There exist three numbers in the ratio of 3 : 2 : 7 such that the sum of their squares is equal to 558. The three numbers will be respectively:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 9, 6, 21
Numbers in the ratio 3 : 2 : 7 can be written as \(3k\), \(2k\) and \(7k\) for some number \(k\); the sum of their squares then gives \(k\).
\((3k)^2+(2k)^2+(7k)^2=9k^2+4k^2+49k^2=62k^2\)
\(62k^2=558\), so \(k^2=9\) and \(k=3\)
The numbers are \(3\times3=9\), \(2\times3=6\) and \(7\times3=21\)