Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 8160, the money borrowed by him (in Rupees) was?

Answer: Option C

Let the money borrowed by Nitin = Rs. P
According to the question,
$$\begin{array}{rl}& \frac{\text{P}\times 6\times 3}{100}+\frac{\text{P}\times 9\times 5}{100}+\frac{\text{P}\times 13\times 3}{100}=\text{Rs}\text{. }8160\\ & \Rightarrow \frac{\text{18P}}{100}+\frac{\text{45P}}{100}+\frac{\text{39P}}{100}=\text{Rs}\text{. 1860}\\ & \Rightarrow \frac{\text{102P}}{100}=\text{Rs}\text{. 8160}\\ & \Rightarrow \text{P = Rs}\text{.}\frac{8160\times 100}{102}\\ & \text{ = Rs}\text{. 8000}\end{array}$$

Alternate
Note : In such type of questions to save your valuable time follow the given below method.
Let principal = Rs. 100
Total Interest
$$\frac{100\times 6\times 3}{100}+\frac{100\times 9\times 5}{100}+$$      $$\frac{100\times 13\times 3}{100}$$
$$\begin{array}{rl}& =18+45+39\\ & =102\text{ units}\\ & \text{According to the question,}\\ & \text{102 units = Rs}\text{. 8160}\\ & \text{1 unit = Rs}\text{. }\frac{8160}{102}\\ & \text{ = Rs}\text{. 80}\\ & \text{100 units = Rs}\text{. 8000}\\ & \text{Hence sum = Rs}\text{. 8000}\end{array}$$

Alternate
Total rate of interest in 11 years
$$\begin{array}{rl}& \left(6\times 3\right)\%+\left(5\times 9\right)\%+\left(3\times 13\right)\%\\ & \Rightarrow 102\%=8160\\ & \Rightarrow 100\%=8000\\ & \therefore \text{Sum = Rs}\text{. 8000}\end{array}$$