SZF IOM 2025 previous year paper practice · PYQ-pattern set

SZF IOM 2025 PYQ-pattern questions

Showing 10 of the 16 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 System

   What is the smallest four-digit number that is divisible by both 12 and 15?

   • A. 1020
   • B. 1040
   • C. 1060
   • D. 1080
   Answer and solution

   Answer: A

   LCM(12,15)=60. Smallest four-digit multiple of 60 is 17×60=1020, which is divisible by both.

2. Q2 · Sequence and Series

   The sum of the first n terms of a series is given by Sₙ = 3n² + 2n. What is the 8th term of this series?

   • A. 45
   • B. 47
   • C. 49
   • D. 51
   Answer and solution

   Answer: B

   The nth term aₙ = Sₙ - Sₙ₋₁. Thus a₈ = S₈ - S₇ = [3(64) + 16] - [3(49) + 14] = 208 - 161 = 47.

3. Q3 · Triangles

   In triangle ABC, AB = 8 cm, BC = 15 cm, and AC = 17 cm. What type of triangle is ABC?

   • A. Acute-angled triangle
   • B. Right-angled triangle
   • C. Obtuse-angled triangle
   • D. Equilateral triangle
   Answer and solution

   Answer: B

   Since 8² + 15² = 64 + 225 = 289 = 17², the triangle satisfies the Pythagorean theorem and is a right-angled triangle.

4. Q4 · Trigonometry

   If sin θ = 3/5, what is the value of cos θ for an acute angle θ?

   • A. 4/5
   • B. 3/4
   • C. 5/4
   • D. 2/5
   Answer and solution

   Answer: A

   Using sin²θ + cos²θ = 1, we get cos²θ = 1 - (9/25) = 16/25, so cos θ = 4/5 for acute angle.

5. Q5 · Logarithms

   If log₂ 64 + log₅ 125 = x, then what is the value of x?

   • A. 8
   • B. 9
   • C. 10
   • D. 12
   Answer and solution

   Answer: B

   log₂ 64 = log₂ 2⁶ = 6, and log₅ 125 = log₅ 5³ = 3. Therefore, x = 6 + 3 = 9.

6. Q6 · Mathematical Reasoning

   The sum of three consecutive odd numbers is 147. What is the largest among them?

   • A. 47
   • B. 49
   • C. 51
   • D. 53
   Answer and solution

   Answer: C

   Let the three consecutive odd numbers be x, x+2, x+4. Their sum is x + (x+2) + (x+4) = 147, so 3x+6=147, 3x=141, x=47. The largest is x+4 = 47+4 = 51.

7. Q7 · Advanced Geometry

   A regular hexagon is inscribed in a circle of radius 12 cm. Another circle is inscribed inside the hexagon. What is the ratio of the area of the outer circle to the area of the inner circle?

   • A. 3 : 2
   • B. 4 : 3
   • C. 2 : 1
   • D. 5 : 3
   Answer and solution

   Answer: B

   For a regular hexagon inscribed in a circle of radius R, the inradius (radius of inscribed circle) is R√3/2. Here R = 12, so inradius = 6√3. Area ratio = (12²)/(6√3)² = 144/108 = 4/3.

8. Q8 · Advanced Algebra

   If x² + 1/x² = 47, what is the value of x³ + 1/x³?

   • A. 322
   • B. 327
   • C. 336
   • D. 343
   Answer and solution

   Answer: A

   First find x + 1/x: (x + 1/x)² = x² + 2 + 1/x² = 47 + 2 = 49, so x + 1/x = 7. Then x³ + 1/x³ = (x + 1/x)³ - 3(x + 1/x) = 7³ - 3(7) = 343 - 21 = 322.

9. Q9 · Circles

   A chord of length 24 cm is at a distance of 5 cm from the center of a circle. Find the radius of the circle.

   • A. 12 cm
   • B. 13 cm
   • C. 14 cm
   • D. 15 cm
   Answer and solution

   Answer: B

   Half the chord = 12 cm. Using Pythagoras: r² = 12² + 5² = 144 + 25 = 169. So r = 13 cm.

10. Q10 · Trigonometry

    If sin θ + cos θ = √2, then what is the value of sin θ · cos θ?

    • A. 1/2
    • B. 1/4
    • C. 1
    • D. √2/2
    Answer and solution

    Answer: A

    Squaring both sides: (sin θ + cos θ)² = 2, which gives sin²θ + cos²θ + 2sinθcosθ = 2. Since sin²θ + cos²θ = 1, we get 1 + 2sinθcosθ = 2, so sinθcosθ = 1/2.