Question

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

Answer 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

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf Amount = Rs. \: 1081.6$

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

$\sf \implies 1081.6 - 1000 = 81.6$

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