A, B and C can individually finish a job in 10, 15 and 6 days, respectively. If all of them work together, in how many days will they finish the job ?
- (a)2 days
- (b)3 days
- (c)4 days
- (d)5 days
Correct — B, (b) 3 days. The reliable method is to give the job a convenient size instead of working in fractions. Take the total work as the LCM of 10, 15 and 6, which is 30 units. Then: • A finishes 30 units in 10 days, so A does 3 units a day. • B finishes 30 units in 15 days, so B does 2 units a day. • C finishes 30 units in 6 days, so C does 5 units a day. Working together they do 3 + 2 + 5 = 10 units a day, and 30 ÷ 10 = 3 days. The fraction method gives the same thing: 1/10 + 1/15 + 1/6 = 3/30 + 2/30 + 5/30 = 10/30 = 1/3 of the job per day, so the whole job takes 3 days. The LCM method is worth preferring under time pressure because it replaces addition of unlike fractions with addition of whole numbers. Check the answer before moving on, since it costs five seconds. In three days A does 3/10 of the job, B does 3/15 = 1/5, and C does 3/6 = 1/2. Adding, 3/10 + 2/10 + 5/10 = 1 — the whole job, exactly. There is also a bracketing argument that eliminates an option without any computation. Three people working together must finish faster than the fastest of them working alone, so the answer is under 6 days. And they cannot beat the time three copies of the fastest worker would take, which is 6 ÷ 3 = 2 days; since A and B are slower than C, the answer must be strictly more than 2 days. The answer is therefore strictly between 2 and 6.
- (a)2 days — Two days is the theoretical floor, achievable only if all three worked at C's rate — three workers each finishing in 6 days would together take 6 ÷ 3 = 2 days. A takes 10 days and B takes 15, both slower than C, so the true figure must exceed 2. In work-rate terms, 2 days would require 15 units a day and the three together manage only 10.
- (c)4 days — Four days corresponds to a combined rate of 30 ÷ 4 = 7.5 units a day, which is less than the 10 units the three actually produce. It is the answer of a candidate who has dropped or mis-scaled one of the three rates — B and C alone, for instance, do 7 units a day and would take about 4.3 days, which rounds towards this option.
- (d)5 days — Five days implies a combined rate of only 6 units a day, barely more than C manages alone. It is the sort of figure produced by averaging the three times rather than adding the three rates — a tempting shortcut that is always wrong, because time and rate are inversely related and only rates may be added.
Every time-and-work problem rests on one idea: times do not add, rates do. If a worker finishes a job in t days his rate is 1/t of the job per day, and when several people work together their rates add while their times do not. The practical technique that follows is the LCM method — set the total work equal to the LCM of the given times, convert each person's time into a whole-number daily output, add the outputs, and divide the total work by the sum. This converts every problem of this family, including the harder variants where workers join or leave partway or where one of them is inefficient, into arithmetic on integers. The same inverse relation is what powers pipes-and-cisterns problems, where an outlet pipe simply contributes a negative rate.
The quantitative block of the EO/AO paper always contains one or two items of this family, and they are among the fastest marks available — this one is a single line of arithmetic once the rates are written down. Because the numbers are chosen to come out exactly, an answer that is not a whole number is usually a signal that a rate has been mis-copied. The habit rewarded is converting to rates immediately and verifying the total at the end.
- A worker who finishes a job in t days works at a rate of 1/t of the job per day.
- Rates of people working together are added; times are never added.
- LCM method: total work = LCM of the individual times; each person's daily output = total work ÷ his time.
- Here LCM(10, 15, 6) = 30, so A does 3, B does 2 and C does 5 units a day, totalling 10 units a day.
- 30 ÷ 10 = 3 days.
- By fractions: 1/10 + 1/15 + 1/6 = 1/3 of the job per day, hence 3 days.
- A group always finishes faster than its fastest member alone — a one-line upper bound on the answer.
- A group of n workers cannot beat the time the fastest member would take divided by n — a one-line lower bound.
- Averaging the three times instead of adding the three rates.
- Adding the times and dividing, which has no basis at all.
- Failing to sanity-check against the fastest worker — any answer of 6 days or more is impossible here.
- Choosing a total-work figure that is not a common multiple, which reintroduces the fractions the LCM method exists to avoid.
Time and work appears in most EO/AO quantitative blocks, usually with three workers and times chosen so the LCM is small. The variants worth rehearsing are the partial-work ones — somebody works for a few days and leaves — because they use the same rates with one extra step.
No directly related past PYQ was found.
- practice — not a real PYQ
A can complete a piece of work in 12 days and B can complete the same work in 24 days. Working together, in how many days will they complete it ?
- (a)6 days
- (b)8 days
- (c)9 days
- (d)18 days
Answer(b) 8 days
- practice — not a real PYQ
A alone can do a piece of work in 12 days, and A and B working together can do it in 8 days. In how many days can B alone do the work ?
- (a)16 days
- (b)20 days
- (c)24 days
- (d)36 days
Answer(c) 24 days