Question

The S.I. on ₹ 1200 at 15% per annum for 4½ years​

Answer 1

$\begin{gathered}\begin{gathered}\begin{gathered} \begin{gathered} \begin{gathered}\begin{gathered}\\\;\underbrace{\underline{\sf{Question:-}}}\end{gathered}\end{gathered} \end{gathered} \end{gathered} \end{gathered} \end{gathered}$

The S.I. on ₹ 1200 at 15% per annum for 4½ years

$\begin{gathered}\begin{gathered}\begin{gathered} \begin{gathered} \begin{gathered}\begin{gathered}\\\;\underbrace{\underline{\sf{understand \: the \: concept :-}}}\end{gathered}\end{gathered} \end{gathered} \end{gathered} \end{gathered} \end{gathered}$

Here we need to find out the S.I. on rupees 1200 for 4½ years at 15% per annum .This can be done by using the formula of S.I

Let's do it !!

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★ Formula Used :-

$\begin{gathered}\begin{gathered}\\\;\boxed{\sf{\pink{simple \: interest\;=\;\bf{\bigg(\;\dfrac{p \times r \times t}{100}\bigg)}}}}\end{gathered} \end{gathered}$

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★ Solution :-

Given,

» Principal = ( P ) = Rs. 1200

» Rate = 15%

» Years = 4½ years = 9/2 years

To find ,

» Simple Interest

Now , by using the formula of simple interest

$\begin{gathered}\begin{gathered}\\\; \longrightarrow{\sf{simple \: interest\;=\;\bf{\bigg(\;\dfrac{p \times r \times t}{100}\bigg)}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\; \longrightarrow{\sf{simple \: interest\;=\;\bf{\bigg(\;\dfrac{1200 \times 15 \times 9}{100 \times 2}\bigg)}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\; \longrightarrow{\sf{simple \: interest\;=\;\bf{\bigg(\;\dfrac{12 \cancel{00 }\times 15 \times 9}{ \cancel{100 }\times 2}\bigg)}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\; \longrightarrow{\sf{simple \: interest\;=\;\bf{\bigg(\;\dfrac{ \cancel{12 } \times 15 \times 9}{ \cancel{ 2}}\bigg)}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\; \longrightarrow{\sf{simple \: interest\;=\;\bf{\bigg(\;{ 6 \times 15 \times 9} \bigg)}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\; \longrightarrow{\sf{simple \: interest\;=\;\bf{\;{rs \: 810} }}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\underline{\boxed{\tt{\odot\;\;Hence,\;\;simple \: interest\;=\;\bf{\blue{Rs\;\;810}}}}}\end{gathered} \end{gathered}$

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★ More to know :-

if ,

• Principal = P

• Rate = R % per annum

• Time = T years

• Simple interest = SI

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; {simple \: interest\;=\;\bf{\bigg(\;\dfrac{p \times r \times t}{100}\bigg)}}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; {principal\;=\;\bf{\bigg(\;\dfrac{si \times 100 }{r \times t}\bigg)}}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; {rate\;=\;\bf{\bigg(\;\dfrac{si \times 100 }{p \times t}\bigg)}}}}\end{gathered} \end{gathered}$

$\begin{gathered}\begin{gathered}\\\;\sf{\gray{\leadsto\;\; {time\;=\;\bf{\bigg(\;\dfrac{si \times 100 }{p \times r}\bigg)}}}}\end{gathered} \end{gathered}$

Answer 2

QUESTION -

The S.I. on ₹ 1200 at 15% per annum for 4½ years

ANSWER -

rs 810