SSC CGL Preparation – Day 19

Quantitative Aptitude – Time and Work (Repeated – Advanced)

✅ 1. Basic Concept Recap

Time and Work problems involve calculating the amount of work, time taken, or number of workers involved in completing a task.

• Work and Time are inversely proportional.
  More men = less time. More work = more time.

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🔹 Basic Formulae

1. Work = Time × Efficiency
2. If A can do a piece of work in x days, then A’s 1 day work = 1/x
3. If A is ‘n’ times as efficient as B, then:
   • Time taken by A = 1/n × time taken by B
   • Ratio of work done in 1 day = A:B = n:1
4. If A and B together can do a work in X days and A alone can do it in Y days, then:
   • B’s time alone = $$\frac{XY}{Y – X}$$
5. LCM Method: Assume total work = LCM of the number of days of individuals
   (Useful when dealing with multiple people or machines)

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✅ 2. Common Problem Types

🔸 A. Individual Work

“A can do a work in 10 days. How much work does he do in one day?”
✅ Answer: 1/10

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🔸 B. Combined Work

“A can do it in 10 days, B in 15 days. How many days will they take together?”

Solution:
1-day work = 1/10 + 1/15 = $$\frac{5}{30} = \frac{1}{6}$$
✅ So, total days = 6

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🔸 C. Man–Day Concept

If 12 men can build a wall in 15 days, how many men are needed to do it in 10 days?

Solution:
12 × 15 = x × 10 → x = 18 men

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🔸 D. Alternate Work / Replacement

A works for 5 days, then B completes the rest. Total time = ?

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🔸 E. Pipes and Cisterns

• Inlet fills → +ve work
• Outlet empties → −ve work

If a pipe fills in 10 min (1/10 work/min) and another empties in 15 min (−1/15),
Net 1-min work = $$\frac{1}{10} – \frac{1}{15} = \frac{1}{30}$$

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✅ 3. Important Shortcuts

1. If A can do a work in ‘a’ days and B in ‘b’ days, together:

$$\text{Work in 1 day} = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$$

2. Time to finish together =

$$\frac{ab}{a + b}$$

3. If A is x% more efficient than B:
   Let B’s efficiency = 1, A’s = 1 + x/100
   Then A will take: $$\frac{1}{1 + x/100}$$​ of B’s time
4. If A and B together can do a work in X days and A alone can do it in Y days:
   Then B alone = $$\frac{XY}{Y – X}$$

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✅ 4. Example Questions

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Q1. A can do a work in 10 days, B in 15 days. Working together, how many days will they take?

Solution:
1-day work = 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6
✅ Total time = 6 days

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Q2. A is 25% more efficient than B. If B alone does the work in 20 days, how much time will A take?

Solution:
A’s efficiency = 125%, or 1.25 × B’s
A’s time = $$\frac{20}{1.25} = 16$$ days

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Q3. A and B can complete a task in 12 days. B alone can complete it in 20 days. In how many days can A alone do the work?

Solution:
1-day work of A + B = 1/12,
1-day work of B = 1/20
1-day work of A = 1/12 − 1/20 = (5 − 3)/60 = 2/60 = 1/30
✅ A alone takes 30 days

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Q4. 8 men can complete a task in 24 days. How many men are needed to finish it in 12 days?

Solution:
Work = Men × Days → 8 × 24 = x × 12 → x = 16 men

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Q5. A pipe fills a tank in 12 minutes. Another pipe can empty it in 20 minutes. If both are opened together, how long will the tank take to fill?

Solution:
Net work = 1/12 − 1/20 = (5 − 3)/60 = 2/60 = 1/30
✅ Time = 30 minutes

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✅ 5. SSC CGL Focused Tips

• Always assume total work = LCM when needed.
• Use the unit work method (work per day) to simplify multi-person problems.
• Time taken is inversely proportional to efficiency.
• Work = Men × Days × Hours – Useful for shift-based questions.