13, a, b, c are four distinct numbers and the HCf of each pair of numbers ( 13, a): ( 13, b): (13,c), where a,b,c are each less than 60 and a?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 2
HCF is multiple of given no so here we have
13a , 13b, 13c and a
now we will find out the value of a,b,c
a= 13×2 = 26
b= 13×3= 39
c= 13×4 = 52
put a = 26, b = 39, and c= 52 because the value of this should be less than 60
these values satisfy so put the value of a,b,c in (a+c)/b
= (26+52)/39
= 78/39 = 2