Alligation — Part 1 MCQs with Explanations

CP of first tea = Rs. 192 per kg. CP of Second tea = Rs. 150 per kg. Mixture is to be sold in Rs. 194.40 per kg, which has included 20% profit. So, SP of Mixture = Rs. 194.40 per kg. Let the CP of Mixture be Rs. X per kg. Therefore, X + 20% of X = SP $$\frac{{6{\text{X}}}}{5}$$ = 194.40 6X = 194.40 *5 X = Rs. 162 per kg. Let N kg of first tea and M kg of second tea to be added. Now, Using Alligation, We get, $$\frac{{\text{N}}}{{\text{M}}} = \frac{{12}}{{30}}$$ N : M = 2 : 5

Ratio of Zinc, Copper and Tin is given as, Z : C : T = 2 : 3 : 1. Now, let the first alloy be 12 kg (taken as 4 kg Zinc, 6 kg Copper and 2 Kg Lead). Weight of second alloy = 12 kg as, C : T : L = 5 : 4 : 3. (taken as 5 kg Copper, 4 kg Tin and 3 Kg Lead.) Alloys are mixed together to form third alloy. Then the ratio of content in it, Z : C : T : L = 4 : (6 + 5) : (2 + 4) : 3 Weight of third alloy = 12 + 12 = 24 Kg. So, weight of the Lead = $$\frac{{3}}{{24}}$$ = $$\frac{{1}}{{8}}$$ kg.

Quantity of milk in 729 litre of mixture = 7 × $$\frac{{729}}{{9}}$$ = 567 litre Quantity of water = 729 - 567 = 162 litres Let x litre of water be added to become ratio 7 : 3 Or, $$\frac{{7}}{{3}}$$ = $$\frac{{567}}{{162 + {\text{x}}}}$$ Or, 162 × 7 + 7x = 567 × 3 Or, 7x = 1701 - 1134 = 567 Or, x = $$\frac{{567}}{{7}}$$ = 81 litres water is to be added.

1 : 3 is the correct answer to “In what ratio must a mixture of 30% alcohol strength be mixed with that of 50% alcohol strength so as to get a mixture of 45% alcohol strength”. This question is standard in competitive examination syllabi for Alligation.

1 st Method (Method of Alligation): Let the quantity of water to be mixed is x kg. Let cost of milk be Rs. 1 per kg. Then SP of 50 kg of milk with gain 10% = Rs. 55 [As Cost of 50 kg milk = Rs. 50, then SP = (50 + 10% of 50)] Then, water to be mixed is 5 kg, As the selling price of the milk is Rs. 1 per kg. Seller has to have 10% gain on 50 kg milk, he must have to add 5 kg water to 50 kg milk. 2 nd Method (Simple Method): Let the quantity of water mixed be x kg. Let CP of 1 kg of pure milk = Rs. 1. Hence, % gain = x × $$\frac{{100}}{{50}}$$ 10 = $$\frac{{100{\text{x}}}}{{50}}$$ Or, 2x = 10 or, x = 5 kg.

Let pure milk was 100L. So, Water is replaced 20% in per process = 20% of 100 = 20L. Now, we use short-cut formula for it. Quantity of Milk reduced to, $$\eqalign{ & = {\text{X}} \times {\left[ {1 - {\frac{{\text{Y}}}{{\text{X}}}} } \right]^{\text{n}}} \cr & = 100 \times {\left[ {1 - {\frac{{20}}{{100}}} } \right]^3} \cr & = \frac{{100 \times 64}}{{125}} \cr & = 51.2\,{\text{%}} \cr} $$ Here, X = Initial quantity of milk. Y = Replaced water in per process. n = No. of process repeated. Note: The formula used in above problem is quite similar to depreciation formula or Compound interest formula. Alternatively, Let pure milk be 100 litres initially. After third operation, milk will be 100 == 20%↓(- 20L) ⇒ 80 == 20%↓(- 16L) ⇒ 64 == 20%↓(-12.8L) ⇒ 51.2 %

Average cost: Rs. 1.30 per kg. And, Hence, ratio of 1.27, 1.29 and 1.32 ⇒ n 1 : n 2 : n 3 = 2 : (2 × 3) : 3 ⇒ n 1 : n 2 : n 3 = 2 : 6 : 3

Average legs per head = $$\frac{{448}}{{180}}$$ Deers : Ducks = 88 : 272 Hence, the number of Deers = $$\frac{{180 \times 88}}{{360}}$$ = 44

Now, $$\frac{{680}}{x} = \frac{{2.5}}{{1.5}}$$ Or, $$x = \frac{{680 \times 3}}{5}$$ Hence, x = Rs. 408

Hence, ratio of quantity sold at 14% profit and 6% loss, = 2 : 18 = 1 : 9 Hence, pulses sold at 14% profit, = $$\frac{{50 \times 1}}{{10}}$$ = 5 kg. At 6% Wheat sold at loss = 45 kg.

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

By the rule of alligation, we have: ∴ Required rate = 3.50 : 1.50 = 7 : 3

Let C.P. of 1 litre milk be Rs. 1 Then, S.P. of 1 litre of mixture = Rs. 1, Gain = 25% C.P. of 1 litre mixture = Rs. $$\left( {\frac{{100}}{{125}} \times 1} \right)$$ = $$\frac{4}{5}$$ By the rule of alligation, we have: ∴ Ratio of milk to water = $$\frac{4}{5}$$ : $$\frac{1}{5}$$ = 4 : 1 Hence, percentage of water in the mixture = $$\left( {\frac{1}{5} \times 100} \right)$$ % = 20%

$$\eqalign{ & {\text{Amount}}\,{\text{of}}\,{\text{milk}}\,{\text{left}}\,{\text{after}}\,{\text{3}}\,{\text{operations}} \cr & = \left[ {40{{\left( {1 - \frac{4}{{40}}} \right)}^3}} \right]\,{\text{litres}} \cr & = \left( {40 \times \frac{9}{{10}} \times \frac{9}{{10}} \times \frac{9}{{10}}} \right)\,{\text{litres}} \cr & = 29.16\,{\text{litres}} \cr} $$

By the rule of alligation, we have: So, ratio of 1 st and 2 nd quantities = 7 : 14 = 1 : 2 ∴ Required quantity replaced = $$\frac{2}{3}$$

Let C.P. of 1 litre milk be Rs. 1 S.P. of 1 litre of mixture = Rs. 1, Gain = $$\frac{50}{3}$$ % ∴ C.P. of 1 litre of mixture = $$ {100 \times \frac{3}{{350}} \times 1} $$ = $$\frac{6}{7}$$ BY the rule of alligation, we have: ∴ Ration of water and milk = $$\frac{{1}}{{7}}$$ : $$\frac{{6}}{{7}}$$ = 1 : 6

By the rule of alligation: ∴ Required ratio = 60 : 90 = 2 : 3

S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10% C.P. of 1 kg of the mixture = Rs. $$\left( {\frac{{100}}{{110}} \times 68.20} \right)$$ = Rs. 62 By the rule of alligation, we have: ∴ Required ratio = 3 : 2

Let the price of the mixed variety be Rs. x per kg. By rule of alligation, we have: $$\eqalign{ & \therefore \frac{{20 - x}}{{x - 15}} = \frac{2}{3} \cr & \Rightarrow 60 - 3x = 2x - 30 \cr & \Rightarrow 5x = 90 \cr & \Rightarrow x = 18 \cr} $$

Let the quantity of the wine in the cask originally be x litres. Then, quantity of wine left in cask after 4 operations $$\eqalign{ & = \left[ {x{{\left( {1 - \frac{8}{x}} \right)}^4}} \right]\,{\text{litres}} \cr & \therefore \left( {{{\frac{{x\left( {1 - \left( {\frac{8}{x}} \right)} \right)}}{x}}^4}} \right) = \frac{{16}}{{81}} \cr & \Rightarrow {\left( {1 - \frac{8}{x}} \right)^4} = {\left( {\frac{2}{3}} \right)^4} \cr & \Rightarrow {\frac{{x - 8}}{x}} = \frac{2}{3} \cr & \Rightarrow 3x - 24 = 2x \cr & \Rightarrow x = 24 \,\text{litres} \cr} $$

By the rule of alligation, we have: Ratio of 1 st and 2 nd parts = 4 : 6 = 2 : 3 ∴ Quantity of 2 nd kind $$\eqalign{ & = \left( {\frac{3}{5} \times 1000} \right)\,{\text{kg}} \cr & = 600\,{\text{kg}} \cr} $$

Total quantity of mixture $$\eqalign{ & = \frac{7}{{7 + 3}} \times 30{\text{ litres}} \cr & = 21{\text{ litres}} \cr} $$ And quantity of water in the mixture $$\eqalign{ & = \frac{3}{{7 + 3}} \times 30{\text{ litres}} \cr & = 9{\text{ litres}} \cr} $$ Let water to be mixed 'a' litre Then, $$\eqalign{ & \Rightarrow \left( {30 + a} \right) \times \frac{{40}}{{100}} = 9 + a \cr & \Rightarrow 120 + 4a = 90 + 10a \cr & \Rightarrow 120 - 90 = 10a - 4a \cr & \Rightarrow 30 = 6a \cr & \Rightarrow a = 5 \cr} $$ Hence, 5 litres water mixed in the mixture.

Let the Cost price of spirit be Rs. 1 per litre Spirit in 1 litre mixture in A = $$\frac{{5}}{{7}}$$ litre; Cost price of 1 litre mixture in A = Rs. $$\frac{{5}}{{7}}$$ Spirit in 1 litre mixture of B = $$\frac{{7}}{{13}}$$ litre; Cost price of 1 litre mixture in B = Rs. $$\frac{{7}}{{13}}$$ Spirit in 1 litre mixture of C = $$\frac{{8}}{{13}}$$ litre; Mean price = Rs. $$\frac{{8}}{{13}}$$ By the rule of alligation, we have: ∴ Request ratio = $$\frac{{1}}{{13}}$$ : $$\frac{{9}}{{91}}$$ = 7 : 9

Let the quantity of milk and water in initial mixture be 4x and 3x litres. Quantity of water = 3x litres On adding 6 litres of water, we get $$\eqalign{ & \Rightarrow \frac{{4x}}{{3x + 6}} = \frac{8}{7} \cr & \Rightarrow 28x = 24x + 48 \cr & \Rightarrow 28x - 24x = 48 \cr & \Rightarrow 4x = 48 \cr & \Rightarrow x = \frac{{48}}{4} \cr & \Rightarrow x = 12 \cr} $$ ∴ Required quantity of milk = 4x = 4 × 12 = 48 litres

Cost price of mixture $$\eqalign{ & = \frac{{{\text{Sale price}}}}{{\left( {100 + {\text{gain }}\% } \right)}} \times 100 \cr & = \frac{{767}}{{118}} \times 100 \cr & = {\text{Rs}}{\text{. 650}} \cr} $$ Ratio = 5 : 4 ∴ Quantity of A type of sandal is 35 kg ∴ 5x = 35 kg ∴ x = 7 kg Thus B type sandal = 7 × 4 = 28 kg

Let cost of 1 litre milk be Rs. 1 Milk in 1 litre mixture in A = $$\frac{8}{13}$$ litre; Cost price of 1 litre mixture in A = Rs. $$\frac{8}{13}$$ Milk in 1 litre mixture in B = $$\frac{5}{7}$$ litre; Cost price of 1 litre mixture in B = Rs. $$\frac{5}{7}$$ Milk in 1 litre of final mixture $$\eqalign{ & = {\frac{{900}}{{13}} \times \frac{1}{{100}} \times 1} \cr & = \frac{9}{{13}}{\text{ litre}}{\text{}} \cr} $$ Mean price = Rs. $$\frac{9}{{13}}$$ By the rule of alligation, we have: ∴ Required ratio = $$\frac{2}{{91}}$$ : $$\frac{1}{{13}}$$ = 2 : 7

In 20 litres of mixture Quantity of milk $$ \Rightarrow \frac{3}{4} \times 20 = 15{\text{ litres}}$$ Quantity of water $$ \Rightarrow \frac{1}{4} \times 20 = 5{\text{ litres}}$$ Let the quantity of milk added be x litres. According to the question, $$\eqalign{ & \Rightarrow \frac{{15 + x}}{5} = \frac{4}{1} \cr & \Rightarrow 15 + x = 4 \times 5 \cr & \Rightarrow x = 20 - 15 \cr & \Rightarrow x = 5{\text{ litres}} \cr} $$

By the rule of alligation : Ratio of water to milk = 4 : 8 = 1 : 2

Cost price of 1 litres of milk = Rs. 100 ∴ Mixture sold for Rs. 125 = $$\frac{125}{100}$$ = $$\frac{5}{4}$$ litres ∴ Quantity of mixture = $$\frac{5}{4}$$ litres ∴ Quantity of milk = 1 litre ∴ Quantity of water = $$\frac{5}{4}$$ - 1 = $$\frac{1}{4}$$ litre ∴ Required ratio = $$\frac{1}{4}$$ : 1 = 1 : 4

Let quantity of grape, pineapple and banana juice in the mixture be 4x, 6x and 5x respectively Total initial quantity of juice in the vessel = 4x + 6x + 5x = 15x litres In 15 litres of juice, Grape's juice = 4 litres Pineapple's juice = 6 litres Banana's juice = 5 litres Resultant quantity of grape juice is 10 litres less than the resultant quantity of pineapple juice $$\eqalign{ & \Rightarrow \left( {6x - 6 + 2} \right) - \left( {4x - 4 + 8} \right) = 10 \cr & \Rightarrow 6x - 4 - 4x - 4 = 10 \cr & \Rightarrow 2x - 8 = 10 \cr & \Rightarrow 2x = 10 + 8 \cr & \Rightarrow x = 9 \cr} $$ ∴ Initial quantity of mixture = 15x = 15 × 9 = 135 litres

By the rule of alligation : ∴ Required ratio = 750 : 250 = 3 : 1