Question

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 1

$\sf\small\underline{Let:-}$

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

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

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

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

$\sf\small\underline{Solution:-}$

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):-]

$\tt{\longrightarrow Pr=420000}$

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

$\tt{\longrightarrow P=70000}$

$\sf\large{Hence,}$

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

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

Answer 2

$\LARGE\boxed{\textsf{\textbf{\pink{Given\::-}}}}$

$\:$

• Simple interest and compound interest of a certain sum of money for two years are Rs. 8400 and Rs. 8652.

$\:$

$\LARGE\boxed{\textsf{\textbf{\green{To\:Find\::-}}}}$

$\:$

• Sum of money and rate of interest?

$\:$

$\LARGE\boxed{\textsf{\textbf{\blue{Solution\::-}}}}$

$\:$

$\underline{\sf{\bigstar\:Formulae\:used\::-}}$

$\:$

$\quad\odot\:{\underline{\boxed{\bf{\red{S.I = \dfrac{P\:\times\:R\:\times\:T}{100}}}}}}$

$\:$

$\quad\odot\:{\underline{\boxed{\bf{\purple{C.I = {P\bigg(1 + \dfrac{R}{100}\bigg)}^{T} - P}}}}}$

$\:$

Where,

$\:$

• P = Principal (sum of money)

• R = Rate of interest

• T = Time

• S.I = Simple interest

• C.I = Compound interest

$\:$

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$\:$

• Let sum of money be P

• And rate of interest be R

$\:$

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$\:$

$\underline{\sf{\bigstar\:According\:to\:the\:given\:Question\::-}}$

$\\ \longrightarrow \:\sf S.I = 8400$

$\\ \longrightarrow \:\sf \dfrac{P\:\times\:R\:\times\:T}{100} = 8400$

$\\ \longrightarrow \:\sf \dfrac{P\:\times\:R\:\times\:2}{100} = 8400$

$\\ \longrightarrow \:\sf \dfrac{\cancel{2}PR}{\cancel{100}} = 8400$

$\\ \longrightarrow \:\sf \dfrac{PR}{50} = 8400$

$\\ \longrightarrow \:\sf PR = 8400\:\times\:50$

$\\ \longrightarrow \:\sf PR = 420000 \:\dashrightarrow\:(1)$

$\:$

Now,

$\\ \longrightarrow \:\sf C.I = 8652$

$\\ \longrightarrow \:\sf {P\bigg(1 + \dfrac{R}{100}\bigg)}^{T} - P = 8652$

$\\ \longrightarrow \:\sf {P(1 + 0.01R)}^{2} - P = 8652$

$\:$

Using identity, (a + b)² = a² + b² + 2ab on (1 + 0.01R)² ::

$\\ \longrightarrow \:\sf P[(1)^2 + (0.01R)^2 + 2(1)(0.01R)] - P = 8652$

$\\ \longrightarrow \:\sf P(1 + 0.0001R^2 + 0.02R - P = 8652$

$\\ \longrightarrow \:\sf \cancel{P} + 0.0001PR^2 + 0.02PR - \cancel{P} = 8652$

$\\ \longrightarrow \:\sf 0.0001(PR)R + 0.02PR = 8652$

$\:$

Putting value of PR from (1) ::

$\\ \longrightarrow \:\sf 0.0001(420000)R + 0.02(420000) = 8652$

$\\ \longrightarrow \:\sf 42R + 8400 = 8652$

$\\ \longrightarrow \:\sf 42R = 8652 - 8400$

$\\ \longrightarrow \:\sf 42R = 252$

$\\ \longrightarrow \:\sf R = {\cancel{\dfrac{252}{42}}}$

$\\ \longrightarrow \:\bf \orange{R = 6}$

$\:$

• Therefore, Rate of interest is 6 % per annum.

$\:$

Putting value of R in (1) ::

$\\ \longrightarrow \:\sf P(6) = 420000$

$\\ \longrightarrow \:\sf P = {\cancel{\dfrac{420000}{6}}}$

$\\ \longrightarrow \:\bf\orange{ P = 70000}$

$\:$

• Therefore, sum of money is Rs. 70,000.

$\:$

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