A and B can do a work in 4 hours. B and C can complete the same work in 6 hours, while A and C take 8 hours to complete the work. C can complete the work independently in:

Correct option is B

Given:

A + B can do the work in 4 hours → $$\frac{1}{A} + \frac{1}{B} = \frac{1}{4}$$​​

B + C can do the work in 6 hours → $$\frac{1}{B} + \frac{1}{C} = \frac{1}{6}$$​​

A + C can do the work in 8 hours →$$\frac{1}{A} + \frac{1}{C} = \frac{1}{8}$$​​

Solution:

​$$\frac{1}{A} + \frac{1}{B} = \frac{1}{4}$$    ...(1)

$$\frac{1}{B} + \frac{1}{C} = \frac{1}{6}$$     .....(2)
Add equations (1) and (2):

​$$\left( \frac{1}{A} + 2\cdot\frac{1}{B} + \frac{1}{C} \right) = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}$$​

​$$\frac{1}{A} + \frac{1}{C} = \frac{1}{8}$$    ....(3)​

Now subtract (3) from the above sum:

​$$\left( \frac{1}{A} + 2\cdot\frac{1}{B} + \frac{1}{C} \right) - \left( \frac{1}{A} + \frac{1}{C} \right) = \frac{5}{12} - \frac{1}{8}$$​​

$$\frac{2}{B} = \frac{10 - 3}{24} = \frac{7}{24}$$​

Now substitute into (2):

​$$\frac{7}{48} + \frac{1}{C} = \frac{1}{6}$$​

​$$\frac{1}{C} = \frac{1}{6} - \frac{7}{48} = \frac{8 - 7}{48} = \frac{1}{48}$$​

C alone can do the work in 48 hours

Alternate Method:​

Efficiency 

(A + B) + (B + C) + (A + C) = 6 + 4 + 3 = 13

A + B + C =$$\frac{13}{2}$$​= 6.5 units/hour

Now subtract:

(A + B + C) – (A + B) = 6.5 – 6 = 0.5 → C = 0.5 units/hour

Time taken by C alone =​$$\frac{24 \text{ units}}{0.5 \text{ units/hour}} =48 \text{ hours}$$​