Question

A and B can finish a work in 16 days while a alone can do the same work in 24 days. In how many days B alone will complete the work?​

Answer 1

Answer:

A+B=16 days

A=24 days

So, let's take LCM of 16 and 24. That's 48.

 This 48 unit is the total work to be done.

So, A and B do (48/16) = 3 unit work/day …(1)

And, A does (48/24) = 2 unit work/day …(2)

Equation(1)-(2)

{A+B-A=3–2=1}

B does (3–2)= 1 unit/day

So, Total days required = (Total work)/(unit per day)

 (48/1)=48 days.

Please mark as brainliest!!!!    

Hope it will help you!!!!

Answer 2

Explanation:

A + B ---- 16 days

A ---- 24 days

Work done by A + B in 1 day = 1/16

Work done by A in one day = 1/24

Let the number of days taken by B is X

then, work done by B in one day = 1/X

1/x + 1/24 = 1/36

1/x = 1/36 - 1/24

1/x = 5/72

Therefore, number of days in which B alone can complete the work is 72/5 ( reciprocal of 5/72)