Let the total work = 100 units Efficiency of (A + B) = 100/20 = 5 units/day Efficiency of ‘A’ = 100/25 = 4 unit/day Therefore, efficiency of ‘B’ = 5 – 4 = 1 units/day Efficiency of ‘C’ = 1.6 × 1 = 1.6 units/day Required time taken = 100/1.6 = 62.5 days
? ÷ [35% of 379 - 34(4/5)] = 0.4
(25 × 12 + 30 × 8 – 22 × 10) = ?
25.6% of 250 + √? = 119
12 % of 72 × 25 – (x ÷ 20) × (16 ÷ 24) × 36 + 1/5 × x = (4 ÷ 12) × 36 ÷ 1/4
(75 + 0.25 × 10) × 4 = ?2 - 14
1540 ÷ 7 - 184 ÷ 8 = ?
Simplify the following expression.
(8.25)3 - 3× (8.25)2 × 0.25 +24.75 × (0.25) ² - (0.25)3 / (4×4×4)
...5555 ÷ 11 ÷ 5 = 100 + ?
(12 × 48 ÷ 6) ÷ 2 + ? = 106