Episode 8 — Aptitude and Reasoning / 8.9 — Work and Time
8.9.b Tips, Tricks and Shortcuts -- Work and Time
Table of Contents
- The LCM Method -- The Ultimate Shortcut
- Shortcut Formulas for 2-Person and 3-Person Work
- Efficiency Comparison Shortcuts
- Common Time-Work Relationships
- Alternating Work -- Quick Pattern Recognition
- Man-Days Speed Tricks
- Wages -- Quick Ratio Method
- Commonly Tested Traps and How to Avoid Them
1. The LCM Method -- The Ultimate Shortcut
The LCM method is the single most powerful technique for Work and Time problems. It eliminates fractions entirely and converts every problem into simple integer arithmetic.
Why LCM Works
Instead of treating total work as "1" (which creates fractions), we assign total work a concrete number of units equal to the LCM of all given times. This ensures every worker's rate comes out as a whole number.
Step-by-Step Process
Step 1: Identify all the "days" values given in the problem.
Step 2: Calculate LCM of those values = Total Work (in units).
Step 3: Each worker's rate = Total Work / Their individual days.
Step 4: Add or subtract rates as needed.
Step 5: Time = Total Work / Combined Rate.
Full Comparison: Fraction Method vs LCM Method
Problem: A does a job in 12 days, B in 18 days, C in 36 days. How long together?
Fraction Method (slow):
Combined rate = 1/12 + 1/18 + 1/36
= 3/36 + 2/36 + 1/36
= 6/36
= 1/6
Time = 6 days
LCM Method (fast):
LCM(12, 18, 36) = 36 units
A = 36/12 = 3 units/day
B = 36/18 = 2 units/day
C = 36/36 = 1 unit/day
Combined = 6 units/day
Time = 36/6 = 6 days
Both give the same answer, but the LCM method uses only whole numbers -- no fraction addition needed.
LCM Method for "A and B Together, Then A Leaves" Type Problems
Problem: A and B can do a job in 20 and 30 days. They start together. After 4 days, A leaves. How many more days for B to finish?
LCM(20, 30) = 60 units
A = 60/20 = 3 units/day
B = 60/30 = 2 units/day
Together for 4 days: (3 + 2) x 4 = 20 units done
Remaining: 60 - 20 = 40 units
B alone: 40 / 2 = 20 more days
Answer: 20 days
LCM Method for "Finding Individual from Pairs" Problems
Problem: A+B finish in 10 days, B+C in 12 days, A+C in 15 days. Find each alone and all three together.
LCM(10, 12, 15) = 60 units
(A+B) rate = 60/10 = 6 units/day
(B+C) rate = 60/12 = 5 units/day
(A+C) rate = 60/15 = 4 units/day
Sum of all three equations:
2(A + B + C) = 6 + 5 + 4 = 15
(A + B + C) = 7.5 units/day
Individual rates:
A = 7.5 - 5 = 2.5 units/day => alone: 60/2.5 = 24 days
B = 7.5 - 4 = 3.5 units/day => alone: 60/3.5 = 120/7 days
C = 7.5 - 6 = 1.5 units/day => alone: 60/1.5 = 40 days
Together: 60/7.5 = 8 days
2. Shortcut Formulas for 2-Person and 3-Person Work
Two-Person Shortcut
If A takes 'a' days and B takes 'b' days:
Together = (a x b) / (a + b) days
This formula should be MEMORIZED. It is the single most-used formula.
Quick mental math examples:
a = 10, b = 15: (10 x 15)/(10 + 15) = 150/25 = 6 days
a = 12, b = 12: (12 x 12)/(12 + 12) = 144/24 = 6 days
a = 6, b = 12: (6 x 12)/(6 + 12) = 72/18 = 4 days
a = 20, b = 30: (20 x 30)/(20 + 30) = 600/50 = 12 days
"One Helps, One Hinders" Shortcut
When A builds and B destroys (or a pipe fills while another leaks):
If A completes in 'a' days and B destroys in 'b' days (b > a for net positive):
Net time = (a x b) / (b - a) days
Three-Person Shortcut
If A, B, C take a, b, c days:
Together = (a x b x c) / (b.c + a.c + a.b) days
This is harder to use mentally but useful for clean numbers.
"B Joins After Some Days" Shortcut
A starts alone. After 'd' days, B joins. Total time?
Let A take 'a' days, B take 'b' days.
Total time T satisfies:
T/a + (T - d)/b = 1
Solving: T = (a.b + d.a) / (a + b)
T = a(b + d) / (a + b)
Example: A=20 days, B=30 days, B joins after d=4 days:
T = 20(30 + 4) / (20 + 30) = 20 x 34 / 50 = 680/50 = 13.6 days
Verification: A works 13.6 days = 13.6/20 = 0.68
B works 13.6-4 = 9.6 days = 9.6/30 = 0.32
Total = 0.68 + 0.32 = 1.00 (correct)
3. Efficiency Comparison Shortcuts
Ratio-Based Efficiency
Shortcut: If A is n times as efficient as B, and B takes 'b' days:
A takes b/n days
Together they take b/(n+1) days [since A takes b/n, Together = (b/n x b)/(b/n + b)]
Simplified:
Together = b / (n + 1) <-- VERY useful shortcut
Example: A is 3 times as efficient as B. B takes 24 days alone.
A takes 24/3 = 8 days alone
Together = 24/(3+1) = 24/4 = 6 days
(No formula needed beyond the shortcut!)
Percentage Efficiency
If A is p% more efficient than B, and they together take T days:
A alone = T x (200 + p) / (100 + p) ... (derived from the ratio)
B alone = T x (200 + p) / 100
This is less commonly needed but saves time in specific problem types.
Quick Efficiency Table
If A:B efficiency = 2:1 => A:B time = 1:2 => Together = B_time/3
If A:B efficiency = 3:1 => A:B time = 1:3 => Together = B_time/4
If A:B efficiency = 3:2 => A:B time = 2:3 => Together = (2/5) x B_time
If A:B efficiency = 4:3 => A:B time = 3:4 => Together = (3/7) x B_time
If A:B efficiency = 5:3 => A:B time = 3:5 => Together = (3/8) x B_time
4. Common Time-Work Relationships
Memorize These Common Pairs
These come up repeatedly in exams. Knowing them by heart saves crucial time.
Two workers, individual times -> time together:
(2, 2) -> 1
(3, 6) -> 2
(4, 12) -> 3
(5, 20) -> 4
(6, 30) -> 5
(4, 6) -> 2.4
(6, 12) -> 4
(8, 24) -> 6
(10, 15) -> 6
(12, 15) -> 60/9 = 6.67
(10, 20) -> 20/3 = 6.67
(12, 18) -> 36/5 = 7.2
(15, 20) -> 60/7 = 8.57
(20, 30) -> 12
(15, 45) -> 45/4 = 11.25
(24, 36) -> 14.4
(30, 40) -> 120/7 = 17.14
(36, 45) -> 20
Pattern: When One Time is a Multiple of the Other
If B = n x A (i.e., B takes n times longer):
Together = A x n/(n+1)
Examples:
A=5, B=10 (n=2): Together = 5 x 2/3 = 10/3
A=6, B=18 (n=3): Together = 6 x 3/4 = 18/4 = 4.5
A=4, B=20 (n=5): Together = 4 x 5/6 = 20/6 = 10/3
5. Alternating Work -- Quick Pattern Recognition
The 2-Day Cycle Method
For A and B alternating (A on odd days, B on even days):
Step 1: Work per 2-day cycle = Rate(A) + Rate(B)
Step 2: Full cycles possible = Total Work / Work per cycle (take integer part)
Step 3: Remaining work after full cycles
Step 4: Next person in sequence finishes remaining
CRITICAL: Check if the remaining work can be done by the next person in
exactly 1 day or if it takes a fraction of a day.
Quick Check: Does It Finish Evenly?
Using LCM method, if Total Work is divisible by (Rate_A + Rate_B):
=> Finishes in an exact number of cycles
=> Total days = 2 x (Total Work / (Rate_A + Rate_B))
If not:
=> There will be a fractional last day
Three Workers Alternating
A, B, C work on days 1, 2, 3, 4(A), 5(B), 6(C), ...
Work per 3-day cycle = Rate(A) + Rate(B) + Rate(C)
Full cycles = Total Work / Work per cycle
Remaining work: check who works next in sequence
6. Man-Days Speed Tricks
The Chain Rule Shortcut
For problems with multiple changing variables:
Original: M1 men, D1 days, H1 hours/day, for W1 work
New: M2 men, D2 days, H2 hours/day, for W2 work
Write as a proportion chain:
M2 = M1 x (D1/D2) x (H1/H2) x (W2/W1)
Each ratio flips depending on the relationship:
- More days available -> fewer men needed (inverse)
- More hours/day -> fewer men needed (inverse)
- More work -> more men needed (direct)
Quick Setup
Ask yourself for each variable:
"If this increases, do I need MORE or FEWER of what I'm solving for?"
MORE -> direct ratio (new/old)
FEWER -> inverse ratio (old/new)
Example: 15 men, 8 hrs/day finish job in 10 days. How many men for 12 days, 5 hrs/day?
M2 = 15 x (10/12) x (8/5)
= 15 x 10 x 8 / (12 x 5)
= 1200 / 60
= 20 men
7. Wages -- Quick Ratio Method
The Fast Approach
Step 1: Find rate ratio of all workers using LCM method
Step 2: If all work same days: wage ratio = rate ratio
Step 3: If different days: wage ratio = rate x days for each worker
Step 4: Split total wage in that ratio
Shortcut for "A and B work together" wage split
If A takes 'a' days and B takes 'b' days, and they complete the job together:
A's share = Total Wage x b/(a+b) (NOTE: b in numerator, not a!)
B's share = Total Wage x a/(a+b)
Why? Because efficiency is inversely proportional to time.
A's efficiency : B's efficiency = 1/a : 1/b = b : a
Example: A=10 days, B=15 days. Total wage = Rs. 5000.
A's share = 5000 x 15/(10+15) = 5000 x 15/25 = 5000 x 3/5 = Rs. 3000
B's share = 5000 x 10/(10+15) = 5000 x 10/25 = 5000 x 2/5 = Rs. 2000
8. Commonly Tested Traps and How to Avoid Them
Trap 1: "A and B can do a work in 10 days" vs "A can do in 10 days and B can do in 10 days"
"A and B can do a work in 10 days"
=> Combined rate = 1/10 (they take 10 days TOGETHER)
"A can do in 10 days AND B can do in 10 days"
=> Individual rates: A = 1/10, B = 1/10
=> Combined = 1/5, Together = 5 days
Always read carefully whether the time given is for the pair or for each individual.
Trap 2: "A is twice as good as B" vs "A takes twice as long as B"
"A is twice as good as B"
=> A's rate = 2 x B's rate
=> A takes HALF the time of B
"A takes twice as long as B"
=> A's time = 2 x B's time
=> A is HALF as efficient as B
These are OPPOSITE situations. Read twice.
Trap 3: Alternating Work -- Who Starts?
The person who starts matters because:
- If A starts and work finishes on an odd day, A finishes it
- If A starts and work finishes on an even day, B finishes it
The total time can differ by up to 1 day depending on who starts.
Trap 4: Negative Work
"A can build in 10 days, B can destroy in 15 days"
If both work simultaneously:
Net rate = 1/10 - 1/15 = 1/30 per day (net positive, work progresses)
Time = 30 days
Do NOT accidentally ADD the rates. The destroyer SUBTRACTS.
Trap 5: "Remaining Work" Does Not Mean "Remaining Time"
If 1/3 of work is done, 2/3 remains.
Remaining TIME depends on WHO does the remaining work and at what rate.
If A did the first 1/3 and now B takes over, the time is:
(2/3) / Rate(B) -- NOT simply 2/3 of the original total time.
Trap 6: Adding Workers Mid-Way
"5 men start a job. After 4 days, 3 more men join."
WRONG: Calculating as if 8 men worked the entire time.
RIGHT: 5 men x 4 days + 8 men x remaining days = total work
Quick Reference: When to Use Which Method
| Problem Type | Best Method |
|---|---|
| Simple two-person together | Formula: ab/(a+b) |
| Three or more workers | LCM method |
| Worker joins/leaves mid-way | LCM method |
| Alternating work | LCM + cycle analysis |
| Efficiency comparison | Ratio method |
| Man-days / scaling | Chain rule: M1.D1.H1/W1 = M2.D2.H2/W2 |
| Wages distribution | LCM + ratio |
| Finding individuals from pairs | LCM + simultaneous equations |
Next: 8.9.c Solved Examples