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Let total work = 200 {L.C.M of 40 and 25}
Efficiency of ‘R’ = 200 ÷ 40 = 5 units/day
Combined efficiency of ‘R’ and ‘S’ = 200 ÷ 25 = 8 units/day
So, efficiency of ‘S’ alone = 8 – 5 = 3 units/day
Work done by ‘S’ in 8 days = 3 × 8 = 24 units
Remaining work = 200 – 24 = 176 units
Time taken by ‘R’ to finish remaining work = 176 ÷ 5 = 35.2 days
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