Question

in what time will 5400rs amount to 6773.76rs at 12% per annum compounded anually​

Answer 1

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

$\green{\tt{\therefore{Time=2\:years}}}$

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

$\green{ \underline \bold{Given :}} \\ \tt: {\implies Principal(p) = 5400 \: rupees} \\ \\ \tt: \implies Amount(A) = 6773.76 \: rupees \\ \\ \tt: \implies Rate\% = 12\% \\ \\ \red{ \underline \bold{To \: Find:}} \\ \tt: \implies Time(t) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies A = p(1 + \frac{r}{100} )^{t} \\ \\ \tt: \implies 6773.76 = 5400(1 + \frac{12}{100} )^{t} \\ \\ \tt: \implies \frac{6773.76}{5400} =( 1 + 0.12)^{t} \\ \\ \tt: \implies 1.2544 = (1.12)^{t} \\ \\ \tt: \implies (1.12)^{2} = {(1.12)}^{t} \\ \\ \text{Both \: side \: bases \: are \: same} \\ \text{So, \: powers \: are \: also \: same} \\\\ \tt: \implies 2 = t \\ \\ \green{\tt: \implies t = 2 \: years}$

Answer 2

Hi.

Increasing Rate :- 12%

Starting Money :- 5400

Formula :- $5400\times (1+\frac{12}{100}) ^x$

$5400\times(1+\frac{12}{100} )^x=6773.76$

$(1+\frac{12}{100} )^x=\frac{6773.76}{5400}$

$(\frac{28}{25}) ^x=\frac{784}{625}$

log$_\frac{28}{25}$ $(\frac{28}{25} )^x$ = log $_\frac{28}{25}$ $\frac{28^2}{25^2}$

x = 2 (years)