Number system

Answer: (a) 47

The mean of five consecutive odd numbers is the middle (third) one, so the third number is 39 and the list starts at 35. The mean of thirteen consecutive odd numbers is the 7th number: 35 + 6 × 2 = 47.

1. Mean of 5 consecutive odd numbers = the 3rd number = 39, so the first number is 39 − 4 = 35.
2. The 13 numbers are 35, 37, …, 59 (35 + 12 × 2 = 59), all 2-digit.
3. Mean of an evenly spaced list = middle (7th) term = 35 + 6 × 2 = 47.
4. Check: (35 + 59) ÷ 2 = 47.

Remember · For an evenly spaced list, the mean equals the middle term, or the average of the first and last terms.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 4

If a number and its reverse are both multiples of 7, so is their difference, 99 × (first digit − last digit). Since 99 is not a multiple of 7, the first and last digits must differ by 7 — giving just 168, 861, 259 and 952.

1. Let the number be 100a + 10b + c; its reverse is 100c + 10b + a.
2. Their difference is 99(a − c). Both are multiples of 7 and 99 is not, so a − c must be a multiple of 7.
3. The digits are all different, so a − c = ±7: (a, c) = (1, 8), (8, 1), (2, 9), (9, 2) or (7, 0). The last fails: 7b0 is divisible by 7 only if b = 0 or 7, which repeats a digit.
4. Find the middle digit in each case: 1b8 → 168 = 7 × 24; 8b1 → 861 = 7 × 123; 2b9 → 259 = 7 × 37; 9b2 → 952 = 7 × 136.
5. Each case has exactly one suitable middle digit, so there are 4 such numbers.
6. Check: 861 and 952 are the reverses of 168 and 259, and all four have three different digits.

Remember · A 3-digit number and its reverse differ by 99 × (first − last digit); this turns 'both divisible by 7' into an end-digit condition.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (c) 90

Numbers between 99 and 1000 are the 3-digit numbers 100 to 999. With 8 fixed in the units place, the hundreds digit has 9 choices and the tens digit 10 choices, giving 9 × 10 = 90.

1. Numbers strictly between 99 and 1000: 100 to 999.
2. Units digit fixed as 8.
3. Hundreds digit: 1–9 (9 ways); tens digit: 0–9 (10 ways).
4. Count = 9 × 10 = 90.
5. Check: 108, 118, …, 998 rise in steps of 10, so the count is (998 − 108) ÷ 10 + 1 = 90.

Remember · Fix the given digit, then multiply the free choices for the other places — the leading digit cannot be 0.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 38

Next year's age is 2 more than last year's, so we need a square that is 2 less than a cube: 25 + 2 = 27. Mr. X was 25, is 26 now and turns 27 next year. The next cube is 64, which is 38 years away from his present age.

1. If last year's age is n² and next year's is m³, then m³ − n² = 2.
2. Try cubes: 8 − 2 = 6 (not a square), 27 − 2 = 25 = 5² ✓, 64 − 2 = 62, 125 − 2 = 123, 216 − 2 = 214 — only 27 works for a human age.
3. So he was 25 last year, is 26 now and will be 27 next year.
4. After 27, the next cube is 4³ = 64.
5. Years to wait from his present age: 64 − 26 = 38.

Remember · 'Square last year, cube next year' means the two numbers differ by 2 — list squares and cubes side by side.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) 45

A number and its reverse differ by 9 × (difference of the digits). When the larger is less than twice the smaller, the quotient is 1 and the remainder is that difference; the biggest such case is 94 ÷ 49, which leaves 45.

1. Let the larger number be 10a + b and the smaller 10b + a, with a > b.
2. Their difference is 9(a − b). If the larger is less than twice the smaller, the quotient is 1 and the remainder is exactly 9(a − b).
3. Quotient 1 needs 10a + b < 2(10b + a), i.e. 8a < 19b. With a = 9 this needs b ≥ 4, so the best pair is 94 and 49: remainder 9 × 5 = 45.
4. If the quotient is 2 or more, then 8a ≥ 19b forces b ≤ 3, so the smaller number is at most 39 and the remainder is less than 39.
5. Largest possible remainder = 45 (94 = 1 × 49 + 45).

Remember · A number and its reverse differ by 9 × (difference of digits); when the quotient is 1, that difference is the remainder.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (d) None of the above

A 2-digit number and its reverse differ by 9 × (difference of the digits), so the digits must differ by 3. That gives 41, 52, 63, 74, 85, 96 and 14, 25, 36, 47, 58, 69 (and 30, if 03 is allowed) — at least 12 numbers, far more than 3, 4 or 5.

1. (10a + b) − (10b + a) = 9(a − b), which is 27 in size when the digits differ by 3.
2. Tens digit 3 more than units: 41, 52, 63, 74, 85, 96 (and 30, if its reverse 03 = 3 is accepted).
3. Tens digit 3 less than units: 14, 25, 36, 47, 58, 69.
4. That is 12 numbers (13 counting 30); even one direction alone gives 6. None of 3, 4 or 5 fits.

Remember · Number − reverse = 9 × (digit difference). Divide the given difference by 9 to get the digit gap, then list the pairs.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error

Answer: (b) 342

Pages 1–9 use 9 digits, pages 10–99 use 90 × 2 = 180 digits, and pages 100–150 use 51 × 3 = 153 digits. The total is 342.

1. Pages 1–9: 9 pages × 1 digit = 9.
2. Pages 10–99: 90 pages × 2 digits = 180.
3. Pages 100–150: 51 pages × 3 digits = 153.
4. Total = 9 + 180 + 153 = 342.

Remember · Count digits band by band: pages in a band = last − first + 1, times the digits on each page.

Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink · Report an error