Question

Kindly answer no. 5

5. Find the sum of money when.
(i) simple interest at $7\frac{1}{4}\%\mathrm{p}.\mathrm{a}.\mathrm{for}2\frac{1}{2}\mathrm{years}\mathrm{is}\mathrm{Rs}.2356.25$
(ii) Final amount is Rs. 11,300 at 4% p.a. for 3 years 3 months.

Answer 1

Dear student,

$$\begin{gathered} i)LetPbetheprincipal \\ Weknow, \\ SI=\frac{PRT}{100} \\ \Rightarrow 2356.25=\frac{P\times 29\times 5}{100\times 4\times 2} \\ \Rightarrow P=\frac{2356.25\times 100\times 4\times 2}{29\times 5} \\ \Rightarrow P=\frac{235625\times 100\times 4\times 2}{29\times 5\times 100} \\ \Rightarrow P=\frac{235625\times 4\times 2}{29\times 5} \\ \Rightarrow P=\frac{8125\times 4\times 2}{5} \\ \Rightarrow P=1625\times 8 \\ \Rightarrow P=Rs13000 \\ ii)LetPbetheprincipal \\ Weknow, \\ Amount=SI+P \\ \Rightarrow 11300=\frac{PRT}{100}+P \\ \Rightarrow 11300=\frac{P\times 4\times 13}{100\times 4}+P..[3yrs3months=3+\frac{3}{12}=3+\frac{1}{4}=\frac{12+1}{4}=\frac{13}{4}years] \\ \Rightarrow 11300=\frac{P\times 13}{100}+P \\ \Rightarrow 11300=\frac{13P+100P}{100} \\ \Rightarrow 1130000=113P \\ \Rightarrow P=\frac{1130000}{113} \\ \Rightarrow P=Rs10000 \end{gathered}$$

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