Aman and Varun together can complete a task in 120 days, while Varun and Sameer together can complete the same task in 80 days. Aman starts working on the task and works for 20 days. Then Varun takes over and works for 120 days. Finally, Sameer takes over and completes the remaining task in 20 days. If instead Aman had worked on every day, Varun on every alternate day and Sameer on every third day, with all three starting on the first day, then on which numbered day would the task be completed?
Answer and solution
Correct answer: C
Solution
Answer: C — Day 155
1. Take the total work as 1.
2. Aman and Varun together finish 1/120 of the work in a day.
Varun and Sameer together finish 1/80 of the work in a day.
3. From the first condition:
4. A + V = 1/120
5. From the second condition:
6. V + S = 1/80
7. In the first arrangement, Aman works for 20 days, Varun for 120 days and Sameer for 20 days. So:
8. 20A + 120V + 20S = 1
9. Solving these three relations gives:
10. Aman's one-day work = 1/960
Varun's one-day work = 7/960
Sameer's one-day work = 5/960
11. Now consider the new arrangement. In every 6 days:
Aman works all 6 days.
Varun works on days 1, 3 and 5.
Sameer works on days 1 and 4.
12. Work done in 6 days:
13. Aman: 6/960
Varun: 21/960
Sameer: 10/960
14. Total = 37/960.
15. In 25 such cycles, or 150 days, the work completed is 925/960.
16. The remaining work is 35/960.
17. During the next five days, the work done is:
Day 151: 13/960
Day 152: 1/960
Day 153: 8/960
Day 154: 6/960
Day 155: 8/960
18. Total work in these five days = 36/960.
19. The remaining work is completed on the 155th day.
20. Answer: 155 days.
21. Precision check: by the end of day 154, 953/960 is complete. Remaining work is 7/960; on day 155 Aman and Varun work at 8/960 per day. They need 7/8 day. Exact elapsed time is 154⅞ days, so the job finishes during day 155.
Why this method works:
The six-day cycle is the least common multiple of the two attendance periods, 2 and 3. Because everyone starts on day 1, Varun’s days are 1,3,5 and Sameer’s are 1,4. Starting either schedule a day late would change the result.
Independent check and common trap:
Daily rates can be checked directly: A+V=8/960=1/120, V+S=12/960=1/80, and 20A+120V+20S=(20+840+100)/960=1. Use whole cycles first, then daily work to distinguish elapsed time from the numbered completion day.
Apply: Use 960 work units: A, V, S rates 1,7,5. Six-day cycle 37. After 150 days 925 done; next four days 28; 7 remain on day 155 at rate 8 →154⅞ days, completion during day 155.
Check: Daily rates can be checked directly: A+V=8/960=1/120, V+S=12/960=1/80, and 20A+120V+20S=(20+840+100)/960=1. Use whole cycles first, then daily work to distinguish elapsed time from the numbered completion day.
1. Take the total work as 1.
2. Aman and Varun together finish 1/120 of the work in a day.
Varun and Sameer together finish 1/80 of the work in a day.
3. From the first condition:
4. A + V = 1/120
5. From the second condition:
6. V + S = 1/80
7. In the first arrangement, Aman works for 20 days, Varun for 120 days and Sameer for 20 days. So:
8. 20A + 120V + 20S = 1
9. Solving these three relations gives:
10. Aman's one-day work = 1/960
Varun's one-day work = 7/960
Sameer's one-day work = 5/960
11. Now consider the new arrangement. In every 6 days:
Aman works all 6 days.
Varun works on days 1, 3 and 5.
Sameer works on days 1 and 4.
12. Work done in 6 days:
13. Aman: 6/960
Varun: 21/960
Sameer: 10/960
14. Total = 37/960.
15. In 25 such cycles, or 150 days, the work completed is 925/960.
16. The remaining work is 35/960.
17. During the next five days, the work done is:
Day 151: 13/960
Day 152: 1/960
Day 153: 8/960
Day 154: 6/960
Day 155: 8/960
18. Total work in these five days = 36/960.
19. The remaining work is completed on the 155th day.
20. Answer: 155 days.
21. Precision check: by the end of day 154, 953/960 is complete. Remaining work is 7/960; on day 155 Aman and Varun work at 8/960 per day. They need 7/8 day. Exact elapsed time is 154⅞ days, so the job finishes during day 155.
Why this method works:
The six-day cycle is the least common multiple of the two attendance periods, 2 and 3. Because everyone starts on day 1, Varun’s days are 1,3,5 and Sameer’s are 1,4. Starting either schedule a day late would change the result.
Independent check and common trap:
Daily rates can be checked directly: A+V=8/960=1/120, V+S=12/960=1/80, and 20A+120V+20S=(20+840+100)/960=1. Use whole cycles first, then daily work to distinguish elapsed time from the numbered completion day.
G Strategy shortcut
Spot: Time and Work – Combined Efficiency.Apply: Use 960 work units: A, V, S rates 1,7,5. Six-day cycle 37. After 150 days 925 done; next four days 28; 7 remain on day 155 at rate 8 →154⅞ days, completion during day 155.
Check: Daily rates can be checked directly: A+V=8/960=1/120, V+S=12/960=1/80, and 20A+120V+20S=(20+840+100)/960=1. Use whole cycles first, then daily work to distinguish elapsed time from the numbered completion day.
G Strategy: Spot: Time and Work – Combined Efficiency. Apply: Use 960 work units: A, V, S rates 1,7,5. Six-day cycle 37. After 150 days 925 done; next four days 28; 7 remain on day 155 at rate 8 →154⅞ days, completion during day 155. Check: Daily rates can be checked directly: A+V=8/960=1/120, V+S=12/960=1/80, and 20A+120V+20S=(20+840+100)/960=1. Use whole cycles first, then daily work to distinguish elapsed time from the numbered completion day.