Alligation or Mixture Questions and Answers

Answer: A

Selling at cost price means the entire gain comes from the free water.

A gain of \( 16\tfrac{2}{3}\% = \tfrac{1}{6} \) means the profit is one sixth of the cost, and the profit is exactly the value of the water:

\[ \frac{\text{water}}{\text{milk}} = \frac{1}{6} \]

Checking with 6 litres of milk costing Rs. 6: adding 1 litre of water gives 7 litres sold at Rs. 7, a profit of Re. 1 on a cost of Rs. 6 - exactly \( 16\tfrac{2}{3}\% \).

Answer: D

The alligation rule gives the ratio in which two ingredients must be mixed:

\[ \frac{\text{cheaper quantity}}{\text{dearer quantity}} = \frac{\text{dearer} - \text{mean}}{\text{mean} - \text{cheaper}} \]

Work with the fraction of milk in each vessel.

\[ A = \frac{4}{7}, \qquad B = \frac{2}{5}, \qquad \text{mixture} = \frac{1}{2} \]

\[ \frac{A}{B} = \frac{\tfrac{1}{2} - \tfrac{2}{5}}{\tfrac{4}{7} - \tfrac{1}{2}} = \frac{\tfrac{1}{10}}{\tfrac{1}{14}} = \frac{14}{10} = \frac{7}{5} \]

Vessel A is richer in milk and is further from the target than B, so it contributes the larger share - 7 parts to 5.

Answer: C

Water is free, so the total value of the bucket never changes; only the volume does.

\[ \text{value} = 40 \times 3.50 = \text{Rs. } 140 \]

At Rs. 2 per litre that value must spread over

\[ \frac{140}{2} = 70 \text{ litres} \]

\[ \text{water added} = 70 - 40 = 30 \text{ litres} \]

By alligation, water is priced at 0 and \( \tfrac{\text{milk}}{\text{water}} = \tfrac{2-0}{3.50-2} = \tfrac{4}{3} \), so 40 litres of milk needs 30 of water.

Answer: B

Selling at cost price means the whole gain comes from the free water.

A gain of 20% is \( \tfrac{1}{5} \) of the cost, and that profit is exactly the value of the water added:

\[ \frac{\text{water}}{\text{spirit}} = \frac{1}{5} \]

Checking with 5 litres of spirit costing Rs. 5: adding 1 litre of water gives 6 litres sold for Rs. 6, a profit of Re. 1 on Rs. 5 - exactly 20%.

Answer: B

Correct answer: 63 kg

Apply alligation between the two prices with a mean of 8.40: \[ \frac{\text{quantity at } 7}{\text{quantity at } 9} = \frac{9 - 8.40}{8.40 - 7} = \frac{0.60}{1.40} = \frac{3}{7}. \] The cheaper and dearer sugars are therefore in the ratio 3 : 7. The cheaper quantity is 27 kg, so one part is \( 27/3 = 9 \) kg, and the dearer quantity is \[ 7 \times 9 = 63 \text{ kg}. \] Checking: 27 kg at 7 and 63 kg at 9 cost \( 189 + 567 = 756 \) for 90 kg, which is 8.40 per kg. Note that the differences are taken diagonally: the quantity of each ingredient is proportional to the distance of the mean from the other price. Since the mean lies nearer 9 than 7, more of the dearer sugar is needed - which is a useful check before any arithmetic.

Answer: A

Correct answer: ₹20.40

The price of a mixture is the total cost divided by the total quantity - a weighted average rather than a simple one. The total cost is \[ 20 \times 24 + 30 \times 18 = 480 + 540 = 1020, \] and the total quantity is 50 kg, so \[ \text{price} = \frac{1020}{50} = 20.40 \text{ per kg}. \] The distractor 21 is the simple average of 24 and 18, which would be correct only if equal quantities were mixed. Because more of the cheaper wheat is used, the mixture price must lie below the midpoint - and noticing that rules out both 21 and 22 at once. This is the direct problem; alligation solves the inverse one, finding the ratio when the mixture price is given. The two are the same relation read in opposite directions, and it is worth being able to move either way: \[ \text{mean} = \frac{q_{1}p_{1} + q_{2}p_{2}}{q_{1} + q_{2}}. \]

Answer: A

Correct answer: 100 ml

The 400 ml contains \( 400 \times 0.15 = 60 \) ml of alcohol and 340 ml of water. The water is unchanged by adding alcohol, so anchor on it. If the alcohol is to be 32%, the water must be 68% of the new total: \[ 340 = 0.68 \times \text{new total} \;\Rightarrow\; \text{new total} = \frac{340}{0.68} = 500 \text{ ml}. \] The alcohol to add is therefore \[ 500 - 400 = 100 \text{ ml}. \] Checking: 500 ml containing \( 60 + 100 = 160 \) ml of alcohol is \( 160/500 = 32\% \). The alternative is alligation between the 15% solution and pure alcohol at 100%, with a mean of 32: \[ \frac{\text{solution}}{\text{alcohol}} = \frac{100 - 32}{32 - 15} = \frac{68}{17} = \frac{4}{1}, \] so for 400 ml of solution, 100 ml of alcohol - the same answer by a shorter route.

Answer: B

Correct answer: 17 : 18

Take equal quantities from each vessel - a convenient amount is the LCM of the ratio totals, 7 and 5, which is 35 litres from each. From the first vessel, 35 litres in the ratio 4 : 3 gives 20 of milk and 15 of water. From the second, 35 litres in the ratio 2 : 3 gives 14 of milk and 21 of water. Combining, \[ \text{milk} = 20 + 14 = 34, \qquad \text{water} = 15 + 21 = 36, \] so the ratio is \( 34 : 36 = \mathbf{17 : 18} \). Choosing the LCM as the common quantity is what keeps every figure a whole number; taking 1 litre from each would give fractions and the same answer more painfully. The check that costs nothing: the first vessel is richer in milk than the second, so the mixture must lie between \( 4/7 \approx 0.57 \) and \( 2/5 = 0.40 \) as a fraction of milk - and \( 17/35 \approx 0.49 \) does.

Answer: C

Correct answer: 120 litres

Let the capacity be \( V \) litres. The milk initially is \( \frac{3}{5}V \). Drawing off 20 litres of the mixture removes milk in the same proportion, that is \( \frac{3}{5} \times 20 = 12 \) litres, leaving \( \frac{3}{5}V - 12 \). Adding 20 litres of water returns the total to \( V \). For the final ratio to be 1 : 1 the milk must be half the total: \[ \frac{3}{5}V - 12 = \frac{V}{2} \;\Rightarrow\; \frac{3V}{5} - \frac{V}{2} = 12 \;\Rightarrow\; \frac{V}{10} = 12 \;\Rightarrow\; V = 120. \] Checking: 120 litres in the ratio 3 : 2 is 72 of milk and 48 of water. Drawing 20 litres removes 12 and 8, leaving 60 and 40; adding 20 litres of water gives 60 and 60, which is 1 : 1. The key step is that the portion drawn off carries both components in their current proportion.

Answer: C

Correct answer: 1 : 3

By alligation the ratio of the two quantities is the inverse ratio of their distances from the mean. Taking Basmati first: \[ \frac{\text{Basmati}}{\text{ordinary}} = \frac{90 - 80}{120 - 90} = \frac{10}{30} = \frac{1}{3}. \] The ratio is 1 : 3 - one part Basmati to three parts ordinary rice. Checking: 1 kg at 120 and 3 kg at 80 cost \( 120 + 240 = 360 \) for 4 kg, which is 90 per kg. Note carefully which difference belongs to which ingredient. The quantity of an ingredient is proportional to the distance of the mean from the other price - the differences are taken diagonally across the alligation diagram. The sanity check settles it regardless: the mean of 90 is much nearer 80 than 120, so the ordinary rice must dominate, and 3 : 1 can be rejected without calculation.