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.

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Worker A

Efficiency: 3 units/day

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Worker B

Efficiency: 2 units/day

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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

Work = Efficiency × Time
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

Q: A can complete a work in 20 days and B can complete it in 30 days. In how many days will they complete the work if they work together?
Method 1: Using basic formula
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.

Situation Relationship Formula Example A is n times efficient as B Efficiency ratio = n:1 E_A : E_B = n : 1 If A=2×B, ratio = 2:1 Time ratio given Efficiency ∝ 1/Time E_A : E_B = T_B : T_A A=6 days, B=3 days → E_A:E_B = 3:6 = 1:2 Work ratio in same time Efficiency ratio = Work ratio E_A : E_B = W_A : W_B In 1 hr, A does 3 units, B does 2 → 3:2 Combined efficiency Sum of individual efficiencies E_total = E_A + E_B + ... E_A=3, E_B=2 → E_total=5 units/day

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

Q: A is twice as efficient as B. If B can complete a work in 30 days, in how many days can A and B together complete the work?
Step 1: Find individual efficiencies
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

Q: A and B can complete a work in 12 days, B and C in 15 days, C and A in 20 days. In how many days will A, B, and C together complete the work?
Method 1: Using equation method
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.

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Inlet

Fills in 6 hrs

Rate: +1/6 per hr

Tank
Capacity: 1 unit
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Outlet

Empties in 12 hrs

Rate: -1/12 per hr

Pipe Type Work Rate Time Formula Example Inlet Pipe Positive (+) If fills in 't' hours: Rate = +1/t Fills in 4 hrs → Rate = +1/4 per hr Outlet Pipe Negative (-) If empties in 't' hours: Rate = -1/t Empties in 6 hrs → Rate = -1/6 per hr Net Rate Sum of rates Net = Σ(Inlets) - Σ(Outlets) Inlet:1/4, Outlet:1/6 → Net=1/4-1/6=1/12 Time to Fill/Empty 1/Net Rate Time = 1/Net Rate Net=1/12 → Time=12 hrs to fill Leakage Problems Outlet with time Effective rate = Fill rate - Leak rate Pipe fills in 3 hrs, leak empties in 6 hrs → Effective=1/3-1/6=1/6

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

Q: Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. There is also an outlet pipe C. If all three are opened together, the tank fills in 15 minutes. How much time will pipe C alone take to empty the full tank?
Step 1: Find filling rates
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

Individual's share = (Individual's work/Total work) × Total wage

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

Q: A, B, and C can complete a work in 10, 12, and 15 days respectively. They start together but A leaves after 2 days and B leaves 3 days before completion. If total wage is ₹3000, find C's share.
Step 1: Calculate efficiencies using LCM
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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Includes all types: basic work, efficiency, combined work, pipes-cisterns, work-wages

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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