Question

Ashok borrowed Rs. 12,000 at some rate percent C.I. After a year he paid back Rs. 4,000. If C.I. for 2nd year be Rs.920 Find:

(i) rate of interest

(ii) amount of debt at the end of the second year

Answer 1

$$\begin{gathered} \left(i\right)Principal=Rs12000 \\ After1yearhepaidbackRs4000 \\ Lettherateofinterestber \\ Hencethepricipalafter1year12000(1+\frac{r}{100})-4000 \\ NowitisgiventhatC.Ifor2ndyear=920 \\ Hence\left\{12000(1+\frac{r}{100})-4000\right\}(1+\frac{r}{100})=\left\{12000(1+\frac{r}{100})-4000+920\right\} \\ (1+\frac{r}{100})=\frac{\left\{12000(1+\frac{r}{100})-4000+920\right\}}{\left\{12000(1+\frac{r}{100})-4000\right\}} \\ \Rightarrow (1+\frac{r}{100})=\frac{\left\{8920+120r\right\}}{\left\{8000+120r\right\}} \\ \Rightarrow \frac{r}{100}=\frac{\left\{8920+120r\right\}}{\left\{8000+120r\right\}}-1 \\ \Rightarrow \frac{r}{100}=\frac{920}{8000+120r} \\ \Rightarrow 120r^2+8000r-92000=0 \\ \Rightarrow 3r^2+200r-2300=0 \\ \Rightarrow 3r^2-30r+230r-2300=0 \\ \Rightarrow 3r(r-10)+230(r-10)=0 \\ \Rightarrow (3r+230)(r-10)=0 \\ Sinceratecannotbenegative \\ Hencer=10 \\ \therefore Rateofinterest=10\% \\ \left(ii\right)Nowinitialprincipal=12000 \\ C.Ifor1styear=12000(1+\frac{10}{100})-12000=13200-12000=1200 \\ C.Ifor2ndyear=920 \\ Hencesumofprincipalandinterestofbothyears=12000+1200+920=14120 \\ Buthepaidback4000 \\ Henceamountofdebtattheendof2ndyear=14120-4000=Rs10120 \end{gathered}$$