Question 6

E and F can do a work in 10 days. If E alone can do it in 30 days, F alone can do it in days.

Solution

E and F can do a work =10 days

E alone can do it = 30 days

total work = lcm (10,30) = 30

(E+ F)'s efficiency = $$\frac{work}{time}$$ = $$\frac{30}{10}$$ = 3

E' s efficiency = $$\frac{work}{time}$$ = $$\frac{30 }{30}$$ = 1

so

F's efficiency = 3 - 1 = 2

so , F time taken = $$\frac{work}{time}$$ =$$\frac{30}{2}$$ = 15 days

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