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.
Profit (%) = 9.09 % = 1/11 Since the ratio of water and milk is 1 : 11, Therefore the ratio of water is to mixture = 1:12 Thus the quantity of water in mixture of 1 liter = 1000 x (1/12) = 83.33 ml
General Formula: Final or reduced concentration = initial concentration x 1 - a m o u n t b e i n g r e p l a c e d i n e a c h o p e r a t i o n t o t a l a m o u n t n where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time. Therefore, 50 × 1 - 5 50 3 = 36.45 L
Rate of rice of quantity 280 kg = Rs. 15.60/kg Rate of rice of quantity 120 kg = Rs. 14.40/kg He want to earn a profit of Rs. 10.45/kg Rate of Mix to sell to get profit of 10.45 = 280 x 15 . 60 + 120 x 14 . 40 280 + 120 + 10 . 45 4368 + 1728 400 + 10 . 45 = > 15 . 24 + 10 . 45 = 25 . 69
By the rule of alligation: Cost of 1 kg pulses of 1st kind Cost of 1 kg pulses of 2nd kind Rs. 15 Mean Price Rs. 16.50 Rs. 20 3.50 1.50 ∴ Required rate = 3.50 : 1.50 = 7 : 3.
Ratio of milk and water = 2 : 1Quantity of milk = 60 X 2/3 = 40 litreQuantity of water = 20 litreTo make ratio, 1: 2, we have to double the water that of milkSo, water should be 80 litre.That means 80 ? 20 = 60 litre water to be added.