Example 6: A can complete a job in 10 days. B in 12 days and C in 15 days. They all started the work together but A had to leave the work after 2 days and B left the work 3 days before the job was completed. How long did the work last?
Answers
Answer:
Step-by-step explanation:
All of them worked for the first 2 days, so work done was 1/2. ( 2 * Sum of reciprocals of 10, 12 and 15 )
Let the total time taken be 'x'
B and C worked together for x-2-3 = x-5
So, total work done by them was [ 9(x-5) / 60 ].
C worked alone for the last 3 days
So, worked done was 1/5.
Add all of them and equate them to one ( 1 )
( 1/2 ) + ( 9x-45/60 ) + ( 1/5 ) = 1
( 30 + 9x - 45 + 12 ) / 60 = 1
9x - 3 = 60
Therefore, 9x = 63
Therefore x= 7
Therefore, total time taken to complete the work was : 7 days
HOPE THIS HELPED YOU : )
1/10+1/12+1/15=15/60=4
It means if all work together they can complete the work in 4 days. But they worked together for 2 days only. So half the work done ie 1/2.
C alone worked last 3 days which means 3/15=1/5 work was completed by him alone.
So in these 5 days they actually completed
1/2 + 1/5 =7/10
Remaining work 3/10 were done by b and c together in middle. Here itself you can come to a conclusion that answer is either 5 plus 1 or 5 plus 2. Check options. If both 6 and 7 are there then take next step
1/12+1/15 = 9/60 is one day work of b and c
Total time taken to complete full work by both is 60/9
But only 3/10 work is needed to be done
Hence 60/9*3/10=2 days
Add this to already found 5 days then you will get answer ie 7 days.