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CSAT

CSAT · 132 questions

Arithmetic: percentage, ratio, averages, time & work

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Arithmetic: percentage, ratio, averages, time & work questions per year: 2016: 16, 2017: 11, 2018: 8, 2019: 15, 2020: 15, 2021: 16, 2022: 13, 2023: 2, 2024: 11, 2025: 9, 2026: 16 Asked in 11 of 11 years · most in 2026 (16)

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” See the full syllabus →

Showing 1–30 of 132, newest first.

CSAT 2026 · Q8

Medium Provisional key

The weight of X, in kg, is denoted by X. The weights of A, B, C, D, P, Q, R and S are measured. Given:

A + B + C + D = 17

A + C = 6

P + Q + S + D = 15

P + Q + R + B = 17

P = R and Q = S

Which one of the following statements is correct?

Answer & explanation

Answer: (b) P and Q together weigh more than the total weight of A and C.

Adding the last two equations after substituting R = P and S = Q gives 3(P + Q) + B + D = 32, and B + D = 11 from the first two. So P + Q = 7, which is more than A + C = 6.

  1. From A + B + C + D = 17 and A + C = 6: B + D = 11.
  2. Put S = Q: P + 2Q + D = 15. Put R = P: 2P + Q + B = 17.
  3. Add them: 3P + 3Q + B + D = 32, so 3(P + Q) = 32 − 11 = 21 and P + Q = 7.
  4. (a) B + D = 11 is not less than 7 — false. (b) P + Q = 7 > A + C = 6 — true.
  5. P and Q individually are not fixed (for example P = 3, Q = 4 or P = 4, Q = 3 with suitable B, D), so (c) and (d) cannot be asserted.
  • ✗ (a) B + D = 11 is more than P + Q = 7.
  • ✓ (b) P + Q = 7 exceeds A + C = 6.
  • ✗ (c) Only P + Q = 7 is fixed; P can be larger or smaller than Q.
  • ✗ (d) Same reason: the data do not fix which of P and Q is heavier.

Remember · When single values can't be found, add the equations to get the combined quantity the options ask about.

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

CSAT 2026 · Q13

Easy Provisional key

A toy T jumps forward or backward. In each forward jump, it moves 5′ forward whereas in each backward jump, it moves 2′ backward. If in 31 jumps, T moves exactly 15′ forward, then what is the difference of the number of forward and backward jumps?

Answer & explanation

Answer: (d) 9

Let f forward and b backward jumps: f + b = 31 and 5f − 2b = 15. This gives f = 11 and b = 20, so the difference is 9.

  1. f + b = 31 and 5f − 2b = 15.
  2. Substitute b = 31 − f: 5f − 62 + 2f = 15, so 7f = 77 and f = 11.
  3. b = 31 − 11 = 20.
  4. Difference = 20 − 11 = 9 (more backward jumps than forward).
  5. Check: 11 × 5 − 20 × 2 = 55 − 40 = 15′ forward.

Remember · Two unknowns, two facts (count and net result): write both equations and eliminate one variable.

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

CSAT 2026 · Q20

Easy Provisional key

The speed of a train T is 100 km per hour and the speed of a person P is 4 km per hour. T crosses P in 15 seconds, if P travels along the direction of motion of T. If P travels along the opposite direction of T, then in how much time does T cross P, in seconds, approximately?

Answer & explanation

Answer: (c) 13.85

The train's length is fixed, so crossing time is inversely proportional to relative speed. Relative speed changes from 96 km/h to 104 km/h, so the new time is 15 × 96/104 ≈ 13·85 seconds.

  1. Same direction: relative speed = 100 − 4 = 96 km/h. Opposite direction: 100 + 4 = 104 km/h.
  2. Length of T = 96 × (5/18) m/s × 15 s = 400 m.
  3. Time in the opposite case = 400 ÷ (104 × 5/18) = 3600/260 ≈ 13·85 s.
  4. Check: 15 × 96/104 = 13·846 s.

Remember · Same length, different relative speed: new time = old time × old relative speed ÷ new relative speed.

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

CSAT 2026 · Q25

Easy Provisional key

The class average x in a test increases by 4 when the score of a student is rectified, whose corrected score is 100 instead of 0. Later, the score of another student was found to have been recorded as 81 in place of 56. If there are no other corrections and the final corrected average is y, then y − x is

Answer & explanation

Answer: (b) 3

Adding 100 marks raised the average by 4, so the class has 25 students. The second correction removes 25 marks, lowering the average by 1, so the net rise is 4 − 1 = 3.

  1. First correction adds 100 − 0 = 100 marks and raises the average by 4, so number of students = 100 ÷ 4 = 25.
  2. Second correction: 81 recorded instead of 56 means the total falls by 25.
  3. Average falls by 25 ÷ 25 = 1.
  4. y − x = 4 − 1 = 3.

Remember · Change in average = change in total ÷ number of items; find the count from the first change.

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

CSAT 2026 · Q30

Easy Provisional key

Suppose x, y and z are variables taking positive real numbers as their possible values. It is given that y is directly proportional to x² and x is inversely proportional to z. For z = 7/25, the values of x and y are 5 and 50, respectively. If y = 98, what is z equal to?

Answer & explanation

Answer: (b) 1/5

From y = kx², k = 50/25 = 2, so y = 98 gives x = 7. From xz = constant = 5 × 7/25 = 7/5, z = (7/5)/7 = 1/5.

  1. y = kx²: 50 = k × 25, so k = 2.
  2. y = 98: 2x² = 98, x² = 49, x = 7 (x is positive).
  3. xz = constant = 5 × 7/25 = 7/5.
  4. z = (7/5) ÷ 7 = 1/5.
  5. Check: x rose from 5 to 7, so z must fall in the ratio 5 : 7 — (7/25) × (5/7) = 1/5.

Remember · Direct variation: ratio stays constant; inverse variation: product stays constant. Find each constant from the given pair.

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

CSAT 2026 · Q37

Hard Provisional key

X and Y are two runners who run for the same duration of time on the same circular track. They started running at the same time in the same direction with uniform speeds. When X completed 7 rounds, Y did exactly 5. After completing 5 rounds, Y changed his direction and started running in the opposite direction with speed which is double of his earlier speed. On the other hand, X continued to run with the same speed. They stopped running when X completed exactly 21 rounds. How many times did X and Y meet after they had started and before they finally stopped?

Answer & explanation

Answer: (a) 35

On a circular track, runners meet once for every full round of difference (same direction) or every full round of combined distance (opposite directions). That gives 2 meetings in the first phase and 34 in the second, but the last one happens exactly at the stop, so 35 are counted.

  1. Speeds are in the ratio 7 : 5. Phase 1 (same direction): X runs 7 rounds, Y runs 5, so X gains 2 rounds → 2 meetings (the second at the start point when Y turns).
  2. Phase 2: X runs 21 − 7 = 14 more rounds. In the same time Y, at double speed (10 per 7 of X), runs 20 rounds the other way.
  3. Opposite directions: they meet once per round of combined distance: 14 + 20 = 34 meetings.
  4. The 34th meeting is exactly when they stop (X at 21 rounds, Y at 25 — both at the start point), which is not 'before they finally stopped'.
  5. Meetings counted = 2 + 34 − 1 = 35.

Remember · Circular track meetings = relative rounds covered (difference if same direction, sum if opposite). Check whether the final meeting falls at the stop.

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

CSAT 2026 · Q38

Easy Provisional key

In an objective type question paper, 5 marks are awarded for a correct answer and 2 marks are deducted for a wrong answer. A student attempted all the questions and got a score of 69. Had he been awarded 4 marks for a correct answer and 1 mark deducted for a wrong answer, he would have scored 84. How many questions were there in the question paper?

Answer & explanation

Answer: (b) 81

With c correct and w wrong, 5c − 2w = 69 and 4c − w = 84. Solving gives c = 33 and w = 48, so the paper had 81 questions.

  1. 5c − 2w = 69 and 4c − w = 84.
  2. From the second: w = 4c − 84.
  3. Substitute: 5c − 8c + 168 = 69, so 3c = 99 and c = 33.
  4. w = 132 − 84 = 48.
  5. Total = 33 + 48 = 81.
  6. Check: 5 × 33 − 2 × 48 = 165 − 96 = 69.

Remember · Two scoring schemes for the same answers give two equations; solve for right and wrong counts, then add.

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

CSAT 2026 · Q41

Medium Provisional key

An explosion takes place at a certain distance from an army camp. As soon as the sensor in the camp receives the sound of the explosion, a drone starts flying towards the spot of explosion. The drone clicks a picture from the spot and the camp receives it at the same time. Immediately another drone starts flying to the spot and it also sends a picture as soon as it reaches the spot. The two pictures were received at 5:02 PM and 5:05 PM, respectively. If the speed of the drones is 30 m/s, at what time did the explosion take place? Assume that the speed of sound is 300 m/s.

Answer & explanation

Answer: (c) 4:58:42 PM

The second drone's one-way trip takes the 3 minutes between the two pictures, so the spot is 180 × 30 = 5400 m away. The first drone left 3 minutes before 5:02, and the sound had taken 5400 ÷ 300 = 18 s to arrive, so the explosion was 3 min 18 s before 5:02 PM.

  1. The second drone starts at 5:02 PM and its picture arrives at 5:05 PM, so one trip takes 180 s.
  2. Distance = 30 m/s × 180 s = 5400 m.
  3. The first drone also needed 180 s, so it left the camp at 4:59:00 PM — the moment the sound arrived.
  4. Sound took 5400 ÷ 300 = 18 s, so the explosion happened at 4:59:00 − 18 s = 4:58:42 PM.
  5. Check: 4:58:42 + 18 s + 180 s = 5:02:00 PM.

Remember · Work backwards on a timeline: find the distance from the cleanest interval, then subtract each leg's travel time.

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

CSAT 2026 · Q43

Easy Provisional key

A person saves 10% of his salary every month. If his salary increases by 12% and the expenditure increases by 10%, then what will be the change in his saving per month?

Answer & explanation

Answer: (b) 30% increase

Take the salary as ₹100: saving ₹10, spending ₹90. After the changes, salary is ₹112 and spending ₹99, so saving becomes ₹13 — a 30% rise on ₹10.

  1. Let salary = ₹100. Saving = ₹10, expenditure = ₹90.
  2. New salary = ₹112; new expenditure = 90 × 1.1 = ₹99.
  3. New saving = 112 − 99 = ₹13.
  4. Change = (13 − 10)/10 × 100 = 30% increase.

Remember · Percentage problems with several parts: assume a base of 100 and compute each part directly.

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

CSAT 2026 · Q50

Easy Provisional key

X travels 6 km on a bicycle with average speeds of 5 km per hour, 10 km per hour and 4 km per hour during the first 1 km, the next 2 km and the remaining 3 km, respectively. Y travels the same distances with average speeds of 4 km per hour, 10 km per hour and 5 km per hour, respectively. How many minutes early will Y complete the journey if both X and Y start at the same time?

Answer & explanation

Answer: (d) 6

Add the time for each stretch. X takes 12 + 12 + 45 = 69 minutes; Y takes 15 + 12 + 36 = 63 minutes, so Y finishes 6 minutes earlier.

  1. X: 1/5 h = 12 min; 2/10 h = 12 min; 3/4 h = 45 min. Total 69 min.
  2. Y: 1/4 h = 15 min; 2/10 h = 12 min; 3/5 h = 36 min. Total 63 min.
  3. Y is earlier by 69 − 63 = 6 minutes.

Remember · Average speed over a journey is never the average of the speeds; add time = distance ÷ speed for each stretch.

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

CSAT 2026 · Q58

Easy Provisional key

The ratio of male to female workers in two companies A and B is 13: 10 and 7: 5, respectively. If both the companies have the same number of female workers, then what is the ratio of the total number of workers in A to those in B?

Answer & explanation

Answer: (b) 23: 24

Make the female part equal: write B's 7 : 5 as 14 : 10. Then A has 13 + 10 = 23 parts and B has 14 + 10 = 24 parts.

  1. A: male : female = 13 : 10.
  2. B: 7 : 5 = 14 : 10, so females are 10 parts in both.
  3. Totals: A = 23 parts, B = 24 parts.
  4. Ratio = 23 : 24.

Remember · To compare two ratios sharing a common quantity, scale them so that the common term is equal.

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

CSAT 2026 · Q65

Easy Provisional key

An alloy P contains 20% copper and 80% zinc by weight. Another alloy Q contains 60% copper and 40% zinc by weight. A third alloy R is to be prepared from P and Q so that it contains equal amount of copper and zinc. In what ratio, amounts of P and Q be mixed in order to get R?

Answer & explanation

Answer: (a) 1: 3

R needs 50% copper. By alligation, P (20%) and Q (60%) must be mixed in the ratio (60 − 50) : (50 − 20) = 10 : 30 = 1 : 3.

  1. Equal copper and zinc means 50% copper in R.
  2. P : Q = (60 − 50) : (50 − 20) = 10 : 30 = 1 : 3.
  3. Check: 1 kg P + 3 kg Q has 0.2 + 1.8 = 2 kg copper out of 4 kg = 50%.

Remember · Alligation: ratio of cheaper to dearer = (dearer − mean) : (mean − cheaper).

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

CSAT 2026 · Q69

Medium Provisional key

A train has to complete a journey of 800 km. If it meets a minor accident, its speed becomes half of the existing speed. If there is a mechanical defect, the speed becomes one-fourth of the existing speed. On its way, the train meets with a minor accident after 200 km; and 400 km thereafter, it develops a mechanical defect. Had the train developed the mechanical defect after 200 km and met the minor accident 400 km thereafter, it would have taken 4 more hours to reach its destination. What was the original speed of the train in km per hour?

Answer & explanation

Answer: (a) 200

With original speed v, the actual trip takes 200/v + 400/(v/2) + 200/(v/8) = 2600/v hours, and the alternative takes 200/v + 400/(v/4) + 200/(v/8) = 3400/v hours. The gap 800/v = 4 hours gives v = 200 km/h.

  1. Actual: first 200 km at v; next 400 km at v/2; last 200 km at (v/2)/4 = v/8.
  2. Time = 200/v + 800/v + 1600/v = 2600/v.
  3. Alternative: 200 km at v; 400 km at v/4; last 200 km at (v/4)/2 = v/8.
  4. Time = 200/v + 1600/v + 1600/v = 3400/v.
  5. 3400/v − 2600/v = 800/v = 4, so v = 200 km/h.
  6. Check: 13 h and 17 h — a 4-hour difference.

Remember · Write each scenario's time as distance ÷ speed with the speed in terms of v, then equate the given difference.

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

CSAT 2026 · Q78

Medium Provisional key

Three partners A, B and C entered into a business. A invested one-third of the capital for one-third duration. B invested one-fourth of the capital for one-fourth duration. C invested the remaining capital for the whole duration. Out of a profit of ₹ 17,000, how much profit will C get?

Answer & explanation

Answer: (a) ₹ 12,000

Profit is shared in the ratio of capital × time. A: 1/3 × 1/3 = 1/9, B: 1/4 × 1/4 = 1/16, C: 5/12 × 1 = 5/12, i.e., 16 : 9 : 60 out of 85. C gets 60/85 of ₹ 17,000 = ₹ 12,000.

  1. C's capital = 1 − 1/3 − 1/4 = 5/12.
  2. Capital × time: A = 1/9, B = 1/16, C = 5/12.
  3. Multiply by 144: A = 16, B = 9, C = 60; total 85.
  4. C's share = 17,000 × 60/85 = ₹ 12,000.
  5. Check: A gets ₹ 3,200 and B ₹ 1,800; 3,200 + 1,800 + 12,000 = 17,000.

Remember · Partnership: profit share ∝ capital × time. Clear fractions with a common denominator before dividing.

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

CSAT 2026 · Q79

Easy Provisional key

There are two chemicals which do not react with each other. A container contains 10 litres of the chemical A. One litre of this chemical is removed from it and one litre of the chemical B is poured. Then one litre of the mixture is removed from the container and one litre of B is poured. If this process of replacing one litre of the mixture by one litre of B is performed once more, then what is the volume of B that is present in the container approximately (in percentage)?

Answer & explanation

Answer: (b) 27

Each replacement keeps 9/10 of whatever A is present. After three replacements A = 10 × (0·9)³ = 7·29 litres, so B = 2·71 litres, about 27% of the 10 litres.

  1. Each step removes 1/10 of the mixture, so A falls by a factor 9/10 each time.
  2. After 3 steps, A = 10 × 0·9³ = 10 × 0·729 = 7·29 L.
  3. B = 10 − 7·29 = 2·71 L ≈ 27%.

Remember · Repeated replacement: amount left = original × (1 − removed/total)ⁿ.

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

CSAT 2026 · Q80

Medium Provisional key

A shopkeeper employs a delivery boy and gives him a motorcycle for home delivery. For every delivery, the boy is given ₹ 5. At the end of the day, he also gets ₹ 2 for every kilometre of the distance covered in the day. The boy wants to earn more than ₹ 500 a day, but does not want to travel more than 100 km. Which of the following numbers of deliveries would definitely meet his target?

Answer & explanation

Answer: (d) The question cannot be answered due to insufficient data

Earnings are 5 × deliveries + 2 × kilometres. Even 90 deliveries give only ₹ 450, so the target depends on the distance travelled, which is not given. No option guarantees more than ₹ 500.

  1. Earning = 5d + 2k, where d = deliveries and k ≤ 100 km.
  2. d = 80, 85, 90 give ₹ 400, ₹ 425, ₹ 450 from deliveries alone.
  3. Whether he crosses ₹ 500 depends on k: 90 deliveries need more than 25 km; 80 need more than 50 km.
  4. The distance is not given, so none of the numbers definitely meets the target.

Remember · 'Definitely' means it must hold in every allowed case; test the worst case (here, the shortest distance).

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

CSAT 2025 · Q9

Medium

Team X scored a total of N runs in 20 overs. Team Y tied the score in 10% less overs. Had team Y’s average run rate (runs per over) been 50% higher, the scores would have been tied in 12 overs. How many runs were scored by team X?

Answer & explanation

Answer: (d) Cannot be determined

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 ·

CSAT 2025 · Q10

Medium

The price (p) of a commodity is first increased by k%; then decreased by k%; again increased by k%; and again decreased by k%. If the new price is q, then what is the relation between p and q?

Answer & explanation

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 ·

CSAT 2025 · Q17

Easy

The petrol price shot up by 10% as a result of the hike in crude oil prices. The price of petrol before the hike was ₹ 90 per litre. A person travels 2200 km every month and his car gives a mileage of 16 km per litre. By how many km should he reduce his travel if he wants to maintain his expenditure at the previous level?

Answer & explanation

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 ·

CSAT 2025 · Q37

Medium

P and Q walk along a circular track. They start at 5:00 a.m. from the same point in opposite directions. P walks at an average speed of 5 rounds per hour and Q walks at an average speed of 3 rounds per hour. How many times will they cross each other between 5:20 a.m. and 7:00 a.m.?

Answer & explanation

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 ·

CSAT 2025 · Q39

Medium

A tram overtakes 2 persons X and Y walking at an average speed of 3 km/hr and 4 km/hr in the same direction and completely passes them in 8 seconds and 9 seconds respectively. What is the length of the tram?

Answer & explanation

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 ·

CSAT 2025 · Q45

Medium

A set (X) of 20 pipes can fill 70% of a tank in 14 minutes. Another set (Y) of 10 pipes fills 3/8th of the tank in 6 minutes. A third set (Z) of 16 pipes can empty half of the tank in 20 minutes. If half of the pipes of set X are closed and only half of the pipes of set Y are open, and all pipes of the set (Z) are open, then how long will it take to fill 50% of the tank?

Answer & explanation

Answer: (d) 16 minutes

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 ·

CSAT 2025 · Q59

Medium

X can complete one-third of a certain work in 6 days, Y can complete one-third of the same work in 8 days and Z can complete three-fourth of the same work in 12 days. All of them work together for n days and then X and Z quit and Y alone finishes the remaining work in 8 2/3 days. What is n equal to?

Answer & explanation

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 ·

CSAT 2025 · Q75

Medium

In a T20 cricket match, three players X, Y and Z scored a total of 37 runs. The ratio of number of runs scored by X to the number of runs scored by Y is equal to ratio of number of runs scored by Y to number of runs scored by Z.

  1. Value-I: Runs scored by X
  2. Value-II: Runs scored by Y
  3. Value-III: Runs scored by Z

Which one of the following is correct?

Answer & explanation

Answer: (d) Cannot be determined due to insufficient data

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 ·

CSAT 2025 · Q78

Easy

The average of three numbers p, q and r is k. p is as much more than the average as q is less than the average. What is the value of r?

Answer & explanation

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 ·

CSAT 2024 · Q7

Easy

A certain number of men can complete a piece of work in 6k days, where k is a natural number. By what percent should the number of men be increased so that the work can be completed in 5k days?

Answer & explanation

Answer: (c) 20%

For a fixed job, the number of men is inversely proportional to the number of days. Cutting the days from 6k to 5k needs 6/5 times the men, which is a 20% increase.

  1. Work = men × days, so M × 6k = M′ × 5k.
  2. M′ = M × 6k/5k = (6/5)M; the k cancels.
  3. Increase = (6/5 − 1) × 100% = (1/5) × 100% = 20%.
  4. Check: with k = 1, 5 men for 6 days = 30 man-days = 6 men for 5 days, and 6 is 20% more than 5.

Remember · For fixed work, new men = old men × old days ÷ new days. Common factors like k cancel out.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2024 · Q8

Hard

X, Y and Z can complete a piece of work individually in 6 hours, 8 hours and 8 hours respectively. However, only one person at a time can work in each hour and nobody can work for two consecutive hours. All are engaged to finish the work. What is the minimum amount of time that they will take to finish the work?

Answer & explanation

Answer: (c) 6 hours 45 minutes

X is the fastest, so X should take the first hour and every alternate hour after it. In six hours 21 of the 24 units are done, and X finishes the last 3 units in 45 minutes of the seventh hour.

  1. Take the work as 24 units (LCM of 6 and 8): X does 4 units per hour, Y 3 and Z 3.
  2. Nobody may work two hours in a row, so X can work at most every other hour; start with X so that he gets as many hours as possible.
  3. Hours 1 to 6: X, Y, X, Z, X, Y gives 3 × 4 + 3 × 3 = 21 units, and all three have worked.
  4. Hour 7 is X’s again: the remaining 3 units at 4 units per hour take 3/4 hour = 45 minutes.
  5. Total = 6 hours 45 minutes.
  6. Check: if Y or Z starts, X still gets only three of the first six hours (21 units), but hour 7 then goes to Y or Z at 3 units per hour, finishing at exactly 7 hours.

Remember · In alternate-hour work problems, give the fastest worker the first slot and every second slot after it.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2024 · Q16

Medium

What percent of water must be mixed with honey so as to gain 20% by selling the mixture at the cost price of honey?

Answer & explanation

Answer: (a) 20%

Selling at the honey’s cost price and still gaining 20% means 120 litres of mixture must be sold for what 100 litres of honey cost. So 20 litres of water go into every 100 litres of honey, i.e. 20% of the honey.

  1. Take 100 litres of honey at ₹1 per litre: cost = ₹100.
  2. A 20% gain means sales must bring ₹120; at ₹1 per litre that is 120 litres of mixture.
  3. Water added = 120 − 100 = 20 litres, which is 20% of the honey.
  4. Check: water is 20/120 = 16.67% of the mixture, which is not an option, so the question means water as a percentage of honey.

Remember · When free water is sold at the price of the pure item, the gain percent equals water as a percent of the pure item.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2024 · Q25

Medium

Two persons P and Q enter into a business. P puts ₹14,000 more than Q, but P has invested for 8 months and Q has invested for 10 months. If P’s share is ₹400 more than Q’s share out of the total profit of ₹2,000, what is the capital contributed by P?

Answer & explanation

Answer: (a) ₹30,000

Out of ₹2,000, P gets ₹1,200 and Q ₹800, a ratio of 3 : 2. Profit is shared in the ratio of capital × time, so 8(Q + 14,000) : 10Q = 3 : 2, which gives Q = ₹16,000 and P = ₹30,000.

  1. Shares add to 2,000 and differ by 400, so P gets ₹1,200 and Q ₹800; ratio 3 : 2.
  2. Let Q’s capital be x; then P’s capital is x + 14,000.
  3. 8(x + 14,000) : 10x = 3 : 2 → 16(x + 14,000) = 30x → 14x = 2,24,000 → x = 16,000.
  4. P’s capital = 16,000 + 14,000 = ₹30,000.
  5. Check: 30,000 × 8 = 2,40,000 and 16,000 × 10 = 1,60,000, a ratio of 3 : 2.

Remember · Profit share is proportional to capital × time. Turn the profit clue into a ratio first, then solve one equation.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2024 · Q26

Easy

P’s salary is 20% lower than Q’s salary which is 20% lower than R’s salary. By how much percent is R’s salary more than P’s salary?

Answer & explanation

Answer: (b) 56.25%

Take R’s salary as 100: Q gets 80 and P gets 64. R’s salary is 36 more than P’s, and 36 out of 64 is 56.25%.

  1. Let R’s salary be 100. Q’s is 20% lower: 80. P’s is 20% lower than Q’s: 64.
  2. R − P = 100 − 64 = 36.
  3. R is more than P by 36/64 × 100 = 56.25%.
  4. Check: 64 × 1.5625 = 100.

Remember · Take 100 as the base. ‘More than P’ means divide the difference by P’s value, not by 100.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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