Alligation or Mixture Questions and Answers

Answer: C

Correct answer: 120 litres

Let the capacity be \( V \) litres. The milk initially is \( \frac{3}{5}V \). Drawing off 20 litres of the mixture removes milk in the same proportion, that is \( \frac{3}{5} \times 20 = 12 \) litres, leaving \( \frac{3}{5}V - 12 \). Adding 20 litres of water returns the total to \( V \). For the final ratio to be 1 : 1 the milk must be half the total: \[ \frac{3}{5}V - 12 = \frac{V}{2} \;\Rightarrow\; \frac{3V}{5} - \frac{V}{2} = 12 \;\Rightarrow\; \frac{V}{10} = 12 \;\Rightarrow\; V = 120. \] Checking: 120 litres in the ratio 3 : 2 is 72 of milk and 48 of water. Drawing 20 litres removes 12 and 8, leaving 60 and 40; adding 20 litres of water gives 60 and 60, which is 1 : 1. The key step is that the portion drawn off carries both components in their current proportion.

Answer: A

Remove the profit first to get the cost price of the mixture.

\[ \text{CP} = \frac{68.20}{1.10} = \text{Rs. } 62 \]

\[ \frac{\text{Rs. 60 tea}}{\text{Rs. 65 tea}} = \frac{65 - 62}{62 - 60} = \frac{3}{2} \]

The mean must lie between the two prices; 68.20 does not, which is why it cannot be used directly.

Answer: D

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

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

\[ \frac{\text{Rs. 8.70 sugar}}{\text{Rs. 9.70 sugar}} = \frac{9.70 - 9}{9 - 8.70} = \frac{0.70}{0.30} = \frac{7}{3} \]

The mean of 9 lies much nearer 8.70, so the cheaper sugar dominates - consistent with 7 parts to 3.

Answer: D

Take one unit from each glass and add the parts.

\[ \text{water} = \frac{2}{7} + \frac{4}{11} = \frac{22 + 28}{77} = \frac{50}{77} \]

\[ \text{juice} = \frac{5}{7} + \frac{7}{11} = \frac{55 + 49}{77} = \frac{104}{77} \]

\[ \text{water} : \text{juice} = 50 : 104 = 25 : 52 \]

The question asks for water first, so the ratio is not the reciprocal 52 : 25 - which is offered as a distractor.

Answer: A

Let the amount taken from S1 be 7x
And amount taken from S2 be 13y

(2x + 6y)/(5x + 7y) = 5/8
16x + 48y = 25x + 35y
9x = 13y
x/y = 13/9

Actual ratios of amounts 
               = 7x/13y
               = (7/13) * (13/9)
               = 7/9

Answer: C

Take the vessel as 8 units: 3 of water and 5 of syrup. Drawing off a fraction \( f \) removes both in proportion, and the water poured back keeps the total at 8.

\[ \text{water after} = 3(1 - f) + 8f = 3 + 5f \]

Half water means 4 of the 8 units:

\[ 3 + 5f = 4 \quad\Longrightarrow\quad f = \frac{1}{5} \]

The replacement adds water twice over - it removes syrup and puts water in its place - which is why only a fifth needs drawing off to move the water from \( \tfrac{3}{8} \) to \( \tfrac{1}{2} \).

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: A

Quantity of milk in the mixture = 90/100 (70) = 63 litres.

After adding water, milk would form 87 1/2% of the mixture.

Hence, if quantity of mixture after adding x liters of water, (87 1/2) / 100 x = 63 => x = 72

Hence 72 - 70 = 2 litres of water must be added.

Answer: A

The 80 L contains \( 45\% \times 80 = 36 \) L of milk. Adding \( x \) L of pure milk raises both the milk and the total.

\[ \frac{36 + x}{80 + x} = 0.75 \]

\[ 36 + x = 60 + 0.75x \quad\Longrightarrow\quad 0.25x = 24 \]

\[ x = 96 \text{ litres} \]

The non-milk part stays fixed at 44 L, and it must end up as 25% of the total - so the total becomes 176 L, confirming the 96 L added.

Answer: D

Convert each solution to litres of honey and water.

\[ 28 \text{ L at } 4:3 \rightarrow 16 \text{ honey}, \ 12 \text{ water} \]

\[ 21 \text{ L at } 2:1 \rightarrow 14 \text{ honey}, \ 7 \text{ water} \]

\[ 51 \text{ L at } 9:8 \rightarrow 27 \text{ honey}, \ 24 \text{ water} \]

\[ \text{total: } 100 \text{ L}, \ 57 \text{ honey}, \ 43 \text{ water} \]

Removing 10 L takes away 10% of each, then 10 L of pure honey is added:

\[ \text{honey} = 0.9(57) + 10 = 61.3, \qquad \text{water} = 0.9(43) = 38.7 \]

\[ \text{water} : \text{honey} = 387 : 613 \]

The convenient total of exactly 100 L makes the 10% removal easy to apply.