Find the compound interest on Rs. 1000 for two years at 4% per annuall​

Question

Find the compound interest on Rs. 1000 for two years at 4% per annuall​

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Faith 5 months 2021-06-30T07:45:47+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-06-30T07:46:58+00:00

    Answer:

    ₹81.60

    Step-by-step explanation:

    Principal (P) = ₹1000

    Rate (R) = 4%

    Time (n) = 2 years

    Amount (A) = P (1 + R/100)ⁿ

    => A = 1000 (1 + 4/100)²

    => A = 1000 (26/25)²

    => A = ₹1081.60

    Compound Interest = A-P = ₹1081.60 – ₹1000 = ₹81.60

    0
    2021-06-30T07:47:18+00:00

    \bf \underline{ \underline{\maltese\:Given} }

     \sf \implies Principal  \: (P) = Rs.  \: 1000

    \sf \implies Time  \: (n)=2  \: years

    \sf \implies Rate  \: of  \: interest  \: (R)  = 4 \:  \%

    \bf \underline{ \underline{\maltese \: To  \: find } }

    \sf \implies Compound \:  interest = \:  ?

    \bf \underline{ \underline{\maltese \: Solution } }

     \sf To  \: find  \: the  \: C.I, \:  first \:  of \:  all  \: we  \: have  \: to \:  find  \: the  \: amount.

     \underline{ \boxed{ \sf Amount = Principal  \:  \bigg(1 +  \dfrac{Rate }{100}  \bigg)^{Time } }}

    \sf  \implies  Amount =1000   \: \bigg(1 +  \dfrac{4}{100}  \bigg)^{2 }

    \sf  \implies  1000  \:  \bigg(1 +  \dfrac{1}{25}  \bigg)^{2 }

    \sf  \implies  1000   \: \bigg(\dfrac{25 + 1}{25}  \bigg)^{2 }

    \sf  \implies 1000   \: \bigg (\dfrac{26}{25}  \bigg)^{2 }

    \sf  \implies  1000   \times  \dfrac{676}{625}

    \sf  \implies \dfrac{8 \times 676}{5}

    \sf  \implies  \dfrac{5408}{5}  = 1081.6

     \bf Amount = Rs.  \: 1081.6

     \sf  Now, \:  compound  \: interest = Amount - Principal

    \sf  \implies   1081.6 - 1000 = 81.6

     \underline{ \boxed{ \bf \red{Therefore,  \:compound  \: interest = Rs. \:  81.6}}}

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