A, B and C can finish a work in 20, 25 and 30 days, respectively. They start working together but B quits after working for 3 days. In how many days from the start shall the work be completed?
- (a)9 8/15 days
- (b)10 1/15 days
- (c)10 14/25 days
- (d)11 1/10 days
Correct — C, 10 14/25 days. Take the work as 300 units, the lowest common multiple of 20, 25 and 30. Then A does 15 units a day, B does 12 and C does 10. For the first three days all three work, at 15 + 12 + 10 = 37 units a day, completing 111 units. B then leaves, and A and C continue at 15 + 10 = 25 units a day with 300 − 111 = 189 units left. That takes 189 ÷ 25 = 7.56 days, which is 7 and 14/25 days. Counting from the start, the job ends at 3 + 7 14/25 = 10 14/25 days.
- (a)9 8/15 days — This is roughly what the job would take if all three worked throughout, since 300 ÷ 37 is about 8.1 days. B leaves after three days, so the finish must be later.
- (b)10 1/15 days — Close to the answer but arrived at by mishandling the remaining work; 189 units at 25 units a day is 7.56 days, not 7.07.
- (d)11 1/10 days — This overshoots. It would need about 202 units to be left for A and C, which happens only if the first three days are miscounted at 33 units a day.
Time-and-work questions are cleanest in units rather than fractions. Fix the total job as the lowest common multiple of the individual times, and each worker's daily output becomes a whole number. Rates then add: workers together do the sum of their daily units, and any change in the team changes only the rate from that point on.
The phrase that decides the arithmetic is 'from the start'. The question asks for the total elapsed time, so the three days B worked must be added back at the end — a step candidates skip after computing the 7 14/25 days that A and C need. The unit method also keeps the fractions honest: 189/25 reduces to 7 14/25, and option (c) prints exactly that fraction, which is a useful confirmation that the reading of the stem is right.
- Taking the job as the LCM of 20, 25 and 30, that is 300 units, gives A 15, B 12 and C 10 units a day.
- Working together the three complete 37 units a day, so three days give 111 units.
- A and C together do 25 units a day, and the remaining 189 units take 189 ÷ 25 = 7 14/25 days.
- Total elapsed time is 3 + 7 14/25 = 10 14/25 days.
- Rates add; times do not — two workers who each take 20 days finish together in 10, not 40.
The lowest common multiple removes every fraction until the last division.
- Reporting 7 14/25 days, the time after B leaves, instead of the total from the start.
- Averaging the three individual times instead of adding rates.
- Rounding 189 ÷ 25 to 7.5 and choosing the nearest option.
A two-phase work item: a team that changes partway through, with the answer demanded from the start of the job rather than from the change.
Suppose A and B can complete a work together in 10 days. If B alone can complete the work in 15 days, then in how many days can A alone finish the work?
- (a) 20 days
- (b) 24 days
- (c) 25 days
- (d) 30 days
Answer(d) 30 days
The same rule that rates add and times do not, used in reverse. That item subtracts one worker's rate from a combined rate; this one adds three rates and then removes one partway through the job.
- practice — not a real PYQ
A and B can do a piece of work in 12 and 18 days respectively. Working together, how long do they take?
- (a)6.8 days
- (b)7.2 days
- (c)7.5 days
- (d)8 days
Answer(b) 7.2 days — on a 36-unit job A does 3 and B does 2 units a day, so together 5 a day, and 36 ÷ 5 = 7.2.
- practice — not a real PYQ
P can finish a job in 15 days. He works for 5 days and leaves. Q finishes the rest in 8 days. How long would Q alone take for the whole job?
- (a)10 days
- (b)12 days
- (c)14 days
- (d)16 days
Answer(b) 12 days — P completes one-third in 5 days, so Q does two-thirds in 8 days and the whole in 12.