Arithmetic: percentage, ratio, averages, time & work

Answer: (d) निर्धारित नहीं किया जा सकता

The '12 overs' condition follows automatically from the first one: a run rate 1.5 times N/18 is N/12, which reaches N in exactly 12 overs whatever N is. With no independent equation, N cannot be found.

1. 10% fewer than 20 overs is 18 overs, so team Y's run rate was N/18 runs per over.
2. A 50% higher run rate is 1.5 × N/18 = N/12 runs per over.
3. At N/12 runs per over, N runs take N ÷ (N/12) = 12 overs — for every value of N.
4. The last condition is always true and gives no new information, so N cannot be determined.
5. Check: N = 72 and N = 144 both satisfy every condition, so the data cannot pick one value.

Remember · Test whether a second condition is independent; if it reduces to an identity, the value cannot be determined.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) p(10⁴ − k²)² = q × 10⁸

Each k% rise followed by a k% fall multiplies the price by (100 + k)(100 − k)/10⁴ = (10⁴ − k²)/10⁴. Two such pairs give q = p(10⁴ − k²)²/10⁸, i.e. p(10⁴ − k²)² = q × 10⁸.

1. A k% increase multiplies the price by (100 + k)/100; a k% decrease multiplies it by (100 − k)/100.
2. One increase and one decrease: (100 + k)(100 − k)/10⁴ = (10⁴ − k²)/10⁴.
3. This pair happens twice, so q = p × (10⁴ − k²)²/10⁸.
4. Cross-multiplying: p(10⁴ − k²)² = q × 10⁸.
5. Check with k = 10: q = p × 1.1 × 0.9 × 1.1 × 0.9 = 0.9801p, and (10⁴ − 100)²/10⁸ = 9900²/10⁸ = 0.9801.

Remember · Successive percentage changes multiply. A k% rise and k% fall together leave the factor (10⁴ − k²)/10⁴ — always a net loss.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 200 km

With a fixed budget, the distance he can drive is inversely proportional to the petrol price. A 10% price rise brings the distance down to 100/110 of 2200 km, i.e. 2000 km, so he must travel 200 km less.

1. New price = ₹90 × 1.1 = ₹99 per litre.
2. Petrol used now: 2200 ÷ 16 = 137.5 litres, costing 137.5 × 90 = ₹12,375.
3. At ₹99 a litre, ₹12,375 buys 12,375 ÷ 99 = 125 litres, enough for 125 × 16 = 2000 km.
4. Reduction = 2200 − 2000 = 200 km.
5. Check (faster route): distance falls in the ratio 100 : 110, so 2200 × 100/110 = 2000 km; the mileage figure is not even needed.

Remember · Fixed spending: quantity varies inversely with price. A p% price rise cuts quantity by p/(100 + p) — here 10/110 = 1/11.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 13

Why not the tempting option · UPSC's key is 13. Counting the meeting exactly at 7:00 a.m. would give 14, but 'between 5:20 a.m. and 7:00 a.m.' excludes the end points, and 5:20 itself is not a meeting time. In the exam, read 'between' as excluding the end points, as UPSC does.

Walking in opposite directions, together they cover 8 rounds an hour, so they meet every 7.5 minutes. Between 5:20 and 7:00 they meet at 5:22:30, 5:30, … 6:52:30 — 13 times; the meeting exactly at 7:00 falls on the end point and is not 'between' the two times.

1. In opposite directions, they close the gap at 5 + 3 = 8 rounds per hour, so they meet once every 1/8 hour = 7.5 minutes.
2. Meetings happen 7.5, 15, 22.5, 30, … minutes after 5:00 a.m.
3. 5:20 is 20 minutes and 7:00 is 120 minutes after 5:00. The first meeting after 5:20 is at 22.5 minutes (5:22:30).
4. Meetings from 22.5 to 112.5 minutes: (112.5 − 22.5) ÷ 7.5 + 1 = 13. The next one, at 120 minutes, is exactly 7:00 a.m. — the end point, not between the two times.
5. So they cross each other 13 times.

Remember · Opposite directions on a circular track: meetings per hour = sum of rounds per hour. Then check the end points: 'between' excludes them.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 1 Oct 2026 (how we verify). Permalink · Report an error

Answer: (c) 20 m

To pass a walker, the tram covers its own length at the relative speed, so (v − 3) × 8 = (v − 4) × 9, giving v = 12 km/h. Relative to X it moves at 9 km/h = 2.5 m/s, so in 8 seconds it covers 20 m, which is its length.

1. Let the tram's speed be v km/h. Passing a walker completely means covering the tram's length at the relative speed.
2. Length = (v − 3) × 8 = (v − 4) × 9 (same units), so 8v − 24 = 9v − 36 and v = 12 km/h.
3. Relative speed past X = 12 − 3 = 9 km/h = 9 × 5/18 = 2.5 m/s.
4. Length = 2.5 m/s × 8 s = 20 m.
5. Check with Y: (12 − 4) × 5/18 = 20/9 m/s, and 20/9 × 9 s = 20 m.

Remember · Overtaking a moving person: length = speed difference × time; equate two such lengths to find the speed. 1 km/h = 5/18 m/s.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 16 मिनट

Each X-pipe fills 1/400, each Y-pipe fills 1/160 and each Z-pipe empties 1/640 of the tank per minute. With 10 X-pipes, 5 Y-pipes and 16 Z-pipes open, the X and Z effects cancel (1/40 each), leaving 1/32 per minute, so half the tank takes 16 minutes.

1. Take the tank as 1. Set X: 20 pipes fill 0.7 in 14 minutes, so one pipe fills 0.7 ÷ (14 × 20) = 1/400 per minute.
2. Set Y: 10 pipes fill 3/8 in 6 minutes, so one pipe fills (3/8) ÷ 60 = 1/160 per minute.
3. Set Z: 16 pipes empty 1/2 in 20 minutes, so one pipe empties (1/2) ÷ 320 = 1/640 per minute.
4. Open pipes: 10 of X fill 10/400 = 1/40; 5 of Y fill 5/160 = 1/32; all 16 of Z empty 16/640 = 1/40 per minute.
5. Net rate = 1/40 + 1/32 − 1/40 = 1/32 of the tank per minute.
6. Time for 50% = (1/2) ÷ (1/32) = 16 minutes.

Remember · With sets of pipes, find the rate of one pipe first, then multiply by the number open; watch for rates that cancel.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 4

Alone, X, Y and Z would take 18, 24 and 16 days, i.e. 8, 6 and 9 units a day out of 144 units. Y's last 8 2/3 days cover 52 units, leaving 92 units done together at 23 units a day, so n = 4.

1. Full-work times: X = 6 × 3 = 18 days, Y = 8 × 3 = 24 days, Z = 12 × 4/3 = 16 days.
2. Take the work as 144 units (LCM of 18, 24, 16). Daily work: X = 8, Y = 6, Z = 9 units; together 23 units a day.
3. Y alone for 8 2/3 = 26/3 days does 6 × 26/3 = 52 units.
4. Work done by all three together = 144 − 52 = 92 units, so n = 92 ÷ 23 = 4 days.
5. Check: 4 × 23 + 52 = 92 + 52 = 144.

Remember · Turn partial-work data into full-work days, take the LCM as total units, then work in units per day.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) अपर्याप्त आँकड़ों के कारण निर्धारित नहीं किया जा सकता

The condition only says X, Y and Z are in geometric progression with a total of 37. Both 16, 12, 9 and 9, 12, 16 satisfy it, giving opposite orders, so the comparison cannot be determined.

1. X : Y = Y : Z means Y² = X × Z, with X + Y + Z = 37.
2. Try Y = 12: then X × Z = 144 and X + Z = 25, so X and Z are 16 and 9.
3. Both (X, Y, Z) = (16, 12, 9) and (9, 12, 16) fit: 16 × 9 = 144 = 12², and the total is 37.
4. The first gives Value-III < Value-II < Value-I; the second gives Value-I < Value-II < Value-III.
5. The order cannot be fixed from the data.

Remember · For a 'cannot be determined' option, try to build two valid cases with different orders; one such pair settles it.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) k

p is above the average by exactly the amount q is below it, so p and q together total 2k. Since all three total 3k, r must equal k.

1. Let p = k + d. Then q is below k by the same amount: q = k − d.
2. p + q = 2k, and p + q + r = 3k because the average is k.
3. So r = 3k − 2k = k.
4. Check: p = 7, q = 3, r = 5 has average 5 and r equals the average.

Remember · Deviations from an average always sum to zero; equal and opposite deviations leave the third number at the average.

Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error