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.
Answers
Answered by
1
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.
Answered by
3
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.
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