Question

find the simple interest on Rs 12000 and 2years at 6℅per annum​

Answer 1

Answer:

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Answer 2

$given \:$

$\\principle \: \: \: \: \: \: (p) \: \: \: = \: r s \: 12000 \\ simple \: interest \: rate \: = \: 6\%p.a \\ time \: \: \: period \: \: \: n \: \: = \: 2 \: \: \: years$

$\: therefore \:$

$simple \: interest \: si \: at \: 6\% \: for \: 2 \\ years \: = 2 \times \frac{12000 \times 6}{100} \\ =rs .1440$

$if \: he \: would \: have \: borrowed \: it \: at \\ a \: compound \: interest \: rate \: (r \: ) = \\ 6\%p.a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$we \: know \: amount \: when \: interest \: \\ is \: compounded \: anually \: (a) = \: \: \: \: \: \ \: \: \: \\ a = p(1 + \frac{r}{100} ) ^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ therefore \: a = 1200(1 + \frac{6}{100} ) ^{2} \: \: \: \: \: \: \\ \: \: \: \: = rs \: 13483.20$

$therefore \: compound \: interest \:ci = \\ a - p \: = r \: (13483.20 - 12000) \: \: \: \: \: \: \: \: \: \\ = rs.1483.20 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ therefore \: \: he \: would \: have \: to \: pay \: \: \: \: \: \: \: \\ rs(1483.20 - 1440 = rs \: 43.20 \: extra \: \:$