Question

$A$ can do a piece of work in $40$ days. He worked at it for $5$ days, then $B$ finished it in $21$ days. The number of days, that $A$ and $B$ take together to finish the work, are

• A. $15$ days
• B. $13$ days
• C. $10$ days
• D. $14$ days

Answer 1

Answer: (A)

Part of work done by $A$ in $1day:\frac{1}{40}$

$\because A$ worked for $5$ days.

$\therefore$ Part of work done by $A$ in $5day:\frac{1}{40}\times 5=\frac{1}{8}$

$\because A$ left after $5$ days and remaining work was completed by $B$ in $21$ days.

Remaining work $=1-\frac{1}{8}=\frac{8-1}{8}=\frac{7}{8}$

$\because \frac{7}{8}$ th part of work completed by $B$ in $21$ days.

$\therefore 1$ complete work was completed by $B$ in $\frac{21}{\frac{7}{8}}$ days.

$=\frac{21\times 8}{7}$ days

$=24$ days.

$\because A$ complete the whole work in $40$ days and $B$ complete the whole work in $24$ days.

$\therefore$ if $A$ and $B$ work together

$\therefore$ time taken by $A$ and $B$ togethere to complete the work

$=\frac{40\times 24}{40+24}$ days (Formula $A$ complete work in $X$ days. $B$ complete work in $Y$ days. $A$ and $B$ can complete work in $\left(\frac{X\times Y}{X+Y}\right)$

$=\frac{40\times 24}{64}$ days

$15$ days.