Question

there are 315 420 and 1555 boys of different Schools taking part in an essay competition to the minimum number of rooms required such that all the rooms have equal number of student and each room has students of same school only​

Answer 1

Given,

Boys of first school = 315

Boys of second school = 420

Boys of third school = 1155

To Find,

Minimum rooms required such that all the rooms have an equal number of students and each room has students of the same school only​ =?

Solution,

Factorizing all three numbers,

315 = 3*3*5*7

420 = 2*2*3*5*7

1155 = 3*5*7*11

HCF of these 3 numbers = 3 * 5 * 7 = 105

The rooms required for school 1 = 315 / 105 = 3

The rooms required for school 2 = 420 / 105 = 4

The rooms required for school 3 = 1155 / 105 = 11

Total room required = 11 + 3 + 4 = 18

Hence, the  minimum number of rooms required such that all the rooms have an equal number of students and each room has students of the same school only​ is 18