Math, asked by Neelam27, 1 year ago

A and B can do it piece of work in 14 days and 21 days respectively. They start work together but 3days before completely work A leave it. In How many days work will be finish?


SreekanthK: Put my answer as best
abhi001: (A+B) 1 day work =1/14+1/21=5/42 now, 3 days work 3*5/42,now remaining work =1-15/42=42-15/42=17/42 so b I day work =1/21 hence remaining days =21*17/42=17/2 now total no of days= 3+17/2=so final answer is 23/2 days i.e 11.5 days only
Neelam27: thx to all

Answers

Answered by Shravani83
2
Suppose A & B together could do work in n days, (n = 8.4).
A left after n-3 days, ie., 5.4 days.
Work left by A is 3/14.
B can do that in 4.5 days.
Total Time is n-3 days + 4.5 = 12.9.

Answered by kvnmurty
2
In 1 day (A+B) can do work = 1/14 + 1/21  = 5/42
Now, 3 days before Completion the work remaining is = 3 days work = 3 * 5/42 = 5 / 14.  This is the work remained when A left the project.
So 1 -  5/14 = 9/ 14 work was done by A & B together.  The time taken for that would be :  (9/14) / (5/42)  = 27 / 5 days = 5.4 days

Now the remaining work 5/14 is done by B alone.
1 day work by B = 1/21
B took (5 /14)  / (1/21) = 15/2 = 7.5 days
Total time to complete project is  5.4 + 7.5  = 12.9 days

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