Alligation - Aptitude MCQ Questions and Solutions 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^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
⇒ n₁ : n₂ : n₃ = 2 : (2 × 3) : 3
⇒ n₁ : n₂ : n₃ = 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.