Alligation or Mixture Questions and Answers

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

Correct answer: 25 litres

Let the milk be \( 5x \) and the water \( x \). The milk is unchanged by adding water, which is what makes the problem tractable. After adding 5 litres the water becomes \( x + 5 \), and \[ \frac{5x}{x + 5} = \frac{5}{2} \;\Rightarrow\; 10x = 5x + 25 \;\Rightarrow\; x = 5. \] The milk is therefore \( 5x = 25 \) litres. Checking: 25 litres of milk and 5 of water is 5 : 1; adding 5 litres of water gives 25 and 10, which is 5 : 2. A shortcut is available and worth knowing. Since the milk is constant, write both ratios with the same first term - here both are already 5 - and the water has gone from 1 part to 2 parts, an increase of 1 part. That 1 part equals the 5 litres added, so each part is 5 litres and the milk is 25.

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

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

Correct answer: 1 : 4

The gain arises entirely from the water, which costs nothing but is sold at the price of milk. Let the cost of one litre of milk be 1. If the milkman sells \( 1 + x \) litres of mixture containing 1 litre of milk and \( x \) of water, his cost is 1 and his receipt is \( 1 + x \), so \[ \text{gain} = \frac{x}{1} = 25\% = \frac{1}{4} \;\Rightarrow\; x = \frac{1}{4}. \] The ratio of water to milk is therefore \( \frac{1}{4} : 1 = \mathbf{1 : 4} \). The useful shortcut is that the ratio of water to milk equals the gain expressed as a fraction: a 25% gain means 1 : 4, a 20% gain means 1 : 5, and a 50% gain means 1 : 2. The step that trips people is reading the ratio the right way round - the question asks for water to milk, and reversing it gives 4 : 1, which is also offered in many versions of this question.

Answer: D

Correct answer: 56.89 litres

Each operation leaves the same fraction of wine behind, so after \( n \) operations the quantity remaining is \[ Q\left(1 - \frac{x}{Q}\right)^{n} = 81\left(1 - \frac{9}{81}\right)^{3} = 81 \times \left(\frac{8}{9}\right)^{3}. \] Computing, \[ 81 \times \frac{512}{729} = \frac{41472}{729} = 56.89 \text{ litres}. \] The reason a formula is needed rather than repeated subtraction is that after the first replacement the cask holds a mixture, so each later withdrawal removes water as well as wine. Subtracting 9 litres three times would give 54, which is the distractor and the commonest error. Following it through confirms the result: 72 litres of wine remain after the first replacement, 64 after the second, and \( 64 \times \frac{8}{9} = 56.89 \) after the third.

Answer: D

Total solution quantity 57+43 = 100

Out of 100 we are removing 10 liter (i.e. 10%)

so proportionately from honey we will remove = remove 5.7

& proportionately from water we will remove 10% remove = 4.3

Also, we add 10 liter pure honey

So now honey 57-5.7+10 = 61.3

& Water 43-4.3 = 38.7

water: Honey = 38.7 : 61.3 = 387:613

Answer: A

Let Type-2 acid quantity be T liters.

Final mixture has '60+T' liters

There is no profit no less

So, (Type-I quantity × rate) + (Type-2 quantity × rate) - Mix quality × rate

60 × 32 +T × 23 × 28

T = 48 liters = Type-2 acid quantity

Answer: D

Correct answer: 729 : 271

Each operation leaves the fraction \( 1 - \frac{3}{30} = \frac{9}{10} \) of the milk, so after three operations the milk is \[ 30 \times \left(\frac{9}{10}\right)^{3} = 30 \times \frac{729}{1000} = 21.87 \text{ litres}. \] The total is still 30 litres, so the water is \( 30 - 21.87 = 8.13 \) litres, and the ratio is \[ 21.87 : 8.13 = 729 : 271. \] The ratio can be written down directly without computing the volumes at all: the milk is \( (9/10)^{3} = 729/1000 \) of the whole, so the water is \( 271/1000 \), and the ratio is 729 : 271. That shortcut is the useful one, since the ratio depends only on the fraction replaced and the number of repetitions, never on the actual volume. Subtracting 3 litres three times would give 21 : 9, or 7 : 3, and is wrong because each later withdrawal removes a mixture rather than pure milk.

Answer: B

Correct answer: 3 : 2

By alligation the quantities are in the inverse ratio of their distances from the mean: \[ \frac{\text{cheaper}}{\text{dearer}} = \frac{280 - 220}{220 - 180} = \frac{60}{40} = \frac{3}{2}. \] The ratio is 3 : 2. Checking: 3 kg at 180 and 2 kg at 280 cost \( 540 + 560 = 1100 \) for 5 kg, which is 220 per kg. The rule can be sanity-checked without arithmetic: the mean of 220 is nearer 180 than 280, so more of the cheaper tea must be used - which immediately rules out 2 : 3. That check catches the commonest error in this topic, which is writing the ratio the wrong way round. The alligation diagram makes the direction visible: place the mean in the middle, the two prices at the upper corners, and take the differences diagonally. Each difference belongs to the ingredient on the opposite side.