Railway Exams Maths Arithmetic Time & Work: Formulas, Concepts & Solved Questions

Time & Work is a high-weightage topic in Railway Exams and is frequently asked in RRB NTPC, RRB Group D, ALP, Technician, and other competitive examinations. Questions from this topic often involve concepts such as efficiency, LCM-based work calculations, alternate work patterns, and work-time relationships. Many candidates find these questions challenging because a small mistake in understanding efficiency or work cycles can lead to incorrect answers.

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• Railway Exams Maths Arithmetic Partnership

To solve Time & Work questions quickly and accurately, candidates must understand the concepts of total work, efficiency, and alternate working arrangements. Here, we explained the important time and work problems to help students prepare for railway examinations.

Alternate Work Problems

Alternate work problems involve individuals performing a task by taking turns, typically for one day each. This pattern repeats until the job is completed. The crucial step is to calculate the work done and time taken for one complete "round" or cycle of turns.

Worked Example 1: P and Q Alternate Work (Q Starts)

Problem: P and Q can complete the work in 9 and 12 days, respectively. How many days will it take to complete the work if they work on alternate days, starting with Q?

Given: P: 9 days, Q: 12 days.

Solution Steps:

1. Total Work (LCM of 9, 12): 36 units.

2. Efficiency: P = 4 units/day (36/9), Q = 3 units/day (36/12).

3. Work per Round (2 days): Q (Day 1) = 3 units, P (Day 2) = 4 units. Total = 7 units in 2 days.

4. Number of Rounds: 36 units / 7 units/round = 5 full rounds (35 units completed).

5. Time for 5 Rounds: 5 rounds * 2 days/round = 10 days.

6. Remaining Work: 36 - 35 = 1 unit.

7. Complete Remaining Work: Q starts the next turn. Q does 1 unit in 1/3 day (1 unit / 3 units/day).

8. Total Time: 10 + 1/3 = 10 1/3 days.

Worked Example 2: P, Q, R Alternate Work

Problem: P, Q, and R can complete a work in 9, 18, and 12 days, respectively. They start the work with P on the first day, Q on the second day, and R on the third day. This cycle continues. How many days will it take to complete the work?

Given: P: 9 days, Q: 18 days, R: 12 days.

Solution Steps:

1. Total Work (LCM of 9, 18, 12): 36 units.

2. Efficiency: P = 4 (36/9), Q = 2 (36/18), R = 3 (36/12) units/day.

3. Work per Round (3 days): P (Day 1) = 4, Q (Day 2) = 2, R (Day 3) = 3. Total = 9 units in 3 days.

4. Number of Rounds: 36 units / 9 units/round = 4 full rounds.

5. Total Time: 4 rounds * 3 days/round = 12 days.

Worked Example 3: P Works Daily, Q and R Assist Every Third Day

Problem: P, Q, and R can complete the work in 12, 24, and 36 days, respectively. P works daily, and Q and R assist him every third day. How many days will it take to complete the work?

Given: P: 12 days, Q: 24 days, R: 36 days.

Solution Steps:

1. Total Work (LCM of 12, 24, 36): 72 units.

2. Efficiency: P = 6, Q = 3, R = 2 units/day.

3. Work per Round (3-day cycle):

• Day 1: P works = 6 units

• Day 2: P works = 6 units

• Day 3: P+Q+R work = 6+3+2 = 11 units

• Total work in 3 days = 6+6+11 = 23 units.

4. Number of Rounds: 72 units / 23 units/round = 3 full rounds (69 units completed).

5. Time for 3 Rounds: 3 rounds * 3 days/round = 9 days.

6. Remaining Work: 72 - 69 = 3 units.

7. Complete Remaining Work: The next day (Day 1 of new cycle), P works. P does 3 units in 1/2 day (3 units / 6 units/day).

8. Total Time: 9 + 1/2 = 9 1/2 days.

Worked Example 4: Ajay and Vijay Alternate Work (Vijay Assists Every Third Day)

Problem: Ajay can complete a job in 60 days. On every third day, he is assisted by Vijay, who can complete the job alone in 84 days. In how many days will the job be completed?

Given: Ajay (A): 60 days, Vijay (V): 84 days.

Solution Steps:

1. Total Work (LCM of 60, 84): 420 units.

2. Efficiency: Ajay = 7 (420/60), Vijay = 5 (420/84) units/day.

3. Work per Round (3-day cycle):

• Day 1: Ajay = 7 units

• Day 2: Ajay = 7 units

• Day 3: Ajay+Vijay = 7+5 = 12 units

• Total work in 3 days = 7+7+12 = 26 units.

4. Number of Rounds: 420 units / 26 units/round = 16 full rounds (416 units completed).

5. Time for 16 Rounds: 16 rounds * 3 days/round = 48 days.

6. Remaining Work: 420 - 416 = 4 units.

7. Complete Remaining Work: The next day (Day 1 of the new cycle), Ajay works. Ajay does 4 units in 4/7 days (4 units / 7 units/day).

8. Total Time: 48 + 4/7 = 48 4/7 days.

Worked Example 5: Sagar and Manish Alternate Work (Manish Alone)

Problem: Sagar and Manish are working on a heavy machine. Sagar can complete the work in 15 days. If both workers work on an alternate basis, they complete the work in 12 days. How many days will Manish alone take to complete the work?

Given: Sagar (S): 15 days. S+M (alternate): 12 days.

Solution Steps:

1. Combined Work Time: If Sagar and Manish complete the work in 12 alternate days, they effectively complete the same work in 6 days if working together simultaneously (12 days / 2).

2. Total Work (LCM of 15, 6): 30 units.

3. Efficiency: Sagar = 2 units/day (30/15). (Sagar+Manish) combined = 5 units/day (30/6).

4. Manish's Efficiency: (S+M) efficiency - Sagar's efficiency = 5 - 2 = 3 units/day.

5. Time for Manish Alone: 30 units / 3 units/day = 10 days.

Worked Example 6: Rathish and Vineet Alternate Work (Find Vineet Alone)

Problem: Rathish and Vineet work on alternate days, starting with Rathish. The entire work is completed in 54 3/4 days. If Rathish alone can finish the work in 48 days, in how many days can Vineet alone do the work?

Given: Rathish (R) alone: 48 days. R+V (alternate, R starts): 54 3/4 days.

Solution Steps (Logical Derivation):

1. Let Rathish's efficiency be R_eff and Vineet's efficiency be V_eff. Total work = 48 * R_eff.

2. In 54 3/4 days: Rathish worked for (54/2) + 3/4 = 27 + 3/4 = 111/4 days. Vineet worked for (54/2) = 27 days.

3. Total work done can be expressed as: (111/4)R_eff + 27V_eff.

4. Equating total work: 48R_eff = (111/4)R_eff + 27V_eff.

5. Rearranging for V_eff: 27V_eff = (48 - 111/4)R_eff = ((192 - 111)/4)R_eff = (81/4)R_eff.

6. Ratio V_eff / R_eff = (81/4) / 27 = 3/4. So, V_eff:R_eff = 3:4.

7. Assume R_eff = 4 units/day. Total work = 48 days * 4 units/day = 192 units.

8. From the ratio, V_eff = 3 units/day.

9. Time for Vineet alone = 192 units / 3 units/day = 64 days.

Efficiency-Based Problems

Key Concepts for Efficiency Problems

• Relationship between Efficiency and Time: Efficiency is inversely proportional to Time. This means if someone is more efficient, they take less time to complete the same work, and vice versa.

• Total Work Formula: Total Work = Efficiency × Time

• Time Formula: Time = Total Work / Efficiency

Worked Example 7: A and B Efficiency (B is 60% More Efficient than A)

Problem: A can complete a job in 32 days. B is 60% more efficient than A. How many days will B alone take to complete the work?

Given: A: 32 days. B is 60% more efficient than A.

1. Efficiency Ratio: Assume A's efficiency = 100 units/day. B's efficiency = 100 + 60 = 160 units/day. Ratio A:B = 100:160 = 5:8.

2. Total Work: A's efficiency × A's time = 5 units/day × 32 days = 160 units.

3. Time for B Alone: 160 units / B's efficiency (8 units/day) = 20 days.

Worked Example 8: A Does Half Work in 3/4 Time of B

Problem: A does half as much work as B in 3/4 of the time taken by B. If both together complete the work in 12 days, how many days will B alone take to complete the work?

Solution Steps:

1. Time Ratio (for full work):

• Let B's time for full work = T.

• A does half work in (3/4)T. Therefore, A does full work in 2 * (3/4)T = (3/2)T.

• Ratio of A:B time = (3/2)T : T = 3:2.

2. Efficiency Ratio (Inverse of Time): A:B efficiency = 2:3.

3. Total Work: Combined efficiency of A and B (2+3=5 units/day) × Combined time (12 days) = 5 × 12 = 60 units.

4. Time for B Alone: 60 units / B's efficiency (3 units/day) = 20 days.

Worked Example 9: X Does Half Work in 1/6 Time of Y

Problem: X does half as much work as Y in 1/6 of the time taken by Y. If both together complete the work in 10 days, how many days will Y alone take to complete the work?

Solution Steps:

1. Time Ratio (for full work):

• Let Y's time for full work = T.

• X does half work in (1/6)T. Therefore, X does full work in 2 * (1/6)T = (1/3)T.

• Ratio of X:Y time = (1/3)T : T = 1:3.

2. Efficiency Ratio (Inverse of Time): X:Y efficiency = 3:1.

3. Total Work: Combined efficiency of X and Y (3+1=4 units/day) × Combined time (10 days) = 4 × 10 = 40 units.

4. Time for Y Alone: 40 units / Y's efficiency (1 unit/day) = 40 days.

Worked Example 10: A, B, C Efficiency (Complex Percentage)

Problem: A is only 40% as efficient as B. C is 50% as efficient as A and B together. C alone can complete the task in 30 days. In how many days will A, B, and C together complete the work?

Solution Steps:

1. Efficiency Ratios:

• Assume B's efficiency = 100 units. Then A's efficiency = 40 units (40% of B).

• Combined (A+B) efficiency = 40 + 100 = 140 units.

• C's efficiency = 50% of 140 = 70 units.

• Final A:B:C efficiency ratio = 40:100:70, simplified to 4:10:7.

2. Total Work: C's efficiency (7 units/day) × C's time (30 days) = 7 × 30 = 210 units.

3. Time for A, B, C Together:

• Combined efficiency of A+B+C = 4 + 10 + 7 = 21 units/day.

• Time = 210 units / 21 units/day = 10 days.

Combined Work with Positive and Negative Contributions

Worked Example 11: Builder and Destroyer

Problem: A builder can build a wall in 20 hours, while a destroyer can demolish such a wall in 50 hours. Both were initially set to work together. After 30 hours, the destroyer was taken out. What is the total time to build the wall?

Given: Builder (B): 20 hours (positive work), Destroyer (D): 50 hours (negative work).

Solution Steps:

1. Total Work (LCM of 20, 50): 100 units.

2. Individual Efficiencies: Builder = +5 units/hour (100/20). Destroyer = -2 units/hour (100/50).

3. Combined Efficiency (Both together): +5 - 2 = +3 units/hour.

4. Work Done in First 30 Hours: 3 units/hour × 30 hours = 90 units.

5. Remaining Work: 100 - 90 = 10 units.

6. Work Done by Builder Alone: After 30 hours, only the builder works. Time taken for the builder to do 10 units = 10 units / 5 units/hour = 2 hours.

7. Total Time: 30 hours (first phase) + 2 hours (second phase) = 32 hours.

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