Alligation or Mixture problems
The average age of 6 members of a family is 20 years. If the age of the servant is included, then the average age increase by 25%. What is the age (in years) of the servant?
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.
Given rate of wheat at cheap = Rs. 2.90/kg Rate of wheat at cost = Rs. 3.20/kg Mixture rate = Rs. 3/kg Ratio of mixture = 2.90 3.20 3 (3.20 - 3 = 0.20) (3 - 2.90 = 0.10) 0.20 : 0.10 = 2:1 Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.
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.
Let the amount of juice and water in original mixture '4x' litre and '3x' litre respectively. According to given data, 4x/3x+6 =8/7 28x=24x+48 28x?24x=48 4x = 48 x = 12 Amount of juice = 4x = 4×12 = 48 litre.
Ratio of Milk and water in a vessel A is 4 : 1 Ratio of Milk and water in a vessel B is 3 : 2 Ratio of only milk in vessel A = 4 : 5 Ratio of only milk in vessel B = 3 : 5 Let 'x' be the quantity of milk in vessel C Now as equal quantities are taken out from both vessels A & B => 4/5 : 3/5 x 3/5-x x - 4/5 => 3 5 - x x - 4 5 = 1 1 (equal quantities) => x = 7/10 Therefore, quantity of milk in vessel C = 7 => Water quantity = 10 - 7 = 3 Hence the ratio of milk & water in vessel 3 is 7 : 3
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