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ATQ, Let total amount of work be 525 units Amount of work done by ‘A’ in one day (with increased efficiency) = 525/25 = 21units Original efficiency of ‘A’ = 21/1.40 = 15 units per day Amount of work done by ‘B’ in one day (with increased efficiency) = 525/21 = 25units Original efficiency of ‘B’ = 25/1.25 = 20 units per day Amount of work completed by ‘A’ and ‘B’ together in 9 days = 35 × 9 = 315 units Remaining work = 525 – 315 = 210 units Desired time = 210/15 = 14 day
(342 – 20% of 5280) = ? ÷ 3
324² × 36 ÷ 18⁵ × 1120 =?
1.25 × 36 + 2.75 × 40 = ? × 3.1
172 - 92 + 121 - 74 = ?
30% of 2200 +’?’ x 50 – 1020 = 11x 306 + (√ 2250 ÷ 0.1)
(7/4×18/21)+ (51/7× 28/17) + (25/2 × 48/10) =?
9 × 4 ? ÷ 16 = 144