Time Work

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Time and Work problems test your ability to calculate how long tasks take when workers, pipes, or machines work together or separately. The fundamental principle is that work done is inversely proportional to time taken — more workers means less time. Understanding the relationship between work, time, and worker capacity is essential for SSC CGL.

Basic Formulas:

If a person can complete a work in $n$ days, their work rate is $\frac{1}{n}$ work per day.

If person A can complete work in $a$ days and person B can complete it in $b$ days: $$A\text{‘s rate} = \frac{1}{a}, \quad B\text{‘s rate} = \frac{1}{b}$$ $$\text{Combined rate} = \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$$ $$\text{Time to complete together} = \frac{ab}{a+b} \text{ days}$$

Work Units:

Work = Rate × Time If rate = $\frac{1}{n}$ work/day, then in t days, work done = $\frac{t}{n}$ of the total work.

When total work = 1 (complete job):

• Person A finishes $\frac{1}{a}$ per day
• Together they finish $\frac{1}{a} + \frac{1}{b}$ per day
• Time to finish = $\frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{ab}{a+b}$

⚡ Exam Tip: Always set the total work as 1 unit. If a pipe fills a tank in 6 hours, its rate is $\frac{1}{6}$ tank per hour. If it fills in 10 hours, its rate is $\frac{1}{10}$ tank per hour. Together: $\frac{1}{6} + \frac{1}{10} = \frac{5+3}{30} = \frac{8}{30} = \frac{4}{15}$ tank per hour. Time = $\frac{15}{4} = 3.75$ hours.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding and problem-solving practice.

Extension to Multiple Workers:

If A, B, C can complete work in $a$, $b$, $c$ days respectively: $$\text{Time together} = \frac{abc}{ab + bc + ca} \text{ days}$$

For $n$ workers with rates $r_1, r_2, …, r_n$: $$\text{Time together} = \frac{1}{r_1 + r_2 + … + r_n}$$

Pipes and Cisterns:

Filling pipe (inlet): Positive contribution Emptying pipe (outlet): Negative contribution

If inlet pipe fills in $a$ hours and outlet pipe empties in $b$ hours: $$\text{Net rate} = \frac{1}{a} - \frac{1}{b} \text{ (if } a < b\text{, otherwise negative)}$$ $$\text{Time to fill} = \frac{1}{\frac{1}{a} - \frac{1}{b}} = \frac{ab}{b-a}$$

Work-Time Equivalence:

If A works for $x$ days, then B completes the rest in $y$ days: $$\frac{x}{a} + frac{y}{b} = 1$$ (assuming total work = 1)

Percentage Work:

If a machine does $p$% of work in one day, it does $\frac{p}{100}$ of total work per day. To complete P% of work: Time = $\frac{P}{p}$ days.

Worker Efficiency Variations:

If workers have different efficiencies, express as fractions: Example: A is twice as efficient as B. If B completes work in 12 days, A would take 6 days.

If A and B together complete work in $d$ days and A alone would take $a$ days: $$B\text{‘s time} = \frac{ad}{a-d}$$

Combined Work with Different Hours:

If workers work different hours per day:

• Calculate daily work: $\text{hours per day} \times \text{daily rate}$
• Sum rates and find total time

⚡ SSC CGL-Specific Tip: In questions where A starts and works alone for some days, then B joins, find the fraction of work done by A in that time, subtract from 1, and find how long B takes for the remaining work. If they work for different durations or at different rates, carefully track work fractions.

Common Student Mistakes:

• Confusing addition with multiplication when combining rates
• Forgetting to subtract outlet rates from inlet rates (or vice versa)
• Using wrong total work value when it isn’t 1 unit

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Man-Days Concept:

Work = Number of workers × Number of days × Efficiency factor

If work requires $M$ man-days and $x$ men work for $d$ days, then:

• Work done = $x \times d$ man-days
• Remaining work = $M - x \times d$ man-days
• Time needed with $y$ additional men: $\frac{M - x \times d}{y}$ days

Fraction of Work Done:

If A can complete work in $a$ days, fraction done by A in $t$ days: $$\text{Fraction} = \frac{t}{a} \text{ (if } t \le a\text{)}$$

When Workers Join/Leave Mid-Project:

Example: A starts working alone. After 3 days, B joins. A and B work for 2 more days, then A leaves. B finishes remaining work in 5 days. Find total time.

Let total work = 1. A’s rate = $\frac{1}{8}$ (if A alone takes 8 days) First 3 days: A works → work done = $3 \times \frac{1}{8} = \frac{3}{8}$ Next 2 days: A + B work together → work done = $2 \times (\frac{1}{8} + \frac{1}{b})$ Remaining work = $1 - [\frac{3}{8} + 2(\frac{1}{8} + \frac{1}{b})] = \frac{5}{8} - \frac{2}{b}$ B finishes this in 5 days: $5 \times \frac{1}{b} = \frac{5}{8} - \frac{2}{b}$ Solve for b: $\frac{5}{b} = \frac{5}{8} - \frac{2}{b}$ → $\frac{7}{b} = \frac{5}{8}$ → $b = \frac{56}{5} = 11.2$ days

Alternative Approach:

Let B’s time alone = $b$ days (rate = $\frac{1}{b}$) Work after 5 days = $1 - \frac{5}{b}$ Work done in first 5 days (3 by A alone + 2 by A+B) = $\frac{3}{8} + 2(\frac{1}{8} + \frac{1}{b}) = \frac{5}{8} + \frac{2}{b}$

Chain Rule in Work:

If $M_1$ workers can do $W_1$ work in $D_1$ days, working $h_1$ hours per day: $$M_1 \times D_1 \times h_1 \propto W_1$$

If $M_2$ workers need to do $W_2$ work in $D_2$ days, working $h_2$ hours per day: $$M_2 \times D_2 \times h_2 = \frac{M_1 \times D_1 \times h_1 \times W_2}{W_1}$$

Sample SSC CGL Questions:

Q1: A and B can do a work in 12 days. B and C can do it in 15 days. C and A can do it in 20 days. How long will A, B, C take together? Solution: A + B rate = $\frac{1}{12}$ B + C rate = $\frac{1}{15}$ C + A rate = $\frac{1}{20}$

Adding all three: $2(A+B+C) = \frac{1}{12} + \frac{1}{15} + \frac{1}{20}$ LCM of 12, 15, 20 = 60 $\frac{1}{12} = \frac{5}{60}$; $\frac{1}{15} = \frac{4}{60}$; $\frac{1}{20} = \frac{3}{60}$ Sum = $\frac{12}{60} = \frac{1}{5}$ So $2(A+B+C) = \frac{1}{5}$ → $A+B+C = \frac{1}{10}$ Time together = 10 days

Q2: A pipe fills a tank in 20 minutes. B pipe fills it in 30 minutes. A pipe is opened for 5 minutes, then B is also opened. How much more time to fill? Solution: A’s 5 minutes: $5 \times \frac{1}{20} = \frac{1}{4}$ filled Remaining = $\frac{3}{4}$ A+B together rate = $\frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$ Time to fill remaining $\frac{3}{4}$ = $\frac{3}{4} \times 12 = 9$ minutes

Q3: If 12 men can do a work in 18 days, how many days will 8 men take? Solution: Work = 12 × 18 = 216 man-days With 8 men: 216/8 = 27 days

⚡ Advanced Tip: When solving Q3, we assume all men work equally. But if efficiency varies, we must account for individual rates. More importantly, if 8 men work slower (less efficient per man), the formula M₁D₁ = M₂D₂ assumes constant efficiency per man. If some men are more efficient than others, express their rates as fractions of a “standard” worker’s rate before solving.

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