Question

A an dB can do a piece of work in 16 days and A along can do it in 24 day, how long will how long will' 'B' alone take to complete the work​

Answer 1

Answer:

A very easy and simple trick.

A+B=16 days

A=24 days

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

Step-by-step explanation:

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)

That is B can do it in (48/1)=48 days.

Answer 2

Answer:

48 days

Step-by-step explanation:

Time taken by A and B = 16 days

A and B together one day work = $\frac{1}{16}$

Time taken by A alone = 24 days

A alone one day work = $\frac{1}{24}$

B alone one day work

= (A and B together one day work) - (A alone one day work)

= $\frac{1}{16} - \frac{1}{24}$

= $\frac{3-2}{48}$

= $\frac{1}{48}$

Time taken by B alone

= 1 ÷ One day work

= 1 ÷ $\frac{1}{48}$

= 48 days