Alligation - Aptitude MCQ Questions and Solutions with Explanations

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