Time and Work: Notes और MCQ Practice

Time & Work – आसान भाषा में

काम = Rate × Time. अगर दो लोग साथ में काम करते हैं तो उनका combined rate = sum of individual rates.

Important Formulas / Rules

• Work done = Rate × Time
• Rate = Work ÷ Time
• If A does a work in a days, his rate = 1/a work per day.
• n workers each with rate r → combined rate = n × r.
• Efficiency % = (Actual work ÷ Standard work) × 100.
• Pipe‑and‑cistern: Fill rate – Drain rate = Net rate.

Quick Tricks

1. Reciprocal Method: For two persons A (a days) & B (b days), total time = (a × b) ÷ (a + b).
2. Work‑in‑%: If a person works 75% efficiency, effective days = given days ÷ 0.75.

Common Mistakes

• Forgetting to convert % efficiency to decimal before using in formulas.
• Adding times of workers instead of adding their rates.

Time and Work

1. Concept

Time and Work problems deal with how long it takes for one or more workers (or machines) to complete a job when they work together or separately. The basic idea is that Work = Rate × Time. A worker’s rate is the fraction of the whole job he/she can finish in one unit of time.

2. Important formulas / rules / facts

• Work done by a person = (Rate) × (Time)
• Rate of a person = (Work done) ÷ (Time)
• If a person can finish a job in X days, his/her daily rate = 1 ÷ X (job per day).
• When workers work together, combined rate = sum of individual rates.
• If two workers A and B work together for t days and finish the whole job, then (1/ A_days + 1/ B_days) × t = 1.
• Efficiency = (Actual work done) ÷ (Time taken) × 100 %.
• For a group of workers, total work = sum of individual works.
• If a worker’s efficiency changes by p %, new rate = old rate × (1 + p/100).

3. Solved examples (exam style)

Example 1

Problem: A can finish a piece of work in 12 days and B can finish the same work in 15 days. How many days will they take to complete the work if they work together?

Solution:

• Rate of A = 1/12 job per day.
• Rate of B = 1/15 job per day.
• Combined rate = 1/12 + 1/15 = (5 + 4) / 60 = 9/60 = 3/20 job per day.
• Time required = 1 ÷ (3/20) = 20/3 = 6.67 days (≈ 6 days 8 hours).

Example 2

Problem: A and B together can complete a work in 10 days. B alone can finish it in 15 days. Find the time taken by A alone.

Solution:

• Let A’s time = x days → rate of A = 1/x.
• Rate of B = 1/15.
• Combined rate = 1/x + 1/15 = 1/10.
• 1/x = 1/10 – 1/15 = (3 – 2) / 30 = 1/30.
• Hence x = 30 days.

Example 3

Problem: C can do a job in 20 days. He works for 5 days, then leaves and D finishes the remaining work in 8 days. Find D’s time to complete the whole job alone.

Solution:

• Work done by C in 5 days = 5 × (1/20) = 5/20 = 1/4 of the job.
• Work left = 1 – 1/4 = 3/4.
• D finishes 3/4 of the job in 8 days, so D’s daily rate = (3/4) ÷ 8 = 3/32 job per day.
• Time for D to do whole job = 1 ÷ (3/32) = 32/3 = 10.67 days (≈ 10 days 16 hours).

Example 4

Problem: A, B and C can complete a work in 18, 20 and 30 days respectively. They work together for 4 days, then A leaves. How many more days will B and C need to finish the remaining work?

Solution:

• Rates: A = 1/18, B = 1/20, C = 1/30.
• Combined rate for first 4 days = 1/18 + 1/20 + 1/30 = (10 + 9 + 6) / 180 = 25/180 = 5/36 job per day.
• Work done in 4 days = 4 × 5/36 = 20/36 = 5/9.
• Work left = 1 – 5/9 = 4/9.
• Rate of B + C = 1/20 + 1/30 = (3 + 2) / 60 = 5/60 = 1/12 job per day.
• Additional days = (4/9) ÷ (1/12) = (4/9) × 12 = 48/9 = 5.33 days (≈ 5 days 8 hours).

Example 5

Problem: A can do a piece of work in 24 days. He works for 6 days, then his efficiency drops by 25 % for the remaining period. The work is completed in 30 days total. Find the original work amount (in job units) if the final work completed equals 1 whole job.

Solution:

• Original daily rate = 1/24.
• Work done in first 6 days = 6 × 1/24 = 6/24 = 1/4.
• Remaining work = 1 – 1/4 = 3/4.
• New rate after 25 % drop = (1/24) × (1 – 0.25) = 1/24 × 0.75 = 3/96 = 1/32.
• Remaining days = 30 – 6 = 24 days.
• Work done in remaining days = 24 × 1/32 = 24/32 = 3/4, which matches the remaining work. Hence the data is consistent; the whole job = 1 unit.

4. Shortcuts and tricks

• Reciprocal method: Instead of converting to rates, directly use reciprocals. If A finishes in a days, treat his contribution as 1/a of the job per day.
• Common denominator shortcut: When adding rates like 1/12 + 1/15, use LCM of denominators (60) to avoid fractions.
• Work left = 1 – work done is quicker than calculating percentages.
• Use proportion: If two workers together finish in T days, then (A_days × B_days) ÷ (A_days + B_days) = T (useful for quick mental check).
• Efficiency change: New time = Old time ÷ (1 – % change/100).

5. Common mistakes to avoid

• Forgetting to convert percentages to decimals when adjusting efficiency.
• Adding times instead of rates; always add rates (1/ days), not the days themselves.
• Ignoring the fact that work is a dimensionless quantity; mixing units (hours with days) without conversion leads to errors.
• Overlooking the remaining work after a partial period; always compute remaining work = 1 – work already done.
• Misplacing the denominator when using the reciprocal method; ensure the denominator is the individual time, not the combined time.