Max’s efficiency is 2/5th Kendrick’s efficiency. Max starts off three pieces of work and works for a day alone and Kendrick joins him for the next half a day. As Max leaves after that, Kendrick bucks up and doubles his efficiency. If he can finish the remaining work alone in 4 hours, then in how much time could Max and Kendrick together do one piece of work at initial efficiency?
Solution
Correct Answer: Option E
Let Kendrick be able to complete one piece of work in x days.
⇒ In one day, work done by Kendrick alone = 1/x ---(1)
∴ Work done by Max in 1 day = (2/5x) ---(2) (given)
Work done by Max and Kendrick in 1 day = (1/x) + (2/5x) = 7/5x ---(3)
∴ Work done by Max and Kendrick in half a day = 7/10x ---(4)
From 2 and 4,
Total work completed at the end of 1.5 days = (2/5x) + (7/10x) = 11/10x
∴ Remaining work =3- (11/10x) = (30x – 11)/ 10x
If Kendrick doubles his efficiency in a day,
Work done by Kendrick in 1 day = 2/x
If it takes 4 hours (i.e. 1/6th of a day) for him to finish remaining work,
(1/6) × (2/x) = (30x – 11)/ 10x
⇒ x = 43/90
∴ Work done by Kendrick and Max together in 1 day = 7/5x = (7 × 90)/(5 × 43) = 126/43
∴ if it takes n days for them to finish a single piece of work together,
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