Advanced Analytical Skills β II
Unit 1: Advanced Efficiency & Pipes
Master efficiency-based Time & Work, Wages, Chain Rule, Alternate Day problems, and Pipes & Cisterns β the most frequently asked topics in TCS, Infosys, and Wipro placement exams.
β±οΈ 6 hrs | π― TCS / Infosys / Wipro | π° Placement Essential | π 30 MCQs + 8 Short + 3 Long
Opening Hook β Why Efficiency & Pipes Rule Placement Exams
π’ The TCS NQT Question That Stumped 72% of Test-Takers
In the 2024 TCS National Qualifier Test (NQT), this question appeared: "A is 25% more efficient than B. If B alone can complete a work in 30 days, in how many days can A and B together complete the work?" Out of 3.8 lakh candidates who attempted the test, 72% got this wrong β not because the math is hard, but because they lacked a systematic approach.
Here's the secret: every efficiency, wages, and pipes problem follows the same core logic β convert everything to "work per day" (or "fraction of tank per hour"), then add or subtract rates. This chapter gives you that one framework that solves 95% of all questions in under 90 seconds.
Companies like TCS, Infosys, Wipro, Cognizant, and Accenture ask 3β5 questions from this chapter in every placement exam. Mastering this unit alone can add 15β20 marks to your score.
Learning Outcomes β Bloom's Taxonomy Mapped
| Bloom's Level | Learning Outcome |
|---|---|
| π΅ Remember | Recall the formulas for efficiency ratio, wages distribution, and pipe filling/emptying rates |
| π΅ Understand | Explain why "A is 20% more efficient than B" means their work-rate ratio is 6:5, and how wages are proportional to work done |
| π’ Apply | Solve standard efficiency, wages, chain rule, and pipe problems using the GivenβFindβFormulaβSolution framework |
| π’ Analyze | Break down complex scenarios β workers joining/leaving midway, alternate-day work, and pipes with leaks β into sub-problems |
| π Evaluate | Choose the fastest approach (LCM method vs. fraction method vs. shortcut) for a given problem type under exam time pressure |
| π Create | Formulate and solve original multi-step problems combining efficiency, pipes, and wages in a single scenario |
Efficiency-Based Problems
Core Concept: What is Efficiency?
Plain English: Efficiency is simply how much work someone does per unit time. If A is "20% more efficient than B," it means in the same amount of time, A does 20% more work than B. Think of it like two students writing an exam β if student A writes 12 answers per hour and student B writes 10 answers per hour, A is 20% more efficient.
π Efficiency Formula Framework
Efficiency β 1/Time β If A takes fewer days, A is more efficient.
If A is x% more efficient than B, then:
β’ Ratio of efficiencies β A : B = (100 + x) : 100
β’ Ratio of time taken β A : B = 100 : (100 + x) (inverse of efficiency)
LCM METHOD (RECOMMENDED)Step 1: Take LCM of the days given β This = Total Work (in units)
Step 2: Each person's per-day work = Total Work Γ· Their Days
Step 3: Combined rate = Sum of individual rates
Step 4: Time together = Total Work Γ· Combined rate
EXAMPLEIf A takes 15 days and B takes 20 days:
LCM(15, 20) = 60 units of work
A's rate = 60/15 = 4 units/day, B's rate = 60/20 = 3 units/day
Together = 7 units/day β Time = 60/7 = 8β΄ββ days
Quick Reference: Common Efficiency Ratios
| Statement | Efficiency Ratio (A:B) | Time Ratio (A:B) |
|---|---|---|
| A is 20% more efficient than B | 6 : 5 | 5 : 6 |
| A is 25% more efficient than B | 5 : 4 | 4 : 5 |
| A is 33β % more efficient than B | 4 : 3 | 3 : 4 |
| A is 50% more efficient than B | 3 : 2 | 2 : 3 |
| A is twice as efficient as B | 2 : 1 | 1 : 2 |
| A is thrice as efficient as B | 3 : 1 | 1 : 3 |
Wages-Based Problems
Core Concept: Wages β Work Done
Plain English: If three workers paint a house together and one of them did half the painting, that person gets half the payment. Wages are always distributed in the ratio of work done, not in the ratio of days spent or people involved.
π° Wages Distribution Formula
Wage of A : Wage of B : Wage of C = Work done by A : Work done by B : Work done by C
And since all work for the same number of days (in standard problems):
Wage Ratio = Efficiency Ratio = (1/Time_A) : (1/Time_B) : (1/Time_C)
FORMULAIf A, B, C can individually complete work in a, b, c days respectively, and total wage is βΉW:
A's share = W Γ (1/a) / (1/a + 1/b + 1/c)
SHORTCUTIf A can do in 6 days, B in 8 days β Efficiency ratio = 1/6 : 1/8 = 4:3
If total wage = βΉ2,100 β A gets (4/7)Γ2100 = βΉ1,200, B gets (3/7)Γ2100 = βΉ900
Chain Rule (MenβDaysβHours)
Core Concept: Mβ Γ Dβ Γ Hβ = Mβ Γ Dβ Γ Hβ
Plain English: The Chain Rule says that the total "man-hours" of work stays constant. If you have more workers, you need fewer days. If workers work longer hours, the job finishes sooner. It's like pouring water into a bucket β more taps = faster filling.
π Chain Rule Master Formula
Mβ Γ Dβ Γ Hβ = Mβ Γ Dβ Γ Hβ
Where M = Men (workers), D = Days, H = Hours per day
EXTENDED FORM (with work amount & efficiency)(Mβ Γ Dβ Γ Hβ) / Wβ = (Mβ Γ Dβ Γ Hβ) / Wβ
Where Wβ, Wβ = amount of work done (if different)
WITH EFFICIENCY FACTOR(Mβ Γ Dβ Γ Hβ Γ Eβ) / Wβ = (Mβ Γ Dβ Γ Hβ Γ Eβ) / Wβ
Where Eβ, Eβ = individual efficiency of each worker type
Chain Rule Variation Table
| What Changes | Relationship | Effect |
|---|---|---|
| More Men | Inversely proportional to Days | β Men β β Days |
| More Hours/Day | Inversely proportional to Days | β Hours β β Days |
| More Work | Directly proportional to Days | β Work β β Days |
| Higher Efficiency | Inversely proportional to Days | β Efficiency β β Days |
Alternate Day Work Problems
Core Concept: Work in 2-Day Cycles
Plain English: "A and B work on alternate days" means A works on Day 1, B works on Day 2, A on Day 3, B on Day 4, and so on (or vice versa). The trick is to calculate how much work gets done in a 2-day cycle, then figure out how many full cycles are needed.
π Alternate Day Work Framework
Step 1: Find individual per-day rates (use LCM method)
Step 2: Calculate work done in one 2-day cycle = Rate_A + Rate_B
Step 3: Find number of complete cycles: Total Work Γ· Work per cycle
Step 4: Check remaining work β who works on the final day?
CRITICAL POINTβ οΈ The answer depends on who starts first! "A and B work on alternate days starting with A" β "starting with B". Always check the question carefully.
EXAMPLEA completes in 12 days, B in 18 days. Working alternately starting with A.
LCM(12,18) = 36. A's rate = 3 units/day, B's rate = 2 units/day.
2-day cycle: 3 + 2 = 5 units. Cycles needed: 36/5 = 7 full cycles + 1 unit remaining.
7 cycles = 14 days. Day 15: A works β does 3 units (but only 1 needed).
A finishes remaining 1 unit in 1/3 day. Total = 14β days.
Advanced Time & Work β Join/Leave Problems
Core Concept: Workers Joining or Leaving Midway
Plain English: Real projects don't have everyone working from start to finish. Sometimes a contractor leaves after 3 days, or extra workers join after 5 days. These problems test your ability to track work done in phases.
πΆ Join/Leave Problem Framework
Phase 1: Calculate work done before the change (person leaving/joining)
Phase 2: Calculate remaining work
Phase 3: Calculate time for remaining work with the new team
Total Time = Phase 1 time + Phase 2 time
"A LEAVES AFTER x DAYS" PATTERN1. Find rates: A = 1/a per day, B = 1/b per day
2. Work done in x days (both working) = x Γ (1/a + 1/b)
3. Remaining work = 1 β [work done in x days]
4. Time for B alone to finish remaining = Remaining Γ· (1/b)
"B JOINS AFTER y DAYS" PATTERN1. A works alone for y days β Work done = y/a
2. Remaining = 1 β y/a
3. Time for A+B together = Remaining Γ· (1/a + 1/b)
Pipes & Cisterns β Inlet, Outlet, and Leaks
Core Concept: Pipes = Time & Work (But with a Twist!)
Plain English: Pipes & Cisterns is exactly like Time & Work, but with water instead of work. An inlet pipe fills a tank (positive work), and an outlet pipe / leak empties it (negative work). The net rate = inlet rate β outlet rate.
π° Pipes & Cisterns Master Framework
Pipe A fills tank in a hours β Rate = 1/a tank/hour (positive)
Pipe B empties tank in b hours β Rate = 1/b tank/hour (negative)
Net rate when both open = 1/a β 1/b
Time to fill = 1 / (1/a β 1/b) = ab / (b β a) [only if b > a, i.e. filling wins]
Net filling rate = (Sum of all inlet rates) β (Sum of all outlet rates)
If net rate is positive β tank fills. If negative β tank empties.
PIPE WITH LEAKWithout leak: pipe fills in a hours.
With leak: pipe fills in b hours (b > a, takes longer).
Leak's rate = 1/a β 1/b β Leak empties tank in ab/(bβa) hours.
If pipe runs for t hours at rate 1/a, part filled = t/a
Key Differences: Time & Work vs. Pipes & Cisterns
| Aspect | Time & Work | Pipes & Cisterns |
|---|---|---|
| Positive contributor | Worker doing work | Inlet pipe filling tank |
| Negative contributor | β | Outlet pipe / leak emptying tank |
| Combined rate | Always additive | Inlets add, outlets subtract |
| Key formula | 1/a + 1/b | 1/a β 1/b (if one empties) |
| LCM method | β Works perfectly | β Works perfectly (outlet = negative units) |
Worked Examples β 15 Problems with Complete Solutions
π Example 1 β Basic Efficiency
A is 25% more efficient than B. B alone can complete a work in 20 days.
FINDIn how many days can A alone complete the work?
FORMULAEfficiency ratio A:B = 125:100 = 5:4. Time ratio A:B = 4:5.
SOLUTIONB takes 20 days. Time ratio A:B = 4:5.
So A's time = (4/5) Γ 20 = 16 days
SHORTCUT25% more efficient β Time ratio = 4:5 β A's time = B's time Γ 4/5 = 20 Γ 4/5 = 16 days. Done in 10 seconds!
π Example 2 β Efficiency + Combined Work
A is 30% more efficient than B. Together they finish a work in 13 days.
FINDIn how many days can B alone complete the work?
FORMULALet B's efficiency = 10 units/day. Then A's = 13 units/day. Combined = 23 units/day.
SOLUTIONTotal work = 23 Γ 13 = 299 units.
B alone = 299/10 = 29.9 β 30 days
(In exam MCQ options, pick 30 days)
SHORTCUTEfficiency ratio = 13:10. Combined rate = 23 parts. Together in 13 days β Total = 299 parts. B alone = 299/10 β 30 days.
π Example 3 β Wages Distribution
A can do a piece of work in 6 days, B in 8 days. They work together and get βΉ2,800.
FINDA's share of the wages.
FORMULAWage ratio = Efficiency ratio = 1/6 : 1/8 = 4 : 3
SOLUTIONTotal parts = 4 + 3 = 7
A's share = (4/7) Γ 2800 = βΉ1,600
B's share = (3/7) Γ 2800 = βΉ1,200
SHORTCUTLCM(6,8) = 24. A's rate = 4, B's rate = 3. Ratio = 4:3. A gets 4/7 of total.
π Example 4 β Wages with Three Workers
A, B, C can complete a work in 10, 15, and 30 days respectively. Total payment = βΉ3,000.
FINDEach person's share.
FORMULAWage ratio = 1/10 : 1/15 : 1/30 = 3 : 2 : 1
SOLUTIONLCM(10,15,30) = 30. Rates: A = 3, B = 2, C = 1. Ratio = 3:2:1.
A gets (3/6) Γ 3000 = βΉ1,500
B gets (2/6) Γ 3000 = βΉ1,000
C gets (1/6) Γ 3000 = βΉ500
SHORTCUTLCM method gives ratio directly: 3:2:1. Total = 6 parts. Divide βΉ3,000 accordingly.
π Example 5 β Chain Rule (Basic)
12 men can complete a work in 18 days working 8 hours/day.
FINDHow many men are needed to complete the same work in 12 days working 9 hours/day?
FORMULAMβ Γ Dβ Γ Hβ = Mβ Γ Dβ Γ Hβ
SOLUTION12 Γ 18 Γ 8 = Mβ Γ 12 Γ 9
1728 = 108 Γ Mβ
Mβ = 1728/108 = 16 men
SHORTCUTPlug and solve. The formula is the shortcut itself β no tricks needed.
π Example 6 β Chain Rule (Different Work Amount)
15 men can build a wall 30 m long in 20 days working 6 hrs/day.
FINDHow many men are needed to build a wall 45 m long in 15 days working 8 hrs/day?
FORMULA(Mβ Γ Dβ Γ Hβ) / Wβ = (Mβ Γ Dβ Γ Hβ) / Wβ
SOLUTION(15 Γ 20 Γ 6) / 30 = (Mβ Γ 15 Γ 8) / 45
1800/30 = 120Mβ/45
60 = 120Mβ/45
Mβ = (60 Γ 45)/120 = 2700/120 = 22.5 β 23 men
SHORTCUTSince we can't have half a man, round up to 23. In MCQs, look for the option closest to 22.5.
π Example 7 β Alternate Day Work
A can do a work in 10 days, B can do it in 15 days. They work on alternate days starting with A.
FINDIn how many days will the work be completed?
FORMULALCM(10,15) = 30 units. A's rate = 3/day, B's rate = 2/day. Per cycle (2 days) = 5 units.
SOLUTIONComplete cycles: 30/5 = 6 cycles = 12 days.
Work is completed exactly in 12 days.
SHORTCUTCheck if Total Work is exactly divisible by 2-day cycle work. If yes, answer = 2 Γ (Total/Cycle). Here 30/5 = 6 β 12 days exactly.
π Example 8 β Alternate Day (Fractional Finish)
A can do a work in 12 days, B in 18 days. They work alternately, starting with A.
FINDWhen is the work finished?
FORMULALCM(12,18) = 36. A = 3 units/day, B = 2 units/day. Cycle = 5 units in 2 days.
SOLUTIONFull cycles: 36/5 = 7 cycles with 1 unit remaining.
7 cycles = 14 days. Day 15 is A's turn. A does 3 units/day, only 1 unit needed.
A finishes in 1/3 day. Total = 14β days
SHORTCUTAfter complete cycles, check whose turn it is. Remaining work Γ· that person's rate = fractional day.
π Example 9 β A Leaves After 3 Days
A can do a work in 12 days, B in 15 days. They start together but A leaves after 3 days.
FINDTotal days to complete the work.
FORMULALCM(12,15) = 60 units. A = 5/day, B = 4/day.
SOLUTIONFirst 3 days (A+B): (5+4) Γ 3 = 27 units done.
Remaining: 60 β 27 = 33 units.
B alone: 33/4 = 8.25 days.
Total = 3 + 8.25 = 11.25 days = 11ΒΌ days
SHORTCUTPhase 1 (together) + Phase 2 (alone). LCM method avoids all fraction work in the intermediate steps.
π Example 10 β B Joins After 5 Days
A can do a work in 20 days. B can do it in 30 days. A starts alone and B joins after 5 days.
FINDTotal days to complete the work.
FORMULALCM(20,30) = 60. A = 3/day, B = 2/day.
SOLUTIONPhase 1: A alone for 5 days = 3 Γ 5 = 15 units.
Remaining: 60 β 15 = 45 units.
Phase 2: A + B together = 3 + 2 = 5 units/day.
Time for 45 units = 45/5 = 9 days.
Total = 5 + 9 = 14 days
SHORTCUTA alone for 5 days covers 15/60 = ΒΌ of work. Remaining ΒΎ at combined rate (1/20+1/30 = 1/12) = ΒΎ Γ 12 = 9 days. Total = 14.
π Example 11 β Basic Pipe Filling
Pipe A can fill a tank in 12 hours. Pipe B can fill it in 18 hours.
FINDTime to fill the tank if both pipes are opened simultaneously.
FORMULACombined rate = 1/12 + 1/18 = (3+2)/36 = 5/36 tank/hr.
SOLUTIONTime = 36/5 = 7.2 hours = 7 hrs 12 min
SHORTCUTLCM(12,18) = 36. A = 3 units/hr, B = 2 units/hr. Together = 5 units/hr. Time = 36/5 = 7.2 hrs.
π Example 12 β Inlet + Outlet Pipe
Pipe A fills a tank in 20 min. Pipe B empties it in 30 min. Both are opened together.
FINDTime to fill the tank.
FORMULANet rate = 1/20 β 1/30 = (3β2)/60 = 1/60 tank/min.
SOLUTIONTime = 60 minutes
SHORTCUTLCM(20,30) = 60. Inlet = +3, Outlet = β2. Net = +1 unit/min. Time = 60/1 = 60 min.
π Example 13 β Pipe with Leak
A pipe can fill a tank in 10 hours. Due to a leak at the bottom, it takes 12 hours to fill.
FINDHow long will the leak take to empty a full tank?
FORMULAFilling rate β Leak rate = Effective rate β 1/10 β 1/L = 1/12
SOLUTION1/L = 1/10 β 1/12 = (6β5)/60 = 1/60
L = 60 hours
SHORTCUTPipe alone = 10 hrs, with leak = 12 hrs. Leak rate = 1/10 β 1/12 = 1/60. Leak empties in 60 hrs.
π Example 14 β Three Pipes (2 Inlet + 1 Outlet)
Pipe A fills in 20 min, Pipe B fills in 30 min, Pipe C empties in 15 min. All opened together.
FINDTime to fill the tank.
FORMULANet rate = 1/20 + 1/30 β 1/15
SOLUTIONLCM(20,30,15) = 60. A = +3, B = +2, C = β4. Net = +1 unit/min.
Time = 60/1 = 60 minutes
SHORTCUTLCM method instantly: 3 + 2 β 4 = 1. Time = 60/1 = 60 min. Done in 15 seconds.
π Example 15 β Part of Tank + Time-Based Opening
Pipe A fills a tank in 24 hours. Pipe B fills it in 32 hours. A is opened first. After 8 hours, B is also opened.
FINDTotal time to fill the tank.
FORMULALCM(24,32) = 96. A = 4 units/hr, B = 3 units/hr.
SOLUTIONPhase 1: A alone for 8 hrs = 4 Γ 8 = 32 units.
Remaining: 96 β 32 = 64 units.
Phase 2: A + B = 4 + 3 = 7 units/hr.
Time = 64/7 = 9.14 hours β 9 hrs 8Β½ min.
Total = 8 + 64/7 = 8 + 9ΒΉββ = 17ΒΉββ hours
SHORTCUTPhase approach: A alone fills 32/96 = 1/3 of tank in 8 hrs. Remaining 2/3 at combined rate 7/96 per hour = (2/3)Γ(96/7) = 64/7 hours.
MCQ Assessment Bank β 30 Questions (Bloom's Mapped)
Remember / Recall (Q1βQ5)
If A is 20% more efficient than B, what is the ratio of their efficiencies (A:B)?
- 5:4
- 6:5
- 4:5
- 5:6
In the formula Mβ Γ Dβ Γ Hβ = Mβ Γ Dβ Γ Hβ, what does 'H' represent?
- Height of the wall
- Hours worked per day
- Horsepower of the machine
- Head count of supervisors
Wages in a Time & Work problem are distributed in the ratio of:
- Time taken by each worker
- Number of days each worker works
- Work done by each worker (efficiency)
- Age of each worker
An outlet pipe in a Pipes & Cisterns problem contributes:
- Positive work (fills the tank)
- Negative work (empties the tank)
- Zero work
- Depends on the inlet pipe
If A can complete a work in 10 days, what fraction of work does A complete in 1 day?
- 1/5
- 10
- 1/10
- 1/100
Understand / Explain (Q6βQ10)
Why does "A is 25% more efficient than B" imply A takes less time, not more?
- Because efficiency and time are directly proportional
- Because efficiency and time are inversely proportional β higher efficiency means less time
- Because A works fewer hours per day
- Because A gets paid more
In a pipe problem, if Pipe A fills in 10 hrs and Pipe B empties in 15 hrs, why does the tank eventually fill up?
- Because B is not working
- Because A's filling rate (1/10) is greater than B's emptying rate (1/15)
- Because both pipes fill the tank
- Because the tank has a lid
In the chain rule, if the amount of work doubles, what happens to the number of days (all else constant)?
- Days halve
- Days double
- Days remain same
- Days become zero
Why is the LCM method preferred over fractions in competitive exams?
- It gives different answers
- It eliminates fractions, making calculations faster and less error-prone
- It only works for pipe problems
- Exams require LCM notation
In an alternate-day problem, why does the answer depend on who starts first?
- It doesn't β the answer is always the same
- Because the person who starts first may complete the remaining work on a different day, changing the total fractional day
- Because the person starting first is always faster
- Because odd-numbered days are longer
Apply / Solve (Q11βQ18)
A is 50% more efficient than B. If B can do a work in 24 days, how many days will A take?
- 12
- 16
- 18
- 20
A can do a work in 10 days, B in 12 days. They work together for 3 days. What fraction of work remains?
- 11/20
- 9/20
- 1/2
- 7/20
8 men can finish a work in 15 days. How many men are needed to finish it in 10 days?
- 10
- 12
- 15
- 16
A pipe fills a tank in 15 min. A leak empties it in 20 min. How long to fill with both active?
- 45 min
- 50 min
- 55 min
- 60 min
A can do work in 8 days, B in 12 days. Total wage βΉ4,000. What is B's share?
- βΉ1,200
- βΉ1,600
- βΉ2,000
- βΉ2,400
A is 33β % more efficient than B. If A takes 18 days, how many days does B take?
- 20
- 22
- 24
- 27
A and B work on alternate days starting with A. A does 1/10 per day, B does 1/15 per day. Total days?
- 11
- 12
- 12β
- 13
15 men working 8 hrs/day build a wall in 10 days. 20 men working 6 hrs/day will take how many days?
- 8
- 9
- 10
- 12
Analyze / Break Down (Q19βQ23)
A does work in 12 days, B in 15 days. They start together but A leaves after 4 days. How many total days to complete?
- 9
- 10
- 10ΒΎ
- 11ΒΌ
Two pipes A (fills in 24 min) and B (empties in 36 min) are opened. After the tank is full, only B is opened. How long to empty?
- 24 min
- 30 min
- 36 min
- 72 min
A does work in 10 days, B in 15 days, C in 30 days. They work together for 2 days, then C leaves. How many more days for A and B?
- 3
- 4
- 5
- 6
A pipe fills in 10 hrs. Due to a leak, it fills in 14 hrs. The leak empties a full tank in:
- 25 hrs
- 30 hrs
- 35 hrs
- 40 hrs
A is twice as efficient as B. Together they finish in 14 days. How many days for A alone?
- 18
- 21
- 28
- 42
Evaluate / Compare (Q24βQ27)
Which method is faster for solving "A takes 12 days, B takes 18 days, together?"
- Fraction method: 1/12 + 1/18
- LCM method: LCM = 36, rates 3+2 = 5, time = 36/5
- Both take the same time
- Trial and error
A worker is 40% more efficient but charges 60% more. Is this worker cost-effective compared to a standard worker?
- Yes, because they finish faster
- No, because cost increases more than efficiency
- They're exactly equivalent
- Cannot determine
For filling a tank faster, which is more effective: adding another inlet of the same capacity or closing a leak that empties at half the inlet rate?
- Adding another inlet β doubles the filling rate
- Closing the leak β removes the negative rate
- Both have the same effect
- Depends on tank size
In a chain rule problem with 3 variables changing, what is the most common student error?
- Forgetting to use LCM
- Mixing up direct and inverse proportion
- Using wrong units
- Not reading the question
Create / Design (Q28βQ30)
A, B, C can do a work in 12, 18, 24 days. A works for 2 days alone, then B joins for 3 days, then C also joins. Total days?
- 8βΈβββ
- 9
- 9β΅βββ
- 10
Pipe A fills in 12 hrs, B fills in 16 hrs, C empties in 24 hrs. A is opened for 4 hrs alone, then B and C are also opened. Total time to fill?
- 10
- 12
- 10β
- 11
10 men working 7 hrs/day can complete Β½ of a work in 12 days. How many men working 5 hrs/day can complete the remaining Β½ in 14 days?
- 10
- 12
- 14
- 16
Short Answer Questions (8 Questions)
SA-1: Efficiency Ratio Derivation
Q: A is 40% more efficient than B. If A can do a work in 15 days, find the time B takes. Show your working.
Ans: Efficiency ratio A:B = 140:100 = 7:5. Time ratio = 5:7 (inverse). A takes 15 days β 15 corresponds to 5 parts β 1 part = 3. B takes 7 Γ 3 = 21 days.
SA-2: Wages for Partial Work
Q: A can do a job in 5 days, B in 10 days. A works for 2 days then leaves. B finishes the rest. Total wage is βΉ5,000. Find each person's share.
Ans: A's 2-day work = 2/5. B finishes remaining 3/5, taking 3/5 Γ 10 = 6 days. Work ratio = 2/5 : 3/5 = 2:3. A gets (2/5) Γ 5000 = βΉ2,000. B gets (3/5) Γ 5000 = βΉ3,000.
SA-3: Chain Rule with Efficiency
Q: 5 men can paint a fence in 8 days working 6 hrs/day. How many hours/day must 4 men work to finish it in 10 days?
Ans: MβDβHβ = MβDβHβ β 5 Γ 8 Γ 6 = 4 Γ 10 Γ Hβ β 240 = 40Hβ β Hβ = 6 hours/day.
SA-4: Alternate Day Completion
Q: A takes 6 days, B takes 9 days. Working alternately starting with A, when is the work completed?
Ans: LCM(6,9) = 18. A = 3 units/day, B = 2 units/day. 2-day cycle = 5 units. 18/5 = 3 full cycles (15 units in 6 days). Remaining = 3 units. Day 7: A's turn β A does 3 units. Done! Total = 7 days.
SA-5: Pipe with Leak Detection
Q: A pipe fills a tank in 8 hours. With a leak, it fills in 10 hours. Find the time the leak takes to empty a full tank.
Ans: Leak rate = 1/8 β 1/10 = (5β4)/40 = 1/40. The leak empties the tank in 40 hours.
SA-6: Part of Tank Filled
Q: Pipes A and B can fill a tank in 16 and 24 hours respectively. If both are opened together, what part of the tank is filled in 6 hours?
Ans: Combined rate = 1/16 + 1/24 = (3+2)/48 = 5/48. In 6 hours = 6 Γ 5/48 = 30/48 = 5/8 of the tank.
SA-7: Worker Leaves Midway
Q: A and B together can do a work in 12 days. A alone can do it in 20 days. After working together for 4 days, A leaves. How many more days does B take?
Ans: B alone: 1/12 β 1/20 = (5β3)/60 = 1/30 β B takes 30 days alone. Together for 4 days: 4/12 = 1/3 done. Remaining = 2/3. B alone: (2/3) Γ 30 = 20 more days.
SA-8: Efficiency Comparison
Q: A takes 20 days to do a work. B takes 25 days. Who is more efficient and by what percentage?
Ans: A's rate = 1/20, B's rate = 1/25. A is more efficient. Percentage = [(1/20 β 1/25) / (1/25)] Γ 100 = [(5β4)/100 / (4/100)] Γ 100 = (1/4) Γ 100 = 25% more efficient.
Long Answer Questions (3 Questions)
LA-1: Multi-Phase Complex Problem
Q: A, B, and C can individually complete a work in 10, 15, and 20 days respectively. They start working together. After 2 days, C leaves. After 2 more days, B also leaves. In how many total days is the work completed? Also find the share of each if total wage is βΉ9,000.
Solution:
Step 1: Set up LCM
LCM(10, 15, 20) = 60 units (Total Work)
A = 60/10 = 6 units/day, B = 60/15 = 4 units/day, C = 60/20 = 3 units/day
Step 2: Phase-wise calculation
Phase 1 (Days 1β2): A + B + C work together
Rate = 6 + 4 + 3 = 13 units/day. Work in 2 days = 26 units.
Phase 2 (Days 3β4): A + B work (C has left)
Rate = 6 + 4 = 10 units/day. Work in 2 days = 20 units.
Phase 3 (Day 5 onwards): A works alone
Remaining = 60 β 26 β 20 = 14 units. A's rate = 6/day. Time = 14/6 = 2β days.
Total time = 2 + 2 + 2β = 6β days
Step 3: Wage distribution
A's total work = (6Γ2) + (6Γ2) + (6Γ2β ) = 12 + 12 + 14 = 38 units
B's total work = (4Γ2) + (4Γ2) = 8 + 8 = 16 units
C's total work = (3Γ2) = 6 units
Total = 38 + 16 + 6 = 60 β
Wage ratio = 38 : 16 : 6 = 19 : 8 : 3
A = (19/30) Γ 9000 = βΉ5,700
B = (8/30) Γ 9000 = βΉ2,400
C = (3/30) Γ 9000 = βΉ900
LA-2: Pipes with Multiple Phases and a Leak
Q: A tank has two inlet pipes A and B, and one outlet pipe C. A fills the tank in 12 hours, B fills in 18 hours, and C empties in 36 hours. Pipe A is opened first. After 3 hours, B is also opened. After 3 more hours, C is accidentally opened. From that point, how much longer does it take to fill the tank? What is the total time?
Solution:
Step 1: LCM and rates
LCM(12, 18, 36) = 36 units. A = +3/hr, B = +2/hr, C = β1/hr.
Step 2: Phase-wise
Phase 1 (Hours 1β3): Only A
Work = 3 Γ 3 = 9 units.
Phase 2 (Hours 4β6): A + B
Work = (3+2) Γ 3 = 15 units.
Phase 3 (Hour 7+): A + B + C
Net rate = 3 + 2 β 1 = 4 units/hr.
Remaining = 36 β 9 β 15 = 12 units.
Time = 12/4 = 3 hours.
Total time = 3 + 3 + 3 = 9 hours
From when C opened, it takes 3 more hours to fill.
LA-3: Comprehensive Efficiency + Chain Rule + Wages
Q: A is 20% more efficient than B, and B is 25% more efficient than C. If C alone takes 30 days to complete a work:
(a) Find the days each person takes individually.
(b) If 4 workers of type A and 5 workers of type B work 6 hrs/day, how many days to finish?
(c) If A, B, C all work together and the total contract payment is βΉ18,600, find each person's share.
Solution:
(a) Individual days
C takes 30 days. B is 25% more efficient than C.
Eff ratio B:C = 125:100 = 5:4. Time ratio = 4:5. B's time = (4/5) Γ 30 = 24 days.
A is 20% more efficient than B.
Eff ratio A:B = 120:100 = 6:5. Time ratio = 5:6. A's time = (5/6) Γ 24 = 20 days.
(b) Chain Rule with multiple worker types
A's daily rate = 1/20, B's daily rate = 1/24.
4A's hourly rate = 4 Γ (1/20) Γ (1/H_std). Let's work with a standard 1-day unit.
Per day (6 hrs): 4 workers of A do 4 Γ (1/20) = 1/5 work per standard day.
But they work 6 hrs/day. If standard is full day (assumed), then directly: 4A's per day = 4/20 = 1/5 of work.
5 workers of B per day = 5 Γ 1/24 = 5/24 of work.
Combined daily = 1/5 + 5/24 = 24/120 + 25/120 = 49/120.
Days = 120/49 β 2.45 days β 2 days 11 hours (at 6 hrs/day = about 2.45 working days).
(c) Wage distribution
A = 20 days, B = 24 days, C = 30 days.
LCM(20,24,30) = 120. A = 6, B = 5, C = 4. Ratio = 6:5:4. Total = 15 parts.
A gets (6/15) Γ 18600 = βΉ7,440
B gets (5/15) Γ 18600 = βΉ6,200
C gets (4/15) Γ 18600 = βΉ4,960
Verification: 7440 + 6200 + 4960 = βΉ18,600 β
Formula Sheet β Quick Revision Card
π Complete Formula Reference
Print this page or screenshot this section for last-minute revision before your placement exam.
A. Basic Time & Work
Work Rate Formula
If A can do a work in n days β A's 1 day work = 1/n
A and B together β 1/a + 1/b per day β Time together = ab/(a+b)
LCM Method
Total Work = LCM(days of all workers)
Each worker's rate = Total Work Γ· Their Days
Combined Time = Total Work Γ· Sum of Rates
B. Efficiency
Efficiency Ratio
If A is x% more efficient than B:
Efficiency ratio A:B = (100+x) : 100
Time ratio A:B = 100 : (100+x) (inverse)
If B takes d days β A takes d Γ 100/(100+x) days
C. Wages
Wage Distribution
Wages β Work Done β Efficiency β 1/Time
Wage ratio of A, B, C = 1/a : 1/b : 1/c
A's share = Total Wage Γ (A's efficiency / Sum of all efficiencies)
D. Chain Rule
Men-Days-Hours Formula
Mβ Γ Dβ Γ Hβ / Wβ = Mβ Γ Dβ Γ Hβ / Wβ
M = Men, D = Days, H = Hours/day, W = Amount of work
β Men or β Hours β β Days (inverse)
β Work β β Days (direct)
E. Alternate Day Work
Two-Day Cycle Method
Work per cycle (2 days) = Rate_A + Rate_B
Full cycles = Total Work Γ· Work per cycle (take integer part)
Remaining work β check whose turn β remaining Γ· their rate = fractional day
β οΈ Answer depends on who starts first!
F. Join/Leave Problems
Phase Method
"A leaves after x days": Phase 1 = x days (both), Phase 2 = remaining Γ· B's rate
"B joins after y days": Phase 1 = y days (A alone), Phase 2 = remaining Γ· (A+B rate)
Total = Phase 1 time + Phase 2 time (+ Phase 3 if needed)
G. Pipes & Cisterns
Pipe Rates
Inlet fills in a hrs β Rate = +1/a
Outlet empties in b hrs β Rate = β1/b
Net rate = Ξ£(inlet rates) β Ξ£(outlet rates)
Time to fill = Total capacity Γ· Net rate
Pipe with Leak
Without leak: fills in a hrs. With leak: fills in b hrs (b > a)
Leak rate = 1/a β 1/b
Leak empties full tank in = ab/(bβa) hrs
Part Filled / Emptied
Part of tank filled in t hours = t Γ (net rate)
If part filled = 1, tank is full. If > 1, it overflows.
H. Quick Percentage β Fraction Table
| Percentage | Fraction | Use In |
|---|---|---|
| 10% | 1/10 | Efficiency increase/decrease |
| 12.5% | 1/8 | Efficiency problems |
| 16β % | 1/6 | Efficiency problems |
| 20% | 1/5 | Very common in TCS |
| 25% | 1/4 | Very common in Infosys |
| 33β % | 1/3 | Efficiency problems |
| 50% | 1/2 | Twice as efficient |
| 66β % | 2/3 | Less common |
| 75% | 3/4 | Less common |
β Unit 1 Complete β Advanced Efficiency & Pipes Mastered!
Ready for Unit 2. Keep practising β every placement exam will reward you for mastering this chapter.