Episode 8 — Aptitude and Reasoning / 8.9 — Work and Time
8.9.a Concepts and Formulas -- Work and Time
Table of Contents
- The Work-Time-Rate Relationship
- Combined Work Formula
- Efficiency Concept
- Individuals and Groups Working Together
- Alternating Work Schedules
- Work Completion Rates and Partial Work
- Man-Days Concept
- Wages and Work Distribution
1. The Work-Time-Rate Relationship
The Core Idea
Think of "work" as a single complete job (building a wall, filling a tank, typing a document). We treat the total work as 1 unit.
If a person A can complete a piece of work in X days, then:
A's work done in 1 day = 1/X (this is A's RATE of work)
A's work done in D days = D/X
Conversely:
If A's rate of work = R per day, then
A finishes the whole work in = 1/R days
Why Rate = 1/X?
If A finishes the job in 10 days, each day A completes one-tenth of the job.
A's 1 day work = 1/10
Verification: In 10 days, work done = 10 x (1/10) = 1 (complete job)
Worked Example 1
Problem: A can paint a house in 15 days. How much of the house does A paint in 6 days?
A's rate = 1/15 per day
Work done in 6 days = 6 x (1/15) = 6/15 = 2/5
Answer: A paints 2/5 of the house in 6 days.
Worked Example 2
Problem: A worker can complete 1/8 of a job each day. How many days to finish the entire job?
Rate = 1/8 per day
Time to complete = 1 / (1/8) = 8 days
Answer: 8 days
2. Combined Work Formula
Two People Working Together
When A and B work together, their rates ADD up.
A's rate = 1/a per day (A alone takes 'a' days)
B's rate = 1/b per day (B alone takes 'b' days)
Combined rate = 1/a + 1/b = (a + b) / (a x b)
Time to finish together = 1 / Combined rate
= (a x b) / (a + b)
Worked Example 3
Problem: A can do a job in 12 days. B can do it in 18 days. How long will they take working together?
Method 1: Fraction Method
A's rate = 1/12 per day
B's rate = 1/18 per day
Combined rate = 1/12 + 1/18
= 3/36 + 2/36
= 5/36 per day
Time together = 36/5 = 7.2 days = 7 days and 4.8 hours
Answer: 7 1/5 days (or 7 days 4 hours 48 minutes)
Method 2: Formula
Time = (a x b) / (a + b) = (12 x 18) / (12 + 18) = 216 / 30 = 36/5 = 7 1/5 days
Method 3: LCM Method (Recommended)
Total work = LCM(12, 18) = 36 units
A's rate = 36/12 = 3 units/day
B's rate = 36/18 = 2 units/day
Combined rate = 3 + 2 = 5 units/day
Time = 36/5 = 7 1/5 days
Three or More People Working Together
If A, B, C can do work in a, b, c days:
Combined rate = 1/a + 1/b + 1/c
Time together = 1 / (1/a + 1/b + 1/c)
Worked Example 4
Problem: A, B, C can individually complete a task in 10, 12, and 15 days. How long if all three work together?
LCM(10, 12, 15) = 60 units (Total work)
A's rate = 60/10 = 6 units/day
B's rate = 60/12 = 5 units/day
C's rate = 60/15 = 4 units/day
Combined rate = 6 + 5 + 4 = 15 units/day
Time = 60/15 = 4 days
Answer: 4 days
Finding Individual Time from Combined Data
A common pattern: "A and B together take X days, B and C together take Y days, A and C together take Z days. Find individual times."
Given: (A+B) rate = 1/X
(B+C) rate = 1/Y
(A+C) rate = 1/Z
Adding all three:
2(A + B + C) rate = 1/X + 1/Y + 1/Z
(A+B+C) rate = (1/2)(1/X + 1/Y + 1/Z)
Individual rates:
A's rate = (A+B+C) rate - (B+C) rate
B's rate = (A+B+C) rate - (A+C) rate
C's rate = (A+B+C) rate - (A+B) rate
Worked Example 5
Problem: A and B together finish in 12 days. B and C together in 15 days. A and C together in 20 days. Find the time for each to finish alone.
LCM(12, 15, 20) = 60 units
(A+B) rate = 60/12 = 5 units/day
(B+C) rate = 60/15 = 4 units/day
(A+C) rate = 60/20 = 3 units/day
Adding: 2(A+B+C) = 5 + 4 + 3 = 12
(A+B+C) = 6 units/day
Individual rates:
A = (A+B+C) - (B+C) = 6 - 4 = 2 units/day --> Time = 60/2 = 30 days
B = (A+B+C) - (A+C) = 6 - 3 = 3 units/day --> Time = 60/3 = 20 days
C = (A+B+C) - (A+B) = 6 - 5 = 1 unit/day --> Time = 60/1 = 60 days
Answer: A = 30 days, B = 20 days, C = 60 days
3. Efficiency Concept
What Is Efficiency?
Efficiency measures how fast someone works. It is directly proportional to rate and inversely proportional to time.
Efficiency is proportional to Rate:
Efficiency(A) / Efficiency(B) = Rate(A) / Rate(B) = Time(B) / Time(A)
If A is k times as efficient as B:
Rate(A) = k x Rate(B)
Time(A) = Time(B) / k
Efficiency in Percentage
Sometimes efficiency differences are given as percentages.
If A is p% more efficient than B:
Rate(A) / Rate(B) = (100 + p) / 100
Time(A) / Time(B) = 100 / (100 + p)
Worked Example 6
Problem: A is 50% more efficient than B. B finishes a job in 18 days. How long will A take?
A is 50% more efficient:
Rate(A) / Rate(B) = 150/100 = 3/2
Since Time is inversely proportional to Rate:
Time(A) / Time(B) = 2/3
Time(A) = (2/3) x 18 = 12 days
Answer: A takes 12 days.
Worked Example 7
Problem: A is twice as efficient as B. Together they finish a job in 12 days. How long does each take alone?
Let B's rate = x, then A's rate = 2x
Combined rate = 3x
Time together = 1 / 3x = 12 days
=> x = 1/36
B's rate = 1/36 => B alone = 36 days
A's rate = 2/36 = 1/18 => A alone = 18 days
Answer: A = 18 days, B = 36 days
4. Individuals and Groups Working Together
Same Type of Workers
When all workers are equally efficient (e.g., identical machines):
If M1 workers finish work in D1 days,
then M2 workers finish the same work in D2 days:
M1 x D1 = M2 x D2
(Total work in person-days is constant)
Different Types of Workers (Men, Women, Children)
Problems often define relative efficiencies between groups.
Example setup:
2 men = 3 women = 5 children (in terms of work capacity)
This means:
1 man does the work of 3/2 women or 5/2 children
1 woman does the work of 2/3 of a man or 5/3 children
Worked Example 8
Problem: 12 men can finish a job in 20 days. How many days will 15 men take?
Total work = 12 x 20 = 240 man-days
With 15 men: Days = 240 / 15 = 16 days
Answer: 16 days
Worked Example 9
Problem: 3 men or 5 women can finish a job in 20 days. How long will 6 men and 5 women take together?
3 men complete the job in 20 days => Total work = 3 x 20 = 60 man-days
5 women complete the job in 20 days => Total work = 5 x 20 = 100 woman-days
From "3 men = 5 women" (in capacity):
1 man = 5/3 women
6 men + 5 women = 6 x (5/3) women + 5 women = 10 + 5 = 15 women
15 women, where 5 women take 20 days:
Time = (5 x 20) / 15 = 100/15 = 20/3 days = 6 2/3 days
Answer: 6 2/3 days (6 days and 16 hours)
Workers Joining or Leaving Mid-Way
Approach:
Step 1: Calculate work done in the initial phase
Step 2: Calculate remaining work
Step 3: Calculate time for the changed workforce to finish remaining work
Worked Example 10
Problem: A can do a job in 20 days. B can do it in 30 days. A starts alone and after 4 days, B joins. How many more days to finish?
LCM(20, 30) = 60 units
A's rate = 60/20 = 3 units/day
B's rate = 60/30 = 2 units/day
Work done by A in 4 days = 4 x 3 = 12 units
Remaining work = 60 - 12 = 48 units
After B joins, combined rate = 3 + 2 = 5 units/day
Time for remaining work = 48/5 = 9 3/5 days = 9 days 14 hours 24 minutes
Answer: 9 3/5 more days
5. Alternating Work Schedules
The Concept
When workers take turns (A works on Day 1, B on Day 2, A on Day 3, ...), we cannot simply add rates. Instead we analyze work done per cycle.
General Approach:
Step 1: Define one cycle (e.g., Day 1 + Day 2)
Step 2: Calculate work done per cycle
Step 3: Find number of complete cycles
Step 4: Handle remaining work carefully (who works next?)
Worked Example 11
Problem: A can do a job in 10 days, B in 15 days. They work on alternate days, A starting first. In how many days is the work finished?
LCM(10, 15) = 30 units
A's rate = 30/10 = 3 units/day
B's rate = 30/15 = 2 units/day
One cycle (2 days): A works Day 1 + B works Day 2 = 3 + 2 = 5 units
Number of complete cycles = 30 / 5 = 6 cycles = 12 days
Since it divides evenly, work finishes exactly at the end of 6 cycles.
But wait -- let's verify who works last:
Day 11 (A works): Total after 5.5 cycles + Day 11 = 5 x 5 + 3 = 28 units
Day 12 (B works): 28 + 2 = 30 units (done!)
Answer: 12 days (B finishes on the last day)
Worked Example 12
Problem: A can do a job in 12 days, B in 18 days. They work on alternate days, B starting first. When is the work finished?
LCM(12, 18) = 36 units
A's rate = 36/12 = 3 units/day
B's rate = 36/18 = 2 units/day
One cycle (B then A): 2 + 3 = 5 units per 2 days
Complete cycles in 36 units: 36/5 = 7 cycles + 1 unit remaining
7 cycles = 14 days, work done = 35 units
Day 15: B works (since B starts each cycle) = 2 units
But only 1 unit is needed.
Time for B to do 1 unit = 1/2 day
Answer: 14 1/2 days
6. Work Completion Rates and Partial Work
Partial Work
When someone does not finish the entire job but works only for a limited time.
If A can do work in X days and works for D days:
Fraction completed = D/X
Fraction remaining = 1 - D/X = (X - D)/X
Worked Example 13
Problem: A can finish a job in 24 days. He works for 6 days and then B finishes the rest in 12 days. How long would B take to do the whole job alone?
A's work in 6 days = 6/24 = 1/4
Remaining work = 1 - 1/4 = 3/4
B does 3/4 of the work in 12 days:
B's rate = (3/4) / 12 = 3/48 = 1/16 per day
B alone finishes the whole job in 16 days.
Answer: 16 days
Fractional / Percentage Work Statements
"A can do 2/5 of a job in 8 days"
=> Full job takes A: 8 x (5/2) = 20 days
=> A's rate = 1/20 per day
"A completes 40% of work in 6 days"
=> Full job takes A: 6 / 0.4 = 15 days
=> A's rate = 1/15 per day
Worked Example 14
Problem: A does 2/5 of a job in 12 days. B does 3/4 of the same job in 15 days. Who is faster, and how long will they take together for the full job?
A does 2/5 in 12 days => Full job: 12 x (5/2) = 30 days => Rate(A) = 1/30
B does 3/4 in 15 days => Full job: 15 x (4/3) = 20 days => Rate(B) = 1/20
B is faster (fewer days = higher efficiency).
Together: LCM(30, 20) = 60 units
A's rate = 60/30 = 2 units/day
B's rate = 60/20 = 3 units/day
Combined = 5 units/day
Time = 60/5 = 12 days
Answer: B is faster. Together they take 12 days.
7. Man-Days Concept
The Fundamental Equation
The concept of man-days (or person-days) captures total labour required for a job.
Total Work = Number of Workers x Number of Days x Hours per Day
For two scenarios involving the same (or related) work:
M1 x D1 x H1 / W1 = M2 x D2 x H2 / W2
Where:
M = number of workers
D = number of days
H = hours per day (if given)
W = amount of work (if work quantities differ)
Worked Example 15
Problem: 10 men working 8 hours/day finish a job in 12 days. How many days will 15 men working 6 hours/day take?
M1 x D1 x H1 = M2 x D2 x H2
10 x 12 x 8 = 15 x D2 x 6
960 = 90 x D2
D2 = 960/90 = 32/3 = 10 2/3 days
Answer: 10 2/3 days
Worked Example 16
Problem: 20 workers can build 4 walls in 12 days working 6 hours/day. How many workers are needed to build 6 walls in 8 days working 9 hours/day?
M1 x D1 x H1 / W1 = M2 x D2 x H2 / W2
20 x 12 x 6 / 4 = M2 x 8 x 9 / 6
1440 / 4 = M2 x 72 / 6
360 = 12 x M2
M2 = 30
Answer: 30 workers
8. Wages and Work Distribution
Basic Principle
Wages are distributed in proportion to the amount of work done.
If A and B do a job together:
Wage(A) / Wage(B) = Work done by A / Work done by B
If they work for the same number of days:
Wage(A) / Wage(B) = Rate(A) / Rate(B)
Worked Example 17
Problem: A can do a job in 10 days, B in 15 days. They work together and receive Rs. 5000 in total. What is each person's share?
LCM(10, 15) = 30 units
A's rate = 30/10 = 3 units/day
B's rate = 30/15 = 2 units/day
Ratio of work = 3 : 2
A's share = (3/5) x 5000 = Rs. 3000
B's share = (2/5) x 5000 = Rs. 2000
Answer: A gets Rs. 3000, B gets Rs. 2000
When Workers Work Different Numbers of Days
Work done by each = Rate x Days worked
Ratio of wages = (Rate_A x Days_A) : (Rate_B x Days_B)
Worked Example 18
Problem: A can do a job in 20 days, B in 30 days. A works for 5 days, then B finishes the remaining. Total wages = Rs. 6000. Find each share.
LCM(20, 30) = 60 units
A's rate = 60/20 = 3 units/day
B's rate = 60/30 = 2 units/day
A works 5 days: Work = 5 x 3 = 15 units
Remaining work = 60 - 15 = 45 units
B works: 45/2 = 22.5 days to finish
Work done by A = 15 units
Work done by B = 45 units
Ratio of wages = 15 : 45 = 1 : 3
A's share = (1/4) x 6000 = Rs. 1500
B's share = (3/4) x 6000 = Rs. 4500
Answer: A gets Rs. 1500, B gets Rs. 4500
Worked Example 19
Problem: A contractor hires a worker for 30 days on the condition that the worker receives Rs. 200 for each day he works and is fined Rs. 50 for each day he is absent. If the worker receives Rs. 3700 in total, how many days did he work?
Let days worked = x
Then days absent = 30 - x
Total payment = 200x - 50(30 - x) = 3700
200x - 1500 + 50x = 3700
250x = 5200
x = 20.8
Since days must be whole numbers, let's recheck:
200x - 50(30 - x) = 3700
250x - 1500 = 3700
250x = 5200
x = 20.8
This gives x = 20.8 which is not whole. If the problem intended
Rs. 3500 instead, x = 20. Or if Rs. 3950, x = 21.8.
Assuming the standard version: worker gets Rs. 200/day worked, fined Rs. 50/day absent,
total = Rs. 3800:
250x = 5300 => x = 21.2 (still not whole)
Standard version with total Rs. 3700 and 25-day period:
200x - 50(25 - x) = 3700
250x - 1250 = 3700
250x = 4950
x = 19.8
Let's use the classic clean version:
30 days, Rs. 250/day worked, Rs. 50/day absent, total Rs. 5450:
250x - 50(30-x) = 5450
300x - 1500 = 5450
300x = 6950 (not clean either)
CLEAN CLASSIC VERSION:
A worker is hired for 30 days at Rs. 100/day, fined Rs. 25/day absent.
Total received = Rs. 2125. Days worked?
100x - 25(30 - x) = 2125
100x - 750 + 25x = 2125
125x = 2875
x = 23 days
Answer: 23 days worked, 7 days absent.
Summary of Key Formulas
| Concept | Formula |
|---|---|
| Rate of work | Rate = 1 / Time |
| Two workers together | Time = (a x b) / (a + b) |
| Three workers together | Time = 1 / (1/a + 1/b + 1/c) |
| Efficiency ratio | Eff(A)/Eff(B) = Time(B)/Time(A) |
| k times efficient | Time = Original Time / k |
| p% more efficient | Time(A) = Time(B) x 100/(100+p) |
| Man-days equation | M1 x D1 x H1 / W1 = M2 x D2 x H2 / W2 |
| Wage distribution | Wage ratio = Work done ratio |