Alligation or Mixture Questions and Answers

Answer: B

As Ram replaces 10% of kerosene with water, the amount of kerosene left is 90%.

As the replacement is done four times,

After 1st replacement, kerosene left is : 0.9 × 100 = 90 liters

After 2nd replacement, kerosene left is : 0.9 × 90 = 81 liters

After 3rd replacement, kerosene left is : 0.9 × 81 = 72.9 liters

After 4th replacement, kerosene left is : 0.9 × 72.9 = 65.61 liters

Answer: C

Correct answer: 1 : 7

Water is free, so treat its price as zero and apply alligation between 0 and 32 with a mean of 28: \[ \frac{\text{water}}{\text{milk}} = \frac{32 - 28}{28 - 0} = \frac{4}{28} = \frac{1}{7}. \] The ratio is 1 : 7 - one litre of water to seven of milk. Checking: 7 litres of milk cost \( 7 \times 32 = 224 \), and the mixture is 8 litres, so the rate is \( 224/8 = 28 \) per litre as required. The whole of the trick with this family of questions is recognising that water costs nothing, which turns a dilution problem into an ordinary alligation. A related form asks for the profit from such adulteration: selling the 8-litre mixture at the milk price of 32 brings in 256 against a cost of 224, a gain of 32 on 224, which is \( 14\frac{2}{7}\% \).

Answer: C

Correct answer: 15 litres

The 45 litres contains \( 45 \times \frac{4}{5} = 36 \) litres of milk and 9 of water. The milk does not change. For a final ratio of 3 : 2 the water must be \( \frac{2}{3} \) of the milk, so it must become \[ \frac{2}{3} \times 36 = 24 \text{ litres}. \] The water to add is \[ 24 - 9 = 15 \text{ litres}. \] Checking: the vessel then holds 36 litres of milk and 24 of water, a total of 60 in the ratio 3 : 2. A quicker route uses parts. Expressing both ratios with the milk fixed - \( 4 : 1 \) becomes \( 12 : 3 \) and \( 3 : 2 \) becomes \( 12 : 8 \) - shows the water rising from 3 parts to 8, an increase of 5 parts. Since 12 parts of milk is 36 litres, one part is 3 litres, and 5 parts is 15.

Answer: B

Correct answer: 7 : 3

By alligation the quantities are in the inverse ratio of the distances from the mean: \[ \frac{\text{cheaper}}{\text{dearer}} = \frac{20 - 16.50}{16.50 - 15} = \frac{3.50}{1.50} = \frac{7}{3}. \] The ratio is 7 : 3. Checking: 7 kg at 15 and 3 kg at 20 cost \( 105 + 60 = 165 \) for 10 kg, which is 16.50 per kg. The direction can be verified without arithmetic. The mean of 16.50 is much closer to 15 than to 20, so the cheaper variety must dominate heavily - which rules out 3 : 7 and 2 : 5 at a glance. Reversing the ratio is the commonest error in this topic, and this check catches it every time. Multiplying both differences by 2 to clear the decimals - \( 7 : 3 \) from \( 3.5 : 1.5 \) - is the practical step that keeps the arithmetic clean.

Answer: D

As vessel A contains 150 liters of mixture of milk and water in the ratio 3:1, the quantity of milk is 3 × 150/4 liters, and

that of water is 150/4 liters.

Similarly, as vessel B contains 200 liters of mixture of milk and water in the ratio 5:2, the quantity of milk is 5 × 200/7,

and that of water is 2 × 200/7 liters.

After mixing the total quantity is 350 liters.

The quantity of milk is {(3 × 150/4) + (5 × 200/7)} = 7150/28

The quantity of water is {(150/4) + (2 × 200/7)} = 2650/28

The ratio of the resulting mixture = 7150/2650 = 143 : 53

Answer: B

Let the jar initially contain 8 liters of mixture:3 liters water and 5 liters alcohol

Let x liters of this mixture is drawn out and is replaced by x liters of water.

Amount of water now in the solution: 3-(3/8)x+x.

Amount of alcohol now in the solution: 5-(5/8)x

Desired ratio: 1/1

=>3-(3/8)x+x=5-(5/8)x

x=8/5 liters

%x=((8/5)/8) × 100=20%

Answer: D

Currently the mixture is in ratio of A/B: 7:5

=> This means fraction of A in the solution: 7/12

=> Fraction of B in the solvent: 5/12

=>New concentration of A= 7-(7/12)9

=>New concentration of B= 5-(5/12)9+x

=> (7-(7/12)9)/(5-(5/12)9+x) = 7/9

=>x=21

Answer: C

Correct answer: 3 : 2

By the rule of alligation, the two quantities are in the inverse ratio of their distances from the mean price: \[ \frac{\text{cheaper quantity}}{\text{dearer quantity}} = \frac{\text{dearer price} - \text{mean}}{\text{mean} - \text{cheaper price}}. \] Here \[ \frac{45 - 36}{36 - 30} = \frac{9}{6} = \frac{3}{2}. \] So the ratio is 3 : 2 - three parts of the cheaper rice to two of the dearer. The result is worth checking directly: 3 kg at 30 and 2 kg at 45 cost \( 90 + 90 = 180 \) for 5 kg, which is 36 per kg as required. The mean must lie between the two prices for the problem to have a solution, and the larger share always goes to the ingredient whose price is nearer the mean - here the cheaper, since 36 is closer to 30 than to 45. That observation alone rules out 2 : 3 without any arithmetic.

Answer: C

Take the vessel as 8 units: 3 of water and 5 of syrup. Drawing off a fraction \( f \) removes both in proportion, and the water poured back keeps the total at 8.

\[ \text{water after} = 3(1 - f) + 8f = 3 + 5f \]

Half water means 4 of the 8 units:

\[ 3 + 5f = 4 \quad\Longrightarrow\quad f = \frac{1}{5} \]

The replacement adds water twice over - it removes syrup and puts water in its place - which is why only a fifth needs drawing off to move the water from \( \tfrac{3}{8} \) to \( \tfrac{1}{2} \).

Answer: B

Cost Price of 1kg of Type 1 sugar = 6200 p.
Cost Price of 1kg of Type 2 sugar = 7200 p.
Mean Price of 1 kg of mixture = 6450 p.
According to the Rule of Alligation,
(Quantity of Cheaper):(Quantity of Dearer) = (CP of dearer - Mean Price):(Mean Price - CP of cheaper)
Therefore, the required ratio = (7200-6450):(6450-6200) = 750:250 = 3:1.