Aptitude · Time and Work · Model 2

They never work on the same day, so stop adding their rates

Q: A takes 16 days and B takes 12. Working alternate days starting with A, how long?

13¾ days. On 48 units A does 3 a day and B does 4, so a two-day cycle moves 7 units. Six cycles give 42 units in 12 days, leaving 6. Day 13 A does 3, leaving 3, and on day 14 B needs three-quarters of a day.

Q: Why can’t you just add their rates?

Because they never work on the same day. Adding rates describes two people working simultaneously, which is not what happens here. The correct unit of progress is the cycle: 7 units per two days, not 7 units per day.

Q: How does the alternate-day answer compare with working together?

Roughly double. Together the pair would take 48/7 ≈ 6.86 days; alternating takes 13.75. Each worker is idle half the time, so progress is about halved. That comparison is a good sanity check on your answer.

Q: Does it matter who starts?

Yes, to the answer, though not to the cycle work. Swapping the starting worker leaves the number of whole cycles the same but changes the final part-cycle, and the tail is where the fraction comes from. Questions are set on exactly this difference.

Q: Why are these answers usually fractional?

Because the leftover work after the last whole cycle rarely equals a whole day’s output for whoever works next. The fraction is the part of that final day actually needed, and it is a legitimate answer — not something to round.

Q: How do you handle three workers rotating?

The same four steps with a three-day cycle. The cycle work is the sum of all three rates over three days, and the tail is walked in the rotation’s order. Nothing new is required, which is why it pays to learn the procedure rather than a two-worker formula.

Q: What if one worker works every day and the other only on alternate days?

The cycle is still two days, but the daily worker appears in both of them. So the cycle work is twice the daily worker’s rate plus once the alternating worker’s. Write the cycle out explicitly rather than trying to remember a rule.

Q: How do you decide how many whole cycles fit?

Take the largest n for which n times the cycle work is still strictly less than the total job. If n cycles exactly completed the job you would stop there, but in general you want to leave a non-zero remainder so the tail is well defined.

Q: A and B both take 12 days and work alternate days. How long?

Twelve days. On 12 units each does 1 a day, so a cycle moves 2 units in 2 days and the job takes 12 days. Equal alternating workers reproduce exactly one worker’s solo time, which makes sense — there is always exactly one person working.

Q: When would this model appear outside time and work?

Pipes and cisterns constantly — taps opened on alternate hours is one of the four standard question sets there. More broadly, any resource used on a rota rather than in parallel has this structure, and recognising it saves setting up the whole thing from scratch.

When A works on odd days and B on even days, they are not a team — they are a rota. The unit of progress is not the day but the cycle, and the answer is almost always decided in the final part-cycle.

Set the two times, choose who starts, and walk the cycles →

Find the work done in one full cycle, count how many whole cycles fit, then walk the last part-cycle day by day.

01 The idea

The cycle is the unit, not the day

A can finish a job in 16 days and B in 12. They work on alternate days, starting with A. The instinct is to add their rates and treat them as a pair, and that is wrong — on any given day only one of them is working, so the combined rate never actually occurs.

What repeats is the two-day pattern. On a job of 48 units, A does 3 a day and B does 4, so a full cycle of two days moves the job forward by 7 units. Seven units per two days is the honest rate of progress, and it is not the same as the 7 units per day they would manage together.

So count cycles. Six cycles is 42 units in 12 days, leaving 6. Now the pattern restarts with A, who contributes 3 on day 13, leaving 3. Then B, who does 4 a day and therefore needs only three-quarters of day 14. Total: 13¾ days.

That final part-cycle is where every mark in this model lives. You cannot shortcut it, because who works next depends on where the cycle restarts, and the leftover work usually does not divide evenly into a whole day. Expect a fractional answer and be suspicious of a round one.

One cycle of two days moves the job by the sum of the two daily rates. Count whole cycles first, then finish the remainder in order.

CycleThe repeating pattern of who works. Usually two days, but a question may set three workers rotating, or one worker every third day.

Cycle workThe units completed in one full cycle. This is the rate of progress, and it is not the combined rate of the workers.

Part-cycle tailThe leftover work after the last whole cycle, finished in the cycle’s order. Almost always produces a fractional day, and it is where the model is failed.

02 Worked example

A in 16 days, B in 12, alternating from A

The source question, worked in units. A can complete a work in 16 days and B in 12 days. Starting with A, they work on alternate days. In how many days is the work completed?

Set up the unitsLCM of 16 and 12 is 48, so both rates are whole numbers.W = 48 units ⇒ A = 3/day, B = 4/day

Find the work in one cycleDay one is A and day two is B, then it repeats. Two days moves the job by seven units — not the eleven a day they would manage together.cycle = 3 + 4 = 7 units per 2 days

Count whole cyclesSix cycles is 42 units. A seventh would overshoot, so stop at six.6 × 7 = 42 units in 12 days, leaving 48 − 42 = 6

Day 13 — A’s turnThe cycle restarts with A, who contributes 3 units. Not enough to finish, so the day is used in full.6 − 3 = 3 units still left

Day 14 — B needs only part of itB does 4 units a day and only 3 remain, so B finishes three-quarters of the way through the day.3 / 4 = 0.75 ⇒ total = 12 + 1 + 0.75 = 13¾ days

Compare with the same pair working together every day: 48 units at 7 a day is about 6.86 days. Alternating takes twice as long, because each worker is idle half the time. That comparison is a useful check — an alternate-day answer should be roughly double the together answer, and if yours is close to the together time you have added the rates somewhere you should not have.

03 The method

The method, and the variants that change only the cycle

One procedure covers every version of this model. What changes between versions is the length of the cycle and who is in it.

1. Take W as the LCM and find each rate. 2. Compute the cycle work. 3. Whole cycles = the largest n with n × cycle < W. 4. Walk the remainder in the cycle’s order.

The order matters to the answer, even though the cycle work does not. Starting with the faster worker generally finishes sooner, because the faster worker gets more of the final part-cycle. Swapping who starts changes the tail but not the whole-cycle count, so recompute only the tail. Never assume the two orders give the same answer — questions are set on precisely that difference.

Variant	Cycle	What changes
A and B alternate	2 days	Standard case
B starts instead of A	2 days	Only the tail
Three workers rotating	3 days	Cycle work is all three rates
A daily, B every 2nd day	2 days	A appears in both days of the cycle
A daily, B and C alternating	2 days	A plus one of the others each day
Adding the rates	`wrong`	they never share a day
Rounding the tail up	`wrong`	a part-day is a real answer

05 Cheat sheet

Model 2 on one page

The four steps, then the comparisons and errors worth having in mind.

Step	Do this	On A = 16, B = 12
1. Units	`W = LCM`	48 units, A = 3, B = 4
2. Cycle work	`sum of the cycle’s rates`	7 units per 2 days
3. Whole cycles	`largest n with n·cycle < W`	6 cycles, 42 units, 12 days
4. Walk the tail	`in the cycle’s order`	A 3, then B needs 3/4 day
Compare with together	`W/(Eᴱ+Eᵇ)`	6.86 days — about half
Adding the rates	`wrong`	would give 6.86, not 13.75
Order of starting	`changes the answer`	recompute the tail

Cycle work is not the combined rateSeven units per two days, not seven per day. The workers never share a day, so their rates never actually add.

Expect a fractionThe leftover work rarely fills a whole day, so alternate-day answers are usually fractional. A suspiciously round answer is worth re-checking.

Who starts mattersThe cycle work is unchanged but the tail is not, so swapping the starting worker can change the total. Recompute the tail rather than assuming.

06 Where & why

Where Model 2 shows up

A reliable mains-paper item, because the wrong method gives a clean answer and the right one gives a fraction.

Bank PO Mains · SSC CGL

Two workers on alternate days

The standard form. The together-time answer is always among the options for students who added the rates.

“Who starts” variants

The same pair, opposite order

Set specifically to check whether you recompute the tail. The cycle count is identical and the answer is not.

Pipes and cisterns

Taps opened alternately

Identical arithmetic, and one of the four question sets in that module. Getting this model solid covers both.

Three-worker rotas

A three-day cycle

The same four steps with a longer cycle. Nothing new to learn, which is the payoff for doing the procedure properly rather than memorising a two-worker result.

The tell for this model is any phrase that puts the workers on different days — “alternate days”, “on odd days”, “beginning with B”. The moment you see it, stop adding rates and start counting cycles.

07 Interview questions

What gets asked

Ten, and the second is the misconception the whole model is built to catch.

A takes 16 days and B takes 12. Working alternate days starting with A, how long?

13¾ days. On 48 units A does 3 a day and B does 4, so a two-day cycle moves 7 units. Six cycles give 42 units in 12 days, leaving 6. Day 13 A does 3, leaving 3, and on day 14 B needs three-quarters of a day.

Why can’t you just add their rates?

Because they never work on the same day. Adding rates describes two people working simultaneously, which is not what happens here. The correct unit of progress is the cycle: 7 units per two days, not 7 units per day.

How does the alternate-day answer compare with working together?

Roughly double. Together the pair would take 48/7 ≈ 6.86 days; alternating takes 13.75. Each worker is idle half the time, so progress is about halved. That comparison is a good sanity check on your answer.

Does it matter who starts?

Yes, to the answer, though not to the cycle work. Swapping the starting worker leaves the number of whole cycles the same but changes the final part-cycle, and the tail is where the fraction comes from. Questions are set on exactly this difference.

Why are these answers usually fractional?

Because the leftover work after the last whole cycle rarely equals a whole day’s output for whoever works next. The fraction is the part of that final day actually needed, and it is a legitimate answer — not something to round.

How do you handle three workers rotating?

The same four steps with a three-day cycle. The cycle work is the sum of all three rates over three days, and the tail is walked in the rotation’s order. Nothing new is required, which is why it pays to learn the procedure rather than a two-worker formula.

What if one worker works every day and the other only on alternate days?

The cycle is still two days, but the daily worker appears in both of them. So the cycle work is twice the daily worker’s rate plus once the alternating worker’s. Write the cycle out explicitly rather than trying to remember a rule.

How do you decide how many whole cycles fit?

Take the largest n for which n times the cycle work is still strictly less than the total job. If n cycles exactly completed the job you would stop there, but in general you want to leave a non-zero remainder so the tail is well defined.

A and B both take 12 days and work alternate days. How long?

Hard

A takes 16 days and B takes 12. (a) Find the completion time starting with A, and starting with B. (b) The two answers differ. Explain, in terms of the tail, exactly why. (c) Construct a pair of solo times for which the starting order makes no difference to the total, and state the general condition on the two rates for that to happen.

Follow-up

Part (c) is the real question. The order stops mattering precisely when the remainder after the whole cycles is cleared by the first worker of the tail in both orderings — and the cleanest sufficient condition is that the two rates are equal, but it is not the only one. Working out the boundary teaches you what the tail actually depends on.

Show the hint

For (c), think about what has to be true of the remainder relative to each worker’s daily output.