Question

A and B can do a piece of work in 12 days, B and C can do it in 15 days and C and A can do the same work in 20 days. Find the number of days in which A alone can do the same work.​

Answer 1

Given:

A and B can do work = 12 days.

B and C can do work = 15 days.

C and A can do work = 20 days.

To find:

In how many days A alone can do the same work.

Solution:

A and B can do the work = $\frac{1}{12}$

B and C can do the work = $\frac{1}{15}$

C and A can do the work = $\frac{1}{20}$

Here, we need to find the work done by A , b and C in One day.

2 ( A + B + C ) = $\frac{1}{12} + \frac{1}{15} + \frac{1}{20}$

Here, we need to take the LCM and find the values.

The LCM of 12 , 15 and 20 is 60.

2 ( A + B + C ) = $\frac{5 + 4 + 3}{60}$    

2 ( A + B + C ) = $\frac{12}{60}$

2 ( A + B + C ) = $\frac{1}{5}$

A + B + C = $\frac{1}{10}$ days.  

Now, we need to find in how many days A alone can do the same work.

Total work done by A, B and C  subtracted by Work done by B and C

= $\frac{1}{10} - \frac{1}{15}$

Now, we need to take LCM of 10 and 15.  

The LCM of 10 and 15 is 30.

$= \frac{3 - 2}{30}\\ \\= \frac{1}{30}$

The same work done by A alone is 30 days.