SSC CHSL 10 Jan 2017 Shift 3 Paper

Three numbers a,b,c are in proportion iff ${\mathit{b}}^2=\mathit{a}\mathit{c}$ where b is the mean proportion and c is the third proportion Mean proportion of two numbers x and y = 6
⇒ $\mathit{x}\mathit{y}={\left(6\right)}^2=36$ -------(i)
Third proportion = 48
⇒ ${\mathit{y}}^2=\mathit{x}\times 48$ ----(ii)
Substituting value of x from equation(i) in equation(ii), we get :
⇒ ${\mathit{y}}^2=\frac{36}{\mathit{y}}\times 48$
⇒ ${\mathit{y}}^3={\left(2\right)}^3\times {\left(6\right)}^3={\left(12\right)}^3$
⇒ $\mathit{y}={\left(12\right)}^{\frac{3}{3}}=12$
Substituting it in equation(i), ⇒ $\mathit{x}=\frac{36}{12}=3$