Question

find the interest when the rate is 6.25% principal is ₹1275 and time is 1 year​

Answer 1

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

$\green{\tt{\therefore{Simple\:Interest=79.6875\:rupees}}}$

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

$\green{\underline\bold{Given :}} \\ \tt: \implies Principal(p) = 1275\: rupees \\ \\ \tt: \implies Rate\%(r) = 6.25\% \\ \\ \tt : \implies Time(t) = 1 \: years \\ \\ \red{\underline\bold{To \: Find :}} \\ \tt: \implies Simple \: Interest = ?$

• According to given question :

$\bold{As \: we \: know \:that} \\ \tt: \implies S.I = \frac{p \times r \times t }{100} \\ \\ \tt: \implies S.I= \frac{1275\times 6.25 \times 1}{100} \\ \\ \tt: \implies S.I =12.75 \times 6.25 \\ \\ \green{\tt: \implies S.I =79.6875 \: rupees} \\ \\ \green{\tt \therefore Interest \: at \: the \: end \: of \: 4th \: year \: is \: 79.6875\: rupees} \\ \\ \blue{ \bold{Some \: related \: formula}} \\ \orange{ \tt \circ \: S = p(1 + \frac{r}{100} )^{t} } \\ \\ \orange{ \tt \circ \: A = p(1 + \frac{ \frac{r}{2} }{100} )^{2t} } \\ \\ \orange{ \tt \circ \: A = p(1 + \frac{ \frac{r}{4} }{100} )^{4t} }$

Answer 2

${ \huge{ \bold{ \underline{ \underline{ \pink{Question:-}}}}}}$

Find the interest when the rate is 6.25% principal is ₹1275 and time is 1 year ..

_______________

${ \huge{ \bold{ \underline{ \underline{ \orange{Answer:-}}}}}}$

✒Given : -

• Rate(r) = 6.25%
• Principal(p) = 1275 Rupee
• Time(t) = 1 year

✒To Find : -

• Simple Interest(S.I.) = ?

${{ \large{ \bold{ \underline{ \underline{ \purple{According\:to\:the\:Question:-}}}}}}}$

✒Formula Used : -

$\leadsto\sf{{ \small{ \bold{ \bold{ \bold{ \red{Simple\:Interest=\dfrac{P\times{R}\times{T}}{100}}}}}}}}$

✒On Substituting Values : -

$\dashrightarrow\sf{S.I.=\dfrac{1275\times{6.25}\times{1}}{100\times{100}}}$

$\dashrightarrow\sf{S.I.=\cancel\dfrac{796875}{10000}}$

$\leadsto\sf{{ \large{ \boxed{ \bold{ \bold{ \green{S.I.=79.6875}}}}}}}$

So , Simple Interest is 79.6875 ..