Time and work is fundamentally about rates. If you do a job in n days, your rate is 1/n of the job per day. Everything else — multiple workers, staggered starts, increasing/decreasing gang sizes — is just accounting: add up the fractions of work done and set the total equal to 1.
Here is the analogy that locks this in. Think of "work" as filling a bucket. Each worker is a tap with a fixed flow rate. The bucket is full when the total water poured in equals 1 bucket. Multiple taps open simultaneously just mean adding flow rates. A tap that drains the bucket (like a pipe emptying a cistern) gets a negative rate. Taps taking turns (alternating workers) means you track the bucket level period by period.
The standard pitfall is treating "men and days" as additive when they are multiplicative. Ten men finishing a job in 10 days does not mean one man finishes in 1 day — it means the total work is 10 × 10 = 100 man-days. This man-day (or worker-day) concept is your universal currency.
Three quantities govern every problem:
W = total work (in man-days or units)R = rate (work done per day by one worker)T = time to completeThe relationship is W = R × T. Change any one, you change the others. Fifty percent more men (1.5R) means T drops to 2/3 of original — and that ratio relationship is the engine behind most CDS questions in this chapter.
The chapter also blends into pipes and cisterns, which is identical in structure except one pipe may "undo" work (a leak or outlet), giving a negative rate. Don't treat it as a separate topic — it is the same bucket-filling logic with a sign change.
CDS specifically tests: (a) combined rates of groups, (b) gang-size changing over time, (c) algebraic work equations (x men in x+1 days), and (d) constraint problems where workers cannot operate consecutively. Each type has a clean mechanical path once you know what to set up.
Let a worker complete a job alone in n days. Their rate is 1/n per day.
When multiple workers work simultaneously, rates add:
If A takes a days and B takes b days, together they do 1/a + 1/b work per day. Time to finish together:
This is just the harmonic mean of a and b, divided by 2 only when there are exactly two workers. Never memorise the formula — derive the rate sum each time, it takes 10 seconds and avoids sign errors.
Total work W = M × D, where M = number of men and D = number of days. This is constant for a given job. So:
If you have efficiency differences (some workers are faster), extend this to:
where E is efficiency per man per day.
When you are told (A+B) finishes in p days, (B+C) in q days, and (C+A) in r days:
The trick is to add all three pair-equations:
So the combined rate of all three is half the sum. Once you have (A+B+C), individual rates are found by subtraction:
C alone = (A+B+C) − (A+B)A alone = (A+B+C) − (B+C)B alone = (A+B+C) − (C+A)This three-pair setup appears frequently in CDS. Always start by adding all three equations.
When workers leave (or join) at regular intervals, you cannot use a single combined rate. Work through it in blocks:
M₀ × D₀ man-days (use original full gang).W.This is arithmetic accumulation, not a shortcut formula. The CDS question with batches of 5 men leaving every 10 days is exactly this type — do it block by block.
When a problem says "x men in (x+1) days equals (x+5) men in (x−2) days," you are equating man-days:
Expand both sides, simplify, solve the linear or quadratic equation. CDS favours cases where the quadratic terms cancel, leaving a linear equation. Always verify your answer by plugging back into both sides.
This is a newer CDS variant. Workers must alternate — no one person works two hours in a row. Strategy: pair the fastest worker with the second fastest and rotate. Compute how much work accumulates per 2-hour cycle, then find how many cycles are needed.
Same setup. Inlet pipes have positive rates; outlet pipes (leaks) have negative rates. If a pipe fills a tank in p hours and a leak empties it in q hours, the net rate is 1/p − 1/q. If q < p, the tank never fills — good to check before computing. CDS occasionally hides this scenario in a word problem.
When three pairs are given — (A+B), (B+C), (C+A) — add all three pair-rates and halve the result to get (A+B+C). Standard method: solve three simultaneous equations (90s). Shortcut: write down the three fractions, add them, halve (15s). Then subtract each pair-rate from the trio to isolate individuals. Works every time this pattern appears.
Whenever the problem says "50% more workers," the time becomes 2q/3. So the reduction in time is q − 2q/3 = q/3. If you are told this reduction equals a number (say, 12 days), then q/3 = 12, so q = 36. No algebra beyond one step. Standard approach: set up pq = 1.5p × (q-12) and solve (60s). Shortcut: q/3 = reduction, write q = 3 × reduction directly (10s).
Instead of working with fractions throughout, pick total work = LCM of all the individual completion times. Each person's rate becomes a whole number. For example, if A takes 3 hours, B takes 6 hours, C takes 9 hours, D takes 12 hours, pick total work = LCM(3,6,9,12) = 36 units. Then A does 12 units/hour, B does 6, C does 4, D does 3. All arithmetic stays in integers. Standard fraction method: 4 fractions with a common denominator throughout (4 steps). LCM method: 4 whole-number additions (2 steps). Time savings: roughly 30s per problem.
Compute total work in man-days upfront: W = M₀ × D₀. Then track cumulative man-days block by block. Stop when the cumulative sum reaches W. You never need to set up equations — it is pure accumulation. Standard method: fraction-based daily rate with variable men (error-prone, 90s). Man-days accumulation: 5 additions on paper margin (30s).
"x men in (x+1) days" equals "some expression in x" — always set up as LHS man-days = RHS man-days, expand, and look for the x² terms to cancel. If they cancel, you have a linear equation solvable in one step. If they do not cancel, use the quadratic formula — but CDS almost always picks values where cancellation happens. Check for cancellation before reaching for the quadratic formula; it saves 45s.
Read the problem and classify it in 10 seconds using this decision tree:
Is total work given implicitly (time + workers)?
→ Yes: compute total man-days = M × D. Use man-days accumulation if gang size changes over time.
Are multiple pairs of workers given? → Three pairs: use the "add all three, halve" trick to get the full-group rate, then isolate individuals by subtraction.
Does it say "X% more workers"? → Use the proportional time reduction directly: 50% more workers → 2/3 time → reduction = original/3.
Is there a constraint (no consecutive shifts, alternating workers)? → Use LCM as work unit, assign integer rates, track work per cycle manually.
Algebraic (x men in x+1 days)?
→ Equate man-days, expand, check if x² cancels. Solve linear equation.
Pipes and cisterns? → Inlet = positive rate, outlet = negative rate. Compute net rate. Check net is positive before finding time.
In every case, write down rates as fractions (or integers if using LCM) before doing any other step. Never try to hold rates in your head — write them first.
Why this question: This is the canonical three-pair group problem. It tests whether you know the "add all three, halve" shortcut and can then compute half the work.
Solving path: Write three rate equations: A+B = 1/10, B+C = 1/12, C+A = 1/15. Add all three: 2(A+B+C) = 1/10 + 1/12 + 1/15. LCM of 10, 12, 15 is 60. Sum = 6/60 + 5/60 + 4/60 = 15/60 = 1/4. So A+B+C = 1/8 per day. Time for whole work = 8 days. Half the work = 4 days.
Why this question: This problem chains multiple constraints — one ratio relationship plus one equality — and requires isolating each individual rate from a group rate. It rewards systematic substitution over trial and error.
Solving path: Total rate: A+B+C = 1/36. Given A+B = 5C, substitute into total: 5C + C = 1/36, so C = 1/216. Given B+C = A, total becomes A + A = 1/36 (since B+C = A), so A = 1/72. Then B = 1/36 − 1/72 − 1/216. LCM of 36, 72, 216 is 216: B = 6/216 − 3/216 − 1/216 = 2/216 = 1/108. Finally: A+C = 1/72 + 1/216 = 3/216 + 1/216 = 4/216. Ratio (A+C)/B = (4/216)/(2/216) = 2. So n = 2.
Why this question: Men-and-women problems test whether you can set up two simultaneous equations from two work-completion conditions, and then apply the result to a scaled-up workforce.
Solving path: Let m = one man's daily output, w = one woman's daily output. Condition 1: 8m + 24w = 1. Condition 2: 12m + 18w = 1. Subtract Condition 1 from Condition 2: 4m − 6w = 0, so m = 1.5w. Substitute back into Condition 1: 8(1.5w) + 24w = 12w + 24w = 36w = 1, so w = 1/36 and m = 1/24. Now compute output of 24 men + 72 women per day: 24 × (1/24) + 72 × (1/36) = 1 + 2 = 3 units per day. Since total work = 1 unit, time = 1/3 of a day.
Why this question: The alternating-worker constraint is the newest pattern in CDS. It forces you to track work period by period rather than using a combined rate, and tests whether you choose the optimal pairing.
Solving path: Use LCM(3,6,9,12) = 36 as total work. Rates: A = 12 units/hr, B = 6 units/hr, C = 4 units/hr, D = 3 units/hr. To minimise total hours, pair A and B on alternate hours (highest and second-highest). Per 2-hour cycle: A contributes 12, B contributes 6 — total 18 units per 2 hours. After 2 cycles (4 hours): 2 × 18 = 36 units = total work. Verify no consecutive-hour rule is violated: Hour 1 → A, Hour 2 → B, Hour 3 → A, Hour 4 → B. No one works two consecutive hours. Minimum time = 4 hours.
Why this question: The percentage-more-workers problem is a direct test of proportional reasoning. CDS repeats this exact structure with different numbers.
Solving path: Total work = pq man-days. With 50% more men, workforce = 1.5p, and they finish in (q − 12) days. Equate: 1.5p × (q − 12) = pq. Cancel p: 1.5(q − 12) = q. Expand: 1.5q − 18 = q. So 0.5q = 18, giving q = 36.
Adding time instead of adding rates. If A takes 10 days and B takes 20 days, their combined time is not 15 days — it is 1/(1/10 + 1/20) = 1/(3/20) ≈ 6.67 days. Always add rates (fractions of work per day), never add days directly.
Forgetting to halve when using the three-pair trick. Adding the three pair-rates gives 2(A+B+C), not (A+B+C). Dividing by 2 is mandatory. Missing this step produces an answer that is exactly double the correct rate — a trap that leads to one of the wrong options in the MCQ.
Using combined rate when gang size changes. If 5 men leave every 10 days, you cannot apply a single average rate over the whole period. Break the timeline into blocks and accumulate man-days block by block.
Misreading "50% more" as "50% of". "50% more workers" means the new count is 1.5 times the original, not 0.5 times. The time becomes q/1.5 = 2q/3, not q/0.5 = 2q. These give wildly different answers.
Ignoring the "half the work" or "twice the work" qualifier. Several CDS questions ask for time to finish half the work, not the full job. After computing the full completion time, read the question again before writing your answer.
Equating men-and-women efficiencies without establishing them first. In problems with two types of workers, you must solve two simultaneous equations for individual efficiencies before computing anything about a new group. Assuming men and women are equally efficient when the problem does not say so is a common and costly error.