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Start learning 50% faster. Sign in nowLet the total work be 100 units. {LCM (20 and 50)}
So, combined efficiency of 'I' and 'J' = 100 ÷ 20 = 5 units/day
Efficiency of 'I' = 100 ÷ 50 = 2 units/day
So, efficiency of 'J' = 5 - 2 = 3 units/day
So, time taken by 'J' to do the work alone = 100 ÷ 3 = (100/3) days
20.11 × 6.98 + 21.03 × 6.12 – 37.95 + 92.9 × 5.02 =?
5 (1/2)% of 3599.67 + {18.11% of 2499.67} + √255.11 - 10% of 1799.99 - (63.01 – 47.08) = ?
630.11 ÷ 20.98 × 5.14 – 125.9 = √?
54.97% of 620.08 + 16.13 × 11.11 – 829.91 ÷ 5 = ?
(18.31)2 – (13.68)2 + (2344.20 + 82.32) ÷ ? = 229.90
116.90 ÷ (77.81 ÷ 6.06) + 6.32 = (15.12% of ?) ÷ 4.652
1242.12 ÷ √530 + 1139.89 ÷ 14.91 = ? + 45.39
? + 157.99 – 101.01 = 25.01 × 5.98