Alligation or Mixture problems
In a class of 30 students , the average weight of boys is 20 kg and the average weight of the girls is 25 kg . The fraction of boys out of the total students of the class is
Let us assume the number of boys = B and number of girls = G.According to question,B + G = 30Lets us assume total weight of boys = W1 and total weight of girls = W2average weight of boys = total weight of boys/number of boystotal weight of boys/number of boys = 20W1/B = 20W1 = 20Baverage weight of girls = total weight of girls/number of girls25 = W2/GW2 = 25GData is not sufficient to solve the equation.since we do not know either the average weight of the whole class or the ratio of no. of boys to girls.
If the two alloys are mixed, the mixture would contain 15 gms of each metal and it would cost Rs. (150 + 120) = Rs. 270.
Cost of (15 gms of metal A + 15 gms of metal B) = Rs. 270
Cost of (1 gm of metal A + 1 gm of metal B) = Rs. (270 / 15) = Rs. 18
Cost of 1 gm of metal B = Rs. (18 ? 6) = Rs. 12
Average cost of original piece of alloy = (150 / 15) = Rs. 10 per gm.
Quantity of metal / A Quantity of metal B = (2 / 4) = (1 / 2)
Quantity of metal B = 2 (1 + 2) × 15 = 10 gms.
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