If Rs. 12000 is divided into two parts such that the simple interest on the first part for 3 years at 12% per annum is equal to the simple interest on the second part for $$4\frac{1}{2}$$ years at 16% per annum, the greater part is = ?

Explanation:-

$$\eqalign{ & {\text{Let the first part = Rs}}{\text{. }}x \cr & \therefore {\text{Hence second part}} \cr & {\text{ = Rs}}{\text{. }}\left( {12000 - x} \right) \cr & {\text{According to the qoestion,}} \cr & \Rightarrow \frac{{x \times 12 \times 3}}{{100}} = \frac{{\left( {12000 - x} \right) \times 9 \times 16}}{{2 \times 100}} \cr & \Rightarrow 36x = 72\left( {12000 - x} \right) \cr & \Rightarrow x = 24000 - 2x \cr & \Rightarrow 3x = 24000 \cr & \Rightarrow x = {\text{Rs}}{\text{. 8000}} \cr & {{\text{1}}^{{\text{st}}}}{\text{ part = Rs}}{\text{. 8000}} \cr & {{\text{2}}^{{\text{nd}}}}{\text{ part}} \cr & {\text{ = Rs}}{\text{. }}\left( {12000 - 8000} \right) \cr & = {\text{Rs}}{\text{. 4000 }} \cr & {\text{Hence maximum part}} \cr & {\text{ = Rs}}{\text{. 8000}} \cr}$$
Alternate
Note : In such type of questions to save your valuable time follow the given below method.
Let two parts P₁ and P₂ respectively
$$\eqalign{ & \Rightarrow {{\text{P}}_1} \times \frac{{36}}{{100}} \times 1 = {{\text{P}}_2} \times \frac{9}{2} \times \frac{{16}}{{100}} \times 1 \cr & \Rightarrow {{\text{P}}_1} \times 4 = 8{{\text{P}}_2} \cr & \Rightarrow \frac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}} = \frac{2}{1} \cr & \Rightarrow {{\text{P}}_1}{\text{:}}{{\text{P}}_2} = 2:1 \cr & {\text{Hence greater part}} \cr & {\text{ = }}\frac{{12000}}{{\left( {2 + 1} \right)}} \times 2 \cr & = {\text{Rs}}{\text{. }}8000 \cr}$$