A can do a piece of work in 12 days. B can do it in 20 days. With the assistance of C, they completed the work in 3 days. C alone can do it in ______________days.

Correct option is C

Given:

A can do the work in 12 days.

B can do the work in 20 days.

A, B, and C together can complete the work in 3 days.

Concept Used:

Work done by a person in a day =$$\frac{1}{\text{Number of days}}.$$​​

Solution:
A can do the work in 12 days, so the work rate of A is:

Rate of A =$$\frac{1}{12}$$​
B can do the work in 20 days, so the work rate of B is:

Rate of B =$$\frac{1}{20}$$​
A, B, and C together can complete the work in 3 days, so their combined work rate is:

Rate of A, B, and C = $$\frac{1}{3}$$​

The combined work rate of A, B, and C is the sum of the individual work rates of A, B, and C.

Thus, we can express C's work rate as:

​$$\frac{1}{3} = \frac{1}{12} + \frac{1}{20} + \text{Rate of C}$$​

​$$\frac{1}{3} = \frac{2}{15} + \text{Rate of C}$$​

Rate of C =$$\frac{1}{3} - \frac{2}{15} = \frac{5}{15} - \frac{2}{15} = \frac{3}{15} = \frac{1}{5}$$​
C alone can do the work in 5 days.

Alternate Method: 

Assume, total work = 60 unit 

Efficiency of A = $$\frac{60}{12} =5$$

​​Efficiency of B = $$\frac{60}{20}=3$$

Efficiency of A + B + C = $$\frac{60}3=20$$

Efficiency of C = 20 - 5 - 3 =12

Time taken by C = $$\frac {60}{12}$$ = 5​