UPSC CSE · CSAT · Numeracy and data interpretation · TOPIC LESSON
Convert completion times to rates, account for changing stages and separate continuous work from finished batches.
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BEFORE YOU START
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01 / UNDERSTAND & APPLY
For a constant-rate model, rate=work/time and time=work/rate when the rate is positive. Normalize the entire task as one job, so a worker who finishes it in 12 days has rate 1/12 job per day. Alternatively, choose a convenient total such as 36 work units and express every worker’s rate in those units.
The same job definition must apply to every rate. One person’s “job” cannot mean one room while another’s means a whole building unless you convert them to a common measure. Rates also need the same time unit before addition.
A completion time is the reciprocal of a one-job rate, not the rate itself. Adding 12 days and 18 days does not describe how much work two people complete together.
ORIGINAL PRACTICE · 1
1/20 job per hour, or 0.05 job per hour.
Divide one job by 20 hours. Multiplying this rate by 20 hours returns one complete job, which checks the units.
02 / UNDERSTAND & APPLY
When workers independently contribute at constant rates to a divisible task without interfering, their work per unit time adds. If A supplies 1/12 job per day and B supplies 1/18, together they supply 5/36 job per day. The completion time is the reciprocal 36/5 days.
These assumptions matter. A job with indivisible stages, a single shared tool or coordination delays may not obey simple rate addition. A numerical problem should stipulate the ideal model rather than pretend that adding workers always scales output perfectly.
Two positive additive rates complete the same job faster than either worker alone. If your answer is slower than both, check whether you added times instead of rates or inverted the sum incorrectly.
ORIGINAL WORKED EXAMPLE
A completes a job alone in 12 days and B in 18 days. Assume constant independent rates, divisible work and no interference. How long do they take together?
Answer 7.2 days under the stated model.
ORIGINAL PRACTICE · 1
6 hours.
Rates add to 3/30+2/30=5/30=1/6 job per hour. One job therefore takes 6 hours.
ORIGINAL PRACTICE · 2
Rate ratio 3:1; completion-time ratio 1:3.
Rates are 1/8 and 1/24, so A works three times as fast in the model. The times 8:24 simplify to 1:3. Inverting times reverses their ratio.
03 / UNDERSTAND & APPLY
When every worker has identical constant productivity, total output is proportional to worker count×hours per day×days. If daily hours change, worker-days alone no longer provide the full exposure. Convert to worker-hours or another consistent unit.
This model holds only when the task can use all those hours productively. More workers do not automatically help if the work cannot be divided or equipment creates idle time. State the fixed-productivity and no-interference assumptions before using proportionality.
ORIGINAL WORKED EXAMPLE
Six workers working 8 hours per day complete a job in 5 days. With identical constant productivity and divisible work, how many days would ten workers need if they work 6 hours per day?
Answer 4 days under the stated productivity model.
04 / UNDERSTAND & APPLY
A staged problem changes the active rates over time. Calculate each stage’s output, subtract it from the job and then divide the remainder by the new rate. A worker who leaves stops contributing; do not keep the original combined rate for the rest of the task.
Keep a running total of completed fractions. If a stage would exceed the remaining work, the job finishes partway through that stage under a divisible-work model. The final duration may therefore be fractional even when earlier stages are whole days.
ORIGINAL WORKED EXAMPLE
A finishes the job alone in 12 days and B in 18 days. They work together for 3 days, then A leaves and B finishes alone. Find the total time, assuming constant additive rates.
Answer 13.5 days in total, including 10.5 days after A leaves.
ORIGINAL PRACTICE · 1
6 days.
Remaining work is 3/5. Divide by 1/10 to get(3/5)×10=6. Do not compute the time for the already completed part again.
05 / UNDERSTAND & APPLY
For alternating workers, calculate the work in one full cycle, take as many complete cycles as fit without finishing the job, then inspect the next active worker’s contribution. The worker who starts matters because the final partial shift may belong to a faster or slower person.
Specify whether a worker may finish partway through a day or whether output appears only at shift completion. Under continuous divisible work, use a fractional final shift. Under discrete batch rules, a partially finished batch may contribute no completed output; that is a different model.
ORIGINAL PRACTICE · 1
After 4⅔days.
One two-day cycle completes 1/4+1/6=5/12. Two cycles take 4 days and complete 5/6, leaving 1/6. A works next and needs(1/6)/(1/4)=2/3 day. Total 4+2/3 days.
06 / UNDERSTAND & APPLY
Measure inflow and outflow in the same units. If capacity is normalized to one tank, an inlet filling in 6 hours has rate 1/6 tank per hour and a drain emptying in 9 hours has magnitude 1/9. While both stated constant rates are active, the net is their difference.
A positive net rate raises the level, zero leaves it unchanged, and a negative net lowers it while water is available. From an empty tank, a stronger drain cannot create negative water or allow filling under the unchanged model. Physical drains may depend on water level; a textbook constant-rate exercise must state the idealization.
Time to fill equals remaining capacity divided by positive net rate. If the tank starts partly full, use only the unfilled fraction. If an outlet stops below a certain level or a pump changes rate, split the work into appropriate stages.
ORIGINAL WORKED EXAMPLE
An ideal inlet fills a tank in 6 hours and a drain empties a full tank in 9 hours. Assume their stated rates are constant whenever water is available, and the inlet provides water immediately. Starting empty with both open, how long does filling take?
Answer 18 hours in the ideal constant-rate model.
ORIGINAL PRACTICE · 1
20 minutes.
Net rate=6 L/min, so 120/6=20 minutes. The model assumes no other losses and that the stated outflow applies as filling proceeds.
ORIGINAL PRACTICE · 2
No.
Their net rate is zero, so the level does not increase. A formula dividing remaining capacity by zero is undefined, not an infinitely fast fill.
07 / UNDERSTAND & APPLY
A continuous average rate can estimate production over long periods, but a machine may release items only when a whole batch finishes. Then count completion events within the stated time window. Include the starting state, batch duration, output per batch, delay between batches and whether an endpoint completion counts.
Starting a batch at time 0 does not normally mean one batch is already complete at time 0. For duration d and uninterrupted sequential production, completions occur at d,2 d,3 d,… . In a closed window ending at T, the completed-batch count is floor(T/d), provided the first batch begins at 0 and no batch was underway earlier.
ORIGINAL WORKED EXAMPLE
A machine starts its first batch at 09:00. Each batch takes 14 minutes, releases 6 finished items only at completion and is immediately followed by the next batch. Count items completed by 10:00, including any completion exactly at 10:00.
Answer 24 finished items. A continuous-rate value of about 25.7 items is not a valid completed-item count here.
ORIGINAL PRACTICE · 1
Three in[0,36], and two in[0,36).
Completions occur at 12,24 and 36. The closed endpoint includes 36; the open endpoint excludes it. There is no completion at 0 because the first batch only starts then.
08 / UNDERSTAND & APPLY
Use units as an error detector: jobs divided by jobs/day gives days; litres divided by litres/minute gives minutes. Adding a time to a rate or a tank fraction to litres without conversion is invalid.
Check whether a stage, capacity or batch boundary changes the model. A worker’s rate may cease when they leave, a tank cannot contain negative water, and an unfinished batch may count differently from partially completed divisible work.
Finally compare the answer with a rough expectation. Two positive simultaneous workers should outperform either alone under the ideal model, while an active leak should lengthen filling time relative to the same inlet alone. A result violating those checks needs explanation or correction.
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