Time and work for RRB NTPC: methods, patterns and practice

Many candidates put off time and work, because the first few questions they try seem to need messy fractions. They do not. With one change of viewpoint, from "days" to "work per day", nearly every question becomes addition and division of whole numbers.

The same method also solves pipes and cisterns, which is time and work with water. Learn it once and you have two topics.

The key idea: work per day

If a person finishes a job in n days, they do 1/n of the job each day.

Why this matters: you cannot add days. If A takes 10 days and B takes 15, together they do not take 25 days, or 12.5. But you can add work done per day, because on any one day the two amounts of work simply pile up. So always convert to rates, add or subtract the rates, then convert back to time.

Method 1: fractions

• A's one-day work = 1/a; B's = 1/b.
• Together, one day = 1/a + 1/b.
• Time together = 1 ÷ (1/a + 1/b) = ab/(a + b).

For A = 12 days and B = 18 days: 12 × 18 ÷ 30 = 7.2 days.

Method 2: total work as the LCM

1. Take the total work as the LCM of the given days. Call the result "units".
2. Each person's units per day = total ÷ their days.
3. Add or subtract units per day, then divide the total by the result.

For A = 12 and B = 18: total = 36 units. A does 3 a day, B does 2, together 5. Time = 36 ÷ 5 = 7 1/5 days.

Why it works: the size of the job is yours to choose, since only the rates matter. Choosing a total that every time divides evenly gives whole-number rates and keeps fractions out until the last step.

Efficiency and time are inversely related

If A is twice as efficient as B, A takes half the time. More generally, if efficiencies are in the ratio x : y, times are in the ratio y : x. A useful way to handle "A is twice as efficient" is to give B 1 unit a day and A 2 units a day.

Men, days and hours

For the same job, M₁ × D₁ × H₁ = M₂ × D₂ × H₂, where M is workers, D is days and H is hours a day. If the amount of work changes, divide each side by its work: M₁D₁H₁/W₁ = M₂D₂H₂/W₂.

Why: M × D × H counts total worker-hours. A fixed job needs a fixed number of worker-hours, however they are arranged.

Work and wages

When people are paid for a job done together, wages are shared in the ratio of work done, which is their efficiency multiplied by the days they worked. If they work the same days, share in the ratio of efficiencies.

Worked questions at NTPC level

Q1. A can finish a job in 20 days. A works for 5 days, then B finishes the rest in 9 days. How long would B take alone?

In 5 days A does 5/20 = 1/4 of the job, leaving 3/4.
B does 3/4 in 9 days, so the whole job takes 9 × 4/3 = 12 days.

Q2. A is twice as efficient as B. Together they finish a job in 12 days. How long will A take alone?

Let B do 1 unit a day and A 2 units. Together 3 units a day, so the job is 3 × 12 = 36 units.
A alone = 36 ÷ 2 = 18 days.

Q3. 12 workers, working 8 hours a day, finish a job in 20 days. How many days will 16 workers take, working 10 hours a day?

12 × 8 × 20 = 16 × 10 × D, so 1,920 = 160D, and D = 12 days.

Q4. A, B and C are three workers. A and B together finish a job in 12 days, B and C in 15 days, and C and A in 20 days. How long will all three take together?

Total work = LCM of 12, 15 and 20 = 60 units.
A + B = 5 a day, B + C = 4 a day, C + A = 3 a day. Adding: 2(A + B + C) = 12, so A + B + C = 6 a day.
All three together = 60 ÷ 6 = 10 days.

Q5. A can do a job in 12 days and B in 18 days. They work on alternate days, starting with A. In how many days will the job be finished?

Total = 36 units. A does 3 a day, B does 2. Each two-day cycle does 5 units.
After 7 cycles (14 days), 35 units are done and 1 unit is left.
Day 15 is A's day. A needs 1/3 of a day for 1 unit. Total = 14 1/3 days.

Q6. A can do a job in 10 days and B in 15 days. Working together they finish it and are paid ₹3,000. How should the money be shared?

Rates: 1/10 and 1/15, in the ratio 3 : 2. They worked the same number of days, so work done is also 3 : 2.
A gets 3/5 of 3,000 = ₹1,800; B gets ₹1,200.

Common mistakes

• Adding days instead of rates. Always convert to work per day first.
• Forgetting that more workers need fewer days. Workers and days are in inverse proportion.
• Ignoring work already done when someone leaves or joins partway.
• Rounding up in alternate-day questions. The last worker may need only part of a day.
• Sharing wages in the ratio of days taken. Share in the ratio of work done, which is the inverse ratio of days when they work together.

Practice set

1. A takes 10 days and B takes 15 days to do a job. How long will they take together?
2. A and B together take 8 days; A alone takes 12. How long does B alone take?
3. 8 workers finish a job in 15 days. How many days will 10 workers take?
4. A is twice as fast as B. B alone takes 30 days. How long will they take together?
5. A takes 20 days and B takes 30 days. They work together for 6 days, then A leaves. How many more days will B take?
6. A and B together can do a job in 10 days, B and C in 12 days, and C and A in 15 days. How long will A, B and C take working together?
7. 20 workers can finish a job in 12 days working 6 hours a day. How many workers are needed to finish it in 10 days working 8 hours a day?
8. A can do a job in 12 days and B in 16 days. They finish it together and are paid ₹5,600. Find A's share.

Answers:

1. 6 days. LCM 30: A 3, B 2 a day; 30 ÷ 5.
2. 24 days. LCM 24: together 3, A 2, so B 1 a day.
3. 12 days. 8 × 15 = 10 × D.
4. 10 days. A takes 15 days. LCM 30: A 2, B 1; 30 ÷ 3.
5. 15 days. LCM 60: A 3, B 2; 6 days × 5 = 30 units; B does the other 30 at 2 a day.
6. 8 days. LCM 60: 6 + 5 + 4 = 15 = 2(A + B + C), so 7.5 a day; 60 ÷ 7.5 = 8.
7. 18 workers. 20 × 12 × 6 = M × 10 × 8, so 1,440 = 80M.
8. ₹3,200. Rates 1/12 : 1/16 = 4 : 3; A gets 4/7 of 5,600.

What to do next

• Solve 20 time and work questions with the LCM method only, until choosing the total is automatic.
• Practise five "pairs together" questions (A + B, B + C, C + A) and five alternate-day questions.
• Move straight on to pipes and cisterns, which uses the same method, and revise LCM and HCF if the LCM step is slow.