Question

A can do a piece of work in 72 days. He works at it for 8 days and then B alone finishes the remaining work
in 48 days. In how many days will A and B finish the work by working together?​

Answer 1

Answer:

Step-by-step explanation:

2(A+B+C)  =12 units/day.

(A+B+C) = 6 units/day

In 3 days.

A+B+C will do

=6×3=18 units

In 3 days the part of work will finish in

=3daysworkT.W=18360

=120

Answer 2

Step-by-step explanation:

Let the total work done is X.

A's 1 day work is = X/72..................(i)

A's 8 days work = 8*X/72 = X/9

Remaining work = X-X/9 = 8X/9

which is completed by B in 48 days so

B's 1 day work will be = 8*X/9*48 = X/54............... (ii)

So by eqn (i) and (ii) it is Clear that A can do X work in 72 days and B can do that in 54 days,

Now the question become simpler as,

We take the lcm of 72 and 54 which is 216 and divide it by their multiples,

So A+B together can do it in = 216/(3+4) = 216/7Which is the required answer