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: B

The new solution is 3% sugar.

The sugar does not change when water is added; only the volume does.

\[ \begin{aligned} \text{sugar} &= 4\% \times 3 = 0.12\ \text{L} \\ \text{new volume} &= 3 + 1 = 4\ \text{L} \\ \text{strength} &= \frac{0.12}{4} \times 100 = 3\% \end{aligned} \]

4% ignores the added water, and 6% would need sugar to be added rather than water.

Answer: C

Take the container as 5 units: 4 of wine and 1 of water. Spilling a fraction \( f \) removes that fraction of the wine as well, and the water poured back keeps the total at 5.

\[ \text{wine left} = 4(1 - f) \]

The new ratio 3 : 2 means the wine is now 3 of the 5 units:

\[ 4(1 - f) = 3 \quad\Longrightarrow\quad 1 - f = \frac{3}{4} \quad\Longrightarrow\quad f = \frac{1}{4} \]

So a quarter of the mixture was replaced by water. Spilling removes wine and water in their existing proportion, which is what makes the wine shrink by the factor \( 1-f \).

Answer: B

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

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

Work with the fraction of wine in each can.

\[ \text{Can 1} = \frac{5}{8} = 0.625, \qquad \text{Can 2} = \frac{2}{5} = 0.4, \qquad \text{mixture} = 0.5 \]

\[ \frac{\text{Can 2}}{\text{Can 1}} = \frac{0.625 - 0.5}{0.5 - 0.4} = \frac{0.125}{0.1} = \frac{5}{4} \]

So Can 1 : Can 2 \( = 4 : 5 \). The can further from the target proportion contributes less, which is why the weaker Can 2 is used in greater quantity.

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

Start from the composition and follow each step.

\[ \text{water} = 30\% \times 140 = 42 \text{ L}, \qquad \text{syrup} = 98 \text{ L} \]

Selling 20 L removes both in proportion, leaving 120 L:

\[ \text{water} = 36, \qquad \text{syrup} = 84 \]

Adding \( x \) of each and requiring water : syrup \( = 1 : 2 \):

\[ 2(36 + x) = 84 + x \quad\Longrightarrow\quad 72 + 2x = 84 + x \]

\[ x = 12 \text{ litres} \]

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.