Alligation or Mixture Questions and Answers

Answer: B

Correct answer: 5 litres

The mixture contains \( 40 \times 0.10 = 4 \) litres of water and 36 litres of milk. The milk is unchanged by adding water, so work from it. If water is to be 20% of the new mixture, milk must be 80%, and \[ 36 = 0.80 \times \text{new total} \;\Rightarrow\; \text{new total} = \frac{36}{0.8} = 45 \text{ litres}. \] The water to add is therefore \[ 45 - 40 = 5 \text{ litres}. \] Checking: 45 litres containing 9 of water is 20% water, and \( 9 - 4 = 5 \) litres were added. The method to carry away is to anchor on the quantity that does not change and express the target percentage in terms of it rather than of the component being added. Working forwards from the water - asking what makes 4 litres into 20% - gives 20 litres total and the wrong answer, because it ignores that the added water also enlarges the total.

Answer: D

Let the quantities of A and B mixed be 3x kg and 7x kg.

Cost of 3x kg of A = 9(3x) = Rs. 27x

Cost of 7x kg of B = 15(7x) = Rs. 105x

Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x

Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66

Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50

Answer: D

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: 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: B

Correct answer: 32.4 litres

Each operation leaves a fixed fraction of the milk behind, so the quantity remaining after \( n \) operations is \[ Q\left(1 - \frac{x}{Q}\right)^{n}, \] where \( Q \) is the original quantity and \( x \) the amount removed each time. Here \[ 40\left(1 - \frac{4}{40}\right)^{2} = 40 \times \left(\frac{9}{10}\right)^{2} = 40 \times \frac{81}{100} = 32.4 \text{ litres}. \] The reason a formula is needed rather than simple subtraction is that the second withdrawal removes a mixture, not pure milk: 4 litres of the diluted contents contains only 3.6 litres of milk. Subtracting 4 twice gives 32, which is the distractor - and the commonest error. Following it through by hand confirms the result. After the first replacement 36 litres of milk remain; the second removes a tenth of that, 3.6 litres, leaving 32.4.

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: D

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

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

Here \( A = 40 \), \( x = 4 \) and \( n = 3 \) (once, then twice more):

\[ 40\left(1 - \frac{4}{40}\right)^{3} = 40(0.9)^{3} \]

\[ = 40 \times 0.729 = 29.16 \text{ litres} \]

Simply subtracting \( 3 \times 4 = 12 \) litres would give 28, but each later removal takes out some water as well as juice.

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.