Question

A B and C can do a work in 20 days 15 days and 12 days respectively they began to work together be left the work after 4 days in what time can a and c complete the remaining work​

Answer 1

Answer:

Step-by-step explanation:

1st day A,B and C can do =1/12+1/15+1/20 = 12/60= 1/5 work

2nd days they do= 2/5 work.

In 1-day A +B can do= 1/12+1/15 =9/60

Then remaining 3/5 of the work can be done by A and B in

(3/5)÷(9/60) = 4 days.

Answer 2

A, B and C can do a work in 20 days, 15 days, and 12 days respectively.

Let's find the no. of days needed for the three to do the work together.

For this, take the reciprocal of the sum of reciprocals of the duration taken by A, B and C each.

(1/20) + (1/15) + (1/12)

=> (3/60) + (4/60) + (5/60)

=> (3 + 4 + 5)/60

=> 12/60

=> 1/5

So they do the work together in 5 days.

But 4 days after starting the work, B left.

In these 4 days, A, B and C together did 4/5 part of the work. So the remaining 1/5 part of the work should be done by A and C together as B has left.

Now let's find the no. of days needed by A and C together for doing the whole work.

(1/20) + (1/12)

=> (3/60) + (5/60)

=> (3 + 5)/60

=> 8/60

=> 2/15

So A and C together could do the whole work in 15/2 days.

No. of days needed for A and C together to complete the whole work = 15/2

No. of days needed for A and C together to complete 1/5 part of the work = (15/2) × (1/5) = 3/2.

So both A and C together can do the remaining work in 3/2 days, i.e., 1 day and 12 hours.

Or we can say that the answer is 36 hours.