Minimalist IAS
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 →

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 ·

From the UPSC Blog

Other CSAT skills