A can do a piece of work in 15 days, B in 20 days. A and B work at it together for 6 days and then C finishes it in 3 days, in how many days could C have done it alon
Answers
Answered by
1
Answer:
10 days
Step-by-step explanation:
A can do in one day: 1/15 of work, B can do in one day: 1/20 of work
A and B completed in 6 days:
6/15+6/20=42/60=21/30 of work
Work completed by C:
1-21/30=9/30=3/10
C completed in one day:
1/3*3/10=1/10
C can do full work in: 1:1/10=10 days
Answered by
0
Answer:
10 days
Step-by-step explanation:
work done by both A and B together in one day
= 1/15+1/20= 7/60
work done by both in 6 days= 7×6/60= 7/10
remaining part= 1-7/10= 3/10
thus 3/10 part of work done by C in 3 days
=> in one day ,work done by C= 3/10÷3= 1/10
=> time for total work done by C alone
= 1÷1/10= 1×10/1= 10 days
hence C finish the total work in 10 days.
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