Alligation or Mixture Questions and Answers

Answer: D

Let the quantities of A and B mixed be 3x kg and 7x kg.

Cost of 3x kg of A = 9(3x) = Rs. 27x

Cost of 7x kg of B = 15(7x) = Rs. 105x

Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x

Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66

Profit made in selling 5 kg of the mixture = 25/100 (cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50

Answer: D

The 180 g mixture is 80% milk.

\[ \text{milk} = 144 \text{ g}, \qquad \text{water} = 36 \text{ g} \]

Adding 36 g of pure milk raises both the milk and the total:

\[ \text{milk} = 180, \qquad \text{total} = 216 \]

\[ \frac{180}{216} \times 100 = 83.33\% \]

That figure appears nowhere on the list, so the answer is none of these. The 84% offered is close but not equal.

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: A

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

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

\[ \frac{\text{Rs. 6 rice}}{\text{Rs. 8.75 rice}} = \frac{8.75 - 7.50}{7.50 - 6} = \frac{1.25}{1.50} = \frac{5}{6} \]

The mean of 7.50 sits slightly nearer the dearer price, so a little more of the expensive rice is needed - consistent with 5 : 6.

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: D

B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.

Quantity of water in B = 8k - 5k = 3k.

Quantity of water in container C = 8k - 3k = 5k

Container: A   B    C

Quantity of water: 8k  3k   5k

It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.

5k - 148 = 3k + 148 => 2k = 296 => k = 148

The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.

Answer: D

Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litres

Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres

Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres

As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5.

Answer: B

Let the initial quantity of milk in vessel be T litres.

Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.

Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T 

For the given problem, T = 90, y = 9 and n = 2.

Hence, quantity of  milk finally in the vessel 

= [(90 - 9)/90]2 (90) =  72.9 litres.

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

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.