Question

At the rate of 8% per annum simple interest a sum of Rs.2000 will earn how much interest by the end of 4 years?

Answer 1

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

$\green{\tt{\therefore{Simple\:Interest=640\:rupees}}}$

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

$\green{\underline\bold{Given :}} \\ \tt: \implies Principal(p) = 2000 \: rupees \\ \\ \tt: \implies Rate\%(r) = 8\% \\ \\ \tt : \implies Time(t) = 4 \: years \\ \\ \red{\underline\bold{To \: Find :}} \\ \tt: \implies Simple \: Interest = ?$

• According to given question :

$\bold{As \: we \: know \:that} \\ \tt: \implies S.I = \frac{p \times r \times t }{100} \\ \\ \tt: \implies S.I= \frac{2000 \times 8 \times 4}{100} \\ \\ \tt: \implies S.I =20 \times 32 \\ \\ \green{\tt: \implies S.I =640 \: rupees} \\ \\ \green{\tt \therefore Interest \: at \: the \: end \: of \: 4th \: year \: is \: 640 \: rupees} \\ \\ \blue{ \bold{Some \: related \: formula}} \\ \orange{ \tt \circ \: S = p(1 + \frac{r}{100} )^{t} } \\ \\ \orange{ \tt \circ \: A = p(1 + \frac{ \frac{r}{2} }{100} )^{2t} } \\ \\ \orange{ \tt \circ \: A = p(1 + \frac{ \frac{r}{4} }{100} )^{4t} }$

Answer 2

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

$\green{\tt{\therefore{Simple\:Interest=640\:rupees}}}$

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

$\green{\underline\bold{Given :}}$

$\tt: \implies Principal(p) = 2000 \: rupees$

$\tt: \implies Rate\%(r) = 8\%$

$\tt : \implies Time(t) = 4 \: years$

$\red{\underline\bold{To \: Find :}}$

$\tt: \implies Simple \: Interest = ?$

• Given Question

$\bold{As \: we \: know \:that}$

$\tt: \implies S.I = \frac{p \times r \times t }{100}$

$\tt: \implies S.I= \frac{2000 \times 8 \times 4}{100}$

$\tt: \implies S.I =20 \times 32$

$\green{\tt: \implies S.I =640 \: rupees}$

$\green{\tt \therefore Interest \: at \: the \: end \: of \: 4th \: year \: is \: 640 \: rupees}$

$\blue{ \bold{Some \: related \: formula}}$

$\orange{ \tt \circ \: S = p(1 + \frac{r}{100} )^{t} }$

$\orange{ \tt \circ \: A = p(1 + \frac{ \frac{r}{2} }{100} )^{2t} }$

$\orange{ \tt \circ \: A = p(1 + \frac{ \frac{r}{4} }{100} )^{4t} }$