Question

Find the principal when simple interest at 16% per annum for 2 years is ₹ 3840.

Answer 1

S.I = p×r×t/100

p = r×t /SI × 100

= 16×2/384000

= 32/384000

=12000

Answer 2

Answer:

Answer:

∴Principal=9,600rupees\begin{lgathered}\green{\therefore{\text{Principal=9,600\:rupees}}}\\\end{lgathered}

∴Principal=9,600rupees

Step−b−stepexplanation:‾‾\orange{\bold{\underline{\underline{Step-b-step\:explanation:}}}}

Step−b−stepexplanation:

Given:‾:⟹Time (t)= 2.5 years:⟹Simple interest(S.I) = 3840rupees:⟹Rate(r) = 16%ToFind:‾:⟹Principal(p) = ?\begin{lgathered}\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) = ?}\end{lgathered}

Given:

:⟹Time (t)= 2.5 years

:⟹Simple interest(S.I) = 3840rupees

:⟹Rate(r) = 16%

ToFind:

:⟹Principal(p) = ?

• According to given question :

Asweknowthat∘Simple interest=Principal × Rate×Time100Puttinggivenvalues:⟹S.I=p×r×t100:⟹3840=p×16×2.5100:⟹p=3840×10016×2.5:⟹p = 9600 rupees∴principal = 9,600 rupeesBasicformularelatedtoC.I∘A=p(1+r100)t\begin{lgathered}\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}\end{lgathered}

Asweknowthat

∘Simple interest=

100

Principal × Rate×Time

Puttinggivenvalues

:⟹S.I=

100

p×r×t

:⟹3840=

100

p×16×2.5

:⟹p=

16×2.5

3840×100

:⟹p = 9600 rupees

∴principal = 9,600 rupees

BasicformularelatedtoC.I

∘A=p(1+

100

r

)

t

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