1.

If simple interest and compound interest of a certain sum of money for two years are RS. 8400 and RS. 8652 ,then find the sum of money and the rate of interest.

Answer»

\sf\small\underline{Let:-}

\tt{\implies The\:sum\:of\: money=P}

\tt{\implies The\:rate\:of\: interest=r}

\sf\small\underline{To\: Find:-}

\tt{\implies The\:sum\:and\:rate=?}

\sf\small\underline{Solution:-}

To CALCULATE the sum and rate at first we have to set up equation by applying formula of simple interest and compound interest with the help of GIVEN clue.

\rm\large\underline{Calculation\:for\:SI:-}

\tt\small\underline{Sum=P\:\:T=2years\:\:R=r\:\:SI=Rs.8400:-}

\tt{\longrightarrow SI=\dfrac{P*T*R}{100}}

\tt{\longrightarrow 8400=\dfrac{P*2*R}{100}}

\tt{\longrightarrow 2Pr=840000}

\tt{\longrightarrow Pr=420000-----(i)}

\rm\large\underline{Calculation\:for\:CI:-}

\tt\small\underline{Sum=P\:\:T=2years\:\:R=r\:\:CI=Rs.8652:-}

\tt{\longrightarrow CI=P\bigg[1+\dfrac{r}{100}\bigg]^n-P}

\tt{\longrightarrow 8652=P\bigg[1+\dfrac{r}{100}\bigg]^2-P}

\tt{\longrightarrow 8652=P\bigg[\dfrac{100+r}{100}\bigg]^2-P}

\tt{\longrightarrow 8652=P\bigg[\dfrac{10000+200r+r^2}{10000}\bigg]-P}

\tt{\longrightarrow 8652=\dfrac{10000P+200Pr+Pr^2}{10000}-P}

\tt{\longrightarrow 8652=\dfrac{10000P+200Pr+Pr^2-10000P}{10000}}

\tt{\longrightarrow 8652*10000=200Pr+Pr^2}

\tt{\longrightarrow 8652*10000=Pr(200+r)----(ii)}

  • Substituting the value of Pr = 420000 :-]

\tt{\longrightarrow 8652*10000=420000(200+r)}

\tt{\longrightarrow 8652=42(200+r)}

\tt{\longrightarrow (200+r)=8652\div\:42}

\tt{\longrightarrow 200+r=206}

\tt{\longrightarrow r=206-200}

\tt{\longrightarrow r=6\%}

  • PUTTING the value of r=6 in eq (i):-]

\tt{\longrightarrow Pr=420000}

\tt{\longrightarrow P(6)=420000}

\tt{\longrightarrow P=70000}

\sf\large{Hence,}

\tt\red{\implies The\:sum\:of\: money=Rs.70000}

\tt\blue{\implies The\:rate\:of\: interest=6\%}



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