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ATQ, Let the total task be 200 units {LCM (25 and 40)} So, efficiency of 'Ajay' = 200 ÷ 25 = 8 units/day And efficiency of 'Bijay' = 200 ÷ 40 = 5 units/day Task done in every 2 days = 8 + 5 = 13 units So, task done in 30 days by 'Ajay' and 'Bijay' = (30/2) × 13 = 195 units Time taken by 'Bijay' to do the remaining work = {(200 - 195)/5} = 1 day So, total time taken = 1 + 30 = 31 days
Find the approximate value of Question mark(?). No need to find the exact value.
59.88% of 419.78 + (24.09 × 5) ÷ 3 – √(80.81) = ?
...15.99% of 549.99 ÷ 11.17 = ? ÷ 20.15
10.10% of 999.99 + 14.14 × 21.21 - 250.25 = ?
33.02% of (143.92 + 7 × ?) + 34.03 = 99.88
33.33% of 110.99 = 19.98% × 244.97 - √?
139.88% of 145.09 + 721.93 ÷ 38.13 +? = 29.14 * 8.18
(14.98% of 319.99) - 7.998 = √?
(78.03 + 116.98) ÷ 13.211 = 89.9 – 25.23% of ?
(3/7 of 1049.88 + 44.95% of 799.79) ÷ (√168.89 + 24.77% of 400.11) = ?