Question

Find 3 no. in the ratio 4:6:10, the sum of whose squares is 608.

Answer 1

Let the no. be 4x, 6x and 10x respectively
According to question
(4x)^2+(6x)^2+(10x)^2 = 608
16x^2+ 36x^2+100x^2= 608
152x^2 = 608
x^2= 608/152
x^2 = 4
x = sq. root of 4 = 2
Therefore the numbers will be 8, 12 and 20.

Answer 2

Suppoze the three numbers=2z,3z and 5z.

A.P.Q.

(2z)²+(3z)²+(5z)²=608

or 4z²+9z²+25²=608

or 38z²=608

or z²=608÷38

or z²=16

or z²=(4)²

or z=4

Hence,the three numbers are

• 2×z=2×4=8
• 3×z=3×4=12
• 5×z=5×4=20