Please answer the given questions Share with your friends Share 0 Ankita Agarwal answered this Dear Student, No. of students in group A = 156No. of students in group B = 208No. of students in group C = 260Total no. of students = 156+208+260 = 624First, we need to find maximum numbers of students in each bus by taking HCF of 156,208 and 260Using prime factorization method,156 = 2×2×3×13208 = 2×2×2×2×13260 = 2×2×5×13HCF of 156, 208 and 260 = 2×2×13 = 52Maximum number of students per bus = 52Minimum number of buses required = 62452 = 12 busesiii Group C and 26 students of group-A not taken i.e. 260+26 = 286 students are not taken for filed tripTotal students = 624-286 = 338 studentsNo. of students in group A = 130No. of students in group B = 208We will find HCF of 130 and 208 using prime factorization,130 = 2×5×13208 = 2×2×2×2×13HCF of 130 and 208 = 2×13 = 26Maximum number of students per bus = 26Minimum number of buses required = 33826 = 13 busesHence, option (b) is correct(iv) No. of students in group B = 208No. of students in group C = 260We will find HCF of 208 and 260 using prime factorization,208 = 2×2×2×2×13260 = 2×2×5×13HCF of 208 and 260 = 2×2×13 = 52Maximum number of students per bus = 52Minimum number of buses required = 46852 = 9 busesIf we represent it in the form of 208x+260y, then value of x and y is 1,1 Regards 0 View Full Answer Kartik Gupta answered this sry but i have no idea 0