Alligation or Mixture Questions and Answers

Answer: C

Bottle B now holds 13.33% alcohol.

Let bottle A hold V and let x be the amount stolen. A keeps V - x at 50% and gains x at 20%, ending at 30%:

\[ \begin{aligned} 0.5(V - x) + 0.2x &= 0.3V \\ 0.2V &= 0.3x \\ x &= \tfrac{2}{3}V \end{aligned} \]

Bottle B, twice as large (2V, full), lost \(\tfrac{2}{3}V\) of its 20% vodka, replaced by water:

\[ \begin{aligned} \text{alcohol} &= 0.2 \times \tfrac{4}{3}V = \tfrac{4}{15}V \\ \text{strength} &= \tfrac{4}{15}V \div 2V = \tfrac{2}{15} \\ &\approx 13.33\% \end{aligned} \]

Forgetting that B is twice the size of A gives 6·67%. 10% and 15% do not follow from the 30% condition on bottle A.

Answer: A

The mixture would cost 29.5/kg (in rupees).

Let the A+B mixture weigh \(q\) kg costing \(p\) per kg, with \(c\) kg of C and \(3c\) kg of D.

\[ \begin{aligned} pq + 20c &= 25(q + c) \Rightarrow pq = 25q + 5c \\ pq + 108c &= 31(q + 3c) \Rightarrow pq = 31q - 15c \\ 6q &= 20c \Rightarrow q = \tfrac{10c}{3} \end{aligned} \]

Then \(pq = \frac{265c}{3}\), and for all four varieties:

\[ \begin{aligned} \text{cost} &= \tfrac{265c}{3} + 20c + 108c = \tfrac{649c}{3} \\ \text{weight} &= \tfrac{10c}{3} + 4c = \tfrac{22c}{3} \\ \text{rate} &= \tfrac{649}{22} = 29.5 \end{aligned} \]

Answer: C

The water to be added is 150 L.

The 150 L mixture in the ratio 2:1 has 100 L of milk and 50 L of water. The milk stays at 100 L, and for a milk to water ratio of 1:2 the water must be twice the milk:

\[ \begin{aligned} \text{water needed} &= 2 \times 100 = 200\ \text{L} \\ \text{to add} &= 200 - 50 = 150\ \text{L} \end{aligned} \]

200 L is the total water required, not the amount to add, and 50 L or 100 L would leave the milk still in excess of half the water.

Answer: D

The vessels should be mixed in the ratio 7 : 2.

Work with the fraction of milk in each. Vessel A has \(\frac{5}{7}\), vessel B \(\frac{8}{13}\), and the mixture must have \(\frac{9}{13}\). By the rule of alligation:

\[ \begin{aligned} A : B &= \left(\tfrac{9}{13} - \tfrac{8}{13}\right) : \left(\tfrac{5}{7} - \tfrac{9}{13}\right) \\ &= \tfrac{1}{13} : \tfrac{2}{91} \\ &= 7 : 2 \end{aligned} \]

Check: 7 litres of A give 5 of milk, 2 litres of B give \(\frac{16}{13}\), so milk is \(\frac{81}{13}\) of 9 litres, which is \(\frac{9}{13}\).

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%.