A, B and C alone can complete a work in
24, 30 and 20 days respectively. They start the
work together. However 6 days and 3 days
before the completion of the work A and B leave
in that order. In how many days the work will be
completed?
Answers
Answer:
The correct answer is 10.8 days.
Step-by-step explanation:
We are going to solve this using the LCM method.
Number of days taken by the three-
A = 24
B = 30
C= 20
Finding their LCM, we get- 120 ( let this be the total work)
Now finding the efficiency of each( Work done in one day) -
A= 120/ 24 = 5
B = 120/ 30 = 4
C= 120/ 20 = 6
A left 6 days and B left 3 days before the completion of the work, hence, in the 3 days after A left, work was done by both B and C
Work was done in 3 days by B+ C = (4+6)*3 = 30
Also, in the last 3 days, work was done only by C so work done in those 3 days = 6*3 = 18
So the amount of work left= 120-( 18+30) = 72
This work was done by all the three so number of days taken by them to do this= 72/15 = 4.8 days
Total number of days required for the entire work = 4.8 + 6 = 10.8 days or 10 4/5 days.
Answer:
10.8 Days
Step-by-step explanation:
A's 1 day work = 1/24
B's 1day work = 1/30
C's 1 Day work = 1/20
A, B & C's 1 day work = 1/24 + 1/30 + 1/20 = (1/120)(5 + 4 + 6) = 15/120 = 1/8
B & C's 1 day work = 1/30 + 1/20 = (1/60)(2 + 3) = 5/60 = 1/12
Let say work is completed in D days
D-6 days A, B & C worked = (D-6)/8
6 -3 = 3 Days B & C worked Together = 3/12 = 1/4
Last 3 Days C worked only = 3/20
(D-6)/8 + 1/4 + 3/20 = 1
=> 5(D-6) + 10 + 6 = 40
=> 5D - 30 = 24
=> 5D = 54
=> D = 54/5 = 10.8 Days