Alligation or Mixture Questions and Answers

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.

Answer: A

Amount of water in solution = 100 % - 30 % = 70 % of total
Amount of water = 70 % x 500 = 350 gm
Amount of water remains same for both the solutions. Therefore, let amount of new solution be x gm.
x = 350 x (100/ 50)
x = 700
Salt added = (700 - 500) = 200 gm

Answer: A

The water is free, so "no loss" means the revenue from the whole adulterated quantity just covers the cost of the pure milk.

\[ 28(m + 1.5) = 36m \]

\[ 28m + 42 = 36m \quad\Longrightarrow\quad 8m = 42 \]

\[ m = 5.25 \text{ litres} \]

Selling at a price below cost is profitable only because the 1.5 L of water is sold as though it were milk.

Answer: C

Correct answer: 15 litres

The 45 litres contains \( 45 \times \frac{4}{5} = 36 \) litres of milk and 9 of water. The milk does not change. For a final ratio of 3 : 2 the water must be \( \frac{2}{3} \) of the milk, so it must become \[ \frac{2}{3} \times 36 = 24 \text{ litres}. \] The water to add is \[ 24 - 9 = 15 \text{ litres}. \] Checking: the vessel then holds 36 litres of milk and 24 of water, a total of 60 in the ratio 3 : 2. A quicker route uses parts. Expressing both ratios with the milk fixed - \( 4 : 1 \) becomes \( 12 : 3 \) and \( 3 : 2 \) becomes \( 12 : 8 \) - shows the water rising from 3 parts to 8, an increase of 5 parts. Since 12 parts of milk is 36 litres, one part is 3 litres, and 5 parts is 15.