Alligation or Mixture Questions and Answers

Answer: A

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

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

\[ \frac{\text{Rs. 15 sugar}}{\text{Rs. 18 sugar}} = \frac{18 - 16}{16 - 15} = \frac{2}{1} \]

So the ratio is \( 2 : 1 \). The mean price of 16 sits closer to 15 than to 18, so the cheaper sugar must dominate - a check on which way round the ratio goes.

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: D

As vessel A contains 150 liters of mixture of milk and water in the ratio 3:1, the quantity of milk is 3 × 150/4 liters, and

that of water is 150/4 liters.

Similarly, as vessel B contains 200 liters 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 liters.

After mixing the total quantity is 350 liters.

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: B

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

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

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

The mixture at 6.30 lies nearer the cheaper price, so the cheaper rice must be the larger share - hence 2 parts of the dearer to 3 of the cheaper.

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: 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.

Answer: C

Let the can hold \( 12k \) litres, with A \( = 7k \).

Drawing off 9 litres removes A in proportion, and adding B keeps the total at \( 12k \):

\[ \text{A left} = 7k - 9 \times \frac{7}{12} = 7k - 5.25 \]

The new ratio 7 : 9 makes A \( \tfrac{7}{16} \) of the total:

\[ \frac{7k - 5.25}{12k} = \frac{7}{16} \quad\Longrightarrow\quad 112k - 84 = 84k \]

\[ 28k = 84 \quad\Longrightarrow\quad k = 3 \]

\[ \text{A initially} = 7 \times 3 = 21 \text{ litres} \]

Answer: B

Start from the composition and follow each step.

\[ \text{water} = 30\% \times 140 = 42 \text{ L}, \qquad \text{syrup} = 98 \text{ L} \]

Selling 20 L removes both in proportion, leaving 120 L:

\[ \text{water} = 36, \qquad \text{syrup} = 84 \]

Adding \( x \) of each and requiring water : syrup \( = 1 : 2 \):

\[ 2(36 + x) = 84 + x \quad\Longrightarrow\quad 72 + 2x = 84 + x \]

\[ x = 12 \text{ litres} \]

Answer: B

The target is 12 L with water : milk \( = 3 : 5 \), so water is \( \tfrac{3}{8} \times 12 = 4.5 \) L.

Let \( x \) litres come from the first can (25% water) and \( 12 - x \) from the second (50% water):

\[ 0.25x + 0.5(12 - x) = 4.5 \]

\[ 0.25x + 6 - 0.5x = 4.5 \quad\Longrightarrow\quad -0.25x = -1.5 \]

\[ x = 6 \quad\Longrightarrow\quad 6 \text{ litres from each can} \]

The target of 37.5% water is exactly midway between 25% and 50%, which is why the two cans contribute equally.

Answer: C

Correct answer: 3 : 2

By the rule of alligation, the two quantities are in the inverse ratio of their distances from the mean price: \[ \frac{\text{cheaper quantity}}{\text{dearer quantity}} = \frac{\text{dearer price} - \text{mean}}{\text{mean} - \text{cheaper price}}. \] Here \[ \frac{45 - 36}{36 - 30} = \frac{9}{6} = \frac{3}{2}. \] So the ratio is 3 : 2 - three parts of the cheaper rice to two of the dearer. The result is worth checking directly: 3 kg at 30 and 2 kg at 45 cost \( 90 + 90 = 180 \) for 5 kg, which is 36 per kg as required. The mean must lie between the two prices for the problem to have a solution, and the larger share always goes to the ingredient whose price is nearer the mean - here the cheaper, since 36 is closer to 30 than to 45. That observation alone rules out 2 : 3 without any arithmetic.