(a) Present ages of a father and his son are in the ratio 10 : 3.
Let the sum of their ages be x.
The son's age = 15 years = $\frac{3}{13}\times x=15$
$x=\frac{15\times 13}{3}=65$
$\therefore$ The Father's present age = 65 - 15 = 50 years.
In z years time, the ratio of their ages = 2 : 1
$\frac{50+z}{15+z}=\frac{2}{1}$
$⟹50+z=2(15+z)$
$50+z=30+2z⟹50-30=2z-z$
$20=z$
Therefore, in 20 years' time, the ratio of their age will be 2 : 1.
(b) $\frac{x+y+z}{3}=6⟹x+y+z=18....(1)$
$\frac{x+y+z+l+u+v+w}{7}=9⟹x+y+z+l+u+v+w=63....(2)$
Putting (1) into (2), we have
$18+l+u+v+w=63$
$\therefore l+u+v+w=63-18=45$
Mean of l, u, v and w = $\frac{45}{4}=11.25$.