Two numbers are such that the square of one is 224 less than 8 times the square of the other. If the numbers are in the ratio of 3 : 4, then their values are = ?

112, 16
26, 8
39, 12
412, 9

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let number be x & y \\ Given, \\ x : y \\ 3 : 4 \\ 3 a : 4 a \\ Now , given that \\ \Rightarrow 8 {(3 a)}^{2} = {(4 a)}^{2} + 224 \\ \Rightarrow 72 a^{2} = 16 a^{2} + 224 \\ \Rightarrow 56 a^{2} = 224 \\ \Rightarrow a^{2} = 4 \\ \Rightarrow a = 2 \\ Numbers are \\ x = 3 \times 2 = 6 \\ y = 4 \times 2 = 8 \end{aligned}

$\eqalign{ & {\text{Let number be }}x\& y \cr & {\text{Given,}} \cr & \,\,x:y \cr & \,\,\,\,3:4 \cr & \,3a:4a \cr & {\text{Now , given that}} \cr & \Rightarrow {\text{8}}{\left( {3a} \right)^2} = {\left( {4a} \right)^2} + 224 \cr & \Rightarrow 72{a^2} = 16{a^2} + 224 \cr & \Rightarrow 56{a^2} = 224 \cr & \Rightarrow {a^2} = 4 \cr & \Rightarrow a = 2 \cr & {\text{Numbers are}} \cr & x = 3 \times 2 = 6 \cr & y = 4 \times 2 = 8 \cr}$