A can do a piece of work in 14 days which B alone can do in 21 days. They begin together but 3 days before the completion of the work, A leaves off. The total number of days for the work to be completed is
Answer:C Explanation:B's one days work=1/21 B's 3 days work=1/21 -3 =1/7 The remaining work finished by both of them=1-1/7 =6/7 Both together can finish it in 14-21/35 days=42/5 days Their, one days work = 5/42 Therefore, the No. of days taken by them together to finish 6/7 work = 6/7/5/42 = 6/7-42/5 = 36/5 =7 1/15 days Therefore, the total No. of days for the work to be completed = 3 + 7 1/5 = 10 1/5 days.
Answer:A Explanation: If A can complete a work in x days and B can complete the same work in y days, then, both of them together can complete the work in x y/ x+ y days. Therefore, here, the required number of days = 12 - 24/ 36 = 8 days.
Anand finishes a work in 7 days, Bittu finishes the same job in 8 days and Chandu in 6 days. They take turns to finish the work. Anand on the first day, Bittu on the second and Chandu on the third day and then Anand again and so on. On which day will the work get over?
Answer:D Explanation:In the 1st day Anand does 1/7th of total work similarly, Bithu does 1/8th work in the 2nd day hence at d end of 3 days, work done = 1/7+1/8+1/6=73/168 remaining work = (168-73)/168 = 95/168 again after 6 days of work, remaining work is = (95-73)/168 = 22/168 and hence Anand completes the work on 7th day