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Let the entire work will be = (LCM of 40 and 25) = 200 units The efficiency of 'B' = (200)/(40) = 5 units/day The efficiency of 'B' and 'C' together = (200)/(25) = 8 units/day Then, efficiency of 'C' = 8 - 5 = 3 units/day Let‚ 'C' worked alone for 'd' days ATQ, we can say that 8 x 10 + (3 x d) = 200 Or, 3d = (200 - 80) = 120 Or, d = 120/3 = 40 Therefore, 'C' can complete the remaining work in 40 days
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