Question

A and B work together and finish a piece of work in 6 days, B alone can do it in 8 days .Supposing B works at it for 5 days,in how many days A alone could finish it?

Answer 1

Answer:

A and B can do the work in 6 days.

They can do 1/6 of work in a day.

Let us write this as A+B=1/6.

A can do it in 8 days.

Let us write this as A=1/8.

Let us solve for B.

B=(1/6)-(1/8)=(8–6)/48=2/48=1/24.

B can do 1/24 work in a day. So, B alone can do the work in 24 days.

Answer 2

Given :

Number of days A and B together takes = 6 days

Number of days in which B can do = 8 days

To find : A alone can do it in = ? days

Solution :

Number of total units of work = LCM = 6 and 8 =24

Number of units A and B together do = $\dfrac{24}{6}=4\ units$

Number of units B does = $\dfrac{24}{8}=3\ units$

Since B works for 5 days,

So, Number of units done by B = $5\times 3=15\ units$

Remaining units = $24-15=9$

Since $4-3=1\ unit$ in 1 day by A

So, 9 units done by A in 9 days alone.

Hence, A takes 9 days to finish it .