11. A, B and C together can complete a piece of work in 10 days. All the three started working at it together and after 4 days A left. Then B and C together complete the work in 10 more days. A alone could complete the work in?

A. 15 days B. 16 days C. 25 days D. 50 days

Answer Answer: Option C Explanation: Work done by A, B and C in 4 days = (1/10*4) = 2/5. Remaining work =(1- 2/5) =3/5.
Now, 3/5 work is done by B and C in 10 days.
whole work will be done by B and C in (10*5/3) = 50/3 days.
(A+B+C)'s 1 day's work = 1/10, (B+C)'s 1 day's work = 3/50.
A's 1 day's work =(1/10 - 3/50) = 2/50 =1/25.
therefore A alone could complete the work in 25 days.

12. A works twice as fast as B.If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work is?

A. 6 days B. 8 days C. 18 days D. 4 days

Answer Answer: Option D Explanation: Ratio of rates of working of A and B = 2: 1.so, ratio of times taken = 1:2.
therefore A's 1 day's work = 1/6; B's 1 day's work = 1/12.
(A+B)'s 1 day's work = (1/6+1/12) = 3/12 = 1/4.
so, A and B together can finish the work in 4 days.

13. A Tyre has two punctures. The first puncture alone would have made the Tyre flat in 9 minutes and the second alone would have done it in 6 minutes.If air leaks out at a constant rate, how long does it take both the punctures together to make it flat?

A. 1 1/2 minutes B. 3 1/2 minutes C. 3 3/5 minutes D. 4 1/4 minutes

Answer Answer: Option C Explanation: 1 minute's work of both the punctures = (1/9+ 1/6) = 5/18.
so, both the punctures will make the Tyre flat in 18/5 = 3 3/5min.

14. A takes twice as much time as B or thrice as much time to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in

A. 4 days B. 6 days C. 8 days D. 12 days

Answer Answer: Option B Explanation: Suppose A,B and C take x, x/2 and x/3 hours respectively to finish the work.
Then, (1/x+ 2/x+3/x) = 1/2 = x = 12.
so, B takes 6 hours to finish the work.