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 ra

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​

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Josephine 1 month 2021-08-16T11:11:57+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-16T11:13:13+00:00

    \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.
    0
    2021-08-16T11:13:32+00:00

    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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    if you liked it makes it brainlist

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