Alligation - Aptitude MCQ Questions and Solutions with Explanations

According to the question,
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Milk : Water}} \cr & {\text{Initial Ratio}}\,\,\,\,\,{{\text{7}}_{ \times 2}}:{3_{ \times 2}} \cr & {\text{Final Ratio}}\,\,\,\,\,\,\,{{\text{2}}_{ \times 7}}:{1_{ \times 7}} \cr} $$
∴ Remember water is added not milk, so make milk equal
\[\begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}\,\,\,\,{\text{:}}\,\,\,\,{\text{W}} \hfill \\ \left. \begin{gathered} {\text{Initial}}\,\,\,\,14\,\,\,\,:\,\,\,\,6 \hfill \\ {\text{Final}}\,\,\,\,\,\,14\,\,\,\,:\,\,\,\,7 \hfill \\ \end{gathered} \right) \hfill \\ \end{gathered} \] $$\eqalign{ & \cr & = \,20\,{\text{unit}} \cr & 1 \cr} $$
$$\eqalign{ & {\text{20 units = 80 litres}} \cr & {\text{1 unit = 4 litres}} \cr & {\text{Water added = 4 litres}} \cr} $$

In 300 kg of solution,
Percentage of sugar = 40%
∴ Sugar = $$\frac{{300 \times 40}}{{100}}$$   = 120 kg
Let x kg of sugar be added.
Now, according to the question
$$\eqalign{ & \frac{{120 + x}}{{300 + x}} = \frac{1}{2} \cr & \Rightarrow 240 + 2x = 300 + x \cr & \Rightarrow 2x - x = 300 - 240 \cr & \Rightarrow x = 60\,{\text{kg}} \cr} $$

According to the question,
$${\text{Milk}}\,\,\,\,:\,\,\,\,{\text{Water}}$$
\[\left. \begin{gathered} \,\,\,\,\,7\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,5 \hfill \\ \,\,\,\,\,7\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,8 \hfill \\ \end{gathered} \right)\,3\,{\text{unit}}\]
∴ Remember water is added and not milk, so make milk equal but here milk is already equal
$$\eqalign{ & {\text{3 units = 15 litres}} \cr & {\text{1 unit = 5 litres}} \cr & {\text{8 units = 40 litres}} \cr} $$
Total quantity of water in the new mixture = 40 litres

According to the question,
By using alligation method :

Ratio of quantity $$ \to \boxed{\,\,1\,\,\,\,\,\,:\,\,\,\,\,\,3\,\,}$$

According to the question,
Average price of mixed tea
$$\eqalign{ & {\text{ = }}\frac{{280 \times 7 + 240 \times 9}}{{16}} \cr & {\text{ = }}\frac{{1960 + 2160}}{{16}} \cr & {\text{ = }}\frac{{4120}}{{16}} \cr & {\text{ = Rs}}{\text{. 257}}{\text{.50}} \cr} $$

Let the price of water = Rs. 0
According to the question,
Cost price of pure milk = Rs. 1.80
Cost price of the mixture = Rs. 0.90
Now using Alligation method.

Therefore the amount of water is equal to the amount of Milk in mixture. i.e. 16 litter Milk in Given Mixture

According to the question,
Cost price of the mixture
= 30 × 9.50 + 40 × 8.50
= 285 + 340
= Rs. 625
Selling price of the mixture
= 8.90 × 70
= Rs. 623
Loss = Cost price - Selling price
Loss = 625 - 623 = Rs. 2

Capacities of vessels
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3:2:1$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}\,{\text{:}}\,{\text{W}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{T}}\,\,{\text{Mixture}} \cr & {\text{V - 1}} \to {{\text{(}}5\,\,\,:2\,\,\, = \,\,\,\,\,7{\text{)}}_{ \times 5}} \cr & {\text{V - 2}} \to {{\text{(4}}\,\,\,:1\,\,\, = \,\,\,\,\,5{\text{)}}_{ \times 7}} \cr & {\text{V - 3}} \to {{\text{(4}}\,\,\,:1\,\,\, = \,\,\,\,\,5{\text{)}}_{ \times 7}} \cr} $$
Equate the mixture
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}\,{\text{:}}\,{\text{W}}\,\,\,\,\,\,\,\,\,\,{\text{T}}\,\,{\text{Mixture}} \cr & \left( {{\text{V - 1}}} \right) \to 25:10\,\,\,\,\,\,\, = 35 \cr & \left( {{\text{V - 2}}} \right) \to 28:7\,\,\,\,\,\,\,\, = 35 \cr & \left( {{\text{V - 3}}} \right) \to 28:7\,\,\,\,\,\,\,\, = 35 \cr} $$
$$\eqalign{ & {\text{Capacities}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{M}}:{\text{W}}\,\,\, = {\text{Total Mix}}{\text{.}} \cr & \left( {{\text{V}} - 1} \right) \times 3\, \to 75:30\,\,\, = 105 \cr & \left( {{\text{V}} - 2} \right) \times 2 \to 56:14\,\,\, = 70 \cr & \left( {{\text{V}} - 1} \right) \times 1\, \to 28:7\,\,\,\, = 35 \cr} $$
Water taken out
$$ \Rightarrow \frac{1}{3}{\text{ of water in (V - 1)}}$$     $$ + \frac{1}{2}{\text{ of water in (V - 2)}}$$     $$ + \frac{1}{7}{\text{ of water in (V - 3)}}$$
$$\eqalign{ & \Rightarrow \frac{1}{3} \times 30 + \frac{1}{2} \times 14 + \frac{1}{7} \times 7 \cr & \Rightarrow 10 + 7 + 1 \cr & \Rightarrow 18 \cr} $$
Similarly mixture will be
$$ \Rightarrow \frac{1}{3} \times 105 + \frac{1}{2} \times 70 + \frac{1}{7}$$     × 35
⇒ 75
$$\eqalign{ & \therefore \% {\text{ of water = }}\frac{{18}}{{75}} \times 100 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 24\% \cr} $$

Let them be mixed in the ratio x : y
Then, in 1^st alloy, Zinc = $$\frac{x}{3}$$ and Copper = $$\frac{{2x}}{3}$$
2^nd alloy, Zinc = $$\frac{{2y}}{5}$$ and Copper = $$\frac{{3y}}{5}$$
Now, we have
$$\eqalign{ & \frac{x}{3} + \frac{{2y}}{5}:\frac{{2x}}{3} + \frac{{3y}}{5} = 5:8 \cr & {\text{or,}}\frac{{5x + 6y}}{{10x + 9y}} = \frac{5}{8} \cr & {\text{or,}}40x + 48y = 50x + 45y \cr & {\text{or,}}10x = 3y \cr & \therefore \frac{x}{y} = \frac{3}{{10}} \cr} $$
Thus, the required ratio = 3 : 10

Alligation Method :
You must know that we can apply this rule over the fractional value of either zinc or copper.
Let us consider the fractional value of zinc.

Therefore, they should be mixed in the ratio
$$\eqalign{ & = \frac{1}{{65}}:\frac{2}{{39}} \cr & {\text{or,}}\frac{1}{{65}} \times \frac{{39}}{2} \cr & = \frac{3}{{10}} \cr & {\text{or,}}\,\,3:10 \cr} $$

Let the quantity of alcohol and water be 4x litres and 3x litres respectively
$$\eqalign{ & \frac{{4x}}{{\left( {3x + 5} \right)}} = \frac{4}{5} \cr & \Rightarrow 20x = 4\left( {3x + 5} \right) \cr & \Rightarrow 8x = 20 \cr & \Rightarrow x = 2.5 \cr} $$
Quantity of alcohol
= (4 × 2.5) litres
= 10 litres.