compound interest on a certain sum of money at rate of 5% become rupees 1537.50 into years find the simple interest on the same time same interest rate.
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Answer:
\underline{\textsf{Given:}}Given:
\textsf{Compound interest for 2 years is Rs.1537.50}Compound interest for 2 years is Rs.1537.50
\mathsf{and\;rate\;of\;interest\;is\;5\%}andrateofinterestis5%
\underline{\textsf{To find:}}To find:
\textsf{Simple interest for the same sum of money}Simple interest for the same sum of money
\underline{\textsf{Solution:}}Solution:
\textsf{We apply copound interest formula}We apply copound interest formula
\boxed{\mathsf{Compound\;interest=P\left(1+\dfrac{r}{100}\right)^n-P}}Compoundinterest=P(1+100r)n−P
\mathsf{Compound\;interest\;for\;2\;years=Rs.1537.50}Compoundinterestfor2years=Rs.1537.50
\mathsf{P\left(1+\dfrac{r}{100}\right)^2-P=1537.50}P(1+100r)2−P=1537.50
\mathsf{P\left(1+\dfrac{5}{100}\right)^2-P}=1537.50}
\mathsf{P\left(1+\dfrac{1}{20}\right)^2-P=1537.50}P(1+201)2−P=1537.50
\mathsf{P\left(1+\dfrac{1}{400}+\dfrac{2}{20}\right)-P=1537.50}P(1+4001+202)−P=1537.50
\mathsf{P+\dfrac{P}{400}+\dfrac{P}{10}-P=1537.50}P+400P+10P−P=1537.50
\mathsf{\dfrac{P}{400}+\dfrac{P}{10}=1537.50}400P+10P=1537.50
\mathsf{\dfrac{P+40P}{400}=1537.50}400P+40P=1537.50
\mathsf{\dfrac{41\,P}{400}=1537.50}40041P=1537.50
\mathsf{P=1537.50{\times}\dfrac{400}{41}}P=1537.50×41400
\mathsf{P=153750{\times}\dfrac{4}{41}}P=153750×414
\mathsf{P=3750{\times}4}P=3750×4
\mathsf{P=Rs.\,15000}P=Rs.15000
\textsf{Simple interest for 2 years}Simple interest for 2 years
\mathsf{=\dfrac{P\,n\,r}{100}}=100Pnr
\mathsf{=\dfrac{15000{\times}2{\times}5}{100}}=10015000×2×5
\mathsf{=\dfrac{15000{\times}10}{100}}=10015000×10
\mathsf{=150{\times}10}=150×10
\mathsf{=Rs.\,1500}=Rs.1500
\underline{\textsf{Answer:}}Answer:
\textsf{Simple interest for 2 years is Rs. 1500}Simple interest for 2 years is Rs. 1500
\underline{\textsf{Find more:}}Find more:
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