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

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

1. 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

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2. [tex]\bf \underline{ \underline{\maltese\:Given} }[/tex]

   [tex] \sf \implies Principal \: (P) = Rs. \: 1000[/tex]

   [tex]\sf \implies Time \: (n)=2 \: years [/tex]

   [tex]\sf \implies Rate \: of \: interest \: (R) = 4 \: \% [/tex]

   [tex]\bf \underline{ \underline{\maltese \: To \: find } }[/tex]

   [tex]\sf \implies Compound \: interest = \: ? [/tex]

   [tex]\bf \underline{ \underline{\maltese \: Solution } }[/tex]

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

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

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

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

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

   [tex]\sf \implies 1000 \: \bigg (\dfrac{26}{25} \bigg)^{2 } [/tex]

   [tex]\sf \implies 1000 \times \dfrac{676}{625} [/tex]

   [tex]\sf \implies \dfrac{8 \times 676}{5} [/tex]

   [tex]\sf \implies \dfrac{5408}{5} = 1081.6[/tex]

   [tex] \bf Amount = Rs. \: 1081.6[/tex]

   [tex] \sf Now, \: compound \: interest = Amount – Principal[/tex]

   [tex]\sf \implies 1081.6 – 1000 = 81.6 [/tex]

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

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