(b) Let the four numbers be a, b, c and d. Then, a : b = c : d =1 : 3 ⇒
a
b
=
c
d
=
1
3
⇒ b = 3a, d = 3c ..........(i) Also, a2+b2+c2+d2 = 50 ⇒ a2+9a2+c2+9c2 = 50 ⇒ 10a2+10c2 = 50 ⇒ a2+c2 = 5 ..............(ii) Also, sum of means = 5 b + c = 5 ⇒b = 5 - c ⇒ a =
5−c
3
‌‌‌‌‌ [using Eq. (i) Putting this value of a in Eq (ii), we get (
5−c
3
)2+c2 = 5 ⇒
25+c2−10c
9
+c2 = 5 ⇒ 10c2 - 10c - 20 = 0 ⇒ c2 - c - 2 = 0 ⇒ c = 2 Therefore, a =
5−c
3
=
5−2
3
= 1 b = 3a = 3 c = 2 d = 3c = 6 Hence, required average =