Arjun is 50% less efficient than the combined efficiency of Bheem and Cherry. If Bheem can complete a task in 12 days and Cherry in 20 days, determine the time required for all three of them working together to complete the task.
ATQ, Let, total work = LCM of (15 and 12) = 60 units Efficiency of ‘Bheem’ = 60/12 = 5 units/day Efficiency of ‘Cherry’ = 60/20 = 3 units/day Efficiency of ‘Arjun’ = (3 + 5) × 0.5 = 4 units/day Combined efficiency of all 3 of them = (3 + 5 + 4) = 12 units/day Required time taken to complete the work = (60/12) = 5 days
(?)2 + 3.113 = 22.92 – 61.03
(15.98% of 399.99) - 6.998 = √?
(124.99)² = ?
64.889% of 399.879 + √? = 54.90% of 799.80 – 44.03% of 400.21
70.008% of 399.98 + ?% of 399.999 = 80.105% of 599.998
? × 32.91 – 847.95 ÷ √16.4 – 13.982 = √24.7 × 24.04
24.11% of 249.99 + √143.97 ÷ 12.02 = ?
√65 of 14.97 + √50 = (12.02)2 - ?
