Question

Find the compound interest correct to the nearest rupee on 10000 for 2 and half years at 10% per annum when interest is annually

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Compound\:Interest=2960.58\:rupees}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Principal(p) = 10000 \: rupees \\ \\ \tt: \implies Time(t) = 2.5\: years \\ \\ \tt: \implies Rate\%(r) = 10\% \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Compound \: Interest(C.I) = ?$

• According to given question :

$\bold{As \: we \: know \:that} \\ \tt: \implies A = p(1 + \frac{r}{100} )^{t} \\ \\ \tt: \implies A = 10000(1 + \frac{10}{100} )^{2.5} \\ \\ \tt: \implies A = 10000 \times (1 + 0.1)^{2.5} \\ \\ \tt: \implies A = 10000 \times {(1.1)}^{2.5} \\ \\ \tt: \implies A = 10000 \times 1.269058 \\ \\ \green{\tt: \implies A = 12690.58 \: rupees} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies C.I= A - p \\ \\ \tt: \implies C.I = 12960.58 - 10000 \\ \\ \green{\tt: \implies C.I=2960.58 \: rupees}$

Answer 2

Answer:

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