PSA_LEVEL-UP_QNR17. A and B can do a job in 20 and 30 days respectively. If both of them work on the job together, when would they finish 75% of it?
1) 8 days
2) 9 days
4) 12 days
Let us assume work to be of 1 unit.
A’s one day work = 1/20
B’s one day work = 1/30
In one day, they finish = 1/20 + 1/30 = 1/12
1/12 units of the work is finished in 1 day.
1 unit of work is finished in 12 days.
3/4 units of work is finished in [3/4] * 12 = 9 days. Option (2).
Let us assume work to be of 60 units [LCM of involved numbers 20 and 30 is 60].
Task is to finish 75% of the work i.e 75% of 60 = 45 units.
A’s one day work = 60/20 = 3 units
B’s one day work = 60/30 = 2 units
Together they finish, 5 units each day. To finish 45 units they would need 9 days. Option (2).
Aspirants can solve most of the Time and Work questions using LCM method to do away with any need of solving it on paper. And they can solve it quickly mentally this way.
Shortcut to calculate number of days required by A and B when both are working together = nA * nB / (nA + nB)
Where, nA = number of days A requires to finish the work alone.
nB = number of days B requires to finish the work alone.
Here, the value comes out to be = 20*30/(20+30) = 600/50 = 12 days.
Since, the question asks about “75% of the total work”, the time taken would be 3/4th of 12 days or 9 days. Option (2).