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Let efficiency of ‘B’ is ‘x’ units per day
Efficiency of ‘A’ = 1.40 × x = 1.4x units per day
Efficiency of ‘C’ = 0.60 × x = 0.6x units per day
Total work = (1.4x + x + 0.6x) × 35 = 105x units
Amount of work done by ‘A’ and ‘B’ together in 40 days = 40 × (1.4x + x) = 96x units
Desired time = (105x – 96x)/0.6x = 15 days
(100.01% of 44.89) ÷ 14.98 = √? - √48.98
25.11% of 199.99 + √143.97 ÷ 6.02 = ?
4.93% of 539.92-48% of 4700=?-8330.33
[(5/9 of 719.87) + (59.73% of 450.31)] ÷ (√168.79 - 3/4 of 63.94) = ?
(20.23% of 780.31) + ? + (29.87% of 89.87) = 283
(363.89% of 224.98 – 319.86% of 134.94) ÷ ? = √(134.88 ÷ 15.25)
? + 96.18 – 15.02 = 118.98 + 31.09
(21.02% of 600.15 ) × 14.95 = ? 2 + 29.99 × 3456 ÷ 1152
(439.98 ÷ 10.99) × 23.98= ? × (23.98) 2 ÷ 47.98
24.99% of 1619.78 + (1259.84 ÷ 12.24) = ? × 16.98