CRACKTARGET • EXAM PRACTICE • 2026
Time and work plus speed-distance: a formula sheet and 30 original MCQs with worked solutions, assumptions and common traps.
Choose a rate model before choosing a shortcut
Time-and-work and speed-distance questions become easier when you define one unit consistently. In work questions, rate means the fraction of a job completed per day or per hour. In movement questions, speed means distance covered per unit of time. Write the unit beside every quantity. Shortcuts fail when a worker changes efficiency, a pipe empties instead of fills, or a traveller changes direction. These thirty questions use explicit assumptions so that each answer can be checked.
Attempt all three variants in a topic before reading the explanations. The first variant establishes the method; the next two test whether you can apply it to different figures. After finishing, mix the question order. A sectional heading reveals the method, whereas the examination will require you to select it. Round only at the end unless the question explicitly asks for an approximation.
Formula sheet with conditions
| Situation | Formula and condition |
|---|---|
| Combined work | If A and B finish alone in a and b days, combined time = ab/(a+b), assuming constant independent rates. |
| Work remaining | Remaining fraction = 1 − rate × elapsed time; use only the rate actually operating in that interval. |
| Efficiency | For the same job, time is inversely proportional to efficiency. |
| Pipes | Add filling rates and subtract emptying rates; a non-positive net rate cannot fill an empty tank. |
| Distance | Distance = speed × time, with consistent units. |
| Conversion | km/h × 5/18 = m/s; m/s × 18/5 = km/h. |
| Equal distances | Average speed = 2uv/(u+v); this is not the formula for equal times. |
| Relative speed | Opposite directions: add speeds. Same direction: subtract speeds. |
| Trains | Passing a pole uses train length; passing a platform uses train length plus platform length. |
| Boats | Downstream = still-water speed + stream speed; upstream = still-water speed − stream speed. |
Set 1: combined work
Question 1. A can finish a job in 12 days and B in 18 days. Both work at constant rates. How many days do they need together?
A. 7.2 · B. 14.4 · C. 3.6 · D. 10.2
Answer and explanation 1
Answer: A (7.2). A completes 1/12 and B completes 1/18 of the job per day. Add these daily fractions, then invert the total rate. Combined time = 12 × 18/(12 + 18). Trap: Adding completion times treats simultaneous work as consecutive work.
Question 2. A can finish a job in 20 days and B in 30 days. Both work at constant rates. How many days do they need together?
A. 24 · B. 12 · C. 6 · D. 15
Answer and explanation 2
Answer: B (12). A completes 1/20 and B completes 1/30 of the job per day. Add these daily fractions, then invert the total rate. Combined time = 20 × 30/(20 + 30). Trap: Adding completion times treats simultaneous work as consecutive work.
Question 3. A can finish a job in 15 days and B in 10 days. Both work at constant rates. How many days do they need together?
A. 12 · B. 3 · C. 6 · D. 9
Answer and explanation 3
Answer: C (6). A completes 1/15 and B completes 1/10 of the job per day. Add these daily fractions, then invert the total rate. Combined time = 15 × 10/(15 + 10). Trap: Adding completion times treats simultaneous work as consecutive work.
Set 2: work already completed
Question 4. A alone finishes a job in 12 days and works for 3 days. B, who could finish the whole job alone in 6 days, completes the remainder. How many days does B work?
A. 9 · B. 2.25 · C. 7.5 · D. 4.5
Answer and explanation 4
Answer: D (4.5). A completes 3/12 of the job. Subtract this fraction from one, then multiply by B’s whole-job time of 6 days. B is responsible only for the remaining fraction, not for the original entire job. Trap: Do not add A’s elapsed time when asked only for B’s working duration.
Question 5. A alone finishes a job in 20 days and works for 4 days. B, who could finish the whole job alone in 10 days, completes the remainder. How many days does B work?
A. 8 · B. 16 · C. 4 · D. 11
Answer and explanation 5
Answer: A (8). A completes 4/20 of the job. Subtract this fraction from one, then multiply by B’s whole-job time of 10 days. B is responsible only for the remaining fraction, not for the original entire job. Trap: Do not add A’s elapsed time when asked only for B’s working duration.
Question 6. A alone finishes a job in 15 days and works for 6 days. B, who could finish the whole job alone in 5 days, completes the remainder. How many days does B work?
A. 6 · B. 3 · C. 1.5 · D. 6
Answer and explanation 6
Answer: B (3). A completes 6/15 of the job. Subtract this fraction from one, then multiply by B’s whole-job time of 5 days. B is responsible only for the remaining fraction, not for the original entire job. Trap: Do not add A’s elapsed time when asked only for B’s working duration.
Set 3: efficiency and time
Question 7. A is 2 times as efficient as B. B completes a job in 18 days. How many days will A require at a constant rate?
A. 18 · B. 4.5 · C. 9 · D. 12
Answer and explanation 7
Answer: C (9). Efficiency measures output per unit time. For a fixed quantity of work, multiply efficiency by time to obtain the same total work. Therefore A’s required time is 18/2 days. Trap: Greater efficiency means less time, so multiplying the time by the efficiency factor reverses the relationship.
Question 8. A is 3 times as efficient as B. B completes a job in 24 days. How many days will A require at a constant rate?
A. 16 · B. 4 · C. 11 · D. 8
Answer and explanation 8
Answer: D (8). Efficiency measures output per unit time. For a fixed quantity of work, multiply efficiency by time to obtain the same total work. Therefore A’s required time is 24/3 days. Trap: Greater efficiency means less time, so multiplying the time by the efficiency factor reverses the relationship.
Question 9. A is 4 times as efficient as B. B completes a job in 40 days. How many days will A require at a constant rate?
A. 10 · B. 20 · C. 5 · D. 13
Answer and explanation 9
Answer: A (10). Efficiency measures output per unit time. For a fixed quantity of work, multiply efficiency by time to obtain the same total work. Therefore A’s required time is 40/4 days. Trap: Greater efficiency means less time, so multiplying the time by the efficiency factor reverses the relationship.
Set 4: filling and emptying pipes
Question 10. A filling pipe fills a tank in 6 hours, while a drain empties a full tank in 12 hours. Both operate at constant rates. How many hours are needed to fill an empty tank?
A. 24 · B. 12 · C. 6 · D. 15
Answer and explanation 10
Answer: B (12). Net filling rate = 1/6 − 1/12 tank per hour. The positive net rate is inverted to obtain the filling time. The drain reduces the useful output of the inlet. Trap: Adding both rates would incorrectly treat an outlet as a second inlet.
Question 11. A filling pipe fills a tank in 8 hours, while a drain empties a full tank in 24 hours. Both operate at constant rates. How many hours are needed to fill an empty tank?
A. 24 · B. 6 · C. 12 · D. 15
Answer and explanation 11
Answer: C (12). Net filling rate = 1/8 − 1/24 tank per hour. The positive net rate is inverted to obtain the filling time. The drain reduces the useful output of the inlet. Trap: Adding both rates would incorrectly treat an outlet as a second inlet.
Question 12. A filling pipe fills a tank in 10 hours, while a drain empties a full tank in 15 hours. Both operate at constant rates. How many hours are needed to fill an empty tank?
A. 60 · B. 15 · C. 33 · D. 30
Answer and explanation 12
Answer: D (30). Net filling rate = 1/10 − 1/15 tank per hour. The positive net rate is inverted to obtain the filling time. The drain reduces the useful output of the inlet. Trap: Adding both rates would incorrectly treat an outlet as a second inlet.
Set 5: wages proportional to work
Question 13. A and B work equal hours. A’s efficiency is 2 times B’s. Total remuneration of 900 is divided strictly in proportion to work done. What amount does A receive?
A. 600 · B. 1200 · C. 300 · D. 603
Answer and explanation 13
Answer: A (600). For equal working time, contributions are in the efficiency ratio 2:1. There are 3 total ratio units. A’s share is 900 × 2/3. Trap: Equal hours do not imply equal output when efficiency differs; the payment rule is expressly contribution-based.
Question 14. A and B work equal hours. A’s efficiency is 3 times B’s. Total remuneration of 1200 is divided strictly in proportion to work done. What amount does A receive?
A. 1800 · B. 900 · C. 450 · D. 903
Answer and explanation 14
Answer: B (900). For equal working time, contributions are in the efficiency ratio 3:1. There are 4 total ratio units. A’s share is 1200 × 3/4. Trap: Equal hours do not imply equal output when efficiency differs; the payment rule is expressly contribution-based.
Question 15. A and B work equal hours. A’s efficiency is 4 times B’s. Total remuneration of 1500 is divided strictly in proportion to work done. What amount does A receive?
A. 2400 · B. 600 · C. 1200 · D. 1203
Answer and explanation 15
Answer: C (1200). For equal working time, contributions are in the efficiency ratio 4:1. There are 5 total ratio units. A’s share is 1500 × 4/5. Trap: Equal hours do not imply equal output when efficiency differs; the payment rule is expressly contribution-based.
Set 6: equal-distance average speed
Question 16. A traveller covers the same distance at 30 km/h and then at 60 km/h. What is the average speed in km/h over both legs?
A. 80 · B. 20 · C. 43 · D. 40
Answer and explanation 16
Answer: D (40). Let each leg have distance d. Total distance is 2d and total time is d/30 + d/60. Divide total distance by total time to obtain 2 × 30 × 60/(30 + 60). Trap: The arithmetic mean is valid for equal times, not for equal distances.
Question 17. A traveller covers the same distance at 40 km/h and then at 60 km/h. What is the average speed in km/h over both legs?
A. 48 · B. 96 · C. 24 · D. 51
Answer and explanation 17
Answer: A (48). Let each leg have distance d. Total distance is 2d and total time is d/40 + d/60. Divide total distance by total time to obtain 2 × 40 × 60/(40 + 60). Trap: The arithmetic mean is valid for equal times, not for equal distances.
Question 18. A traveller covers the same distance at 20 km/h and then at 30 km/h. What is the average speed in km/h over both legs?
A. 48 · B. 24 · C. 12 · D. 27
Answer and explanation 18
Answer: B (24). Let each leg have distance d. Total distance is 2d and total time is d/20 + d/30. Divide total distance by total time to obtain 2 × 20 × 30/(20 + 30). Trap: The arithmetic mean is valid for equal times, not for equal distances.
Set 7: train passing a pole
Question 19. A train 150 metres long travels at 54 km/h. How many seconds does it take to pass a stationary pole?
A. 20 · B. 5 · C. 10 · D. 13
Answer and explanation 19
Answer: C (10). Convert speed into metres per second: 54 × 5/18. The entire train length must pass the pole, so divide 150 metres by this converted speed. The result is measured in seconds. Trap: A pole has negligible length; do not add a platform distance that the question does not supply.
Question 20. A train 200 metres long travels at 72 km/h. How many seconds does it take to pass a stationary pole?
A. 20 · B. 5 · C. 13 · D. 10
Answer and explanation 20
Answer: D (10). Convert speed into metres per second: 72 × 5/18. The entire train length must pass the pole, so divide 200 metres by this converted speed. The result is measured in seconds. Trap: A pole has negligible length; do not add a platform distance that the question does not supply.
Question 21. A train 300 metres long travels at 90 km/h. How many seconds does it take to pass a stationary pole?
A. 12 · B. 24 · C. 6 · D. 15
Answer and explanation 21
Answer: A (12). Convert speed into metres per second: 90 × 5/18. The entire train length must pass the pole, so divide 300 metres by this converted speed. The result is measured in seconds. Trap: A pole has negligible length; do not add a platform distance that the question does not supply.
Set 8: train passing a platform
Question 22. A train is 150 metres long and crosses a 250 metre platform at 72 km/h. How many seconds are needed to clear the platform completely?
A. 40 · B. 20 · C. 10 · D. 23
Answer and explanation 22
Answer: B (20). The front of the train must travel train length plus platform length before the rear clears the platform. Required distance is 400 metres. Divide by 72 × 5/18 metres per second. Trap: Using only platform length measures the wrong interval and understates complete clearance time.
Question 23. A train is 200 metres long and crosses a 400 metre platform at 90 km/h. How many seconds are needed to clear the platform completely?
A. 48 · B. 12 · C. 24 · D. 27
Answer and explanation 23
Answer: C (24). The front of the train must travel train length plus platform length before the rear clears the platform. Required distance is 600 metres. Divide by 90 × 5/18 metres per second. Trap: Using only platform length measures the wrong interval and understates complete clearance time.
Question 24. A train is 180 metres long and crosses a 120 metre platform at 54 km/h. How many seconds are needed to clear the platform completely?
A. 40 · B. 10 · C. 23 · D. 20
Answer and explanation 24
Answer: D (20). The front of the train must travel train length plus platform length before the rear clears the platform. Required distance is 300 metres. Divide by 54 × 5/18 metres per second. Trap: Using only platform length measures the wrong interval and understates complete clearance time.
Set 9: opposite-direction meeting
Question 25. Two vehicles start simultaneously 180 km apart and travel towards each other at 40 and 50 km/h. After how many hours do they meet?
A. 2 · B. 4 · C. 1 · D. 5
Answer and explanation 25
Answer: A (2). The gap closes at 90 km/h because both vehicles reduce the distance between them. Meeting time = 180/(40 + 50). Assume constant speed, the same route and no waiting periods. Trap: Subtracting speeds would be appropriate for some same-direction catch-up questions, not for this approach.
Question 26. Two vehicles start simultaneously 240 km apart and travel towards each other at 60 and 60 km/h. After how many hours do they meet?
A. 4 · B. 2 · C. 1 · D. 5
Answer and explanation 26
Answer: B (2). The gap closes at 120 km/h because both vehicles reduce the distance between them. Meeting time = 240/(60 + 60). Assume constant speed, the same route and no waiting periods. Trap: Subtracting speeds would be appropriate for some same-direction catch-up questions, not for this approach.
Question 27. Two vehicles start simultaneously 300 km apart and travel towards each other at 70 and 80 km/h. After how many hours do they meet?
A. 4 · B. 1 · C. 2 · D. 5
Answer and explanation 27
Answer: C (2). The gap closes at 150 km/h because both vehicles reduce the distance between them. Meeting time = 300/(70 + 80). Assume constant speed, the same route and no waiting periods. Trap: Subtracting speeds would be appropriate for some same-direction catch-up questions, not for this approach.
Set 10: boats and streams
Question 28. A boat’s downstream speed is 18 km/h and upstream speed is 10 km/h. What is the speed of the stream in km/h?
A. 8 · B. 2 · C. 7 · D. 4
Answer and explanation 28
Answer: D (4). Downstream speed adds the stream to still-water speed, while upstream speed subtracts it. Subtract the two equations: twice the stream speed = 18 − 10. Divide the difference by two. Trap: The average of upstream and downstream speeds gives the boat’s still-water speed, not the stream speed.
Question 29. A boat’s downstream speed is 24 km/h and upstream speed is 16 km/h. What is the speed of the stream in km/h?
A. 4 · B. 8 · C. 2 · D. 7
Answer and explanation 29
Answer: A (4). Downstream speed adds the stream to still-water speed, while upstream speed subtracts it. Subtract the two equations: twice the stream speed = 24 − 16. Divide the difference by two. Trap: The average of upstream and downstream speeds gives the boat’s still-water speed, not the stream speed.
Question 30. A boat’s downstream speed is 30 km/h and upstream speed is 18 km/h. What is the speed of the stream in km/h?
A. 12 · B. 6 · C. 3 · D. 9
Answer and explanation 30
Answer: B (6). Downstream speed adds the stream to still-water speed, while upstream speed subtracts it. Subtract the two equations: twice the stream speed = 30 − 18. Divide the difference by two. Trap: The average of upstream and downstream speeds gives the boat’s still-water speed, not the stream speed.
Check the assumptions behind every shortcut
In a work question, confirm that all workers are doing the same job and that their rates remain constant. If someone joins late, leaves early or works on alternate days, split the timeline. If the question includes an efficiency increase, apply it to the rate rather than directly to completion time. For wages, identify whether payment follows time, output or a stated agreement; do not import a contribution rule into a question that uses another basis.
In a movement question, sketch the sequence briefly before calculating. Identify whether the distance between objects is increasing or decreasing. Convert minutes to hours before using kilometres per hour, or convert speed to metres per second when lengths are in metres. A result should pass a rough reasonableness check: two workers together should finish faster than either alone, and a drain should make tank filling slower than the inlet acting alone.
A five-session revision routine
Session one: solve the ten formula types without a time limit and explain each denominator. Session two: repeat only incorrect questions with answers hidden. Session three: mix fifteen questions and record time per question. Session four: practise unfamiliar wording without changing the underlying model. Session five: take a mixed mock and review whether you selected the correct method before doing arithmetic. This sequence builds both recognition and transfer.
Record errors as model, unit, arithmetic or reading mistakes. A unit error requires conversion practice; a model error requires rebuilding the equation. Copying the final number does neither. For a correct but slow solution, compare your setup with the shortest valid method and keep the reasoning that makes the shortcut safe. Speed should follow understanding rather than replace it.
Continue your preparation
Quantitative aptitude formula sheet · Mock-test analysis guide
Sources and editorial notes
Practice questions and worked examples are prepared by CrackTarget. They are not claimed as previous-year questions unless an official paper and question reference are explicitly supplied. Use the recruitment notice applicable to your post for the final syllabus and examination instructions.