Question
A buffalo alone can plough field ‘A’ in 10 days. A
Bull alone can plough the field ‘A’ in 15 days. Find the number of days taken by 3 bulls and 2 buffalo to together plough field ‘B’ that is 20% larger than field ‘A’?Solution
Let the total work done to plough field ‘A’ = 30 units (LCM of 10 and 15) Then, efficiency of a buffalo = (30/10) = 3 units/day Efficiency of a bull = (30/15) = 2 units/day Total work required to plough field ‘B’ = 30 × 1.20 = 36 units Combined efficiency of 3 bulls and 2 buffalo = 2 × 3 + 3 × 2 = 12 units So, number of days required to plough field ‘B’ = 36/12 = 3 days
What approximate value will replace the question mark (?) in the following?
√40...
13.99% of 399.99 ÷ 28.17 = ? ÷ 25.15
245.67 + 20.05² + ?³ = √961.89 * 34.02
? + 39.99 X 3.01 = √624.99 X 10.01% of 49.99
Find the approximate value of Question mark(?) for given equation.
(563.84 ÷ 11.92) × 15.08 + 244.95 = ? × 10.01
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.)...
?2/3 = 33.33% of 107.99 + 45.45
480 ÷ 10 + 18 % of 160 + ? * 9 = 60 * √36
74.99 x 175.01 - 75.12 x 154.98 + + 39.98) x 3.98 = (785.01 + 265.01) x 1.99
- 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.)
Relevant for Exams:
