Question

A and B together can do a piece of work in 8 days. B alone can do it in 12 days. B alone works at it for 4 days. In how many more days after that could A alone complete it.​

Answer 1

Answer:

$\huge\underline\bold {Answer:}$

A alone can do

1/8 – 1/12 = 1/24 of the work in 1 day, i.e., A alone can do the same work in 24 days.

B's days' work = 1/12

=> B's 4 days' work

= 4/12 = 1/3

Therefore remaining 2/3 of the work is done by A alone in 2/3 × 2/4 = 16 days.

Answer 2

$Answer$

A+B=8 days

B=12 days

So, let's take LCM of 16 and 24. That's 24.

Note: 24 unit is the total work

So, A and B do (24/8) = 3 unit work/day …(1)

And, B does (24/12) = 2 unit work/day …(2)

Equation(1)-(2)

{A+B-B=3–2=1}

A =(3–2)= 1 unit/day

Condition: B work 4 days alone and remaining work done by A.

So Work done by B= 2*4=8 units

Now, Left total work = 24–8=16 units

Remaining work done by A= 16/1=16 days

Formula: Total days required = ( left Total work)/(unit per day)

Finally answer = 16 days.