Question

$A$ and $B$ together can build a wall in $10$ days; $B$ and $C$ working together can do it in $15$ days; $C$ and $A$ together can do it in $12$ days. How long will they take to finish the work, working altogether? Also find the number of days taken by each to do the same work, working alone.

Answer 1

Given,

$(A+B)$’s can build a wall in = $10$ day
So, $(A+B)$’s $1$ day's work $=1/10$
$(B+C)$’s can build a wall in $=15$ days
So, $(B+C)$’s $1$ day's work $=1/15$
$(C+A)$’s can build a wall in =$12$ days
So, $(C+A)$’s $1$ day's work $=1/12$
$\therefore \left[\right(A+B)+(B+C)+(C+A\left)\right]$’s $1$ day work $=\frac{1}{10}+\frac{1}{15}+\frac{1}{12}=\frac{6+4+5}{60}=\frac{15}{60}=\frac{1}{4}$
$\Rightarrow 2(A+B+C)$’s $1$ day's work $=\frac{1}{4}$
$\Rightarrow (A+B+C)$’s $1$ day's work $=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$
Hence, $(A+B+C)$ can build the wall in $=8$ days

Now let us find the number of days taken by each to do the same work, working alone:
$\left[\right(A+B+C)-(B+C\left)\right]$’s $1$ day work $=\frac{1}{8}-\frac{1}{15}=\frac{15-8}{120}=\frac{7}{120}$
$\Rightarrow A$’s $1$ day work $=\frac{7}{120}$
So, $A$ can build the wall in $=\frac{120}{7}$ days $=17\frac{1}{7}$ days
Again, $\left[\right(A+B+C)-(A+C\left)\right]$’s $1$ day's work $=\frac{1}{8}-\frac{1}{12}=\frac{3-2}{24}=\frac{1}{24}$
$\Rightarrow B$’s $1$ day's work $=\frac{1}{24}$
So, $B$ can build the wall in $=24$ days
Again, $\left[\right(A+B+C)–(A+B\left)\right]$’s $1$ day work $=\frac{1}{8}-\frac{1}{10}=\frac{5-4}{40}=\frac{1}{40}$
$\Rightarrow C$’s $1$ day's work $=\frac{1}{40}$
So, $C$ can build the wall in $=40$ days
$\therefore$ $A$ can complete the work in $17\frac{1}{7}$ days.
$B$ can complete the work in $24$ days.
$C$ can complete the work in $40$ days.