Episode 8 — Aptitude and Reasoning / 8.9 — Work and Time
8.9.c Solved Examples -- Work and Time
Problems are organized into three difficulty tiers: Basic, Medium, and Advanced.
Basic Level
Problem 1: Simple Individual Work
A can complete a piece of work in 16 days. In how many days will he complete 3/4 of the work?
A's rate = 1/16 per day
Time for 3/4 of work = (3/4) / (1/16) = (3/4) x 16 = 12 days
Answer: 12 days
Problem 2: Two Workers Together (Direct Formula)
A can do a job in 10 days. B can do it in 15 days. How many days will they take working together?
Using the formula: Together = (a x b) / (a + b)
= (10 x 15) / (10 + 15)
= 150 / 25
= 6 days
Answer: 6 days
Problem 3: Two Workers Together (LCM Method)
A finishes a task in 18 days. B finishes the same task in 27 days. Working together, how long do they take?
LCM(18, 27) = 54 units
A's rate = 54/18 = 3 units/day
B's rate = 54/27 = 2 units/day
Combined = 5 units/day
Time = 54/5 = 10 4/5 days = 10 days and 19.2 hours
Answer: 10 4/5 days
Problem 4: Finding Individual Time
A and B together can do a job in 12 days. A alone can do it in 20 days. How long does B take alone?
LCM(12, 20) = 60 units
(A+B) rate = 60/12 = 5 units/day
A's rate = 60/20 = 3 units/day
B's rate = 5 - 3 = 2 units/day
B alone = 60/2 = 30 days
Answer: 30 days
Problem 5: Three Workers Together
A can do a job in 6 days, B in 8 days, C in 12 days. How long if all three work together?
LCM(6, 8, 12) = 24 units
A's rate = 24/6 = 4 units/day
B's rate = 24/8 = 3 units/day
C's rate = 24/12 = 2 units/day
Combined = 9 units/day
Time = 24/9 = 8/3 = 2 2/3 days
Answer: 2 2/3 days
Problem 6: Partial Work and Remaining
A can do a job in 25 days. He works for 10 days and leaves. What fraction of the work is left?
Work done in 10 days = 10/25 = 2/5
Work remaining = 1 - 2/5 = 3/5
Answer: 3/5 of the work is left
Problem 7: Simple Man-Days
12 workers can finish a job in 10 days. How many workers are needed to finish it in 8 days?
Total work = 12 x 10 = 120 man-days
Workers needed = 120 / 8 = 15
Answer: 15 workers
Medium Level
Problem 8: Worker Leaves Mid-Way
A and B can do a job in 15 and 20 days respectively. They start working together, but A leaves 5 days before the work is completed. How many days did the work take in total?
LCM(15, 20) = 60 units
A's rate = 60/15 = 4 units/day
B's rate = 60/20 = 3 units/day
Let total time = T days.
B works all T days. A works (T - 5) days.
Work equation:
4(T - 5) + 3T = 60
4T - 20 + 3T = 60
7T = 80
T = 80/7 = 11 3/7 days
Answer: 11 3/7 days
Problem 9: Efficiency-Based Problem
A is 25% more efficient than B. B can finish a job in 30 days. How long do A and B take working together?
B takes 30 days.
A is 25% more efficient:
Time(A) = 30 x 100/125 = 30 x 4/5 = 24 days
Together = (24 x 30) / (24 + 30) = 720/54 = 40/3 = 13 1/3 days
Answer: 13 1/3 days
Problem 10: Alternating Days (A Starts)
A can do a job in 15 days, B in 20 days. They work on alternate days with A starting. In how many days is the work finished?
LCM(15, 20) = 60 units
A's rate = 60/15 = 4 units/day
B's rate = 60/20 = 3 units/day
One 2-day cycle (A then B): 4 + 3 = 7 units
Full cycles in 60 units: 60/7 = 8 cycles remainder 4
8 cycles = 16 days, work done = 56 units
Remaining = 4 units
Day 17: A works. A's rate = 4 units. A does exactly 4 units.
Total work = 56 + 4 = 60 units. Done!
Answer: 17 days
Problem 11: Alternating Days (B Starts)
Same as Problem 10, but B starts first. In how many days is the work finished?
One 2-day cycle (B then A): 3 + 4 = 7 units (same total per cycle)
8 cycles = 16 days, work done = 56 units
Remaining = 4 units
Day 17: B works (since B starts odd cycles). B's rate = 3 units.
After Day 17: 56 + 3 = 59 units. Still 1 unit left.
Day 18: A works. A needs 1 unit, rate = 4 units/day.
Time = 1/4 day.
Answer: 17 1/4 days
Note: Compared to Problem 10 (17 days), having the slower worker start
results in a slightly longer total time.
Problem 12: Three Workers -- Finding Individuals from Pairs
A and B can do a job in 8 days. B and C in 12 days. A and C in 24 days. How long does each take alone? How long do all three take together?
LCM(8, 12, 24) = 24 units
(A+B) rate = 24/8 = 3 units/day
(B+C) rate = 24/12 = 2 units/day
(A+C) rate = 24/24 = 1 unit/day
Sum: 2(A+B+C) = 3 + 2 + 1 = 6
(A+B+C) = 3 units/day
Individual rates:
A = 3 - 2 = 1 unit/day => A alone = 24/1 = 24 days
B = 3 - 1 = 2 units/day => B alone = 24/2 = 12 days
C = 3 - 3 = 0 units/day => C alone = infinite! C does no work!
Wait -- C's rate = 0? Let's verify:
A+C = 1, and A = 1, so C = 0. This means C contributes nothing.
Actually, let's recheck: if A+C = 1 unit/day and A = 1 unit/day,
then C = 0. This is mathematically valid but practically means
C is not a worker. This indicates the problem data may be
inconsistent for a realistic scenario.
Let us redo with cleaner numbers:
A+B in 10, B+C in 12, A+C in 15.
LCM(10, 12, 15) = 60 units
(A+B) rate = 60/10 = 6
(B+C) rate = 60/12 = 5
(A+C) rate = 60/15 = 4
2(A+B+C) = 15 => (A+B+C) = 7.5 units/day
A = 7.5 - 5 = 2.5 => alone: 60/2.5 = 24 days
B = 7.5 - 4 = 3.5 => alone: 60/3.5 = 120/7 = 17 1/7 days
C = 7.5 - 6 = 1.5 => alone: 60/1.5 = 40 days
All together: 60/7.5 = 8 days
Answer: A = 24 days, B = 17 1/7 days, C = 40 days. Together = 8 days.
Problem 13: Worker Joins Mid-Way
A can do a job in 40 days. He works alone for 8 days. Then B joins and together they finish the remaining job in 16 days. How long would B take alone?
A's work in 8 days (alone) = 8/40 = 1/5
A's work in next 16 days = 16/40 = 2/5
Total work by A = 1/5 + 2/5 = 3/5
Work done by B in 16 days = 1 - 3/5 = 2/5
B's rate = (2/5) / 16 = 2/80 = 1/40 per day
B alone = 40 days
Answer: 40 days
Problem 14: Work and Wages
A can do a job in 6 days, B in 10 days, C in 15 days. They work together and earn a total of Rs. 3000. What is each person's share?
LCM(6, 10, 15) = 30 units
A's rate = 30/6 = 5 units/day
B's rate = 30/10 = 3 units/day
C's rate = 30/15 = 2 units/day
Ratio of work = 5 : 3 : 2 (total parts = 10)
A's share = (5/10) x 3000 = Rs. 1500
B's share = (3/10) x 3000 = Rs. 900
C's share = (2/10) x 3000 = Rs. 600
Answer: A = Rs. 1500, B = Rs. 900, C = Rs. 600
Problem 15: Men and Women
5 men can complete a job in 12 days. 8 women can complete the same job in 15 days. How long will 3 men and 4 women take together?
Total work:
5 men x 12 days = 60 man-days
8 women x 15 days = 120 woman-days
1 man's daily work = 1/60 of the job
1 woman's daily work = 1/120 of the job
3 men + 4 women rate per day:
= 3/60 + 4/120
= 1/20 + 1/30
= 3/60 + 2/60
= 5/60
= 1/12 per day
Time = 12 days
Answer: 12 days
Problem 16: Half the Work
A can do a piece of work in 14 days. B is 40% more efficient than A. How many days does B take to do half the work?
A takes 14 days for full work.
B is 40% more efficient:
Time(B) = 14 x 100/140 = 14 x 5/7 = 10 days for full work
Half the work = 10/2 = 5 days
Answer: 5 days
Advanced Level
Problem 17: Complex Alternating Schedule
A can do a job in 16 days, B in 12 days, C in 24 days. They work in a rotation: A on Day 1, B on Day 2, C on Day 3, A on Day 4, and so on. On which day is the work completed?
LCM(16, 12, 24) = 48 units
A's rate = 48/16 = 3 units/day
B's rate = 48/12 = 4 units/day
C's rate = 48/24 = 2 units/day
One 3-day cycle: 3 + 4 + 2 = 9 units
Full cycles: 48/9 = 5 cycles remainder 3 units
5 cycles = 15 days, work done = 45 units
Remaining = 3 units
Day 16: A works. A does 3 units. 45 + 3 = 48. Done!
Answer: Work is completed on Day 16 (A finishes the job).
Problem 18: Group Efficiency with Men, Women, and Children
2 men or 3 women or 5 children can complete a job in 20 days. How long will 1 man, 1 woman, and 1 child take together?
Total work:
2 men in 20 days => 1 man in 40 days => Man's rate = 1/40
3 women in 20 days => 1 woman in 60 days => Woman's rate = 1/60
5 children in 20 days => 1 child in 100 days => Child's rate = 1/100
Combined rate of 1M + 1W + 1C:
= 1/40 + 1/60 + 1/100
LCM(40, 60, 100) = 600
= 15/600 + 10/600 + 6/600
= 31/600 per day
Time = 600/31 = 19 11/31 days
Answer: 19 11/31 days (approximately 19.35 days)
Problem 19: Worker Leaves, Another Joins
A and B can do a job in 12 days. B and C in 15 days. C and A in 20 days. A, B, and C start working together. After 4 days, A leaves. After 2 more days (i.e., at the end of Day 6), B also leaves. How many more days will C take to finish?
LCM(12, 15, 20) = 60 units
(A+B) rate = 60/12 = 5
(B+C) rate = 60/15 = 4
(A+C) rate = 60/20 = 3
2(A+B+C) = 12 => (A+B+C) = 6 units/day
Individual rates:
A = 6 - 4 = 2 units/day
B = 6 - 3 = 3 units/day
C = 6 - 5 = 1 unit/day
Phase 1: A+B+C work for 4 days = 6 x 4 = 24 units
Phase 2: B+C work for 2 days = (3+1) x 2 = 8 units
Total after 6 days = 24 + 8 = 32 units
Remaining = 60 - 32 = 28 units
Phase 3: C alone at 1 unit/day = 28 days
Answer: C takes 28 more days.
Problem 20: Man-Days with Multiple Variables
24 men working 8 hours/day can complete a project in 10 days. Due to delays, after 4 days only 16 men show up and they work 6 hours/day. How many additional days to complete the project?
Total work = 24 x 8 x 10 = 1920 man-hours
Work done in first 4 days = 24 x 8 x 4 = 768 man-hours
Remaining work = 1920 - 768 = 1152 man-hours
New rate = 16 men x 6 hours/day = 96 man-hours/day
Additional days = 1152 / 96 = 12 days
Answer: 12 additional days
Problem 21: Efficiency Ratio and Wages Combined
A is thrice as efficient as B. A finishes a job 16 days earlier than B. They work together and earn Rs. 4800. Find each person's share and the time they take together.
Let A's time = x days, B's time = 3x days (since A is thrice as efficient).
B takes 16 days more than A:
3x - x = 16
2x = 16
x = 8
A = 8 days, B = 24 days
Together = (8 x 24)/(8 + 24) = 192/32 = 6 days
Efficiency ratio A:B = 3:1
Wage ratio = 3:1
A's share = (3/4) x 4800 = Rs. 3600
B's share = (1/4) x 4800 = Rs. 1200
Answer: Together = 6 days. A gets Rs. 3600, B gets Rs. 1200.
Problem 22: Work with Pipes Crossover
A tap can fill a tank in 12 hours. A leak empties it in 20 hours. If both are open, how long to fill the tank?
(This is a Pipes and Cisterns problem, but uses the same Work and Time logic.)
LCM(12, 20) = 60 units
Filling rate = 60/12 = 5 units/hour
Leak rate = 60/20 = 3 units/hour (negative work)
Net rate = 5 - 3 = 2 units/hour
Time = 60/2 = 30 hours
Answer: 30 hours
Problem 23: Work Done on Different Days
A can do a job in 20 days. B can do it in 25 days. A works on it for 10 days, then B works on it for 5 days. C finishes the remaining work in 12 days. How long would C take to do the entire job alone?
A's work in 10 days = 10/20 = 1/2
B's work in 5 days = 5/25 = 1/5
Total done by A and B = 1/2 + 1/5 = 5/10 + 2/10 = 7/10
Remaining = 1 - 7/10 = 3/10
C does 3/10 in 12 days:
C's rate = (3/10) / 12 = 3/120 = 1/40 per day
C alone = 40 days
Answer: 40 days
Problem 24: Progressive Work (Increasing Workforce)
On Day 1, one man works. On Day 2, two men work. On Day 3, three men work, and so on. If one man can complete the job in 55 days working alone, on which day is the job completed?
Total work = 55 man-days
Work done by end of Day n = 1 + 2 + 3 + ... + n = n(n+1)/2
We need: n(n+1)/2 >= 55
n(n+1) >= 110
Testing: n=10: 10 x 11 = 110. Exactly 110.
n(n+1)/2 = 110/2 = 55 man-days.
Answer: The job is completed on Day 10.
Problem 25: Complex Wage Distribution
A works for 3 days and finishes 1/6 of the job. B then works for 6 days and finishes 1/3 of the job. C finishes the remaining work in 8 days. If total wages are Rs. 7200, how should wages be distributed?
Work done:
A: 1/6
B: 1/3 = 2/6
C: remaining = 1 - 1/6 - 1/3 = 1 - 1/6 - 2/6 = 3/6 = 1/2
Ratio of work done:
A : B : C = 1/6 : 1/3 : 1/2 = 1 : 2 : 3 (multiply all by 6)
Total parts = 6
A's wage = (1/6) x 7200 = Rs. 1200
B's wage = (2/6) x 7200 = Rs. 2400
C's wage = (3/6) x 7200 = Rs. 3600
Answer: A = Rs. 1200, B = Rs. 2400, C = Rs. 3600
Problem 26: "A Can Destroy" Type Problem
A can build a wall in 10 days. B can demolish it in 15 days. A works for 5 days, then both A and B work together. How many more days to complete the wall?
LCM(10, 15) = 30 units
A's rate = +3 units/day (builds)
B's rate = -2 units/day (destroys)
Phase 1: A works alone for 5 days = 3 x 5 = 15 units done
Remaining = 30 - 15 = 15 units
Phase 2: A and B together
Net rate = 3 - 2 = 1 unit/day
Time for remaining 15 units = 15/1 = 15 days
Answer: 15 more days
Problem 27: Multiple Constraints
A group of 10 men can finish a project in 30 days working 8 hours/day. After 10 days, the client doubles the project scope (remaining work doubles). How many extra men must be hired if the deadline remains 30 days from the start and working hours increase to 10 hours/day?
Original total work = 10 x 30 x 8 = 2400 man-hours
Work done in first 10 days = 10 x 10 x 8 = 800 man-hours
Original remaining work = 2400 - 800 = 1600 man-hours
But remaining work doubles: New remaining = 2 x 1600 = 3200 man-hours
Remaining time = 30 - 10 = 20 days at 10 hours/day
Men needed = 3200 / (20 x 10) = 3200/200 = 16 men
Extra men = 16 - 10 = 6 extra men
Answer: 6 extra men must be hired.
Summary of Problem Types Covered
| # | Type | Difficulty |
|---|---|---|
| 1 | Partial work by one person | Basic |
| 2-3 | Two workers together | Basic |
| 4 | Finding individual from combined | Basic |
| 5 | Three workers together | Basic |
| 6 | Fraction remaining | Basic |
| 7 | Man-days | Basic |
| 8 | Worker leaves before completion | Medium |
| 9 | Efficiency percentage | Medium |
| 10-11 | Alternating days | Medium |
| 12 | Individuals from pair data | Medium |
| 13 | Worker joins mid-way | Medium |
| 14 | Work and wages | Medium |
| 15 | Men and women | Medium |
| 16 | Half-work with efficiency | Medium |
| 17 | Three-person rotation | Advanced |
| 18 | Men, women, children | Advanced |
| 19 | Sequential leaving | Advanced |
| 20 | Man-days multi-variable | Advanced |
| 21 | Efficiency + wages combo | Advanced |
| 22 | Constructive + destructive work | Advanced |
| 23 | Three-phase work | Advanced |
| 24 | Progressive workforce | Advanced |
| 25 | Complex wage distribution | Advanced |
| 26 | Build and destroy | Advanced |
| 27 | Mid-project scope change | Advanced |
Next: 8.9 Practice MCQs