Time and Work

Three equivalent ways to solve any time-work problem. Pick the fastest based on the question.

Approach 1: 1/x rate.
If A finishes work in a days, A's 1-day work = 1/a.

A and B together: 1/a + 1/b. Time taken = 1 / (1/a + 1/b) = ab/(a+b).

Worked: A does work in 12 days, B in 18. Together?
Combined rate = 1/12 + 1/18 = 3/36 + 2/36 = 5/36. Time = 36/5 = 7.2 days.

Approach 2: LCM (units of work).
Take LCM of given days as "total work" units. Divide by individual times to get per-day rate.

Same problem with LCM:
LCM(12, 18) = 36 units of work. A's rate = 36/12 = 3 units/day. B's rate = 36/18 = 2 units/day.
Together = 5 units/day. Time = 36/5 = 7.2 days.

LCM approach avoids fractions and is faster for multi-step problems.

Approach 3: Efficiency ratio.
If A is x times more efficient than B, their work-time ratio is reciprocal: t_A : t_B = 1 : x.

Pipes & cisterns (just time-work with positive/negative rates):

• Inlet pipe filling: positive rate.
• Outlet (leak): negative rate.
• Net rate = sum.

Worked example: Pipe A fills tank in 6 hours; pipe B in 4 hours; outlet C drains in 8 hours. With all open, time to fill?

LCM(6, 4, 8) = 24 units (tank capacity).
A: 24/6 = +4. B: 24/4 = +6. C: 24/8 = −3.
Net = +7 per hour. Time = 24/7 ≈ 3.43 hours.

Person-day formula (M₁D₁ = M₂D₂):
M₁ men in D₁ days = M₂ men in D₂ days, for the same work.

Worked example: 18 men complete a job in 24 days. How many days for 27 men?

18 × 24 = 27 × D → D = 432/27 = 16 days.

Trick: alternate-day work.
If A and B work alternately starting with A: time taken depends on pattern. Compute units done in one full cycle (1 day A + 1 day B), then total time.

Worked example: A finishes in 6 days, B in 8 days. Working alternately starting with A, time?

LCM = 24 units. A=4/day, B=3/day. One cycle (2 days) = 4+3 = 7 units. After 3 cycles (6 days) = 21 units. Day 7: A works → 4 more units → 25 units > 24. Actually A only needs 3/4 of a day to finish 3 units.
Total = 6 + 3/4 = 6.75 days.

TIME AND WORK problems involve calculating time to complete jobs given individual or combined efficiencies.

CORE CONCEPTS:

If A completes work in m days:

• A's 1-day work = 1/m.
• In d days, A completes d/m of work.

For two persons A (m days) + B (n days):

• Combined per day = 1/m + 1/n = (m+n)/(mn).
• Together complete work in mn/(m+n) days.

For three: 1/m + 1/n + 1/p; combined = mnp/(np+mp+mn).

EFFICIENCY (compare):

If A:B efficiency = a:b, then time ratio = b:a (inversely proportional).

TYPES OF PROBLEMS:

1. Joint work:

• A & B together. Use 1/m + 1/n.

2. A starts alone, then B joins:

• Calculate A's work for that period, then combined.

3. Substitution / replacement:

• Worker replaced after some days.

4. Pipes and cisterns:

• Fill pipes positive; drain pipes negative.

WORKED EXAMPLES:

Q1. A can do work in 12 days, B in 18 days. Together?

• 1/12 + 1/18 = 3/36 + 2/36 = 5/36.
• Together: 36/5 = 7.2 days.

Q2. A & B can do work in 12 days; A alone in 20 days. B alone?

• A's rate = 1/20. Combined = 1/12.
• B's rate = 1/12 − 1/20 = 5/60 − 3/60 = 2/60 = 1/30.
• B alone: 30 days.

Q3. A is twice as efficient as B. Together they do work in 8 days. B alone?

• Let B's rate = x. A's rate = 2x.
• Combined = 3x = 1/8 → x = 1/24.
• B alone: 24 days.

Q4. A starts alone; B joins after 5 days. A: 20 days; B: 15 days. Total time?

• A alone for 5 days: completes 5/20 = 1/4.
• Remaining: 3/4. Together rate: 1/20 + 1/15 = 7/60.
• Time to do 3/4: (3/4) / (7/60) = 3/4 × 60/7 = 45/7 ≈ 6.43 days.
• Total: 5 + 6.43 = 11.43 days.

Q5. Pipe A fills tank in 6 hr; B fills in 10 hr; C drains in 30 hr. All three open. Time to fill?

• Combined rate: 1/6 + 1/10 − 1/30 = 5/30 + 3/30 − 1/30 = 7/30.
• Time: 30/7 ≈ 4.29 hr.

Q6. A & B together 12 days; B & C together 15 days; A & C together 20 days. All three together?

• Let A, B, C rates = a, b, c.
• a + b = 1/12.
• b + c = 1/15.
• a + c = 1/20.
• Sum: 2(a + b + c) = 1/12 + 1/15 + 1/20 = 5/60 + 4/60 + 3/60 = 12/60 = 1/5.
• a + b + c = 1/10.
• All together: 10 days.

Q7. 30 men do work in 24 days. After 8 days, 6 more join. Total time?

• 30 men in 8 days: 8/24 = 1/3 of work done.
• Remaining: 2/3.
• 36 men's rate per day: 36/(30×24) = 1/20.
• Days needed: (2/3) / (1/20) = 40/3 ≈ 13.33 days.
• Total: 8 + 13.33 = 21.33 days.

Q8. A & B start, but A leaves after 4 days, B finishes alone. A: 10 days; B: 15 days. Total time?

• 4 days A & B: (1/10 + 1/15) × 4 = (3+2)/30 × 4 = 5/30 × 4 = 4/6 = 2/3.
• B alone remaining 1/3: at rate 1/15 → 5 days.
• Total: 4 + 5 = 9 days.

MEN-DAYS-WORK PROBLEMS:

(M₁ × D₁ × H₁) / W₁ = (M₂ × D₂ × H₂) / W₂.

E.g., 10 men work 8 hr/day for 12 days → some work.
20 men work 6 hr/day for x days → same work.
10 × 8 × 12 = 20 × 6 × x → x = 8.

WAGES SHARING:

Workers paid in ratio of work done (or rate × days).

Q9. A works for 6 days at rate 1/10; B works for 4 days at rate 1/15. Total wages ₹100. Each's share?

• A's work: 6 × 1/10 = 3/5.
• B's work: 4 × 1/15 = 4/15.
• Ratio: (3/5)/(4/15) = 9/4.
• A : B = 9:4. Total parts = 13.
• A: 100 × 9/13 = ₹69.23.
• B: 100 × 4/13 = ₹30.77.

EXAM HOOKS:

• Combined time = mn/(m+n) for 2 people.
• Efficiency ↔ time (inverse).
• Pipes: fill positive, drain negative.
• Men-days-work: cross-multiply.
• Wages: in ratio of work done.
• For "A and B alone known, find combined": just add rates.