Question

A and b can do a work in 18 and 24 days respectively. they worked together for 8 days and then a left. the remaining work was finished by b in.

Answer 1

Solution:-
A can do piece of work in 18 days
One day work of A is 1/18
B can do the piece of work in 24 days
One day work of B =1/24
So,
(A+B)'s one day work=
$= > \frac{1}{18} + \frac{1}{24}$
$= > \frac{4 + 3}{72} \\ = > \frac{7}{72}$
They worked for 8 days
So,
Worked complete =
$= > \frac{7}{72} \times 8 \\ = > \frac{7}{9}$
Now,
Remain work=
$= > 1 - \frac{7}{9} \\ = > \frac{2}{9}$
So,
the remain work done by B in
$= > \frac{2}{9} \div \frac{1}{24} \\ = > \frac{2}{9} \times \frac{24}{1} \\ = > \frac{16}{3}$
So,the remain work done by B in 16/3 days

Answer 2

Answer:

In   $\frac{16}{3}$ days B finish the work left by A.

Step-by-step explanation:

No of days A took to complete work = 18 days

Work Done by A in a day = $\frac{1}{18}$

No of days A took to complete work = 24 days

Work Done by A in a day = $\frac{1}{24}$

So,

Work done by A and B both together in a day = $\frac{1}{18}+\frac{1}{24}=\frac{7}{72}$

Work done by A and B both together in 8 day = $8\times\frac{7}{72}=\frac{7}{9}$

Remaining work = $1-\frac{7}{9}=\frac{2}{9}$

So, No of days taken by B to do remaining work = $\frac{2}{9}\times24=\frac{16}{3}$

Therefore, In   $\frac{16}{3}$ days B finish the work left by A.