There are three departments having students 64,58,24 .In an exam they have to be seated in rooms such that each room has equal number of students and each room has students of one type only (No mixing of departments. Find the minimum number rooms required ?
Answers
Answer:73
Step-by-step explanation:
WE NEED TO TAKE GCD WHICH IS 2
THUS ALL THE ROOMS WILL HAVE 2 STUDENTS OF THE SAME DEPT
1) 64/2 = 32
2) 58/2 = 29
3) 24/2 = 12
TOTAL NO. OF MIN ROOMS REQD = 32+12+29 = 73
Answer:
Minimum number of rooms required is 73.
Step-by-step explanation:
Given:
Number of student in 1st department = 64
Number of student in 2nd department = 58
Number of student in 3rd department = 24
To find: Minimum number of rooms required such that each has same number of student and of same department.
First we find number of student in each room of same department.
To find this we find HCF of 64 , 58 , 24
Prime factorization of 64 = 2 × 2 × 2 × 2 × 2 × 2
Prime factorization of 58 = 2 × 29
Prime factorization of 24 = 2 × 2 × 2 × 3
HCF ( 64 , 58 , 24 ) = 2
So, Each room has 2 students of same department.
Rooms required by 1st department = 64 / 2 = 32
Rooms required by 2nd department = 58 / 2 = 29
Rooms required by 3rd department = 24 / 2 = 12
Minimum number of rooms = 32 + 29 + 12 = 73
Therefore, Minimum number of rooms required is 73.