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 41/2 years at 16% per annum, the greater part is = ?

Answer: Option A

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