Alligation or Mixture Questions and Answers

Answer: B

Correct answer: 7 : 3

By alligation the quantities are in the inverse ratio of the distances from the mean: \[ \frac{\text{cheaper}}{\text{dearer}} = \frac{20 - 16.50}{16.50 - 15} = \frac{3.50}{1.50} = \frac{7}{3}. \] The ratio is 7 : 3. Checking: 7 kg at 15 and 3 kg at 20 cost \( 105 + 60 = 165 \) for 10 kg, which is 16.50 per kg. The direction can be verified without arithmetic. The mean of 16.50 is much closer to 15 than to 20, so the cheaper variety must dominate heavily - which rules out 3 : 7 and 2 : 5 at a glance. Reversing the ratio is the commonest error in this topic, and this check catches it every time. Multiplying both differences by 2 to clear the decimals - \( 7 : 3 \) from \( 3.5 : 1.5 \) - is the practical step that keeps the arithmetic clean.

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: C

The water contributes nothing to the value, so all the worth comes from the oil.

\[ 325x = 275(x + 240) \]

\[ 325x = 275x + 66{,}000 \]

\[ 50x = 66{,}000 \quad\Longrightarrow\quad x = 1320 \text{ litres} \]

By alligation, water is priced at 0 and \( \tfrac{\text{oil}}{\text{water}} = \tfrac{275-0}{325-275} = \tfrac{275}{50} = 5.5 \), and \( 5.5 \times 240 = 1320 \).

Answer: C

In regard to the problem let after n times liquid in the container is x and liquid removed is y

=>Liquid in container= x

=>Liquid removed= y

After one trial, quantity of pure liquid= x- (1-y/x)

After n trials quantity of pure liquid = x - (1 -y/x)n

Therefore, Putting the values ( x = 40L y = 4L N=3) and solving

=> 40- (1-4/40)3

= 29.16L

Answer: B

Correct answer: ₹34

The cost of a mixture is the weighted average of the component prices: \[ \text{mean} = \frac{2 \times 25 + 3 \times 40}{2 + 3} = \frac{50 + 120}{5} = \frac{170}{5} = 34. \] The mixture costs 34 per kg. The distractor 32.50 is the simple average of 25 and 40, correct only for equal quantities. Since more of the dearer rice is used, the answer must lie above the midpoint - which rules out 32.50 and 33 immediately, and is worth checking before any calculation. This is the direct form of the problem; alligation solves the inverse, finding the ratio when the mean is given. The relation is the same read either way, and being able to move in both directions is what the topic amounts to: \[ \frac{q_{1}}{q_{2}} = \frac{p_{2} - m}{m - p_{1}}. \] Substituting the answer back confirms it: \( \frac{40-34}{34-25} = \frac{6}{9} = \frac{2}{3} \).

Answer: B

Let the capacity of the can be T litres.

Quantity of milk in the mixture before adding milk = 4/9 (T - 8)

After adding milk, quantity of milk in the mixture = 6/11 T.

6T/11 - 8 = 4/9(T - 8)

10T = 792 - 352 => T = 44.

Answer: B

Correct answer: 63 kg

Apply alligation between the two prices with a mean of 8.40: \[ \frac{\text{quantity at } 7}{\text{quantity at } 9} = \frac{9 - 8.40}{8.40 - 7} = \frac{0.60}{1.40} = \frac{3}{7}. \] The cheaper and dearer sugars are therefore in the ratio 3 : 7. The cheaper quantity is 27 kg, so one part is \( 27/3 = 9 \) kg, and the dearer quantity is \[ 7 \times 9 = 63 \text{ kg}. \] Checking: 27 kg at 7 and 63 kg at 9 cost \( 189 + 567 = 756 \) for 90 kg, which is 8.40 per kg. Note that the differences are taken diagonally: the quantity of each ingredient is proportional to the distance of the mean from the other price. Since the mean lies nearer 9 than 7, more of the dearer sugar is needed - which is a useful check before any arithmetic.

Answer: B

Cost Price of 1kg of Type 1 sugar = 6200 p.
Cost Price of 1kg of Type 2 sugar = 7200 p.
Mean Price of 1 kg of mixture = 6450 p.
According to the Rule of Alligation,
(Quantity of Cheaper):(Quantity of Dearer) = (CP of dearer - Mean Price):(Mean Price - CP of cheaper)
Therefore, the required ratio = (7200-6450):(6450-6200) = 750:250 = 3:1.

Answer: B

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

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

\[ \frac{\text{Rs. 15 wheat}}{\text{Rs. 18 wheat}} = \frac{18 - 17}{17 - 15} = \frac{1}{2} \]

The mean of 17 is nearer the dearer price, so more of the Rs. 18 wheat is needed.

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.