IOQM 2025 previous year paper · official paper and PYQ-pattern practice

IOQM 2025 PYQ-pattern questions

Showing 10 of the 18 PYQ-pattern questions in this set — freshly worded in that year's pattern, not the original questions. Answers and solutions open under each question.

1. Q1 · Number Theory

   Find the sum of all positive divisors of 84.

   • A. 168
   • B. 224
   • C. 192
   • D. 240
   Answer and solution

   Answer: B

   84 = 2² × 3 × 7. Using the divisor sum formula: σ(84) = (1+2+4)(1+3)(1+7) = 7×4×8 = 224.

2. Q2 · Geometry

   A circle is inscribed in a triangle with sides 13, 14, and 15. What is the radius of the inscribed circle?

   • A. 3
   • B. 4
   • C. 5
   • D. 6
   Answer and solution

   Answer: B

   Semi-perimeter s = (13+14+15)/2 = 21. By Heron's formula, Area = √[21×8×7×6] = √7056 = 84. Since r = Area/s, we get r = 84/21 = 4.

3. Q3 · Geometry

   In triangle ABC, the perpendicular from A to BC meets BC at D. If BD = 9, DC = 16, and AD = 12, find the area of triangle ABC.

   • A. 150
   • B. 300
   • C. 225
   • D. 180
   Answer and solution

   Answer: A

   Area of triangle ABC = (1/2) × base × height = (1/2) × BC × AD = (1/2) × (9 + 16) × 12 = (1/2) × 25 × 12 = 150.

4. Q4 · Trigonometry

   Find the value of cos(π/9) × cos(2π/9) × cos(4π/9), expressed as 1/k where k is an integer.

   • A. 02
   • B. 04
   • C. 08
   • D. 16
   Answer and solution

   Answer: C

   Using the product formula for cos(π/9)cos(2π/9)cos(4π/9) with triple angle identities, the result simplifies to 1/8, so k = 08.

5. Q5 · Basic Preliminaries

   Let x be a real number such that x² + 1/x² = 23. What is the value of x + 1/x if x + 1/x is positive?

   • A. 4
   • B. 5
   • C. 6
   • D. 7
   Answer and solution

   Answer: B

   We know (x + 1/x)² = x² + 2 + 1/x². Given x² + 1/x² = 23, we have (x + 1/x)² = 25, so x + 1/x = ±5. Since we want the positive value, the answer is 5.

6. Q6 · Geometry

   In triangle ABC, AB = 13, BC = 14, and CA = 15. Let D be the foot of the altitude from A to BC. Find the length BD.

   • A. 5
   • B. 6
   • C. 7
   • D. 9
   Answer and solution

   Answer: A

   Using the formula in a triangle with sides a, b, c: the foot of altitude from vertex divides opposite side into segments of lengths (c² + a² - b²)/(2a). Here BD = (BC² + AB² - AC²)/(2·BC) = (196 + 169 - 225)/28 = 140/28 = 5.

7. Q7 · Trigonometry

   In a right triangle ABC with right angle at C, if sin A = 5/13, find the value of cos B.

   • A. 5/13
   • B. 12/13
   • C. 13/12
   • D. 13/5
   Answer and solution

   Answer: A

   In a right triangle, angles A and B are complementary (A + B = 90°). Therefore, cos B = cos(90° - A) = sin A = 5/13.

8. Q8 · Counting Principles

   A committee of 5 people is to be formed from 6 men and 4 women such that the committee has at least 2 women. The number of ways to form such a committee is:

   • A. 186
   • B. 192
   • C. 196
   • D. 246
   Answer and solution

   Answer: A

   Total committees with at least 2 women = C(4,2)×C(6,3) + C(4,3)×C(6,2) + C(4,4)×C(6,1) = 6×20 + 4×15 + 1×6 = 120 + 60 + 6 = 186.

9. Q9 · Polynomials

   Let p(x) be a monic polynomial of degree 3 such that p(1) = 1, p(2) = 4, and p(3) = 9. Find p(4).

   • A. 22
   • B. 28
   • C. 34
   • D. 16
   Answer and solution

   Answer: A

   Note p(x) − x² is a monic cubic vanishing at x = 1, 2, 3. So p(x) − x² = (x−1)(x−2)(x−3). Thus p(x) = (x−1)(x−2)(x−3) + x². Hence p(4) = (3)(2)(1) + 16 = 6 + 16 = 22.

10. Q10 · Pigeonhole Principle

    Among any n integers chosen from the set {1, 2, 3, ..., 20}, at least two must have a difference that is a multiple of 6. Find the smallest such n.

    • A. 7
    • B. 8
    • C. 9
    • D. 6
    Answer and solution

    Answer: A

    Partition {1,...,20} by residues mod 6: {6,12,18}, {1,7,13,19}, {2,8,14,20}, {3,9,15}, {4,10,16}, {5,11,17}. We have 6 residue classes. To avoid two numbers with the same residue, we can pick at most one from each class, giving at most 6 numbers. So n = 7 guarantees two with the same residue mod 6.