Alligation - Aptitude MCQ Questions and Solutions with Explanations

Now, given ratio of Ram and Shayam's weight = 4 : 5
Hence, $$\frac{{x - 15}}{{15 - 10}} = \frac{4}{5}$$
Or, x = 19%

Cost of (18g of 1^st metal + 18g of 2^nd metal) = Rs. 165.60
Cost of (1g of 1^st metal + 1g metal of 2^nd metal) = Rs. 9.20
Hence cost of 1g of 2^nd metal,
= 9.20 - 6.70
= Rs. 2.5
Mean price = Rs. $$\frac{{87}}{{18}}$$

Now, $$\frac{{{\text{quantity}}\,{\text{of}}\,{1^{{\text{st}}}}\,{\text{metal}}}}{{{\text{quantity}}\,{\text{of}}\,{2^{{\text{nd}}}}\,{\text{metal}}}}$$     $$ = $$ $$\frac{{14}}{6}:\frac{{56}}{{30}} = 5:4$$
Quantity of 2^nd metal = $$\frac{{18 \times 4}}{9}$$  = 8g

Let the original volume be x.
then, quantity of milk and water,
= $$\frac{{3x}}{5}$$ and $$\frac{{2x}}{5}$$ respectively.
After adding water to it, the volume becomes 150%, the quantity of milk and water,
= $$\frac{{3x}}{5}$$ and $$\frac{{9x}}{10}$$
$$\frac{{\frac{{3x}}{5} - 12}}{{\frac{{9x}}{{10}} + 12}} = \frac{3}{7}$$
14(3x - 60) = 3(9x + 120)
Or, x = 80 L

We use formula for it where
n = number of time process of out going repeated.
y = Out going milk in process.
x = Volume of original solution.
F = Final milk left in mixture.
$$F = x{\left( {1 - \frac{y}{x}} \right)^n}$$
Final quantity of water,
$$F = x{\left( {1 - \frac{3}{x}} \right)^3}$$
$$\frac{F}{x} = {\left( {\frac{{x - 3}}{x}} \right)^3}$$
$$\frac{{27}}{{1728}} = {\left( {\frac{{x - 3}}{x}} \right)^3}$$
$${\left( {\frac{3}{{12}}} \right)^3} = {\left( {\frac{{x - 3}}{x}} \right)^3}$$
$$\frac{3}{{12}} = \frac{{x - 3}}{x}$$
$$x = 4 \,\text{L}$$

Let the rate of second quality be Rs. x per Kg.
C.P of 1 Kg wheat of 1st kind = 930 p
C.P of 1 Kg wheat of 2nd kind = 100x p
Mean price = 1000 p
By rule of alligation we have required ratio 8 : 7
930 x / (Mean Price) (1000) / (x-10) : 0.7
So we get required ratio, (x-10) : 0.7 :: 8 : 7
⇒ x = 10.80 per Kg

Heads Count = 200
Legs count = 580
Average Legs count for per head = $$\frac{{580}}{{200}}$$ = $$\frac{{29}}{{10}}$$

Rabbits : Pigeons = $$\frac{{9}}{{10}}$$ : $$\frac{{11}}{{10}}$$ = 9 : 11
Number of Pigeons = $$200 \times \frac{{11}}{{20}}$$  = 110

Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = $$\left( {3 - \frac{{3x}}{8} + x} \right)$$   litres
Quantity of syrup in new mixture = $$\left( {5 - \frac{{5x}}{8}} \right)$$   litres
$$\eqalign{ & \therefore {3 - \frac{{3x}}{8} + x} = {5 - \frac{{5x}}{8}} \cr & \Rightarrow 5x + 24 = 40 - 5x \cr & \Rightarrow 10x = 16 \cr & \Rightarrow x = \frac{8}{5} \cr} $$
So, part of the mixture replaced
$$\eqalign{ & = {\frac{8}{5} \times \frac{1}{8}} \cr & = \frac{1}{5} \cr} $$

Since first and second varieties are mixed in equal proportions.
So, their average price
$$\eqalign{ & = {\text{Rs}}{\text{.}}\,\,\left( {\frac{{126 + 135}}{2}} \right) \cr & = {\text{Rs}}{\text{.}}\,\,130.50 \cr} $$
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.
By the rule of alligation, we have:

$$\eqalign{ & \therefore \frac{{x - 153}}{{22.50}} = 1 \cr & \Rightarrow x - 153 = 22.50 \cr & \Rightarrow x = 175.50 \cr} $$

$$\eqalign{ & {\text{Suppose}}\,{\text{the}}\,{\text{can}}\,{\text{initially}}\,{\text{contains}}\, \cr & 7x\,{\text{and}}\,5x\,{\text{of}}\,{\text{mixtures}}\,{\text{A}}\,{\text{and}}\,{\text{B}}\,{\text{respectively}}. \cr & {\text{Quantity}}\,{\text{of}}\,{\text{A}}\,{\text{in}}\,{\text{mixture}}\,{\text{left}} \cr & = \left( {7x - \frac{7}{{12}} \times 9} \right)\,{\text{litres}} \cr & = \left( {7x - \frac{{21}}{4}} \right)\,{\text{litres}} \cr & {\text{Quantity}}\,{\text{of}}\,{\text{B}}\,{\text{in}}\,{\text{mixture}}\,{\text{left}} \cr & = \left( {5x - \frac{5}{{12}} \times 9} \right)\,{\text{litres}} \cr & = \left( {5x - \frac{{15}}{4}} \right)\,{\text{litres}} \cr & \therefore \frac{{\left( {7x - \frac{{21}}{4}} \right)}}{{\left( {5x - \frac{{15}}{4}} \right) + 9}} = \frac{7}{9} \cr & \Rightarrow \frac{{28x - 21}}{{20x + 21}} = \frac{7}{9} \cr & \Rightarrow 252x - 189 = 140x + 147 \cr & \Rightarrow 112x = 336 \cr & \Rightarrow x = 3 \cr} $$
So, the can contained 21 litres of A

Let the cost of 1 litre milk be Rs. 1
Milk in 1 litre mixture in 1^st can = $$\frac{{3}}{{4}}$$ litre, C.P. of 1 litre mixture in 1^st can Rs. $$\frac{{3}}{{4}}$$
Milk in 1 litre mixture in 2^nd can = $$\frac{{1}}{{2}}$$ litre, C.P. of 1 litre mixture in 2^nd can Rs. $$\frac{{1}}{{2}}$$
Milk in 1 litre of final mixture = $$\frac{{5}}{{8}}$$ litre, Mean price = Rs. $$\frac{{5}}{{8}}$$
By the rule of alligation, we have:

∴ Ratio of two mixtures = $$\frac{{1}}{{8}}$$ : $$\frac{{1}}{{8}}$$ = 1 : 1
So, quantity of mixture taken from each can = $$\left( {\frac{1}{2} \times 12} \right)$$  = 6 litres