Alligation or Mixture Questions and Answers

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%.

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: D

Let \( x \) kg be the cheaper rice; the rest, \( 15 - x \) kg, is the dearer.

\[ 10.40(15 - x) + 8.80x = 146.40 \]

\[ 156 - 10.40x + 8.80x = 146.40 \]

\[ -1.60x = -9.60 \quad\Longrightarrow\quad x = 6 \text{ kg} \]

The mixture works out at \( \tfrac{146.40}{15} = 9.76 \) per kg, between the two prices as it must be.

Answer: A

The 80 L contains \( 45\% \times 80 = 36 \) L of milk. Adding \( x \) L of pure milk raises both the milk and the total.

\[ \frac{36 + x}{80 + x} = 0.75 \]

\[ 36 + x = 60 + 0.75x \quad\Longrightarrow\quad 0.25x = 24 \]

\[ x = 96 \text{ litres} \]

The non-milk part stays fixed at 44 L, and it must end up as 25% of the total - so the total becomes 176 L, confirming the 96 L added.