Alligation or Mixture problems
A milk man sells the milk at the cost price but he mixes the water in it and thus he gains 9.09%. The quantity of water in the mixture of 1 liter is :
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
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.
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 Here amount being replaced denotes the quantity which is to be withdrawn in each time.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
Quantity of alohol in the mixture = 40 x 5/8 = 25 lit Quantity of water = 40 - 25 = 15 lit According to question, Required ratio = 20 - 40 x 20 100 x 5 8 15 - 40 x 20 100 x 3 8 + 40 x 20 100 = 20 15 - 3 + 8 = 1 : 1
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.
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