Alligation or Mixture problems
The ratio of milk and water in 66 kg of adulterated milk is 5 : 1. Water is added to it to make the ratio 5 : 3. The quantity of water added is ?
Let the milk he bought is 1000 ml Let C.P of 1000 ml is Rs. 100 Here let he is mixing K ml of water He is getting 30% profit => Now, the selling price is also Rs. 100 for 1000 ml => 100 : K% = 100 : 30 10 : 3 is ratio of milk to water => Percentage of milk = 10 x 100/13 = 1000/13 = 76.92%
Average money received by each = 118/50 = Rs. 2.36 Therefore, Ratio of No.of boys and girls = 56 : 24 = 7 : 3 Therefore, Number of boys = 50 x (7/10) = 35 Number of girls = 50 - 35 = 15
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.
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.
EXAMIANS