If 5 men and 9 women can finish a piece of work in 19 days, 3 men and 6 women will do the same work in

The Commission dropped this question. Its final key strikes both the question's number and its answer letter through by hand, and the face of the key sheet records that one item was dropped and 119 taken for scoring. So no option is credited here, and there is nothing on this page for a student to learn as the Commission's answer.

Nothing on the printed booklet says so. The item is set exactly like its neighbours — no note, no asterisk, no cancellation mark — and a candidate sitting the paper answered it in the ordinary way. The withdrawal lives only in the key, it is a fact about how the paper was marked, and the Commission does not publish its reasons for a drop. What follows is therefore the mathematics taught for its own sake, because the topic is a standard one and everything about it is worth knowing.

THE TOPIC IS WORK AND TIME WITH TWO KINDS OF WORKER. Write m for the amount of the job one man completes in a day and w for the amount one woman completes in a day. The stem's single piece of information says that five men and nine women, working together, finish the whole job in nineteen days:

19 × (5m + 9w) = 1 job

and it asks for the number of days t in which three men and six women finish the same job:

t × (3m + 6w) = 1 job

Setting the two equal and rearranging,

t = 19 × (5m + 9w) / (3m + 6w)

TWO UNKNOWNS, ONE EQUATION. That expression still contains m and w, and its value depends on the RATIO between them. Nothing in the stem fixes that ratio. A problem of this kind needs a second, independent piece of information before it has a single numerical answer — a statement such as "two men do as much work as three women", or a second combination of men and women with its own completion time. With one equation and two unknown rates, the problem as printed is under-determined.

Put r = m/w for the ratio of a man's daily output to a woman's. Then

t = 19 × (5r + 9) / (3r + 6)

and different values of r give different values of t. If men and women worked at the same rate, r = 1 and t would be 19 × 14/9, about 29·6 days. If men worked twice as fast, r = 2 and t would be 19 × 19/12, about 30·1 days.

WHAT CAN STILL BE SAID, AND IT IS A GREAT DEAL. Two things follow from the algebra no matter what the ratio turns out to be.

FIRST, THE DIRECTION. Since m and w are both positive, 3m + 6w is always LESS than 5m + 9w — fewer men and fewer women means a smaller daily output. A smaller daily output on the same job means MORE days, so t must be greater than nineteen. The second team is strictly weaker than the first, so it cannot possibly be quicker.

SECOND, THE BOUNDS. The fraction (5r + 9)/(3r + 6) increases steadily as r increases, because the derivative of its numerator times its denominator less the reverse is 15r + 30 − 15r − 27 = 3, a positive constant. As r approaches zero it tends to 9/6 = 1·5, and as r grows without limit it tends to 5/3 ≈ 1·667. So whatever the relative efficiency of the two groups,

19 × 1·5 < t < 19 × 5/3, that is 28·5 days < t < 31·67 days

The problem is under-determined and yet the answer is pinned into a band barely three days wide. That is a genuinely useful lesson: when a question cannot be solved exactly, it is often still worth asking what range the answer must lie in, because the range may be tight enough to be decisive.

Every option printed with this item — 12, 13, 14 and 15 days — is smaller than the nineteen days the first team took, and no option is credited.

WORK AND TIME PROBLEMS ARE RATE PROBLEMS, and the single idea that organises the whole topic is to convert every statement about time into a statement about the FRACTION OF THE JOB DONE PER DAY.

If a worker finishes a job in d days, his rate is 1/d of the job per day. Rates ADD when people work together, so if A finishes a job in 10 days and B in 15, together they complete 1/10 + 1/15 = 1/6 of it per day and finish in 6 days. Inverting once at the start and once at the end is the whole method, and almost every error in the topic is that inversion done in the wrong place.

THE MAN-DAYS SHORT CUT works when all the workers are of ONE kind, that is, when they are interchangeable. Then the total work is measured in man-days, and

M1 × D1 / W1 = M2 × D2 / W2

where M is the number of workers, D the number of days and W the amount of work. That relation is what most candidates reach for first, and it is exactly what CANNOT be applied to a mixed team, because a man-day and a woman-day are different units until something tells us how to convert between them.

WHAT A MIXED-TEAM PROBLEM NEEDS. With two kinds of worker there are two unknown rates, so two independent equations are required. Papers supply the second in one of three standard ways:

A DIRECT EFFICIENCY STATEMENT — "three women do as much work as two men", which gives 3w = 2m at once. A SECOND TEAM WITH ITS OWN TIME — "5 men and 9 women take 19 days, while 6 men and 3 women take 10 days", which gives a second linear equation and lets both rates be found. INDIVIDUAL TIMES — "a man alone would take 40 days and a woman 60", which gives both rates directly.

Given any one of those, the working is mechanical: solve for m and w in terms of the job, add the rates of the new team, and invert.

BOUNDING AN UNDER-DETERMINED PROBLEM is the more advanced idea, and it is worth practising. When an answer depends on an unknown parameter, examine how it varies as that parameter runs over its whole admissible range. Frequently the expression is monotonic — it rises or falls steadily — and then its extreme values occur at the ends of the range, which gives a bracket. Here the two ends correspond to men contributing nothing and to men contributing everything, and the bracket that results is under three days wide out of about thirty.

THE FAMILY OF PROBLEMS this belongs to is large: pipes filling and emptying a cistern, machines producing at different rates, work started by one group and finished by another, and workers who leave part-way through. All of them yield to the same discipline — write each contributor's rate as a fraction of the job per unit time, add, and invert once.

Key facts

• The Commission dropped this question; its key strikes the number and the answer letter through and records 119 items taken for scoring, so no option is credited.
• Nothing on the printed page marks the item as special — it is set exactly like its neighbours, and the withdrawal is recorded only in the key.
• Writing m and w for a man's and a woman's daily output, the stem gives 19 × (5m + 9w) = 1 job, which is a single equation in two unknown rates.
• The quantity asked for is t = 19 × (5m + 9w)/(3m + 6w), and its value depends on the ratio between m and w, which nothing in the stem fixes.
• A mixed-team work problem needs a second independent statement — an efficiency comparison, a second team with its own time, or the individual times — before it has a single numerical answer.
• Because m and w are both positive, 3m + 6w is always less than 5m + 9w, so the second team is strictly weaker and must take MORE than nineteen days.
• Writing r for the ratio m/w, the factor (5r + 9)/(3r + 6) rises steadily from 1·5 as r approaches zero to 5/3 as r grows without limit.
• The time therefore lies between 19 × 1·5 = 28·5 days and 19 × 5/3 ≈ 31·67 days for every admissible ratio — a band under three days wide.
• The man-days relation M1 D1 / W1 = M2 D2 / W2 applies only to interchangeable workers, because a man-day and a woman-day are different units until a conversion is supplied.
• Every option printed with this item is smaller than the nineteen days the larger team took.

Common traps

• Applying the man-days relation to a mixed team. A man-day and a woman-day cannot be added until an efficiency ratio converts one into the other.
• Assuming that a smaller team can finish sooner. Fewer workers of every kind means a smaller daily output and therefore more days on the same job.
• Adding times instead of rates. Two workers who each take 10 days do not take 20 days together; their rates add and the time halves.
• Solving for one unknown and treating the other as though it had cancelled. Two unknown rates need two independent equations.
• Abandoning an intractable problem without extracting the direction and the bounds, which are often available in a few seconds and are frequently decisive.
• Spending several minutes on a single item that will not yield, on a paper where the quantitative strand is the largest and the clock is binding.

• practice — not a real PYQ

  If 5 men and 9 women can finish a piece of work in 19 days, and two women together do as much work in a day as one man does, in how many days will 3 men and 6 women finish the same work ?

  • (a)26 days
  • (b)28 days
  • (c)30 days
  • (d)32 days

  Answer(c) 30 days — with m = 2w the first team's daily output is 10w + 9w = 19w, so the whole job is 19 × 19w = 361w. The second team's daily output is 6w + 6w = 12w, and 361w ÷ 12w is a little over 30 days. Notice that the answer falls inside the band of 28·5 to 31·67 days that holds for every efficiency ratio.

• practice — not a real PYQ

  A alone can finish a job in 12 days and B alone can finish the same job in 24 days. Working together, in how many days will the two of them finish it ?

  • (a)6 days
  • (b)8 days
  • (c)18 days
  • (d)36 days

  Answer(b) 8 days — A does 1/12 of the job a day and B does 1/24, so together they do 1/12 + 1/24 = 3/24 = 1/8 of it a day and finish in 8 days. Rates add and times do not, which is why 36 days and 18 days are both wrong.