Theory & Concepts
SSC CGL Time and Work Questions, Formulas & Short Tricks
Prepare Time and Work for SSC CGL with formulas, short tricks, solved examples, practice questions, PYQs, and free PDF notes for faster exam preparation.
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Practice Notes
24 min readDifficulty: Intermediate
Time and work is one of the most logically satisfying chapters in arithmetic. In Ssc cgl, you can avoid tedious fraction addition entirely by mastering the LCM (Total Work) method.
Whether it's people leaving early, groups of men and women working together, or alternating shifts, every question reduces to a simple efficiency calculation once you define the "Total Work" units.
Learning path
- Basic work-efficiency rule
- The LCM (Unit) method
- Man-days-hours formula
- 10 standard solved problems
- Alternate day logic
1. Core principles
The work equation
Key inversions:
Efficiency and Time are inversely proportional for a constant work.
2. The chain rule formula
For problems involving groups of men working for specific hours and days:
M = Men, D = Days, H = Hours, W = Work quantity.
3. Solved examples
Question 01Standard pattern
A can do a piece of work in 12 days and B can do it in 15 days. How long will they take if they work together?
6(2/3) days
6 days
7 days
5(1/2) days
Correct answer: a) 6(2/3) days
Solution
LCM of \( 12, 15 = 60 \) units (Total Work).
Efficiency of A = \( 60/12 = 5 \text{ units/day} \).
Efficiency of B = \( 60/15 = 4 \text{ units/day} \).
Combined efficiency = \( 5 + 4 = 9 \text{ units/day} \).
Time taken = \( 60 / 9 = 20 / 3 = 6\frac{2}{3} \text{ days} \).
Question 02Standard pattern
A can do a work in 10 days and B in 20 days. They work together for 5 days and then A leaves. How many more days will B take to finish the work?
5 days
10 days
3 days
2 days
Correct answer: a) 5 days
Solution
Total Work (LCM of 10, 20) = \( 20 \text{ units} \).
Efficiencies: \( A = 2, B = 1 \).
Work done in 5 days = \( 5 \times (2 + 1) = 15 \text{ units} \).
Remaining work = \( 20 - 15 = 5 \text{ units} \).
Time for B to finish = \( 5 / 1 = 5 \text{ days} \).
Question 03Standard pattern
A is twice as good a workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work?
27 days
36 days
54 days
24 days
Correct answer: a) 27 days
Solution
Efficiency ratio: \( A : B = 2 : 1 \).
Combined Efficiency = \( 2 + 1 = 3 \).
Total Work = \( 18 \times 3 = 54 \text{ units} \).
Time for A alone = \( 54 / 2 = 27 \text{ days} \).
Question 04Standard pattern
12 men can complete a work in 9 days. After they have worked for 3 days, 6 more men join them. How many days will they now take to complete the remaining work?
4 days
5 days
3 days
6 days
Correct answer: a) 4 days
Solution
Total Man-days = \( 12 \times 9 = 108 \).
Work done in 3 days = \( 12 \times 3 = 36 \).
Remaining work = \( 108 - 36 = 72 \text{ units} \).
Now total men = \( 12 + 6 = 18 \).
Time = \( 72 / 18 = 4 \text{ days} \).
Question 05Standard pattern
A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is:
8/15
7/15
1/4
1/10
Correct answer: a) 8/15
Solution
LCM of 15, 20 = \( 60 \). Efficiencies: \( A = 4, B = 3 \).
Work done in 4 days = \( 4 \times (4 + 3) = 28 \text{ units} \).
Remaining work = \( 60 - 28 = 32 \text{ units} \).
Fraction left = \( 32 / 60 = 8 / 15 \).
Question 06Standard pattern
A can complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at 24 km/hr. Find the total journey in km.
224 km
220 km
230 km
250 km
Correct answer: a) 224 km
Solution
This is a speed-distance problem using average approach.
Average Speed = \( \frac{2 \times 21 \times 24}{21 + 24} = \frac{1008}{45} = 22.4 \text{ km/hr} \).
Total Distance = \( 22.4 \times 10 = 224 \text{ km} \).
Question 07Standard pattern
A group of workers can complete a work in 10 days. But 5 they were absent, and remaining workers finish in 12 days. Find original number of workers.
30
25
20
35
Correct answer: a) 30
Solution
Let original workers be \( x \).
Equation: \( x \times 10 = (x - 5) \times 12 \).
\( 10x = 12x - 60 \).
\( 2x = 60 \implies x = 30 \).
Question 08Standard pattern
A, B and C can do a work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
15 days
12 days
18 days
10 days
Correct answer: a) 15 days
Solution
LCM = \( 60 \). Efficiencies: \( A=3, B=2, C=1 \).
Day 1 (A only): \( 3 \text{ units} \).
Day 2 (A only): \( 3 \text{ units} \).
Day 3 (A+B+C): \( 3+2+1 = 6 \text{ units} \).
Total work in 3 days = \( 3 + 3 + 6 = 12 \text{ units} \).
Number of cycles to reach 60: \( 60 / 12 = 5 \text{ cycles} \).
Total time = \( 5 \times 3 = 15 \text{ days} \).
Question 09Standard pattern
6 men and 8 women can do a work in 10 days. 26 men and 48 women can do it in 2 days. The time taken by 15 men and 20 women in doing the same work is:
4 days
5 days
6 days
3 days
Correct answer: a) 4 days
Solution
Equation: \( 10(6M + 8W) = 2(26M + 48W) \).
\( 60M + 80W = 52M + 96W \).
\( 8M = 16W \implies M = 2W \).
Total work in terms of W: \( 10(6(2W) + 8W) = 10(20W) = 200W \).
Target group in W: \( 15(2W) + 20W = 50W \).
Time = \( 200W / 50W = 4 \text{ days} \).
Question 010Standard pattern
A can do a work in 80 days and B in 100 days. They work together for 20 days and then B goes away. How many days will A take to finish the work?
44 days
40 days
50 days
45 days
Correct answer: a) 44 days
Solution
Total Work (LCM 80, 100) = \( 400 \text{ units} \).
Efficiencies: \( A=5, B=4 \).
Work in 20 days = \( 20 \times 9 = 180 \text{ units} \).
Remaining work = \( 400 - 180 = 220 \text{ units} \).
Time for A = \( 220 / 5 = 44 \text{ days} \).
4. Strategy errors to avoid
Error 01Reciprocal trap: Adding days directly (e.g. 10 + 20 = 30 days total) instead of adding the work rates per day.
Error 02LCM oversight: Not using the Lowest Common Multiple as 'Total Work', leading to messy fractional calculations.
Error 03Efficiency flip: Forgetting that a faster person has a higher efficiency value but needs fewer days.
Error 04Man-days confusion: Forgetting to include the hours per day (H) in the chain rule equation.
