Question

What sum of money, invested at 9%
interest per year, is needed to provide an
income of $45 per year?​

Answer 1

si = simple interest = $45

r = 9%

t = 1

p = ?

$si = \frac{p \times r \times t}{100} \\ \\ 45 = \frac{p \times 9 \times 1}{100} \\ \\ \frac{45 \times 100}{9 \times 1} = p \\ \\ p = 500$

therefore,

p = $500

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