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CSAT · 140 questions

Number system

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

Number system questions per year: 2016: 3, 2017: 7, 2018: 5, 2019: 13, 2020: 19, 2021: 10, 2022: 13, 2023: 21, 2024: 15, 2025: 22, 2026: 12 Asked in 11 of 11 years · most in 2025 (22)

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 2019 · Q9

Easy

The number of times the digit 5 will appear while writing the integers from 1 to 1000 is

Answer & explanation

Answer: (c) 300

Written as three-digit strings from 000 to 999, every digit appears equally often in each place, so 5 fills one-tenth of the 3000 digit places: 300. The number 1000 adds no 5.

  1. Treat 1 to 999 as three-digit strings 001–999 (leading zeros add no 5s).
  2. In each place — units, tens, hundreds — the digit 5 appears in 100 of the 1000 strings 000–999.
  3. So 5 appears 3 × 100 = 300 times from 1 to 999.
  4. 1000 has no 5, so the total is 300.
  5. Check: units place 5, 15, …, 995 = 100; tens place 50–59 in each hundred = 10 × 10 = 100; hundreds place 500–599 = 100.

Remember · From 1 to 10ⁿ − 1, each non-zero digit appears n × 10ⁿ⁻¹ times; check the end number separately.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q15

Easy

In a school every student is assigned a unique identification number. A student is a football player if and only if the identification number is divisible by 4, whereas a student is a cricketer if and only if the identification number is divisible by 6. If every number from 1 to 100 is assigned to a student, then how many of them play cricket as well as football?

Answer & explanation

Answer: (b) 8

A student plays both games only if the number is divisible by both 4 and 6, that is by their LCM 12. There are 8 multiples of 12 up to 100.

  1. Divisible by 4 and by 6 ⇔ divisible by LCM(4, 6) = 12.
  2. Multiples of 12 from 1 to 100: 12, 24, 36, 48, 60, 72, 84, 96.
  3. Count = 8 (100 ÷ 12 = 8 remainder 4).
  4. Check: using 24 (= 4 × 6) would wrongly drop 12, 36, 60 and 84.

Remember · 'Divisible by a and by b' means divisible by LCM(a, b), not a × b.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q33

Easy

The ratio of a two-digit natural number to a number formed by reversing its digits is 4: 7. The number of such pairs is

Answer & explanation

Answer: (b) 4

Writing the number as 10a + b, the ratio 4 : 7 simplifies to b = 2a. The digit a can be 1 to 4, giving 12, 24, 36 and 48 — four pairs.

  1. Let the number be 10a + b; reversed, it is 10b + a.
  2. 7(10a + b) = 4(10b + a) → 70a + 7b = 40b + 4a → 66a = 33b → b = 2a.
  3. b must be a digit, so a = 1, 2, 3, 4.
  4. Pairs: 12 & 21, 24 & 42, 36 & 63, 48 & 84 — each in the ratio 4 : 7.
  5. Number of pairs = 4.

Remember · Turn digit-reversal conditions into 10a + b form; the equation usually gives a simple digit relation to list.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q35

Medium

Seeta and Geeta go for a swim after a gap of every 2 days and every 3 days respectively. If on 1st January both of them went for a swim together, when will they go together next?

Answer & explanation

Answer: (d) 13th January

A gap of 2 days means Seeta swims every 3rd day; a gap of 3 days means Geeta swims every 4th day. They meet again after LCM(3, 4) = 12 days, on 13th January.

  1. Gap of 2 days: Seeta swims on 1, 4, 7, 10, 13 … January (every 3rd day).
  2. Gap of 3 days: Geeta swims on 1, 5, 9, 13 … January (every 4th day).
  3. LCM(3, 4) = 12, so the next common day is 1 + 12 = 13 January.
  4. Check: 13 is in both lists; 7 is only Seeta's and 12 is neither's.

Remember · 'After a gap of n days' means the event repeats every (n + 1) days; then take the LCM.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q37

Easy

If x is greater than or equal to 25 and y is less than or equal to 40, then which one of the following is always correct?

Answer & explanation

Answer: (c) (y − x) is less than or equal to 15

y − x is largest when y is as big as allowed (40) and x as small as allowed (25), giving 15. So y − x can never exceed 15; the other options fail for suitable values.

  1. Maximum of y − x = 40 − 25 = 15, so y − x ≤ 15 always.
  2. (a) fails: x = 25, y = 40 gives x < y.
  3. (b) fails: y − x never goes above 15.
  4. (d) fails: x = 25, y = 10 gives x + y = 35.

Remember · For 'always correct' with bounds, test the extreme values; a single counter-example removes an option.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q48

Easy

What is X in the sequence 132, 129, 124, 117, 106, 93, X?

Answer & explanation

Answer: (c) 76

The terms fall by 3, 5, 7, 11 and 13 — consecutive primes. The next prime is 17, so X = 93 − 17 = 76.

  1. Differences: 132 − 129 = 3, 129 − 124 = 5, 124 − 117 = 7, 117 − 106 = 11, 106 − 93 = 13.
  2. These are consecutive primes; the next one is 17.
  3. X = 93 − 17 = 76.

Remember · When differences grow irregularly, test primes, squares and cubes first.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q50

Easy

If the numerator and denominator of a proper fraction are increased by the same positive quantity which is greater than zero, the resulting fraction is

Answer & explanation

Answer: (b) always greater than the original fraction

For a proper fraction a/b with 0 < a < b, adding the same k > 0 to both parts gives a difference of k(b − a)/[b(b + k)], which is positive. The fraction always increases.

  1. Let the fraction be a/b with 0 < a < b, and add k > 0 to both.
  2. (a + k)/(b + k) − a/b = [b(a + k) − a(b + k)]/[b(b + k)] = k(b − a)/[b(b + k)].
  3. k > 0 and b − a > 0, so the difference is positive: the new fraction is larger.
  4. Example: 1/2 becomes (1 + 1)/(2 + 1) = 2/3, which is greater than 1/2.

Remember · Adding the same positive number to both parts pulls a fraction towards 1: proper fractions rise, improper ones fall.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q51

Easy

What is X in the sequence 4, 196, 16, 144, 36, 100, 64, X?

Answer & explanation

Answer: (b) 64

Every term is a square: 2², 14², 4², 12², 6², 10², 8². Bases in odd places rise 2, 4, 6, 8 and in even places fall 14, 12, 10, so the next even-place base is 8 and X = 64.

  1. 4 = 2², 196 = 14², 16 = 4², 144 = 12², 36 = 6², 100 = 10², 64 = 8².
  2. Odd positions: bases 2, 4, 6, 8 (rising by 2). Even positions: 14, 12, 10, … (falling by 2).
  3. X is in an even position, so its base is 8 and X = 8² = 64.

Remember · In mixed sequences, split the odd and even positions and look for squares or cubes in each.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q53

Easy

A printer numbers the pages of a book starting with 1 and uses 3089 digits in all. How many pages does the book have?

Answer & explanation

Answer: (c) 1049

Pages 1 to 999 use 9 + 180 + 2700 = 2889 digits. The remaining 200 digits give 50 four-digit page numbers, 1000 to 1049.

  1. Pages 1–9: 9 digits; 10–99: 90 × 2 = 180; 100–999: 900 × 3 = 2700. Total 2889.
  2. Digits left = 3089 − 2889 = 200, and 200 ÷ 4 = 50 four-digit pages.
  3. These are pages 1000 to 1049, so the book has 1049 pages.

Remember · Use the digit blocks 9, 180, 2700; divide what remains by the next digit length.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q58

Medium

Consider two statements S1 and S2 followed by a question:

  1. S1.p and q both are prime numbers.
  2. S2.p + q is an odd integer.
  3. Question: Is pq an odd integer?

Which one of the following is correct?

Answer & explanation

Answer: (b) S2 alone is sufficient to answer the question

Two whole numbers have an odd sum only when one is even and the other odd, and then their product is even. So S2 alone gives a definite 'No'. S1 alone fails: 3 × 5 = 15 is odd but 2 × 3 = 6 is even.

  1. S1 alone: p = 3, q = 5 gives pq = 15 (odd); p = 2, q = 3 gives pq = 6 (even) — not sufficient.
  2. S2 alone: for whole numbers, an odd sum means one number is even and the other odd.
  3. Then pq has an even factor, so pq is even — a definite 'No'. S2 is sufficient.
  4. Since S2 works alone, S1 is not needed, so (c) and (d) fail.
  5. Like the question itself, this reading takes p and q to be whole numbers.

Remember · In data sufficiency a definite 'No' is an answer; odd + even = odd, and an even factor makes a product even.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q60

Easy

Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is

Answer & explanation

Answer: (d) 8

The units give 7 + 6 = 13 (carry 1), and the hundreds rise from 5 + 1 to 7, so the tens column must also carry: B + 4 = A + 10, i.e. A = B − 6. For 7A3 to be divisible by 3, 10 + A must be a multiple of 3, so A = 2 and B = 8.

  1. Units: 7 + 6 = 13 → write 3, carry 1.
  2. Hundreds: 5 + 1 + carry = 7, so the tens column must carry 1.
  3. Tens: B + 3 + 1 = A + 10 → A = B − 6; so B is 6, 7, 8 or 9 and A is 0, 1, 2 or 3.
  4. 7A3 divisible by 3 → 7 + A + 3 = 10 + A is a multiple of 3 → A = 2.
  5. B = 8. Check: 587 + 136 = 723, and 7 + 2 + 3 = 12.

Remember · In digit-sum puzzles, go column by column with carries, then use the divisibility rule to fix the digit.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q74

Easy

If $ means ‘divided by’; @ means ‘multiplied by’; # means ‘minus’, then the value of 10#5@1$5 is

Answer & explanation

Answer: (d) 9

Decoded, the expression is 10 − 5 × 1 ÷ 5. Multiplication and division come before subtraction, so 5 × 1 ÷ 5 = 1 and the value is 10 − 1 = 9.

  1. 10#5@1$5 = 10 − 5 × 1 ÷ 5.
  2. 5 × 1 = 5, then 5 ÷ 5 = 1.
  3. 10 − 1 = 9.
  4. Check: working strictly left to right would give (10 − 5) × 1 ÷ 5 = 1, which ignores BODMAS.

Remember · After decoding the symbols, apply BODMAS strictly: division and multiplication before subtraction.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q75

Easy

An 8-digit number 4252746B leaves remainder 0 when divided by 3. How many values of B are possible?

Answer & explanation

Answer: (c) 4

The known digits add to 30, already a multiple of 3, so B itself must be a multiple of 3: 0, 3, 6 or 9 — four values.

  1. 4 + 2 + 5 + 2 + 7 + 4 + 6 = 30.
  2. 30 + B is divisible by 3 exactly when B is divisible by 3.
  3. B can be 0, 3, 6 or 9 → 4 values.

Remember · Divisibility by 3 depends only on the digit sum; if the known digits already give a multiple of 3, the unknown digit must too.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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