Question

find the SI if principal is 15000 rate is 18% and time is one month​

Answer 1

T=1 month=1/12years

SI=P×R×T/100

SI=1500×18×1/12/100

SI=1500×18×1/12×100

SI=₹22.5

Answer 2

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

$\green{\therefore{\text{Simple\:interest=225\:rupees}}}$

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

$\green{ \underline \bold{Given : }} \\ : \implies \text{Time (t)= 1 \: month} \\ \\ : \implies \text{Principal(p) = 15,000\:rupees} \\ \\ : \implies \text{Rate(r) = 18\%} \\ \\ \red{ \underline \bold{To \: Find: }} \\ : \implies \text{Simple\:interest(S.I) = ?}$

• According to given question :

$\bold{As \: we \: know \: that } \\ \circ \: \text{Simple \: interest} = \frac{ \text{Principal }\times \text{ Rate} \times \text{Time}}{100} \\ \\ \bold{Putting \: given \: values} \\ : \implies S.I = \frac{p \times r \times t}{100} \\ \\ : \implies S.I = \frac{15,000 \times 18 \times\frac{1}{12} }{100} \\ \\ : \implies S.I = \frac{15,000 \times 18}{12\times 100 } \\ \\ \green{ : \implies \text{S.I= 225\: rupees}} \\ \\ \green{ \therefore \text{Simple\:interest = 225 \: rupees}} \\ \\ \bold{some \: basic \: formula} \\ \circ \: A = p(1 + \frac{r}{100} )^{t}$