Alligation or Mixture Questions and Answers

Answer: B

Water makes up 20% of the mixture.

Selling the mixture at the cost price of milk and gaining 25% means that for every 100 units of milk he pays for, he sells 125 units. The extra 25 units are water.

\[ \text{water} \% = \frac{25}{125} \times 100 = 20\% \]

The trap is 25%, which is water as a share of the milk, not of the mixture. The question asks for its share of the mixture.

Answer: C

The pure milk to be added is 9.2 litres. Milk in the three mixtures is 3 x 4/5 = 2.4, 4 x 3/4 = 3 and 5 x 5/7 = 25/7 litres, so the total milk is about 8.971 litres in a 12-litre mixture. After adding x litres of pure milk the milk fraction must be 6/7, so (8.971 + x)/(12 + x) = 6/7, which gives 62.8 + 7x = 72 + 6x and x = 9.2. The values 8.2, 8.7 and 9.7 do not satisfy this equation. A check on the final mixture: the milk is 8.971 + 9.2 = 18.171 litres in a total of 21.2 litres, and 18.171 divided by 21.2 is 0.857, which is 6/7 as required.

Answer: B

The mixture has 44 % spirit. Solution A has spirit and water in the ratio 2 : 3, so 6 litres of A contain 6 x 2/5 = 2.4 litres of spirit. Solution B has the ratio 7 : 8, so 9 litres of B contain 9 x 7/15 = 4.2 litres of spirit.

\[ \text{Spirit} = 2.4 + 4.2 = 6.6 \text{ litres} \]

\[ \text{Mixture} = 6 + 9 = 15 \text{ litres} \]

\[ \frac{6.6}{15} \times 100 = 44\% \]

The options 40 %, 52 % and 56 % are not the share of 6.6 litres in the 15 litres of the mixture. As a check, the water in the mixture is 3.6 + 4.8 = 8.4 litres, and 6.6 + 8.4 = 15 litres, the volume of the mixture.

Answer: A

The spirit in 780 mL of the mixture is 285 mL. In solution A the spirit is \(\frac{6}{13}\), so 3 L has \(\frac{18}{13}\) L of spirit. In solution B the ratio of water to spirit is 9 : 4, so spirit is \(\frac{4}{13}\), and 5 L has \(\frac{20}{13}\) L. \[ \begin{aligned} \text{Spirit} &= \frac{18}{13} + \frac{20}{13} \\ &= \frac{38}{13}\text{ L} \end{aligned} \] \[ \text{Total} = 3 + 5 = 8\text{ L} \] \[ \begin{aligned} \text{Fraction of spirit} &= \frac{38}{13\times 8} \\ &= \frac{19}{52} \end{aligned} \] \[ 780\times\frac{19}{52} = 15\times 19 = 285\text{ mL} \] Reading B as spirit to water (4 : 9 swapped) would give a different value, so the order in the stem matters; 304, 270 and 266 do not result.

Answer: C

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

\[ 126 + 135 + 2p = 4 \times 153 = 612 \]

\[ 2p = 612 - 261 = 351 \quad\Longrightarrow\quad p = \text{Rs. } 175.50 \]

The third variety carries half the weight of the mixture, which is why its price does not have to be far above 153 to lift the average.

Answer: A

Correct answer: 12 : 13

The original 50 litres contains \( 50 \times \frac{3}{5} = 30 \) litres of milk and 20 of water. Removing 10 litres of the mixture removes milk and water in their current 3 : 2 proportion - 6 litres of milk and 4 of water - leaving 24 of milk and 16 of water. Adding 10 litres of water brings the water to \( 16 + 10 = 26 \), so the new ratio is \[ 24 : 26 = 12 : 13. \] The essential point is that what is drawn off is a sample of the mixture, carrying both components in proportion. Removing 10 litres of pure milk would be a different problem and would leave 20 : 30, or 2 : 3. As a check, the total stays at 50 litres throughout, and \( 24 + 26 = 50 \).

Answer: A

No profit and no loss means the total cost equals the total revenue.

\[ 32(60) + 23x = 28(60 + x) \]

\[ 1920 + 23x = 1680 + 28x \]

\[ 240 = 5x \quad\Longrightarrow\quad x = 48 \text{ litres} \]

By alligation the same result follows from \( \tfrac{60}{x} = \tfrac{28-23}{32-28} = \tfrac{5}{4} \).

Answer: D

Take each can as 1 unit.

\[ \text{Can 1: water } 1, \text{ wine } 0 \]

\[ \text{Can 2: water } 0.5, \text{ wine } 0.5 \]

\[ \text{total water} = 1.5, \qquad \text{total wine} = 0.5 \]

\[ \text{water} : \text{wine} = 1.5 : 0.5 = 3 : 1 \]

The cans are the same size, which is what lets them be counted as equal units.

Answer: B

Correct answer: 16 litres

Let the original quantity be \( 5x \) litres, containing \( 4x \) of A and \( x \) of B. Removing 10 litres of the mixture removes A and B in the ratio 4 : 1, that is 8 litres of A and 2 of B, leaving \( 4x - 8 \) and \( x - 2 \). Adding 10 litres of B brings B to \( x - 2 + 10 = x + 8 \). The new ratio gives \[ \frac{4x - 8}{x + 8} = \frac{2}{3} \;\Rightarrow\; 12x - 24 = 2x + 16 \;\Rightarrow\; 10x = 40 \;\Rightarrow\; x = 4. \] So A was originally \( 4x = 16 \) litres. Checking: 16 of A and 4 of B; removing 10 litres takes 8 and 2, leaving 8 and 2; adding 10 of B gives 8 and 12, which is 2 : 3. The essential step is that the removed portion carries both liquids in their current proportion.

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%.