Alligation or Mixture Questions and Answers

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.

Answer: D

Currently the mixture is in ratio of A/B: 7:5

=> This means fraction of A in the solution: 7/12

=> Fraction of B in the solvent: 5/12

=>New concentration of A= 7-(7/12)9

=>New concentration of B= 5-(5/12)9+x

=> (7-(7/12)9)/(5-(5/12)9+x) = 7/9

=>x=21

Answer: B

As Ram replaces 10% of kerosene with water, the amount of kerosene left is 90%.
As the replacement is done four times,
After 1st replacement, kerosene left is : 0.9*100 = 90 litres
After 2nd replacement, kerosene left is : 0.9*90 = 81 litres
After 3rd replacement, kerosene left is : 0.9*81 = 72.9 litres
After 4th replacement, kerosene left is : 0.9*72.9 = 65.61 litres

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.