Alligation or Mixture problems
A merchant has 160 kgs of wheat. He sells a part of it at 10% profit and the rest of 6% loss. If he incurs 4% loss on the whole, find the quantity for each part.
According to figure we can find that the ration would be 1 : 7.Quantity sold at 10% profit = 1 / (1 + 7)× 160 = 20 kgs. Quantity sold at 6% loss = (160 ? 20) = 140 kgs.
Cost price of the mixture = 15 × (100 / 180) = Rs. 25/3 per kg
(Quantity of rice @ Rs. 8 per kg) / (Quantity of rice @ Rs.10 per kg) = (5 / 3) / (1/3) = 1/5
Quantity of rice @ Rs. 10 per kg = 25 × (1/ 5) = 5 kgs.
According to figure we find that the ratio will be 3 : 1.Quantity sold at 20% profit = 3 / (3 + 1) × 50 = 37.5 kgs. Quantity sold at 40% profit = (50 ? 37.5) = 12.5 kgs.
pool : kerosene 3 : 2(initially) 2 : 3(after replacement) R e m a i n i n g Q u a n t i t y I n i t i a l Q u a n t i t y = 1 - R e p l a c e d Q u a n t i t y T o t a l Q u a n t i t y (for petrol) 2 3 = 1 - 10 k => K = 30 Therefore the total quantity of the mixture in the container is 30 liters.
Let initially milk and water in container B is 3x liter and x liter respectively
Now, 3x + (8/9) × 18 ? x ? (1/9) × 18 = 30
3x + 16 ? x ? 2 = 30
x = 8
Initial quantity is container B = 8 (3 + 1) = 32 Liter.