CAT 2026 Slot 2 Quant TS&W Questions with Solutions | Expected Paper

2 questions: Q3, Q9. Solve each one here: tap the answer, press Check, then open the solution.

Question 3Time, Speed & Distance

MCQ · +3 / −1 · medium

A bus starts at 8 am and follows a fixed route every day. One day, it travels at a constant speed of 60 km/hr and reaches its destination 3 hours later than its scheduled arrival time. The next day, it travels two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km/hr to reach exactly on time. The scheduled arrival time of the bus is:

Answer and solution

Correct answer: B

Solution

Answer: B — 5:00 pm

1. Let the scheduled journey take T hours and let the total distance be D km.
2. On the first day, the bus travels at 60 km per hour and arrives 3 hours late. Therefore,
3. D = 60(T + 3). ...(1)
4. On the second day, two-thirds of the distance is covered in one-third of the scheduled time. The remaining one-third is covered at 40 km per hour.
5. Time taken for the first part = T/3.
Time taken for the remaining part = D/(3 × 40) = D/120.
6. The bus arrives on time, so:
7. T/3 + D/120 = T
8. This gives D = 80T. ...(2)
9. Putting D = 80T in equation (1):
10. 80T = 60(T + 3)
80T = 60T + 180
20T = 180
T = 9 hours.
11. The bus starts at 8 am and the scheduled journey takes 9 hours.
12. Answer: 5:00 pm.

Why this method works:
The second-day remaining distance is D/3 and its available time is 2T/3. At 40 km/h this gives D/3=40(2T/3), hence D=80T. This relation is about the whole scheduled journey, not an average of the two speeds.

Independent check and common trap:
T=9 gives D=720 km. The first-day duration is 720/60=12 hours, exactly 3 hours late. On day two, 480 km takes 3 hours and the final 240 km takes 6 hours. Both descriptions agree; 8 am plus 9 hours is 5 pm.

G Strategy shortcut

Spot: Arithmetic – Speed, Time and Distance.
Apply: Remaining one-third at 40 takes two-thirds of T: D=80T. Also D=60(T+3), so 20T=180 and T=9; arrival 5pm.
Check: T=9 gives D=720 km. The first-day duration is 720/60=12 hours, exactly 3 hours late. On day two, 480 km takes 3 hours and the final 240 km takes 6 hours. Both descriptions agree; 8 am plus 9 hours is 5 pm.

G Strategy: Spot: Arithmetic – Speed, Time and Distance. Apply: Remaining one-third at 40 takes two-thirds of T: D=80T. Also D=60(T+3), so 20T=180 and T=9; arrival 5pm. Check: T=9 gives D=720 km. The first-day duration is 720/60=12 hours, exactly 3 hours late. On day two, 480 km takes 3 hours and the final 240 km takes 6 hours. Both descriptions agree; 8 am plus 9 hours is 5 pm.

Question 9Time & Work

MCQ · +3 / −1 · hard

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.

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.