Question

principal=50,rate=10%,time=9years,find A and SI​

Answer 1

Answer:

SI=PRT/100 SI=(50×10×9)/100=45rs. Amount =P+SI=50+45=95

Answer 2

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

$\green{\therefore{\text{Simple\:interest=45\:rupees}}}$

$\green{\therefore{\text{Amount=95\:rupees}}}$

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

$\green{ \underline \bold{Given : }} \\ : \implies \text{Time (t)= 9 \:years} \\ \\ : \implies \text{Principal(p) = 50\:rupees} \\ \\ : \implies \text{Rate(r) = 10\%} \\ \\ \red{ \underline \bold{To \: Find: }} \\ : \implies \text{Simple\:interest(S.I) = ?}\\\\ :\implies \text{Amount(A)=?}$

• According to given question :

$\bold{As \: we \: know \: that } \\ \circ \: \text{Simple \: interest} = \frac{ \text{Principal }\times \text{ Rate} \times \text{Time}}{100} \\ \\ \bold{Putting \: given \: values} \\ : \implies S.I = \frac{p \times r \times t}{100} \\ \\ : \implies S.I = \frac{50\times 10 \times9 }{100} \\ \\ : \implies S.I = \frac{500\times 9}{ 100 } \\ \\ \green{ : \implies \text{S.I= 45\: rupees}} \\ \\ \green{ \therefore \text{Simple\:interest = 45 \: rupees}} \\ \\ :\implies \text{Amount=S.I+Principal}\\\\ \green{:\implies \text{Amount=45+50=95\:ruppes}}\\\\ \bold{Basic \: formula\:related\:to\:C.I} \\ \circ \: A = p(1 + \frac{r}{100} )^{t}$