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: B

Let the initial quantity of milk in vessel be T litres.

Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.

Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T 

For the given problem, T = 90, y = 9 and n = 2.

Hence, quantity of  milk finally in the vessel 

= [(90 - 9)/90]2 (90) =  72.9 litres.

Answer: A

Correct answer: 25 litres

Let the milk be \( 5x \) and the water \( x \). The milk is unchanged by adding water, which is what makes the problem tractable. After adding 5 litres the water becomes \( x + 5 \), and \[ \frac{5x}{x + 5} = \frac{5}{2} \;\Rightarrow\; 10x = 5x + 25 \;\Rightarrow\; x = 5. \] The milk is therefore \( 5x = 25 \) litres. Checking: 25 litres of milk and 5 of water is 5 : 1; adding 5 litres of water gives 25 and 10, which is 5 : 2. A shortcut is available and worth knowing. Since the milk is constant, write both ratios with the same first term - here both are already 5 - and the water has gone from 1 part to 2 parts, an increase of 1 part. That 1 part equals the 5 litres added, so each part is 5 litres and the milk is 25.

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: B

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

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

\[ \frac{\text{Rs. 420 rice}}{\text{Rs. 520 rice}} = \frac{520 - 480}{480 - 420} = \frac{40}{60} = \frac{2}{3} \]

So the ratio is \( 2 : 3 \). The mean of 480 lies closer to 520 than to 420, so the dearer variety must be the larger share - which is the check on the direction of the ratio.

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: B

Let rice \( = 5k \) and wheat \( = 3k \). Adding 16 kg of wheat changes only the wheat.

\[ \frac{5k}{3k + 16} = \frac{5}{7} \quad\Longrightarrow\quad 35k = 15k + 80 \]

\[ 20k = 80 \quad\Longrightarrow\quad k = 4 \]

\[ \text{new wheat} = 3(4) + 16 = 28 \text{ kg} \]

The question asks for the wheat in the new mixture, not the original 12 kg.

Answer: D

Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litres

Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres

Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres

As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5.

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.

Answer: A

Take the quantities as 2 kg and 3 kg, five kilograms in all.

\[ \frac{2(15) + 3(20)}{5} = \frac{30 + 60}{5} = \text{Rs. } 18 \]

The result is the weighted average, closer to 20 than to 15 because more of the dearer rice is used.