three persons a ,b and c decide to work in such a way that a and c work together on 1st day ,b alone on 2 nd day , a and c on 3 rd and so on on alternate days .how many days will be required to complete the whole work if c and d alone finished the whole work in 20 days and 34 days respectively and b is 70% more efficient than D but 20% less efficient than A
Answers
Answer:
Step-by-step explanation:
Total work to be done = LCM(6,8,12) = 24 units
A can do 4 units per day
B can do 3 units per day
C can do 2 units per day
B+C in 1 day = 5 units
for 2 days, 10 units will be completed by B and C.
Remaining is 14 units.
Now A and B work together, They can do 4+3 = 7 units per day.
Then 14 units will be completed in 2 days.
Answer:
14 (1/2) days
Step-by-step explanation:
work is done by B in = 100/170 * 34
=20 days
work is done by A in = 100/80 * 20
= 25 days
A=1/25: B=1/20: c=1/20.
1st day - A+C = 1/25 + 1/20 = 9/100
2nd day B = 1/20
so one pair of days work done = 9/100 + 1/20
= 14/100 = 7/50
7 pairs of days (or) 14 days of work done is = 7 * 7/50 = 49/50
Remaining work = 1 - 49/50
= 1/50
so 15 th days is A tern......
A's one day work 1/25
1/50 of work is done by A in = 1/50 * 25 = 1/2 days
total time taken 14 (1/2) days