Question

Find the principal when: simple interest as 16% per annum for 5/2 years is ₹3840

Answer 1

Step-by-step explanation:

P=SI×100/T×R

P=3840×100/(5/2)×16

P=9600

Answer 2

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

$\green{\therefore{\text{Principal=9,600\:rupees}}}$

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

$\green{ \underline \bold{Given : }} \\ : \implies \text{Time (t)= 2.5\: years} \\ \\ : \implies \text{Simple \: interest(S.I) = 3840\:rupees} \\ \\ : \implies \text{Rate(r) = 16\%} \\ \\ \red{ \underline \bold{To \: Find: }} \\ : \implies \text{Principal(p) = ?}$

• According to given question :

$\bold{As \: we \: know \: that } \\ \circ \: \text{Simple \: interest} = \frac{ \text{Principal }\times \text{ Rate} \times \text{Time}}{100} \\ \\ \bold{Putting \: given \: values} \\ : \implies S.I = \frac{p \times r \times t}{100} \\ \\ : \implies 3840= \frac{p \times 16\times2.5}{100} \\ \\ : \implies p = \frac{3840 \times 100}{16\times 2.5 } \\ \\ \green{ : \implies \text{p = 9600\: rupees}} \\ \\ \green{ \therefore \text{principal = 9,600\: rupees}} \\ \\ \bold{Basic \: formula\:related\:to\:C.I} \\ \circ \: A = p(1 + \frac{r}{100} )^{t}$