Question
120 litres of mixture ‘A’ (milk + water) contains 50
litres more water than milk. If 20% of mixture ‘A’, 41.5 litres of milk and 80 litres of water are added in an empty jar, then the quantity of milk is what percentage of the quantity of water in the jar?Solution
Let the quantity of milk in mixture ‘A’ = ‘y’ litres Then, quantity of water in mixture ‘A’ = (y + 50) litres So, y + y + 50 = 120 Or, 2y = 120 – 50 = 70 Or, y = 70/2 = 35 So, quantity of milk in mixture ‘A’ = 35 litres And, quantity of water in mixture ‘A’ = 35 + 50 = 85 litres Quantity of milk in 20% of mixture ‘A’ = 35 × 0.20 = 7 litres Quantity of water in 20% of mixture ‘A’ = 60 × 0.20 = 12 litres So, quantity of milk in the jar = 7 + 41.5 = 48.5 litres Quantity of water in the jar = 12 + 60 = 72 litres So, required percentage = (48.5/72) × 100 ≈ 67%
(124.99)² = ?
25.902 × 78.095 + 999.996% of 200.08 + 20.005 % of 7999.997 = ? × 15.008 × 33.009
?² × 55% of (29 + 32 - 41) = 41.66% of 216 + 9
√65 of 14.97 + √50 = (12.02)2 - ?
(32.15)2 – (255.89)(1/2) – (10648.08)(1/3) = ?
2 (1/4)% of 7999.78 + {49.77% of 899.71} + √144.14 - 20% of 1499.83 = ?
? + 163.99 – 103.01 = 24.01 × 6.98
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
124% of 620.99 + 11.65% of 1279.23 = ?
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
