} \cr} $$
Hence, required part of quantity
$$\eqalign{ & {\text{ = }}\frac{{\frac{4}{3}}}{4} \cr & {\text{ = }}\frac{1}{3} \cr} $$

According to the question,
\[\left. \begin{gathered} {\text{Alloy}}\,{\text{A}} \to 5{ \times _2}:3{ \times _2} = 8{ \times _2} \hfill \\ {\text{Alloy}}\,{\text{B}} \to 5\,\,\,\,\,\,\,\,:11\,\,\,\,\, = 16 \hfill \\ \end{gathered} \right]\]     Equal quantity are mixed

$$\eqalign{ & {\text{Alloy A }} \to 10\,\,\,:\,\,\,6\,\,\,\, = 16 \cr & {\text{Alloy B }} \to {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,5\,\,\,\,\,:\,\,\,11\, = 16 \cr & {\text{Alloy C }} \to {\bf{15}}\,\,\,:\,\,{\bf{17}} \cr} $$

Let first blend is 2 kg and second blend is 3 kg.
$$\eqalign{ & {\text{Total cost price }}{\text{ }} \cr & = \left( {35 \times 2} \right) + \left( {40 \times 3} \right) \cr & = 70 + 120 \cr & = {\text{Rs}}{\text{. 190}} \cr & {\text{Total selling price }} \cr & {\text{ = }}\left( {{\text{ }}1 \times 46} \right) + \left( {4 \times 55} \right) \cr & = 266\left[ {\frac{1}{5}{\text{ of 5kg = 1kg}}} \right] \cr & \therefore {\text{Profit percent }} \cr & \Rightarrow \frac{{{\text{Total profit}}}}{{{\text{Total cost price}}}} \times 100 \cr & \Rightarrow \frac{{\left( {266 - 190} \right)}}{{190}} \times 100 \cr & \Rightarrow \frac{{76}}{{190}} \times 100 \cr & \Rightarrow 40\% \cr} $$

$$\eqalign{ & {\text{Number of girls}} \cr & = \frac{1}{2} \times 12500 \cr & = 6250 \cr} $$

Quantity of tin in 60 kg of A
$$\eqalign{ & = \left( {60 \times \frac{2}{5}} \right)\,{\text{kg}} \cr & = 24\,{\text{kg}} \cr} $$
Quantity of tin in 100 kg of B
$$\eqalign{ & = \left( {100 \times \frac{1}{5}} \right)\,{\text{kg}} \cr & = 20\,{\text{kg}} \cr} $$
∴ Quantity of tin in the new alloy
= (24 + 20) kg
= 44 kg