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All Topics Arithmetic Aptitude Time and Work

Time and Work

Calculate how much work can be done in given time.

Important Formulas

  • [Fundamental model] Work = Rate × Time. When the entire job is treated as 1, rate is measured as job per day or job per hour.
  • [Reciprocal rate] If A alone completes a job in a days, A's one-day work (rate) is 1/a and one-hour work is 1/(total working hours).
  • [Work completed] Work completed in t days by a worker who needs a days = t/a.
  • [Remaining work] Remaining work after t days = 1 - t/a, provided the worker's efficiency stays constant.
  • [Time from work and rate] Required time = Required work ÷ Effective rate.
  • [Combined positive rates] If workers cooperate independently, their rates add: R_total = R1 + R2 + ... + Rn.
  • [Two-person combined time] If A and B take a and b days separately, together they take ab/(a+b) days.
  • [Three-person combined time] If A, B and C take a, b and c days separately, together they take abc/(ab+bc+ca) days.
  • [Unknown partner's rate] If A+B finish in t days and A alone in a days, B's rate = 1/t - 1/a.
  • [Unknown partner's time] If A+B finish in t days and A alone in a days (a>t), B alone takes at/(a-t) days.
  • [Negative or destructive worker] If a person/machine undoes work, subtract its rate: R_net = sum of productive rates - sum of destructive rates.
  • [Efficiency and time] For equal work, efficiency is inversely proportional to time: E1/E2 = T2/T1.
  • [Efficiency ratio to time ratio] If efficiencies A:B = m:n, their times for the same job are n:m.
  • [Time ratio to efficiency ratio] If times A:B = m:n, their efficiencies are n:m.
  • [Percentage more efficient] If A is p% more efficient than B, EA:EB = (100+p):100 and TA:TB = 100:(100+p).
  • [Percentage less efficient] If A is p% less efficient than B, EA:EB = (100-p):100 and TA:TB = 100:(100-p).
  • [Time after efficiency increase] If efficiency increases by p%, new time = old time × 100/(100+p).
  • [Time after efficiency decrease] If efficiency decreases by p%, new time = old time × 100/(100-p).
  • [Combined time from efficiency ratio] If efficiencies A:B = m:n and together they take t days, A alone takes t(m+n)/m and B alone takes t(m+n)/n.
  • [Rate from partial work] If A completes fraction f of a job in t days, A's rate = f/t and full-job time = t/f.
  • [Equivalent work units] Choose total work as the LCM of individual times. A worker taking a days then performs LCM/a units per day.
  • [Worker-day model] For identical workers and fixed hours/efficiency, total work ? number of workers × days.
  • [Worker-hour model] For fixed efficiency, total work ? workers × days × hours per day.
  • [General work proportionality] W1/W2 = (M1×D1×H1×E1)/(M2×D2×H2×E2), where M=workers, D=days, H=hours/day and E=relative efficiency.
  • [Same-work cross multiplication] For the same work: M1D1H1E1 = M2D2H2E2.
  • [Join/leave phase equation] For multiple phases, total work = ?(rate active in phase × duration of phase). Set this sum equal to 1.
  • [Leaves before completion] If A leaves k days before completion time T, A works T-k days while workers who remain work T days.
  • [Joins after a delay] If B joins after k days and total time is T, the first worker works T days while B works T-k days.
  • [Alternate-day cycle] For A and B working on alternate days, two-day cycle work = RA+RB. Count full cycles, then handle the final partial day.
  • [Rotating-team cycle] For a repeating schedule of teams, cycle work = sum of each scheduled day's effective rate.
  • [Periodic absence] For absences every mth, nth, etc. day, use an LCM-length calendar cycle and add only the rates of those present each day.
  • [Work-rest schedule] Calendar duration = required working days + rest days that occur before completion. Do not add a rest after the job is finished.
  • [Equivalent workforce] If x men = y women for the same time and work, one man = y/x women and efficiency M:W = y:x.
  • [Mixed workforce rate] Convert all worker types into one equivalent unit, then use equivalent workers × days × hours.
  • [Wage division] For equal wage rate per unit of work, wages are divided in the ratio of actual work contributions: Ei×Ti.
  • [Deadline workforce] Required workers = remaining work units ÷ (remaining days × work per worker per day).
  • [Minimum whole workers] When a fractional number of workers is obtained, round upward because a partial worker cannot meet the deadline.
  • [Actual-rate correction] If progress differs from plan, calculate actual team rate from completed work/elapsed time before estimating remaining manpower.
  • [Changing workforce] Total work in worker-days = ?(workers active in each phase × phase days × relative efficiency).
  • [Arithmetic-progression workforce] If workforce changes by d each day, total worker-days in n days = n/2[2a+(n-1)d].
  • [Arithmetic-progression productivity] If daily output forms an AP with first output a and increase d, total output after n days = n/2[2a+(n-1)d].
  • [Geometric-progression productivity] If daily output forms a GP with first output a and ratio r, total after n days = a(r^n-1)/(r-1), r?1.
  • [Hourly-rate conversion] If a worker completes in d days at h hours/day, hourly rate = 1/(dh).
  • [Overlapping shifts] Daily work = sum of each worker's hourly rate × actual hours worked that day, including overlap without double-counting time.
  • [Machine productivity] Output = machines × days × hours/day × output per machine-hour × efficiency factor.
  • [Defect/rejection adjustment] Effective productive rate = nominal rate × accepted fraction. If rejection is q%, accepted fraction is 1-q/100.
  • [Rework after rejection] Net completed work = gross completed work × accepted fraction; rejected work must be added back to remaining work.
  • [Efficiency changes by phase] For phase i, time_i = work_fraction_i ÷ adjusted_rate_i. Total time is the sum of all phase times.
  • [Project-scope change] If project size changes, express completed and remaining work in units of the original project before computing the new balance.
  • [Equal-finish work allocation] To make teams with rates R1:R2 finish simultaneously, divide work in the same ratio R1:R2.
  • [Pairwise combined rates] If AB, BC and CA rates are p, q and r, then A+B+C rate = (p+q+r)/2.
  • [Individual rate from pairwise rates] A=(AB+AC-BC)/2, B=(AB+BC-AC)/2, C=(AC+BC-AB)/2.
  • [All-three and pair rate] If ABC rate and AB rate are known, C rate = ABC-AB; similarly for any omitted worker.
  • [Rate chain] Build chained efficiencies on a common base. Example: A=2B and B=1.5C gives A:B:C=6:3:2.
  • [Time-difference equation] If A and B together take t days and their individual times differ by k, solve 1/x + 1/(x+k) = 1/t.
  • [Multiple projects] Keep a separate work ledger for each project. Worker transfers change future rates, not work already completed.
  • [Work share from equal time] If teams work for the same duration, their completed-work ratio equals their effective-rate ratio.
  • [Percentage work in percentage time] If a worker completes x% of work in y% of the usual time, effective efficiency relative to normal = x/y.
  • [Completion within a day] Final fraction of a day = remaining work ÷ that day's active rate.

Shortcut Tricks

  • [Normalize to one job] Treat the whole job as 1 when equations are simple. This prevents unnecessary large numbers.
  • [Use LCM units] When times are integers, take total work as their LCM; daily outputs become whole numbers and hard fractions disappear.
  • [Invert time ratios] Immediately reverse a time ratio to get the efficiency ratio.
  • [Convert percentages to small ratios] 25% more ? 5:4, 20% less ? 4:5, 50% more ? 3:2, 33?% more ? 4:3.
  • [Two-person direct formula] For separate times a and b, use ab/(a+b) instead of repeatedly adding fractions.
  • [Unknown partner direct formula] If together time is t and A alone time is a, B time = at/(a-t).
  • [Pairwise half-sum trick] With AB, BC and CA times, convert to rates and halve their sum to get the all-three rate.
  • [Pairwise cancellation trick] A's rate = ½[(AB rate)+(AC rate)-(BC rate)]; use analogous expressions for B and C.
  • [Cycle before calendar] For alternate or rotating schedules, find work per complete cycle first, then inspect only the leftover days.
  • [Do not average times] Average rates, not completion times. The arithmetic mean of individual times is generally wrong.
  • [Check combined-time sanity] A productive combination must take less time than the fastest individual. Otherwise, recheck signs or data.
  • [Partial-day finish] Never force a full final day. Divide the remaining work by the active rate on that exact day.
  • [Starting worker matters] In alternate-day problems, identify who starts; it can change the answer when completion occurs mid-cycle.
  • [Worker-day ledger] Write total required worker-days, subtract completed worker-days, then divide the balance by the new workforce.
  • [Update from actual progress] If the team finishes only a stated fraction by a checkpoint, ignore the original assumed rate and use actual progress.
  • [Deadline quick ratio] For identical efficiency and unchanged work, workers are inversely proportional to available days.
  • [Equivalent-worker conversion] Convert women, children, machines, or shifts into one base unit before any multiplication.
  • [Wages follow contribution] Use efficiency × actual time, not merely attendance days, to divide payment.
  • [Negative work sign] Use a minus sign for leakage, damage, rework or undoing. A net rate ?0 means the job will never finish.
  • [Rest-day counting] Count only rest days that occur before the final working period; there is no rest after completion.
  • [Hourly conversion first] When daily hours differ, convert every worker to hourly rate before combining.
  • [AP workforce shortcut] For workers increasing/decreasing regularly, use the AP sum instead of writing every day's workforce.
  • [GP output shortcut] For doubling/halving productivity, use the geometric-series sum.
  • [Scope-change ledger] Keep completed work in original-job units, revise total scope, and subtract; do not reset progress.
  • [Rejection shortcut] If q% of output is rejected continuously, multiply that worker's rate by (100-q)/100.
  • [Ceiling rule] For required workers or machines, always take the next integer unless the result is already whole.
  • [Option testing] In MCQs, test answer choices using rate × time; this can avoid solving a quadratic.
  • [Fraction discipline] Keep exact fractions until the final step. Early decimal rounding often changes the final partial day.
  • [Phase table] For join/leave questions, create columns: phase, active workers, rate, time, work done.
  • [Periodic schedule LCM] For 'every 3rd/4th/5th day' conditions, build one LCM-length attendance cycle.
  • [Equal-finish split] Split work in the ratio of team rates so both teams take the same time.
  • [Rate-chain base] For chained comparisons, choose a common base that removes decimals before combining.
  • [Time-difference quadratic] Let faster time be x and slower time x+k; use product/sum after clearing denominators.
  • [Project-transfer rule] Finish calculations for work already done before transferring workers; only future workforce changes.
  • [Same-work cancellation] In M×D×H×E comparisons, cancel common factors before multiplying large numbers.
  • [Fraction-of-work shortcut] If f of a job takes t days, full-job time is t/f instantly.
  • [Percentage-time shortcut] Completing x% work in y% time means efficiency multiplier x/y.
  • [Alternating destructive work] Include the destructive rate on every day it operates, even when different productive workers alternate.
  • [Repeating pattern marker] Write day numbers under one cycle and mark exactly who works; this prevents off-by-one errors.
  • [Feasibility check] Individual rates derived from pairwise data must all be positive; zero or negative values indicate impossible or redundant data.
✏️

Solved Examples

Q1 [Easy | Basic fraction] A can complete a project in 12 days. What fraction is completed in 5 days, and what fraction remains?
Solution: Rate = 1/12 per day. In 5 days, work done = 5/12. Remaining = 1-5/12 = 7/12. Final answer: completed 5/12; remaining 7/12.
Q2 [Easy | Remaining time] A completes 3/8 of a job in 6 days at constant efficiency. How many more days are required?
Solution: Full-job time = 6 ÷ (3/8) = 16 days. Remaining time = 16-6 = 10 days. Final answer: 10 days.
Q3 [Easy | Two workers] A and B can finish a job alone in 15 days and 10 days. How long together?
Solution: Combined rate = 1/15+1/10 = 1/6. Time = 6 days. Shortcut: ab/(a+b)=150/25=6. Final answer: 6 days.
Q4 [Easy | Efficiency ratio] Efficiencies of A and B are 3:5. A needs 20 days. How long will B need?
Solution: Time ratio is inverse, so TA:TB = 5:3. TB = 20×3/5 = 12 days. Final answer: 12 days.
Q5 [Easy | Percentage efficiency] A is 25% more efficient than B. B finishes in 20 days. Find A's time.
Solution: Efficiency ratio A:B = 125:100 = 5:4. Time ratio A:B = 4:5. A's time = 20×4/5 = 16 days. Final answer: 16 days.
Q6 [Easy | Unknown partner] A and B together finish in 8 days. A alone takes 12 days. Find B's time.
Solution: B's rate = 1/8-1/12 = 1/24. Therefore B alone takes 24 days. Direct formula: 12×8/(12-8)=24. Final answer: 24 days.
Q7 [Easy | Worker-days] 18 workers finish a task in 20 days. How many days will 24 equally efficient workers take?
Solution: Total work = 18×20 = 360 worker-days. Required days = 360/24 = 15. Final answer: 15 days.
Q8 [Easy | Worker-hours] 12 workers working 8 hours daily finish in 15 days. How many days will 16 workers working 6 hours daily take?
Solution: Same work: 12×8×15 = 16×6×D. D = 1440/96 = 15 days. Final answer: 15 days.
Q9 [Easy | Men and women] 8 men do the same work in the same time as 12 women. Find the efficiency ratio of one man to one woman.
Solution: 8M = 12W, so M/W = 12/8 = 3/2. Efficiency ratio man:woman = 3:2. Final answer: 3:2.
Q10 [Easy | Three workers] A, B and C alone take 20, 30 and 60 days. How long together?
Solution: Rate = 1/20+1/30+1/60 = 3/60+2/60+1/60 = 1/10. Final answer: 10 days.
Q11 [Easy | Partial-rate comparison] A completes 3/5 of a job in 12 days. B completes 2/3 in 10 days. How long together for the full job?
Solution: A's full time = 12÷(3/5)=20 days. B's full time = 10÷(2/3)=15 days. Combined rate = 1/20+1/15=7/60. Time = 60/7 = 8 4/7 days. Final answer: 8 4/7 days.
Q12 [Easy | Wage sharing] A and B work for 6 days. A alone can finish in 10 days and B in 15 days. Divide ?5,000 in proportion to work.
Solution: Contributions = 6/10 : 6/15 = 3/5 : 2/5 = 3:2. Shares = ?3,000 and ?2,000. Final answer: A ?3,000; B ?2,000.
Q13 [Intermediate | Work then leave] A and B take 18 and 24 days. They work together for 6 days, then A leaves. How long does B need to finish the remainder?
Solution: Together rate = 1/18+1/24=7/72. In 6 days they complete 7/12. Remaining = 5/12. B's time = (5/12)÷(1/24)=10 days. Final answer: 10 more days.
Q14 [Intermediate | Leaves before finish] A alone takes 12 days and B alone 24 days. They start together, but A leaves 3 days before completion. Find total completion time.
Solution: Let total time be T. A works T-3 days and B works T days: (T-3)/12 + T/24 = 1. Multiplying by 24: 2T-6+T=24, so T=10. Final answer: 10 days.
Q15 [Intermediate | Joins later] A can finish in 24 days and B in 16 days. A works alone for 6 days; then B joins. Find total time.
Solution: A completes 6/24=1/4. Remaining = 3/4. Combined rate = 1/24+1/16=5/48. Additional time = (3/4)÷(5/48)=36/5=7 1/5 days. Total = 13 1/5 days.
Q16 [Intermediate | Multiple departures] A, B and C take 12, 18 and 36 days. All start. A leaves after 2 days and B leaves 3 days before completion. Find total time.
Solution: Let total time be T. A works 2 days, B works T-3 days, C works T days: 2/12+(T-3)/18+T/36=1. Multiplying by 36 gives 6+2T-6+T=36, so 3T=36 and T=12. Final answer: 12 days.
Q17 [Intermediate | Alternate days with partial day] A and B take 12 and 18 days. They work on alternate days, starting with A. Find completion time.
Solution: Two-day cycle work = 1/12+1/18=5/36. Seven cycles in 14 days complete 35/36. Remaining 1/36 is done by A on day 15 in (1/36)/(1/12)=1/3 day. Final answer: 14 1/3 days.
Q18 [Intermediate | Alternate days exact finish] A takes 10 days and B 15 days. They work alternately, A first. Find completion time.
Solution: Two-day cycle = 1/10+1/15=1/6. Five cycles complete 5/6 in 10 days. Day 11 A completes 1/10, leaving 1/15; B completes it on day 12. Final answer: 12 days.
Q19 [Intermediate | Rotating pairs] A, B and C alone take 10, 15 and 30 days. Day 1: A+B; day 2: B+C; day 3: C+A; repeat. Find completion time.
Solution: Pair rates are 1/6, 1/10 and 2/15. Three-day cycle = 2/5. Two cycles complete 4/5 in 6 days. Day 7 A+B adds 1/6, reaching 29/30. Remaining 1/30 is completed by B+C at rate 1/10 in 1/3 day. Final answer: 7 1/3 days.
Q20 [Intermediate | Workers leave] 24 workers can finish in 30 days. After 10 days, 6 workers leave. Find total time.
Solution: Total work = 24×30=720 worker-days. Work done = 24×10=240. Remaining=480. With 18 workers, time=480/18=26 2/3 days. Total=36 2/3 days.
Q21 [Intermediate | Workers join] 40 workers can finish in 24 days. After 8 days, 10 more workers join. Find total time.
Solution: Total work=960 worker-days. First 8 days complete 320. Remaining=640. Fifty workers need 640/50=12.8=12 4/5 days. Total=20 4/5 days.
Q22 [Intermediate | Two workforce changes] 10 workers can finish in 24 days. After 6 days, 2 leave; after another 6 days, 4 join. Find total time.
Solution: Total=240 worker-days. First phase: 10×6=60. Second: 8×6=48. Remaining=132. New workforce=12, so 11 days more. Total=6+6+11=23 days.
Q23 [Intermediate | Deadline workforce] 36 workers can finish a project in 25 days. How many workers are needed to finish it in 20 days?
Solution: Total work=36×25=900 worker-days. Required workers=900/20=45. Extra workers=45-36=9. Final answer: 45 total, 9 extra.
Q24 [Intermediate | Actual progress and extra workers] 30 workers were expected to finish in 40 days. After 20 days, only one-third is completed. How many workers at the same actual efficiency are needed to finish in the next 20 days?
Solution: Actual team rate=1/3 ÷20=1/60 job/day. Per-worker rate=1/(60×30)=1/1800. Required rate for remaining 2/3 in 20 days=1/30. Workers needed=(1/30)/(1/1800)=60. Extra=30. Final answer: 60 total, 30 extra.
Q25 [Hard | Mixed workforce] 6 men or 8 women can finish a task in 20 days. How long will 3 men and 4 women take?
Solution: Total work=120 man-days=160 woman-days. Rate of 3 men=3/120=1/40. Rate of 4 women=4/160=1/40. Combined rate=1/20. Final answer: 20 days.
Q26 [Hard | Wages with unequal days] A alone takes 12 days and works 8 days. B alone takes 18 days and works 6 days. Together they complete the job. Divide ?9,000.
Solution: A contributes 8/12=2/3; B contributes 6/18=1/3. Ratio=2:1. Shares: A=?6,000, B=?3,000. Final answer: ?6,000 and ?3,000.
Q27 [Hard | Productive and destructive workers] A, B and C would individually complete, complete and undo a job in 20, 30 and 60 days. All operate together. Find time.
Solution: Net rate=1/20+1/30-1/60=(3+2-1)/60=1/15. Final answer: 15 days.
Q28 [Hard | Efficiency falls] A normally completes in 12 days. After working 3 days, A's efficiency falls by 20%. Find total time.
Solution: First 3 days complete 3/12=1/4. Remaining=3/4. New rate=80% of 1/12=1/15. Remaining time=(3/4)×15=11 1/4 days. Total=14 1/4 days.
Q29 [Hard | Pairwise rates] A+B finish in 12 days, B+C in 15 days and C+A in 20 days. Find each individual time and the all-three time.
Solution: AB=1/12, BC=1/15, CA=1/20. All-three rate=½(1/12+1/15+1/20)=1/10, so together 10 days. A=½(1/12+1/20-1/15)=1/30; B=1/20; C=1/60. Final answer: A 30 days, B 20 days, C 60 days, all three 10 days.
Q30 [Hard | Pairwise data with fractions] A+B finish in 6 days, B+C in 8 days and C+A in 12 days. Find individual times.
Solution: A=½(1/6+1/12-1/8)=1/16. B=½(1/6+1/8-1/12)=5/48, so B's time=48/5=9 3/5 days. C=½(1/8+1/12-1/6)=1/48. Final answer: A 16 days, B 9 3/5 days, C 48 days.
Q31 [Hard | All-three and pair rates] A+B+C finish in 12 days. A+B finish in 15 days and A+C in 20 days. Find individual times.
Solution: C=1/12-1/15=1/60. B=1/12-1/20=1/30. A=1/15-1/30=1/30. Final answer: A 30 days, B 30 days, C 60 days.
Q32 [Hard | Efficiency chain] A is twice as efficient as B, and B is 50% more efficient than C. Together they finish in 8 days. Find individual times.
Solution: Let C=2 units, B=3, A=6. Total=11 units. Individual times are together time × total/individual: A=8×11/6=14 2/3; B=8×11/3=29 1/3; C=8×11/2=44. Final answer: 14 2/3, 29 1/3 and 44 days.
Q33 [Hard | Periodic absence] A takes 30 days and B 45 days. A is absent every 5th day and B every 3rd day. They start together. Find completion time.
Solution: In 15 days A works 12 days and B 10 days, completing 12/30+10/45=28/45. Continue day-wise: by end of day 23 they complete 89/90. On day 24 B is absent and A works at 1/30, so the remaining 1/90 takes (1/90)/(1/30)=1/3 day. Final answer: 23 1/3 days.
Q34 [Hard | Work-rest cycle] A needs 12 actual working days. A works 2 days and rests 1 day repeatedly. How many calendar days are needed?
Solution: Twelve working days form six 2-day work blocks. Only five rest days occur before the final block; no rest is needed after completion. Calendar days=12+5=17. Final answer: 17 days.
Q35 [Hard | Different daily hours] A can finish in 10 days working 6 hours/day. B can finish in 15 days working 8 hours/day. They work together 5 hours/day. Find days.
Solution: A's hourly rate=1/60; B's=1/120. Combined hourly rate=1/40. In 5 hours, daily work=5/40=1/8. Final answer: 8 days.
Q36 [Hard | Scope increase] A team planned to finish in 30 days. After completing 40% in 12 days, project scope increases by 25% of the original job. Find total time at the same rate.
Solution: New total scope=1.25 original. Completed=0.40 original. Remaining=0.85 original. Rate=1/30 original per day. Remaining time=0.85×30=25.5 days. Total=12+25.5=37.5 days.
Q37 [Hard | Rejected work] A team completes 80% of a job in 16 days. Then 10% of the completed work is rejected. At the same gross rate, find total time.
Solution: Gross rate=0.80/16=0.05 job/day. Accepted work after rejection=0.80×0.90=0.72. Remaining=0.28. Extra time=0.28/0.05=5.6 days. Total=21.6 days.
Q38 [Hard | Increasing workforce AP] A project requires 360 worker-days. Ten workers work on day 1, and 2 more workers join each day. In how many days is it completed?
Solution: Workforce sequence is 10,12,14,... Sum after n days=n/2[20+2(n-1)]=n(n+9). Set n(n+9)=360, giving n=15. Final answer: 15 days.
Q39 [Hard | Decreasing workforce AP] A project requires 210 worker-days. Thirty workers begin and 2 leave after each day. Find completion time.
Solution: Daily workforce is 30,28,26,... Sum after n days=n/2[60-2(n-1)]=n(31-n). Set n(31-n)=210. Feasible root n=10 (the other root would extend beyond a valid positive workforce sequence). Final answer: 10 days.
Q40 [Hard | Learning curve AP] Daily output is 5 units on day 1 and rises by 2 units each day. How many days are needed for 221 units?
Solution: AP sum=n/2[10+2(n-1)]=n(n+4). Set n(n+4)=221, so n=13. Final answer: 13 days.
Q41 [Hard | Doubling output GP] A machine produces 1 unit on day 1 and doubles its output every day. When will cumulative output reach 127 units?
Solution: Cumulative output after n days=1+2+...+2^(n-1)=2^n-1. Set 2^n-1=127, so 2^n=128 and n=7. Final answer: 7 days.
Q42 [Hard | Machine-hours] Five machines working 8 hours/day for 12 days produce 4,800 units. How many days will 8 machines working 6 hours/day need?
Solution: Output per machine-hour=4800/(5×8×12)=10 units. New daily output=8×6×10=480. Days=4800/480=10. Final answer: 10 days.
Q43 [Hard | Continuous rejection] A completes in 20 days. B completes in 30 days, but 10% of B's output is rejected continuously. How long together?
Solution: A's effective rate=1/20=0.05. B's accepted rate=0.90/30=0.03. Total=0.08=2/25, so time=25/2=12.5 days. Final answer: 12 1/2 days.
Q44 [Hard | Different teams by stage] A and B take 20 and 30 days. They complete the first 40% together. B and C, where C takes 60 days, complete the rest. Find total time.
Solution: A+B rate=1/12. Time for 40%=0.4×12=4.8 days. B+C rate=1/30+1/60=1/20. Time for remaining 60%=0.6×20=12 days. Total=16.8 days.
Q45 [Hard | Overlapping shifts] A can finish in 18 days at 6 hours/day. B can finish in 24 days at 8 hours/day. Each calendar day A works 6 hours and B works 4 hours. Find time.
Solution: A hourly rate=1/108; B hourly rate=1/192. Daily rate=6/108+4/192=1/18+1/48=11/144. Time=144/11=13 1/11 days. Final answer: 13 1/11 days.
Q46 [Hard | Two periodic absences] A and B take 15 and 30 days. A is absent every 4th day; B is absent every 5th day. They start together. Find completion time.
Solution: Track the 20-day attendance cycle. By the end of day 12, completed work is 14/15. On day 13 both work at 1/15+1/30=1/10. Remaining 1/15 takes (1/15)/(1/10)=2/3 day. Final answer: 12 2/3 days.
Q47 [Hard | Every third day helper] A takes 12 days and works daily. B takes 18 days but works only every third day. Both start on day 1, with B working on days 3,6,9,... Find time.
Solution: In each 3-day cycle A does 3/12=1/4 and B does 1/18, total 11/36. Three cycles complete 11/12 in 9 days. On day 10 A alone completes the remaining 1/12. Final answer: 10 days.
Q48 [Hard | Alternating with damage] A takes 12 days and works on odd days; B takes 18 days and works on even days. C undoes the job at a rate of 1/72 every day. Find completion time.
Solution: Net two-day cycle=1/12+1/18-2/72=(6+4-2)/72=1/9. Nine cycles complete the job in 18 days. Final answer: 18 days.
Q49 [Hard | Three joining times] A, B and C take 24, 36 and 72 days. A starts alone; B joins after 4 days; C joins after another 4 days. Find total time.
Solution: First 4 days: A does 1/6. Next 4 days: A+B rate=5/72, work=5/18. Total completed=4/9. Remaining=5/9. All-three rate=1/24+1/36+1/72=1/12. Extra time=(5/9)×12=6 2/3. Total=14 2/3 days.
Q50 [Very hard | Rotating pair rates] A+B can finish in 8 days, B+C in 15 days and C+A in 20 days. The pairs work in that order, one pair per day, repeating. Find completion time.
Solution: Three-day cycle work=1/8+1/15+1/20=(15+8+6)/120=29/120. Four cycles in 12 days complete 116/120=29/30. On day 13 A+B work at 1/8=15/120; remaining is 4/120, needing (4/120)/(15/120)=4/15 day. Final answer: 12 4/15 days.
Q51 [Very hard | All-three from pair times] A+B take 10 days, B+C 12 days and C+A 15 days. How long do all three take?
Solution: All-three rate=½(1/10+1/12+1/15)=½[(6+5+4)/60]=1/8. Final answer: 8 days.
Q52 [Very hard | Minimum workers after loss] 28 workers can finish in 45 days. After 15 days, only 20 remain. How many additional workers are required to finish by day 35?
Solution: Total work=28×45=1260 worker-days. Done=28×15=420. Remaining=840. Days left=20. Required workforce=840/20=42. Existing=20, so additional=22. Final answer: 22 additional workers.
Q53 [Very hard | Multi-phase efficiency] A's normal full-job time is 20 days. A completes the first 25% at normal efficiency, next 50% at 125% efficiency and final 25% at 80% efficiency. Find total time.
Solution: Phase times: 0.25×20=5; 0.50×20/1.25=8; 0.25×20/0.80=6.25. Total=19.25 days. Final answer: 19 1/4 days.
Q54 [Very hard | Individual times differ] A and B together finish in 12 days. A takes 10 fewer days than B alone. Find their individual times.
Solution: Let A take x and B x+10. 1/x+1/(x+10)=1/12. Clearing denominators: x²-14x-120=0=(x-20)(x+6). Positive root x=20. B=30. Final answer: A 20 days, B 30 days.
Q55 [Very hard | Split work for simultaneous finish] Team A can finish the whole job in 20 days and Team B in 30 days. What fraction should each receive so both finish simultaneously?
Solution: Let A receive x. Completion times: 20x and 30(1-x). Set equal: 20x=30(1-x), so 50x=30 and x=3/5. B receives 2/5. Both take 12 days. Final answer: A 3/5, B 2/5.
Q56 [Very hard | Worker transfer between projects] Project P needs 12 workers for 20 days; Project Q needs 18 workers for 15 days. After 5 days, 4 workers move from P to Q. Find each completion time.
Solution: P total=240 worker-days; done=60; remaining=180 with 8 workers, requiring 22.5 more days. P total=27.5 days. Q total=270; done=90; remaining=180 with 22 workers, requiring 90/11=8 2/11 more days. Q total=13 2/11 days.
Q57 [Very hard | Wages with efficiency and attendance] A, B and C have efficiencies 3:2:5 and work 8, 9 and 6 days. Divide ?14,400.
Solution: Contributions=3×8 : 2×9 : 5×6 =24:18:30=4:3:5. Total parts=12. Shares: A ?4,800; B ?3,600; C ?6,000. Final answer: ?4,800, ?3,600, ?6,000.
Q58 [Very hard | Four-day repeating schedule] A takes 16 days and B 24 days. Schedule: day 1 A, day 2 B, day 3 both, day 4 rest; repeat. Find completion time.
Solution: Four-day cycle work=1/16+1/24+(1/16+1/24)=5/24. Four cycles complete 5/6 in 16 days. Day 17 A ? 43/48; day 18 B ? 45/48; remaining 3/48 on day 19 when both rate=5/48, needing 3/5 day. Final answer: 18 3/5 days.
Q59 [Very hard | Actual productivity improves] 24 workers were estimated to finish in 30 days. After 12 days, only one-third is completed. Productivity per worker then rises by 25%. Minimum workers needed to finish the remaining work in 12 days?
Solution: Actual per-worker rate=(1/3)/(24×12)=1/864. Improved rate=1.25/864=5/3456. Required N satisfies N×12×5/3456=2/3, giving N=38.4. Round up to 39. Final answer: 39 workers total, so 15 more than the current 24.
Q60 [Very hard | Delayed join with combined time] A and B together can finish in 12 days. A works alone for 5 days; then B joins, and the job finishes 9 days later. Find their individual times.
Solution: During the final 9 days, together they complete 9/12=3/4. Therefore A's first 5 days complete 1/4, so A's rate=(1/4)/5=1/20 and A takes 20 days. B's rate=1/12-1/20=1/30, so B takes 30 days. Final answer: A 20 days, B 30 days.
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