Let x,y,x be the ages of 3 persons which lies between 60 and 65

The equation governing the problem $\frac{55n-41-47+(x+y+z)}{n+1}=56$

Also given n is a multiple of 7
55n - 88 +k = 56n + 56  where k=x+y+z, n+ 144=k  
Now k ranges between 3 $\times$ 61 and 3$\times$ 64
183 $\le$ n+44 $\le$ 192
39 $\le$ n $\le$ 48

Thus n =42, the latest number n + 1 =42+1=43

Shortcut

Go from answer options. The initial number of people in the group is a multiple of 7. The answer options give the number of initial people + 1. Thus the correct answer is option (b).