Alligation or Mixture Questions and Answers

Answer: C

Work back from the selling price to the cost of the mixture.

\[ \text{CP} = \frac{9.24}{1.10} = \text{Rs. } 8.40 \text{ per kg} \]

Let \( x \) kg of the Rs. 9 sugar be mixed with 27 kg of the Rs. 7 sugar:

\[ \frac{9x + 7(27)}{x + 27} = 8.40 \]

\[ 9x + 189 = 8.4x + 226.8 \quad\Longrightarrow\quad 0.6x = 37.8 \]

\[ x = 63 \text{ kg} \]

The gain is stated on the mixture, so the selling price must be converted to a cost price before any mixing calculation.

Answer: D

As vessel A contains 150 litres of mixture of milk and water in the ratio 3:1, the quantity of milk is 3*150/4 litres, and that of water is 150/4 litres.
Similarly, as vessel B contains 200 litres of mixture of milk and water in the ratio 5:2, the quantity of milk is 5*200/7, and that of water is 2*200/7 litres.
After mixing the total quantity is 350 litres.
The quantity of milk is {(3*150/4) + (5*200/7)} = 7150/28
The quantity of water is {(150/4) + (2*200/7)} = 2650/28
The ratio of the resulting mixture = 7150/2650 = 143 : 53

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: 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

Correct answer: 60 litres

In 60 litres divided 2 : 1 there are 40 litres of milk and 20 of water. The milk does not change when water is added - that is the key to every question of this kind. For a final ratio of 1 : 2, the water must be twice the milk, so it must become \( 2 \times 40 = 80 \) litres. The water to add is therefore \[ 80 - 20 = 60 \text{ litres}. \] Checking: the vessel then holds 40 litres of milk and 80 of water, a total of 120 in the ratio 1 : 2. The general method is to identify the quantity that stays constant, express the required ratio in terms of it, and subtract what is already present. Working with the total instead invites error, because the total changes while the milk does not. The same approach handles questions where a quantity is removed rather than added.

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

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 \( A = 30 \), \( x = 5 \) and \( n = 3 \):

\[ 30\left(1 - \frac{5}{30}\right)^{3} = 30\left(\frac{5}{6}\right)^{3} \]

\[ = 30 \times \frac{125}{216} = 17.36 \text{ litres} \]

Subtracting \( 3 \times 5 = 15 \) litres would give 15 L, but the later withdrawals remove water as well as milk, so more milk survives than that.