Alligation or Mixture Questions and Answers

Answer: D

When a fraction of a solution is repeatedly removed and replaced, the original liquid left after \( n \) operations is

\[ A\left(1 - \frac{x}{A}\right)^{n} \]

With \( A = 40 \), \( x = 4 \) and \( n = 3 \):

\[ 40(0.9)^{3} = 40 \times 0.729 = 29.16 \text{ litres} \]

Each operation leaves 90% of the milk then present, and three such reductions compound rather than add.

Answer: B

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

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

\[ \frac{\text{Rs. 7.20 rice}}{\text{Rs. 5.70 rice}} = \frac{6.30 - 5.70}{7.20 - 6.30} = \frac{0.60}{0.90} = \frac{2}{3} \]

Note the numerator and denominator: the quantity of the dearer rice is proportional to the gap on the cheaper side, and vice versa.

Answer: A

Convert the selling price to a cost price first.

\[ \text{CP} = \frac{68.20}{1.10} = \text{Rs. } 62 \text{ per kg} \]

\[ \frac{\text{Rs. 60 tea}}{\text{Rs. 65 tea}} = \frac{65 - 62}{62 - 60} = \frac{3}{2} \]

Using 68.20 directly as the mean would put it outside the range 60–65 and give a negative ratio - the signal that the gain must be removed first.

Answer: A

Take the quantities as 2 kg and 3 kg, five kilograms in all.

\[ \frac{2(15) + 3(20)}{5} = \frac{30 + 60}{5} = \text{Rs. } 18 \]

The result is the weighted average, closer to 20 than to 15 because more of the dearer rice is used.

Answer: B

The operation is performed four times in all - once, then three more.

Wine : water \( = 16 : 65 \) means wine is \( \tfrac{16}{81} \) of the cask:

\[ \left(1 - \frac{8}{T}\right)^{4} = \frac{16}{81} \]

\[ \left(1 - \frac{8}{T}\right)^{2} = \frac{4}{9} \quad\Longrightarrow\quad 1 - \frac{8}{T} = \frac{2}{3} \]

\[ \frac{8}{T} = \frac{1}{3} \quad\Longrightarrow\quad T = 24 \text{ litres} \]

The ratio 16 : 65 has to be converted to the fraction \( \tfrac{16}{81} \) of the whole before the formula applies.

Answer: C

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

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

Treat the profit percentages as the two "prices".

\[ \frac{\text{8\% part}}{\text{18\% part}} = \frac{18 - 14}{14 - 8} = \frac{4}{6} = \frac{2}{3} \]

So the 1000 kg splits into 2 and 3 of 5 parts:

\[ \text{18\% part} = \frac{3}{5} \times 1000 = 600 \text{ kg} \]

The overall 14% is nearer 18 than 8, so the higher-profit portion must be the larger - a check on which share is which.

Answer: C

Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters.

Answer: A

Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1.

Answer: D

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

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

\[ \frac{\text{Rs. 18 sugar}}{\text{Rs. 24 sugar}} = \frac{24 - 20}{20 - 18} = \frac{4}{2} = \frac{2}{1} \]

The mean of 20 is close to 18, so the cheaper sugar dominates at twice the quantity.

Answer: D

Let \( x \) litres of the Rs. 40 oil be mixed with 240 litres of the Rs. 60 oil.

\[ \frac{40x + 60(240)}{x + 240} = 52 \]

\[ 40x + 14{,}400 = 52x + 12{,}480 \]

\[ 1920 = 12x \quad\Longrightarrow\quad x = 160 \text{ litres} \]

Alligation confirms it: \( \tfrac{x}{240} = \tfrac{60-52}{52-40} = \tfrac{8}{12} = \tfrac{2}{3} \), and \( \tfrac{2}{3} \times 240 = 160 \).