


22.
Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages? [SCMHRD 2002]
Answer & Solution
Answer: c) 8 hours 15 minutes
Solution: Number of pages typed by Ronald in 1 hour = 32/6 = 16/3 Number of pages typed by Elan in 1 hour = 40/5 = 8. Number of pages typed by both in 1 hour = [(16/3 + 8) = 40/3. Time taken by both to type 110 pages = [110 * (3/40)] hrs = 8 1/4 hrs = 8 hrs 15 minutes. 23.
P can complete a work in 12 days working 8 hours a day. Q can complete the same work in 8 days working 10 hours a day. If both P and Q work together, working 8 hours a day, in how many days can they complete the work? [Bank P.O. 1999]
Answer & Solution
Answer: b) 5 5/11
Solution: P can complete the work in (12 * 8) hrs. = 96 hrs. Q can complete the work in (8 * 10) hrs. = 80 hrs. P's 1 hour's work = 1/96 and Q's 1 hour's work = 1/80 (P + Q)'s 1 hour's work = [(1/96) + (1/80)] = 11/480 So, both P and Q will finish the work in (480/11) hrs. Number of days of 8 hours each = [(480/11) * (1/8)] = 60/11 days = 5 5/11 days. 24.
A and B can do a work in 12 days, B and C in 15 days, C and A in 20 days. If A, B and C work together, they will complete the work in: [S.S.C 1999]
Answer & Solution
Answer: c) 10 days
Solution: (A + B)'s 1 day's work = 1/12 (B + C)'s 1 day's work = 1/15 (A + C)'s 1 day's work = 1/20 Adding, we get, 2(A + B + C)'s 1 day's work = [(1/12) + (1/15) + (1/20)] = 12/60 = 1/5 (A + B + C)'s 1 day's work = 1/10 So, A, B abd C together can complete the work in 10 days. 25.
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in: [R.R.B. 2001]
26.
A and B can do a piece of work in 72 days, B and C can do it in 120 days, A and C can do it in 90 days. In what time can A alone do it?
Answer & Solution
Answer: c) 120 days
Solution: (A + B)'s 1 day's work = 1/72 (B + C)'s 1 day's work = 1/120 (A + C)'s 1 day's work = 1/90 Adding, we get: 2 (A + B + C)'s 1 day's work = (1/72 + 1/120 + 1/90) = 12/360 = 1/30 => (A + B + C)'s 1 day's work = 1/60 So, A's 1 day's work = (1/60  1/120) = 1/120 A alone can do the work in 120 days. 27.
A and B can do a piece of work in 5 days, B and C can do it in 7 days, A and C can do it in 4 days. Who among these will take the least time if put to do it alone?
Answer & Solution
Answer: c) A
Solution: (A + B)'s 1 day's work = 1/5 (B + C)'s 1 day's work = 1/7 (A + C)'s 1 day's work = 1/4 Adding, we get: 2 (A + B + C)'s 1 day's work = (1/5 + 1/7 + 1/4) = 83/140 => (A + B + C)'s 1 day's work = 83/280 A's 1 day's work = [(83/280)  (1/7)] = 43/280 B's 1 day's work = [(83/280)  (1/4)] = 13/280 C's 1 day's work = [(83/280)  (1/5)] = 27/280 Thus time taken by A, B, C is 280/43 days, 280/13 days, 280/27 days respectively. Clearly, the time taken by A is least 28.
A can do a piece of work in 4 hours, B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it? [S.S.C. 2002]
