Alligation or Mixture problems
In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?
S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%. C.P. of 1 kg of the mixture = Rs. ❨ 100 x 68.20 ❩ = Rs. 62. 110 By the rule of alligation, we have: Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind. Rs. 60 Mean Price Rs. 62 Rs. 65 3 2 ∴ Required ratio = 3 : 2.
Ratio of milk and water = 2 : 1
Quantity of milk = 60 X 2/3 = 40 litre
Quantity of water = 20 litre
To make ratio, 1: 2, we have to double the water that of milk
So, water should be 80 litre.
That means 80 ? 20 = 60 litre water to be added.
The given solution has 75% milk.
Milk to be added has 100% milk.
Milk should be added to the given mixture in the ratio 15 : 10 or 3 : 2
Quantity of milk to be added = (3 / 2) × 6 = 9 liters.
Given rate of wheat at cheap = Rs. 2.90/kg Rate of wheat at cost = Rs. 3.20/kg Mixture rate = Rs. 3/kg Ratio of mixture = 2.90 3.20 3 (3.20 - 3 = 0.20) (3 - 2.90 = 0.10) 0.20 : 0.10 = 2:1 Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.
Suppose the can initially contains 7x and 5x of mixtures A and B respectively. Quantity of A in mixture left = ❨ 7x - 7 x 9 ❩ litres = ❨ 7x - 21 ❩ litres. 12 4 Quantity of B in mixture left = ❨ 5x - 5 x 9 ❩ litres = ❨ 5x - 15 ❩ litres. 12 4 ∴ ❨ 7x - 21 ❩ 4 = 7 ❨ 5x - 15 ❩ + 9 4 9 ⟹ 28x - 21 = 7 20x + 21 9 ⟹ 252x - 189 = 140x + 147 ⟹ 112x = 336 ⟹ x = 3. So, the can contained 21 litres of A.
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