A and B can complete a work separately in 10 days and 15 days, respectively. B alone work for 3 days and then he left the work. In how many days will A finish the remaining work?

Correct option is C

Given:
A and B can complete a work separately in 10 days and 15 days
Solution:
B's 1 day work =$$\frac{1}{15}$$​​
B's 3 days work = $$\frac{1}{15}$$​ × 3 = $$\frac{1}{5}$$​​
Remaining work = 1 - $$\frac15 = \frac45$$​​
A's 1 day work = $$\frac{1}{10}$$​​
A can finish the remaining work in $$\frac45$$​ × 10 = 8 days.
Therefore, A can finish the remaining work in 8 days.

Alternate Method:

Total Work = LCM of 10 and 15 = 30 units

B's efficiency = $$\frac{30}{15}$$= 2 unit per day

A's efficiency= $$\frac{30}{10}= 3$$ unit per day
B's work done in 3 days= $$3 \times 2 = 6 \text{unit}$$​​

Remaining work that has to be completed by A = 30-6 = 24 units
Time taken by A to complete that remaining work = $$\frac{24}{3}= 8$$ days​