Question

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

Answer 1

Answer:

to find the sum of simple interest = p×100/t×r

so 8400×100/2 = 16,800

so the simple interest sums = 16,800

then rate = i×100/p×t

8400×100/16800×2 = 50/2

25%

so the rate of interest is 25%

and now compound interest = 16,800

+ interest

16,800+8652

25452

so the compound interest sums = 25,452

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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

• Let sum of money be P
• And rate of interest be R

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

$\begin{gathered}\\ \longrightarrow \:\sf S.I = 8400 \end{gathered}$

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

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

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

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

$\begin{gathered}\\ \longrightarrow \:\sf PR = 8400\:\times\:50\end{gathered}$

$\begin{gathered}\\ \longrightarrow \:\sf PR = 420000 \:\dashrightarrow\:(1) \end{gathered}$

Now,

$\begin{gathered}\\ \longrightarrow \:\sf C.I = 8652\end{gathered}$

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

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

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

$\begin{gathered}\\ \longrightarrow \:\sf P[(1)^2 + (0.01R)^2 + 2(1)(0.01R)] - P = 8652\end{gathered}$

$\begin{gathered}\\ \longrightarrow \:\sf P(1 + 0.0001R^2 + 0.02R - P = 8652\end{gathered}$

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

$\begin{gathered}\\ \longrightarrow \:\sf 0.0001(PR)R + 0.02PR = 8652\end{gathered}$

• Putting value of PR from (1) ::

$\begin{gathered}\\ \longrightarrow \:\sf 0.0001(420000)R + 0.02(420000) = 8652\end{gathered}$

$\begin{gathered}\\ \longrightarrow \:\sf 42R + 8400 = 8652\end{gathered}$

$\begin{gathered}\\ \longrightarrow \:\sf 42R = 8652 - 8400\end{gathered}$

$\begin{gathered}\\ \longrightarrow \:\sf 42R = 252\end{gathered}$

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

$\begin{gathered}\\ \longrightarrow \:\bf \orange{R = 6}\end{gathered}$

Therefore, Rate of interest is 6 % per annum.

• Putting value of R in (1) ::

$\begin{gathered}\\ \longrightarrow \:\sf P(6) = 420000\end{gathered}$

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

$\begin{gathered}\\ \longrightarrow \:\bf\orange{ P = 70000}\end{gathered}$

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