A can do of a work in 6 days and B can do 30% of the same work in 3 days. They worked for 4 days. The remaining work was completed by C alone in 6 days. B and C together can complete 70% of the original work in:
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. 41/2 days.
A can do a work in 6 days and B can do 30% of the same work in 3 days. They worked together for 4 days. The remaining work was completed by C alone in 6 days.
Total work = Work done per day × Total time taken
LCM (3, 4, 6) = 12
Let the total work be five times the LCM of the time taken by different workers.
So. the total work = 12 × 5 = 60 units
The efficiency of A = (2/5 × 60)/6 = 24/6 = 4 units
The efficiency of B = (60 × 30%)/3 = 18/3 = 6 units
Work done by A and B together in 4 days = (6 + 4) × 4 = 40 units
Remaining work = (60 - 40) = 20 units
The efficiency of C = 20/6 = 10/3 units
Now, B and C together can complete 70% of the original work in = (60 × 70%)/(6 + 10/3) = 42/(28/3) = 41/2 days