Problems on H.C.F and L.C.M
In a school, the number of students in Sections A , B and C of 10th class are 70,98 and 126 respectively. Due to overload of student in each class , the administration wants to increase the number roo
The number students in each room = HCF of 70,98 and 126 =14 In each room , maximum 14 students can be seated. Total number students = 70 + 98 +126 =294 Number of rooms required = 294/14 =21
To express any fraction in simple (small form), we have to find the HCF of numerator and denominator. 1095)1168(1 1095 73)1095(15 1095 ——— 0 HCF of 1095 and 1168 is 73Therefore (1168/73)/(1095/73) = 16/15
EXAMIANS