Alligation or Mixture Questions and Answers

Answer: C

The water contributes nothing to the value, so all the worth comes from the oil.

\[ 325x = 275(x + 240) \]

\[ 325x = 275x + 66{,}000 \]

\[ 50x = 66{,}000 \quad\Longrightarrow\quad x = 1320 \text{ litres} \]

By alligation, water is priced at 0 and \( \tfrac{\text{oil}}{\text{water}} = \tfrac{275-0}{325-275} = \tfrac{275}{50} = 5.5 \), and \( 5.5 \times 240 = 1320 \).

Answer: A

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

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

\[ \frac{\text{Rs. 6 rice}}{\text{Rs. 8.75 rice}} = \frac{8.75 - 7.50}{7.50 - 6} = \frac{1.25}{1.50} = \frac{5}{6} \]

The mean of 7.50 sits slightly nearer the dearer price, so a little more of the expensive rice is needed - consistent with 5 : 6.

Answer: A

Take the quantities as 2, 1 and 3 kg, six kilograms in all.

\[ 2(145) + 1(165) + 3p = 6(180) \]

\[ 290 + 165 + 3p = 1080 \]

\[ 3p = 625 \quad\Longrightarrow\quad p = \text{Rs. } 208.33 \]

The third type is the most expensive, which it has to be for the average of 180 to sit above both known prices.

Answer: A

Let the can hold \( T \) litres, with milk \( \tfrac{7T}{12} \).

Spilling 9 litres removes milk in proportion, and the water added keeps the total at \( T \):

\[ \text{milk left} = \frac{7T}{12} \cdot \frac{T-9}{T} = \frac{7(T-9)}{12} \]

The new ratio 9 : 7 makes milk \( \tfrac{7T}{16} \):

\[ \frac{7(T-9)}{12} = \frac{7T}{16} \quad\Longrightarrow\quad 16(T-9) = 12T \]

\[ 4T = 144 \quad\Longrightarrow\quad T = 36 \]

\[ \text{milk} = \frac{7 \times 36}{12} = 21 \text{ litres} \]

Answer: B

Correct answer: 1 : 11

First convert the selling price and profit into a cost price. Selling at 13.75 with a gain of 25% means \[ \text{CP} = \frac{13.75 \times 100}{125} = 11 \text{ per litre}. \] Now apply alligation between water, which costs nothing, and the liquid at 12, with a mean of 11: \[ \frac{\text{water}}{\text{liquid}} = \frac{12 - 11}{11 - 0} = \frac{1}{11}. \] The ratio is 1 : 11. Checking: 11 litres of liquid cost 132, and the 12-litre mixture therefore costs 11 per litre; selling it at 13.75 brings in 165, a gain of 33 on 132, which is 25%. The step most often skipped is the first one - alligating on the selling price rather than converting to cost price gives the wrong ratio, because the profit has not been removed.

Answer: C

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

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

\[ \frac{\text{Rs. 15 pulses}}{\text{Rs. 20 pulses}} = \frac{20 - 16.50}{16.50 - 15} = \frac{3.50}{1.50} = \frac{7}{3} \]

The mean of 16.50 lies much nearer 15, so the cheaper variety dominates - consistent with the 7 : 3 split.

Answer: D

When a fraction of a solution is repeatedly removed and replaced, the original liquid left after \( n \) operations is

\[ A\left(1 - \frac{x}{A}\right)^{n} \]

Here \( A = 40 \), \( x = 4 \) and \( n = 3 \) (once, then twice more):

\[ 40\left(1 - \frac{4}{40}\right)^{3} = 40(0.9)^{3} \]

\[ = 40 \times 0.729 = 29.16 \text{ litres} \]

Simply subtracting \( 3 \times 4 = 12 \) litres would give 28, but each later removal takes out some water as well as juice.

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.