Time, Work & Pipes-Cisterns - Complete SSC CGL Guide
What is Time, Work & Pipes-Cisterns? This topic deals with calculating work efficiency, time taken by individuals/groups to complete tasks, and pipes filling/draining tanks. It's based on the concept that work done is directly proportional to time and efficiency.
Worker A
Efficiency: 3 units/day
Worker B
Efficiency: 2 units/day
Combined
Efficiency: 5 units/day
Pro Tip – Use LCM Method!
For most work problems, take LCM of time taken by individuals as total work. This converts time problems into simple arithmetic. Visit SKY Practice for 500+ work problems with LCM method solutions.
1. Basic Concepts & Formulas
The Fundamental Relationship: Work = Rate × Time. All work problems are variations of this formula.
Core Formulas & Relationships
Understanding Work Concepts
If one person can do a job in 'n' days, their work rate is 1/n per day. Work is usually considered as 1 complete unit unless specified otherwise.
Basic Formulas
Efficiency = Work / Time
Time = Work / Efficiency
Where:
Work = Usually taken as 1 (complete job)
Efficiency = Work done per unit time
Time = Time taken to complete work
Individual Work Rate
- If A completes work in 'a' days: A's rate = 1/a per day
- If B completes work in 'b' days: B's rate = 1/b per day
- Work done in 1 day = 1/a + 1/b (if working together)
- Time together = 1/(1/a + 1/b) = ab/(a+b)
Important Relationships
- More workers → Less time (inverse relationship)
- Higher efficiency → Less time
- If efficiency doubles, time halves
- If work doubles, time doubles (same efficiency)
- Work completed = (Time worked)/(Total time needed)
SSC Shortcut: Quick Formula for Two Workers
A completes in 'a' days, B in 'b' days:
Time together = (a × b) / (a + b) days
Three workers A, B, C: Time = 1/(1/a + 1/b + 1/c)
Example: A=6 days, B=12 days → Together = (6×12)/(6+12) = 72/18 = 4 days
Solved Example: Basic Work Calculation
A's 1 day work = 1/20
B's 1 day work = 1/30
Together 1 day work = 1/20 + 1/30 = (3+2)/60 = 5/60 = 1/12
Time together = 1 ÷ (1/12) = 12 days
Method 2: Using LCM method (recommended)
LCM of 20 and 30 = 60 (Let total work = 60 units)
A's efficiency = 60/20 = 3 units/day
B's efficiency = 60/30 = 2 units/day
Combined efficiency = 3+2 = 5 units/day
Time together = 60/5 = 12 days
Method 3: Using shortcut formula
Time together = (a×b)/(a+b) = (20×30)/(20+30) = 600/50 = 12 days
All methods give same answer: 12 days
2. Work Efficiency & Comparisons
What is Efficiency? The amount of work done per unit time. Higher efficiency means more work in same time.
Efficiency Formulas & Ratios
Understanding Efficiency Ratios
If A is twice as efficient as B, then A does 2 units of work in the time B does 1 unit. Their efficiency ratio is 2:1.
SSC Shortcut: Efficiency from Time
If A takes 'a' days, B takes 'b' days:
Efficiency ratio = b : a (inverse of time ratio)
Example: A=15 days, B=10 days → Efficiency ratio = 10:15 = 2:3
So B is 1.5 times more efficient than A
Solved Example: Efficiency Comparison
Let B's efficiency = 1 unit/day
Since A is twice efficient as B: A's efficiency = 2 units/day
Step 2: Find total work
B completes work in 30 days with efficiency 1 unit/day
Total work = Efficiency × Time = 1 × 30 = 30 units
Step 3: Calculate combined efficiency
A's efficiency = 2 units/day
B's efficiency = 1 unit/day
Combined efficiency = 2 + 1 = 3 units/day
Step 4: Calculate time together
Time = Total work / Combined efficiency
= 30 / 3 = 10 days
Alternative method using ratios:
Efficiency ratio A:B = 2:1
Time ratio A:B = 1:2 (inverse of efficiency)
So A takes 30/2 = 15 days (since B takes 30)
Time together = (15×30)/(15+30) = 450/45 = 10 days
Final Answer: 10 days
3. Combined Work Problems
Multiple Workers: When more than one person works, their efficiencies add up.
Combined Work Formulas
Understanding Combined Work
When multiple workers work together, their work rates add up. But careful: they may work for different time periods or join/leave at different times.
Basic Combined Work
- If A, B, C take a, b, c days individually
- 1 day combined work = 1/a + 1/b + 1/c
- Time together = 1/(1/a + 1/b + 1/c)
- Work done in 't' days = t × (1/a + 1/b + 1/c)
- Remaining work = 1 - Work done
Alternate Day Working
When workers work on alternate days:
- Calculate work done in 2-day cycle
- Divide total work by cycle work
- Multiply cycles by 2 days
- Add remaining days if any
Work After Some Days
When workers join/leave:
- Calculate work done till joining/leaving
- Recalculate with new combination
- Work left = Total - Work done
- Time for remaining = Work left/New efficiency
SSC Shortcut: LCM Method for Combined Work
Step 1: Take LCM of all times as total work
Step 2: Calculate each person's efficiency (Work/Time)
Step 3: Add efficiencies for combined work
Step 4: Time = Total work / Combined efficiency
Example: A=6 days, B=4 days → LCM=12 units, E_A=2, E_B=3, Total E=5, Time=12/5=2.4 days
Solved Example: Complex Combined Work
Let A, B, C's 1 day work be 1/a, 1/b, 1/c
1/a + 1/b = 1/12 ...(1)
1/b + 1/c = 1/15 ...(2)
1/c + 1/a = 1/20 ...(3)
Adding all three equations:
2(1/a + 1/b + 1/c) = 1/12 + 1/15 + 1/20
= (5+4+3)/60 = 12/60 = 1/5
So 1/a + 1/b + 1/c = 1/10
Time together = 10 days
Method 2: LCM method (easier)
LCM of 12, 15, 20 = 60 (Let total work = 60 units)
A+B efficiency = 60/12 = 5 units/day
B+C efficiency = 60/15 = 4 units/day
C+A efficiency = 60/20 = 3 units/day
Adding all three: 2(A+B+C) efficiency = 5+4+3 = 12 units/day
So A+B+C efficiency = 12/2 = 6 units/day
Time for A+B+C = 60/6 = 10 days
Bonus: Find individual efficiencies
A = (A+B+C) - (B+C) = 6-4 = 2 units/day → Time = 60/2=30 days
B = 6-3 = 3 units/day → Time = 60/3=20 days
C = 6-5 = 1 unit/day → Time = 60/1=60 days
Final Answer: A, B, C together take 10 days
4. Pipes & Cisterns Problems
What are Pipes & Cisterns? These are work problems where pipes fill (inlet) or empty (outlet) tanks.
Pipes & Cisterns Formulas
Understanding Pipes as Workers
Inlet pipes are like workers doing positive work (filling). Outlet pipes are like negative workers (emptying). Net work rate = Sum of inlet rates - Sum of outlet rates.
Inlet
Fills in 6 hrs
Rate: +1/6 per hr
Capacity: 1 unit
Outlet
Empties in 12 hrs
Rate: -1/12 per hr
SSC Shortcut: Pipes Problems Formula
Inlet fills in 'a' hours, outlet empties in 'b' hours:
Time to fill when both open = ab/(b-a) if b>a
Example: Inlet:4 hrs, Outlet:6 hrs → Time = (4×6)/(6-4)=24/2=12 hrs
If tank is full and both open: Time to empty = ab/(a-b) if a>b
Solved Example: Complex Pipes Problem
A's rate = 1/20 per minute
B's rate = 1/30 per minute
C's rate = -1/x per minute (negative as outlet)
Step 2: Write equation for combined work
When all three open: 1/20 + 1/30 - 1/x = 1/15
(Since they fill in 15 minutes, combined rate = 1/15)
Step 3: Solve for x
1/20 + 1/30 = (3+2)/60 = 5/60 = 1/12
So: 1/12 - 1/x = 1/15
1/12 - 1/15 = 1/x
(5-4)/60 = 1/x
1/60 = 1/x
x = 60 minutes
Step 4: Verification
A+B rate = 1/12
With C (outlet): 1/12 - 1/60 = (5-1)/60 = 4/60 = 1/15 ✓
So tank fills in 15 minutes when all open
Alternative using LCM method:
LCM of 20,30,15 = 60 (Let tank capacity = 60 units)
A's efficiency = 60/20 = 3 units/min
B's efficiency = 60/30 = 2 units/min
A+B efficiency = 5 units/min
All three efficiency = 60/15 = 4 units/min
So C's efficiency = 5 - 4 = 1 unit/min (negative)
C alone time = 60/1 = 60 minutes to empty
Final Answer: Pipe C alone empties tank in 60 minutes
5. Work & Wages Problems
What are Work & Wages? Problems connecting work done with payment received, based on efficiency.
Work & Wages Formulas
Understanding Wage Distribution
Wages are distributed in proportion to work done, which in turn depends on efficiency and time worked.
Basic Principles
- Wage ∝ Work done
- Work done ∝ Efficiency × Time
- Wage ∝ Efficiency (if same time)
- Wage ∝ Time (if same efficiency)
- Total wage = Sum of individual wages
Distribution Formulas
For two workers A and B:
A's share : B's share = E_A × T_A : E_B × T_B
Where E = Efficiency, T = Time worked
Special Cases
- If workers work equal time: Wage ratio = Efficiency ratio
- If workers have equal efficiency: Wage ratio = Time ratio
- If one leaves early: Calculate work done till leaving
- Daily wages: Divide total by days worked
SSC Shortcut: Quick Wage Calculation
Step 1: Calculate work done by each (Efficiency × Time)
Step 2: Find ratio of work done
Step 3: Distribute total wage in that ratio
Example: A works 3 days at 4 units/day, B works 5 days at 3 units/day
A's work=12, B's work=15, Ratio=12:15=4:5, Total wage=₹1800
A's share=(4/9)×1800=₹800, B's share=(5/9)×1800=₹1000
Solved Example: Work & Wages Problem
LCM of 10,12,15 = 60 (Total work = 60 units)
A's efficiency = 60/10 = 6 units/day
B's efficiency = 60/12 = 5 units/day
C's efficiency = 60/15 = 4 units/day
Step 2: Determine work schedule
Let total time = T days
A works: 2 days
B works: (T-3) days (leaves 3 days before completion)
C works: T days (works till completion)
Step 3: Form equation for total work
Total work = Work by A + Work by B + Work by C
60 = 6×2 + 5×(T-3) + 4×T
60 = 12 + 5T - 15 + 4T
60 = 9T - 3
9T = 63
T = 7 days
Step 4: Calculate work done by each
A's work = 6×2 = 12 units
B's work = 5×(7-3) = 5×4 = 20 units
C's work = 4×7 = 28 units
Total = 12+20+28 = 60 units ✓
Step 5: Distribute wages
Work ratio A:B:C = 12:20:28 = 3:5:7
Sum of ratio = 3+5+7 = 15
C's share = (7/15) × 3000 = 7 × 200 = ₹1400
Verification:
A's share = (3/15)×3000 = ₹600
B's share = (5/15)×3000 = ₹1000
C's share = ₹1400
Total = 600+1000+1400 = ₹3000 ✓
Final Answer: C's share = ₹1400
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Frequently Asked Questions
Q1: How many Time & Work questions in SSC CGL?
Answer: Typically 5-8 questions in Tier I and 10-15 questions in Tier II. This is one of the highest weightage topics with questions ranging from basic to complex combination problems.
Q2: What is the LCM method and why is it better?
Answer: LCM method involves taking LCM of time taken by individuals as total work. It converts fractions to whole numbers, making calculations easier and reducing errors. It's the most recommended method for SSC.
Q3: How to handle negative work/outlet pipes?
Answer: Treat outlet pipes as negative efficiency. For combined rate: Sum of inlet rates - Sum of outlet rates. If result is positive, tank fills; if negative, tank empties; if zero, water level remains constant.
Q4: What if workers join/leave at different times?
Answer: Break problem into phases. Calculate work done in each phase with available workers. Add work from all phases to get total work done. Remaining work divided by new efficiency gives remaining time.
Q5: How to solve alternate day working problems?
Answer: Calculate work done in one cycle (usually 2 days). Divide total work by cycle work to get number of complete cycles. Multiply by cycle days, then add days for remaining work.
Q6: What's the relationship between efficiency and wages?
Answer: Wages are proportional to work done. Work done = Efficiency × Time. So if workers work same time, wage ratio = efficiency ratio. If efficiency same, wage ratio = time ratio.
Final Exam Strategy for Time & Work Problems
Time Allocation: Basic problems: 30-45 seconds, Complex problems: 60-90 seconds.
Priority Method: Always use LCM method for standard problems. For pipes, use positive/negative efficiency approach.
Accuracy Check: Verify that sum of individual works equals total work. For wages, check sum equals total wage.
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