Arithmetic: percentage, ratio, averages, time & work

Answer: (d) 1, 2 and 3

Let a, b, c be the numbers of ₹50, ₹20 and ₹10 notes. The two totals reduce to 4a + b = 134, and none of the three statements gives whole, non-negative numbers of notes, so all three are incorrect.

1. Let the numbers of ₹50, ₹20 and ₹10 notes be a, b and c. Then a + b + c = 50 and 50a + 20b + 10c = 1840, that is 5a + 2b + c = 184.
2. Subtract the first equation from the second: 4a + b = 134.
3. Statement 1: a = 25 gives b = 134 − 100 = 34, and then c = 50 − 25 − 34 = −9. Impossible.
4. Statement 2: b = 35 gives 4a = 99, which is not a whole number. Impossible.
5. Statement 3: c = 20 gives a + b = 30; with 4a + b = 134 this means 3a = 104, not a whole number. Impossible.
6. Check: a valid payment does exist, e.g. a = 30, b = 14, c = 6 — that is 50 notes and 1500 + 280 + 60 = ₹1,840 — so the bill itself is fine; only the three statements fail.

• ✗ 1. a = 25 forces c = −9, a negative number of ₹10 notes.
• ✗ 2. b = 35 forces 4a = 99, so a is not a whole number.
• ✗ 3. c = 20 forces 3a = 104, so a is not a whole number.

Remember · With a count total and a value total, subtract to get one clean equation, then test each statement for whole, non-negative answers.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 3

By the time X finishes 3 km, Y has run 2 km, so X has gained 1000 m — three full laps of 300 m and a little more. Each full lap gained is one pass, so X passes Y 3 times.

1. The race ends when X, the faster runner, completes 3 km = 3000 m.
2. Speeds are 3 : 2, so in that time Y runs 2/3 × 3000 = 2000 m.
3. X gains 3000 − 2000 = 1000 m on Y.
4. X passes Y each time the lead reaches a whole lap: at 300 m, 600 m and 900 m. 1000 ÷ 300 = 3.33, so there are 3 passes.

Remember · On a circular track, passes = lead gained by the faster runner ÷ track length, counting whole laps only.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 10%

A 25% rise followed by a 20% fall brings the price exactly back to where it started (1.25 × 0.8 = 1). Only the last 10% rise remains.

1. Let the price be ₹100. After a 25% increase it is ₹125.
2. A 20% decrease: 125 × 0.8 = ₹100.
3. A 10% increase: 100 × 1.1 = ₹110.
4. The resultant increase is ₹10 on ₹100, i.e. 10%.

Remember · Successive percentage changes multiply: +25% then −20% cancel, because 5/4 × 4/5 = 1.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 40.4 m

With plants at both ends of the road, 101 plants make 100 equal gaps along 1010 m, so each gap is 10.1 m. Five consecutive plants span 4 gaps: 4 × 10.1 = 40.4 m.

1. 101 plants in a line from one end of the road to the other leave 101 − 1 = 100 gaps.
2. Road length = 1.01 km = 1010 m, so each gap = 1010 ÷ 100 = 10.1 m.
3. 5 consecutive plants have 4 gaps between them: 4 × 10.1 = 40.4 m.
4. Trap: dividing 1010 by 101 (counting plants instead of gaps) gives 10 m and the wrong answer 40 m.

Remember · n objects in a line make n − 1 gaps; k consecutive objects span k − 1 gaps.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 52

Work backwards from D’s 10 coins. Each giver handed over half his coins plus 2, so he had 2 × (coins given − 2): C had 16, B had 28 and A had 52.

1. D now has 10 coins (the smallest two-digit number), all received from C.
2. C gave half his coins + 2 = 10, so half = 8 and C had 16 — all received from B.
3. B gave half + 2 = 16, so half = 14 and B had 28 — all received from A.
4. A gave half + 2 = 28, so half = 26 and A had 52.
5. Check: A (52) gives 26 + 2 = 28 to B; B gives 14 + 2 = 16 to C; C gives 8 + 2 = 10 to D.

Remember · For ‘gives half and k more’ chains, reverse each step: original = 2 × (amount given − k).

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 694861x

Convert each unit to seconds: a week is 604800 s, a day 86400 s, an hour 3600 s, a minute 60 s. Adding these and the lone second gives 694861 seconds for every x.

1. 1 week = 7 × 24 × 3600 = 604800 s; 1 day = 24 × 3600 = 86400 s; 1 hour = 3600 s; 1 minute = 60 s.
2. Total = (604800 + 86400 + 3600 + 60 + 1)x = 694861x seconds.
3. Check: the x seconds make the coefficient end in 1, and one week alone is 604800x, so only 694861x fits.

Remember · Use the last digit as a filter: the extra ‘x seconds’ makes the total end in 1.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) ₹2,600

Statements 2 and 3 give P = 13 and Q = 8; then X = 8, Z = 16 and Y = 20. The five bought 65 notebooks, which cost 65 × ₹40 = ₹2,600.

1. From 2 and 3: P + Q = 21 and P − Q = 5, so P = 13 and Q = 8.
2. From 5: X = 13 − 5 = 8. From 1: Z = 8 + 8 = 16. From 4: Y = 28 − 8 = 20.
3. Total notebooks = (P + Q) + (X + Y) + Z = 21 + 28 + 16 = 65.
4. Cost = 65 × ₹40 = ₹2,600.

Remember · Start with the pair of equations that closes on its own (sum and difference), then chain outwards.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 5 minutes fast

Of the 90 minutes he was away by the home clock, 65 were spent driving. The return took 0.625 of the outward time, so the outward drive took 40 minutes and he reached the village at 15:10 home time — the village clock, showing 15:15, is 5 minutes fast.

1. By the home clock he was away from 14:30 to 16:00 = 90 minutes; 25 minutes were spent in the village, so driving took 65 minutes.
2. Let the outward drive take t minutes. The return route is 1.25 times as long at twice the speed, so it takes 1.25 ÷ 2 × t = 0.625t.
3. t + 0.625t = 65, so 1.625t = 65 and t = 40 minutes.
4. He reached the village at 14:30 + 40 min = 15:10 by the home clock, when the village clock showed 15:15.
5. So the village clock is 5 minutes fast.
6. Check: he leaves the village at 15:35 (home time); the return takes 0.625 × 40 = 25 minutes, reaching home at 16:00.

Remember · Use only one clock for elapsed time; compare the other clock’s reading with the true time at that moment.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) Data is inadequate to draw any conclusion

The answer depends on how a man’s work rate compares with a woman’s, and that is not given. Equal rates give exactly 30 days, faster men give more than 30 and faster women fewer — so the data are inadequate.

1. Let a man do m units and a woman w units of work a day. Work = 30 × (24m + 12w).
2. Days for 12 men and 24 women = 30(24m + 12w)/(12m + 24w) = 30(2m + w)/(m + 2w).
3. If m = w this is 30 days; if m > w it is more than 30; if m < w it is less than 30.
4. The relation between m and w is not given, so no conclusion can be drawn. Option (c) fails because it leaves out the possibility of exactly 30 days.

Remember · In time-and-work items with two kinds of workers, you need their relative efficiency before you can swap one for the other.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) z < x < y

The first condition gives 0.7x = 0.65y, so y = 14x/13, just above x. The second gives 0.6x = 0.65z, so z = 12x/13, just below x. Hence z < x < y.

1. 0.7x + y = 1.65y, so 0.7x = 0.65y and y = 70x/65 = 14x/13.
2. 0.6x + z = 1.65z, so 0.6x = 0.65z and z = 60x/65 = 12x/13.
3. For positive x, 12x/13 < x < 14x/13, so z < x < y.
4. Check with x = 13: y = 14 and z = 12; 9.1 + 14 = 23.1 = 1.65 × 14 and 7.8 + 12 = 19.8 = 1.65 × 12.

Remember · Turn each percentage condition into ‘part of x = part of y’ and compare the resulting multiples of x.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) 15000

X got 56% and Y 44% of the votes cast, so the winning margin is 12% of the votes cast: 1440 votes. That makes 12000 votes cast, which is 80% of the list — so the list has 15000 voters.

1. With two candidates and no invalid votes, Y got 100% − 56% = 44% of the votes cast.
2. Margin = 56% − 44% = 12% of the votes cast = 1440, so votes cast = 1440 ÷ 0.12 = 12000.
3. Votes cast are 80% of the voters list, so total voters = 12000 ÷ 0.8 = 15000.
4. Check: 12000 is 80% of 15000; X gets 6720, Y gets 5280, and 6720 − 5280 = 1440.

Remember · In two-candidate elections, the margin is (winner% − loser%) of the votes cast; convert back to the full list last.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (a) m = n

After the two transfers each container again holds 100 ml. Whatever water went into X must be matched by the same amount of milk left in Y, so X’s milk share equals Y’s water share: m = n = 5/6.

1. After the first transfer: X has 80 ml milk; Y has 100 ml water + 20 ml milk = 120 ml.
2. 20 ml of Y’s mixture carries 20 × 20/120 = 10/3 ml milk and 20 × 100/120 = 50/3 ml water back to X.
3. X: milk = 80 + 10/3 = 250/3 ml out of 100 ml, so m = 5/6.
4. Y: water = 100 − 50/3 = 250/3 ml out of 100 ml, so n = 5/6.
5. So m = n.

Remember · Back-and-forth transfers of equal volume: the milk in the water jar always equals the water in the milk jar.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 41 kg

A + B + C = 120 and B + D + E = 126. Since F weighs the same as B, the six weights total 120 + 126 = 246 kg, and the average is 41 kg even though B itself is unknown.

1. A + B + C = 3 × 40 = 120 kg and B + D + E = 3 × 42 = 126 kg.
2. Since F = B, A + B + C + D + E + F = (A + B + C) + (F + D + E) = 120 + (B + D + E) = 120 + 126 = 246 kg.
3. Average = 246 ÷ 6 = 41 kg.

Remember · Before answering ‘cannot be determined’, check whether the unknown cancels — here F = B fills the gap exactly.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error