Time and Work for SSC CGL is one of those chapters that looks complicated when you first see it, but becomes highly mechanical once you understand one simple idea: more work per day means less time, and less work per day means more time.
The chapter is not really about memorising dozens of formulas. Most SSC CGL questions can be reduced to three things:
- How much work is done in one day?
- How much work is completed in the given time?
- How much work is still left?
Once these three questions become automatic, problems involving efficiency, men and women, wages, pipes and even alternate working become much easier.
Golden idea: Time and Work is simply a relationship between Work, Time and Efficiency.
1. The Basic Concept of Time and Work
Suppose A can complete a piece of work in 10 days.
That means A completes the entire work in 10 days. Therefore, the amount of work A completes in one day is:
\[ \text{1 day's work}=\frac{1}{10} \]
If B completes the same work in 20 days:
\[ \text{B's 1 day's work}=\frac{1}{20} \]
So, when A and B work together, their daily work is added:
\[ \frac{1}{10}+\frac{1}{20} =\frac{3}{20} \]
Therefore, the time required is:
\[ \frac{20}{3}=6\frac{2}{3}\text{ days} \]
This fraction method is mathematically straightforward, but for SSC CGL, the LCM method is usually faster.
2. The LCM Method — Your Main SSC Weapon
Suppose:
Take the LCM of 12 and 18:
\[ LCM(12,18)=36 \]
Instead of saying that the total work is 1, imagine that the total work is 36 units.
Together:
\[ 3+2=5\text{ units/day} \]
Therefore:
\[ \text{Time}=\frac{36}{5} \]
\[ \boxed{\frac{36}{5}=7.2\text{ days}} \]
SSC shortcut: Take the LCM of the given days as total work. Divide it by each person's time to get their efficiency.
3. Work, Time and Efficiency
The basic relationship is:
\[ \boxed{\text{Work}=\text{Efficiency}\times\text{Time}} \]
Therefore:
\[ \boxed{\text{Time}=\frac{\text{Work}}{\text{Efficiency}}} \]
and
\[ \boxed{\text{Efficiency}=\frac{\text{Work}}{\text{Time}}} \]
4. Efficiency and Time Are Inversely Proportional
This is one of the most important ideas in the chapter.
If A is twice as efficient as B, then:
\[ A:B=2:1 \]
But efficiency and time move in opposite directions. Therefore their time ratio is:
\[ A:B=1:2 \]
For example, if B takes 30 days, A takes:
\[ 30\div2=15\text{ days} \]
Remember: More efficiency → less time. Less efficiency → more time.
5. “A Is Twice as Efficient as B”
Suppose A is twice as efficient as B.
Then:
\[ A:B=2:1 \]
If they work together, their combined efficiency is:
\[ 2+1=3 \]
If they together complete a work in 12 days, then B alone will take:
\[ 12\times\frac{3}{1}=36\text{ days} \]
A alone will take:
\[ 12\times\frac{3}{2}=18\text{ days} \]
Here, \(T\) represents the time taken by both together.
6. A More Powerful Shortcut
If:
\[ A:B=m:n \]
is the efficiency ratio, and A and B together complete the work in \(T\) days, then:
\[ \boxed{A\text{ alone}=T\times\frac{m+n}{m}} \]
\[ \boxed{B\text{ alone}=T\times\frac{m+n}{n}} \]
Example:
A is 3 times as efficient as B. Together they finish a work in 16 days.
Efficiency ratio:
\[ A:B=3:1 \]
Therefore:
\[ B=16\times\frac{4}{1}=64\text{ days} \]
and:
\[ A=16\times\frac{4}{3}=21\frac{1}{3}\text{ days} \]
7. “A Takes Three Times the Time of B” — Important Trap
Do not confuse this with “A is three times as efficient as B.”
If A takes three times the time taken by B:
\[ T_A:T_B=3:1 \]
Efficiency is reversed:
\[ E_A:E_B=1:3 \]
So:
Exam warning: “Times as efficient” and “times as much time” produce opposite ratios.
8. When One Person Works First and Another Joins
Suppose A can finish a work in 12 days and B in 18 days. A works alone for 4 days, then B joins.
Take:
\[ LCM(12,18)=36 \]
Efficiencies:
\[ A=3,\qquad B=2 \]
A works alone for 4 days:
\[ 3\times4=12 \]
Remaining work:
\[ 36-12=24 \]
Both together complete:
\[ 3+2=5\text{ units/day} \]
Remaining time:
\[ \frac{24}{5}=4.8\text{ days} \]
9. When One Person Leaves
Suppose A and B work together for some time and then A leaves.
The procedure remains the same:
- 1.Find total work.
- 2.Find work completed before A leaves.
- 3.Subtract it from total work.
- 4.Use B's efficiency for the remaining work.
Example:
A takes 12 days and B takes 18 days. They work together for 4 days, then A leaves.
\[ LCM=36 \]
\[ A=3,\quad B=2 \]
Work completed together:
\[ (3+2)\times4=20 \]
Remaining:
\[ 36-20=16 \]
B alone takes:
\[ 16\div2=\boxed{8\text{ days}} \]
10. Men and Days
When all workers have equal efficiency and the total work remains the same:
\[ \boxed{M_1D_1=M_2D_2} \]
Example:
12 men can complete a work in 15 days. How many days will 20 men take?
\[ 12\times15=20\times x \]
\[ x=\frac{180}{20}=\boxed{9\text{ days}} \]
Recognition: More men → fewer days. Fewer men → more days. This is inverse proportion.
11. Men Leave in the Middle of the Work
Example:
15 men can complete a work in 24 days. After working for 8 days, 5 men leave. How many more days are required?
Total work:
\[ 15\times24=360\text{ man-days} \]
Work done in 8 days:
\[ 15\times8=120 \]
Remaining:
\[ 360-120=240 \]
Remaining men:
\[ 15-5=10 \]
Therefore:
\[ \frac{240}{10}=\boxed{24\text{ days}} \]
12. Men and Women
SSC often gives a relationship such as:
\[ 3\text{ men}=5\text{ women} \]
This means:
\[ 1\text{ man}=\frac53\text{ women} \]
Therefore:
\[ 12\text{ men}=12\times\frac53=20\text{ women} \]
This allows us to convert one type of worker into another.
Example
3 men can do as much work as 5 women. If 12 men complete a work in 20 days, how many days will 15 women take?
Since:
\[ 12M=20W \]
12 men working for 20 days are equivalent to 20 women working for 20 days.
Total work:
\[ 20\times20=400\text{ woman-days} \]
Therefore:
\[ \frac{400}{15}=26\frac23\text{ days} \]
13. Wages and Work
Wages are directly related to the amount of work contributed.
The general relationship is:
\[ \boxed{\text{Wage}\propto\text{Efficiency}\times\text{Time}} \]
If two workers work for the same amount of time, their wage ratio is simply their efficiency ratio.
Example:
A is twice as efficient as B and both work for the same time.
\[ A:B=2:1 \]
If their total wage is ₹1500:
\[ A=1500\times\frac23=\boxed{₹1000} \]
When Working Times Are Different
Suppose A is twice as efficient as B.
A works for 6 days and B works for 9 days.
A's work:
\[ 2\times6=12 \]
B's work:
\[ 1\times9=9 \]
Therefore wage ratio:
\[ 12:9=4:3 \]
If total wages are ₹1500:
\[ A=1500\times\frac47 \]
\[ \boxed{A=\frac{6000}{7}\text{ rupees}} \]
14. Efficiency Changes During the Work
This is a slightly more advanced pattern.
Suppose A can complete a work in 20 days. After working for 5 days, his efficiency increases by 50%.
Take total work as 20 units.
Original efficiency:
\[ 1\text{ unit/day} \]
Work completed in 5 days:
\[ 5 \]
Remaining:
\[ 20-5=15 \]
New efficiency:
\[ 1\times1.5=1.5 \]
Remaining time:
\[ 15\div1.5=\boxed{10\text{ days}} \]
For a percentage change:
\[ \boxed{\text{New efficiency}=\text{Old efficiency}\times(1\pm r)} \]
where \(r\) is the percentage expressed as a decimal.
15. Pipes and Cisterns
Pipes and Cisterns is essentially Time and Work with water.
- A pipe that fills a tank does positive work.
- A pipe that empties a tank does negative work.
If A fills a tank in 12 hours:
\[ A=+\frac1{12} \]
If B empties it in 18 hours:
\[ B=-\frac1{18} \]
When both work together:
\[ \frac1{12}-\frac1{18} =\frac1{36} \]
Therefore the tank fills in:
\[ \boxed{36\text{ hours}} \]
LCM Method for Pipes
Suppose three pipes fill or empty a tank in 12, 18 and 36 hours.
Take total work:
\[ LCM(12,18,36)=36 \]
Net efficiency:
\[ 3+2-1=4 \]
Time:
\[ 36\div4=\boxed9\text{ hours} \]
16. Alternate Working
When two people or pipes work alternately, do not simply add their efficiencies and divide the total work by the sum.
You must respect the order in which they work.
Example
A fills a tank in 10 hours and B fills it in 15 hours. They work alternately, starting with A.
Take total work:
\[ LCM(10,15)=30 \]
A:
\[ 30/10=3\text{ units/hour} \]
B:
\[ 30/15=2\text{ units/hour} \]
One complete two-hour cycle produces:
\[ 3+2=5\text{ units} \]
After 5 complete cycles:
\[ 5\times5=25 \]
So after 10 hours, 25 of the 30 units are completed.
Remaining:
\[ 30-25=5 \]
Now the turns continue:
Therefore:
\[ \boxed{12\text{ hours}} \]
Alternate-work trap: After calculating the remaining work, always check whose turn comes next.
17. A Quick Visual Summary
18. The Most Useful SSC CGL Shortcuts
19. Common Mistakes in Time and Work
Mistake 1: Confusing efficiency with time
If A is twice as efficient as B, do not say A takes twice the time.
Correct:
\[ E_A:E_B=2:1 \]
\[ T_A:T_B=1:2 \]
Mistake 2: Forgetting that an outlet does negative work
For pipes:
\[ \text{Net work}=\text{Inlet work}-\text{Outlet work} \]
Mistake 3: Changing total work when workers leave
The work remains the same. Only the rate at which it is completed changes.
Mistake 4: Ignoring turns in alternate work
In alternate-day or alternate-hour questions, you cannot assume that one worker continues indefinitely after a cycle ends.
Mistake 5: Using a complicated fraction method unnecessarily
For most SSC questions involving several whole-number times, the LCM method is faster and reduces calculation mistakes.
20. Exam Decision Tree
When you see a Time and Work question, ask these questions in order:
- 1.Are individual completion times given?
Use LCM to create total work and efficiencies. - 2.Is an efficiency ratio given?
Write the efficiency ratio directly. - 3.Is time ratio given?
Reverse it to obtain efficiency ratio. - 4.Do workers join or leave?
Calculate the work completed before the change and then recalculate the remaining work. - 5.Are men/women involved?
Convert them using the given equivalence. - 6.Are wages involved?
Use efficiency × time. - 7.Are pipes involved?
Treat filling as positive work and emptying as negative work. - 8.Is alternate working mentioned?
Calculate work in a complete cycle and then handle the remaining work according to the next turn.
21. SSC CGL Practice Questions
Try these without looking at the solutions. The aim is not merely to get the answer, but to identify the correct method within a few seconds.
22. Answers
23. Final Revision Sheet
If you have only one minute before the exam, remember these:
\[ \boxed{\text{Work}=\text{Efficiency}\times\text{Time}} \]
\[ \boxed{\text{1-day work}=\frac1{\text{number of days}}} \]
\[ \boxed{\text{Efficiency}\propto\frac1{\text{Time}}} \]
\[ \boxed{M_1D_1=M_2D_2} \]
\[ \boxed{\text{Wage}\propto\text{Efficiency}\times\text{Time}} \]
\[ \boxed{\text{Inlet}=+,\qquad\text{Outlet}=-} \]
And for most standard SSC CGL questions:
LCM → Total Work → Efficiency → Work Done → Remaining Work → Remaining Time.
Conclusion
Time and Work for SSC CGL becomes much easier once you stop treating every question as a separate formula. The same basic relationship keeps appearing in different forms.
A worker becomes more efficient → the time decreases.
More workers join → the work gets completed faster.
A worker leaves → the remaining work takes longer.
A pipe fills → positive work.
A pipe empties → negative work.
People work for different durations → compare their actual work contributions.
Once these relationships become automatic, the chapter turns from a formula-heavy topic into a pattern-recognition chapter—exactly what you want for a time-bound SSC CGL exam.
Best exam habit: Before calculating, identify the pattern. In Time and Work, recognising the correct model is often more important than doing the arithmetic.
