Find the compound interest on 1000 for 9 months at 4% per annum compounded quarterly

asked Aug 14, 2024 7.7k views

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$\sf Compound\:interest = P(1 + (r)/(100) )^(t) - P$

We have,

$\tt \: CI = 1000(1 + (4)/(100) ) ^{ (3)/(4) } - 1000$

$\tt \: CI = 1000(1 + \frac{ \cancel4}{ \cancel{100}} ) ^{ (3)/(4) } - 1000$

$\tt \: CI = 1000(1 + ( 1)/( 25) ) ^{ (3)/(4) } - 1000$

$\tt \: CI = 1000((26)/( 25) ) ^{ (3)/(4) } - 1000$

$\tt \: CI = 1000 \sqrt[4]{ { ((26)/(25) )}^(3) } - 1000$

$\tt \: CI = 1000 (1.02985) - 1000$

$\tt \: CI = 1029.85- 1000$

$\tt \: CI = 29.85$

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