A and B can do a piece of work in 40 days. After working for 10 days they are assisted by "C' and work is finished in 20 days more. If C' does as much work as B does in 3 days, in how many days A alone can do the work?
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7
a+b can do it in 40 days..
task completed in 10 day = 1/4th of the whole
3/4th part of the work done by a+b+c in 20 daysthen these three can complete a whole amount of work in 20 x4/3=80/3 days
now talking of the efficiency of all three
a+b & a+b+c
40 80/3
120 80 lcm 240
2-effe 3-effe
effe of c=1
then b-effe=1/3
1+1/3+a's effe=3
a effe= 5/9
time taken by a=240/5 x 9=48x9=432days.
task completed in 10 day = 1/4th of the whole
3/4th part of the work done by a+b+c in 20 daysthen these three can complete a whole amount of work in 20 x4/3=80/3 days
now talking of the efficiency of all three
a+b & a+b+c
40 80/3
120 80 lcm 240
2-effe 3-effe
effe of c=1
then b-effe=1/3
1+1/3+a's effe=3
a effe= 5/9
time taken by a=240/5 x 9=48x9=432days.
Answered by
4
Answer:
No of day taken by a to do work is 9/5
Step-by-step explanation:
No of days taken by a and b to do work = 40
Work done by both in a day together = 1/40
Work done in 10 day = 1/4
Work left = 1 - 1/4 = 3/4
No of day taken by a , b and c to do 3/4 work = 20 days
No of days taken by a , b and c to complete whole work = 20 × 4/3 = 80/3
let say efficiency of a & b = 2 and a , b & c = 3
efficiency of c = 1
⇒ efficiency of b = 1/3
⇒ efficiency of a = 5/9
No of day taken by a to do work is 9/5
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