A school has 110, 198 and 132 students enrolled for its science, arts and…
2025
A school has 110, 198 and 132 students enrolled for its science, arts and commerce courses. All students have to be seated in rooms for an exam such that each room has students of only the same course and all rooms have an equal number of students. What is the least number of rooms needed?
- A.
20
- B.
23
- C.
25
- D.
29
Attempted by 12 students.
Show answer & explanation
Correct answer: A
Concept: To seat groups of different sizes into equal-sized rooms without mixing courses, the room size must be a common divisor of all the group sizes. The room size that minimizes the total number of rooms is the largest such common divisor — the HCF (highest common factor) of the group sizes. Once the room size is fixed at the HCF, the number of rooms needed for a group equals that group's size divided by the HCF, and the total rooms needed is the sum of these across all groups.
Application:
Find the HCF of 110, 198 and 132 by prime factorisation: 110 = 2 × 5 × 11; 198 = 2 × 32 × 11; 132 = 22 × 3 × 11.
The common prime factors, taken at their lowest powers across all three numbers, are 2 and 11, so HCF = 2 × 11 = 22. Each room therefore holds 22 students of the same course.
Rooms needed = 110 ÷ 22 + 198 ÷ 22 + 132 ÷ 22 = 5 + 9 + 6 = 20.
Cross-check: 22 divides all three enrolments exactly (22 × 5 = 110, 22 × 9 = 198, 22 × 6 = 132), and no larger number divides all three (for example, 44 does not divide 110), confirming 22 is indeed the greatest common room size and 20 is the minimum number of rooms.