Question
A buffalo alone can plough field ‘A’ in 50 days. A
Bull alone can plough the field ‘A’ in 75 days. Find the number of days taken by 1 bulls and 3 buffalo to together plough field ‘B’ that is 12% larger than field ‘A’?Solution
Let the total work done to plough field ‘A’ = 150 units (LCM of 50 and 75) Then, efficiency of a buffalo = (150/50) = 3 units/day Efficiency of a bull = (150/75) = 2 units/day Total work required to plough field ‘B’ = 150 × 1.12 = 168 units Combined efficiency of 1 bulls and 2 buffalo = 1 × 2 + 2 × 3 = 8 units So, number of days required to plough field ‘B’ = 168/8 = 21 days
More Time and work Questions
(30 × 0.80)⁴ ÷ (2160 ÷ 60)⁴ × (54 × 16)⁴ = (6 × 4)?+5
3/7 of 504 ÷ 12 + 17 = √?
20% of 150 + 30 X 4 = ? X 5
52% of 400 + √(?) = 60% of 600 - 25% of 400
135÷ 15 x 19 + 14807 = ? + √3249 - √9604
(22² × 8²) ÷ (92.4 ÷ 4.2) =? × 32
15 * 12 + 35% of 80 + 70% of 130 = ?
60 % of 640 - 57 × 2 - 1520 / 38 = ?
- Find the simplified value of the given expression:
10 of 5 ÷ 4 × 2² + √25 – 8 (168 ÷ 12 + 19 × 64)/(22+1) = ?
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