Breakup | Mixtures + Percentages | HARD | CAT 2025 Slot 1
A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of the resulting solution with water, and finally replaces 10% of this solution with acid, then 15% of the solution obtained with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to
1. 25
2. 27
3. 29
4. 23
Answer
Initial: 200 L solution with 30% acid ⇒ acid = 0.30×200 = 60 L.
Replace 20% with water: remove 40 L (20% of 200). Acid removed = 20% of 60 = 12 L → acid left = 60 − 12 = 48 L. Add 40 L water → volume = 200 L, acid = 48 L.
Replace 10% with acid: remove 20 L. Acid removed = (48/200)×20 = 4.8 L → acid left = 48 − 4.8 = 43.2 L. Add 20 L pure acid → acid = 43.2 + 20 = 63.2 L. Volume = 200 L.
Replace 15% with water: remove 30 L. Acid removed = (63.2/200)×30 = 9.48 L → acid left = 63.2 − 9.48 = 53.72 L. Add 30 L water → acid = 53.72 L, volume = 200 L.
Final percentage acid = (53.72/200)×100 = 26.86% ≈ 26.9% → nearest = 27%.
Answer: 27 (option 2).
Breakup G Strategy Time Speed + APGP | MEDIUM | CAT 2025 Slot 1
Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is
- 86400
- 96000
- 112000
- 76800
Answer
Work in minutes and meters: total time = 3 hr = 180 min, total distance = 224 km = 224000 m.
Times are 30,40,50,60 (sum 180 using AP). If speeds are v, v+d, v+2d, v+3d then total distance = v·180 + d·(40 + 2·50 + 3·60) = v·180 + d·320. With v = 960 m/min, v·180 = 960×180 = 172800, so the extra distance to be made up by the d-terms = 224000 − 172800 = 51200.
So d = 51200/320 = 160 m/min ⇒ v4 = 960 + 3·160 = 1440 m/min ⇒ distance in 4th part = 1440 × 60 = 86,400 m.
Breakup G Strategy | Time & Work | HARD | CAT 2025 Slot 1
Arun, Varun and Tarun, if working alone, can complete a task in 24, 21, and 15 days, respectively. They charge Rs 2160, Rs 2400, and Rs 2160 per day, respectively, even if they are employed for a partial day. On any given day, any workers may or may not be employed to work. If the task needs to be completed in 10 days or less, then the minimum possible amount, in rupees, required to be paid for the entire task is
1. 38400
2. 38880
3. 34400
4. 47040
Answer
Work rates (job/day): Arun = 1/24, Varun = 1/21, Tarun = 1/15.
Daily charges: Arun = 2160, Varun = 2400, Tarun = 2160.
Cost per unit work = daily charge ÷ (work per day) = daily_charge × time:
Arun = 2160×24 = 51,840; Varun = 2400×21 = 50,400; Tarun = 2160×15 = 32,400.
So Tarun is cheapest per unit, then Varun, Arun is most expensive.
Use Tarun for all 10 calendar days to minimize cost: work done = 10/15 = 2/3.
Remaining work = 1 − 2/3 = 1/3. Fill with next cheapest (Varun).
Days of Varun needed: smallest integer d with d/21 ≥ 1/3 ⇒ d ≥ 7. So take d = 7.
Check total work: 10/15 + 7/21 = 2/3 + 1/3 = 1 (done within 10 days since Varun can work on 7 of those 10 days).
Total cost = Tarun: 10×2160 = 21,600; Varun: 7×2400 = 16,800.
Sum = 21,600 + 16,800 = 38,400 rupees.
Hence minimum possible payment = Rs 38,400 (Option 1).
