Question

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

Answer 1

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

We have,

• P = $1000
• R = 4%
• T = 9/12 = 3/4

`\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`