Math, asked by Anonymous, 2 months ago

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\Large{\underline{\underline{\pink{\bf{ \maltese \:  \: Question :-}}}}}
A sum of money put at 11% per annum amounts to Rs. 4491 in 2 years 3 months. What will it amount to in 3 years at the same rate?


anindyaadhikari13: Please use search bar before asking questions to avoid repetitions.
anindyaadhikari13: You can view the answer here -https://brainly.in/question/29204900

Answers

Answered by SƬᏗᏒᏇᏗƦƦᎥᎧƦ
339

 \bold{\underline{\underline{ \Large \mathfrak{Given \:  information  \: to \:  us:-}}}}

  • Rate of interest = 11%
  • Amount = Rs.4491
  • Time period = 2 years 3 months

 \underline{\underline{ \Large \mathfrak{Need \: to \: be \: calculated:-}}}

  • How much will it amount to in 3 years with the same rate of interest.

 \underline{\underline{ \Large \mathfrak{Required \: formulas \: to \: solve \: this \: question:-}}}

Simple Interest :

  •  \leadsto \:  \boxed{\:  \bf{Simple  \: Interest \:  =   \:  \dfrac{Principal \:  \times  \: Rate \:  \times \: Time }{100} }}

Amount:

  • \leadsto  \boxed{ \bf{Amount  \: =  \: Principal  \: +  \: Interest}}

 \underline{\underline{ \Large \mathfrak{Step \: by \: step \: explaination:-}}}

______

 \large {\bf{ \underbrace {\underline{{understanding \: the \: question \: and \: solving \: way}}}}}

»» Here first of all we would be calculating the simple interest taken by using the given formula and applying all the required values. But here we don't know the value of principal so we would assume it be y and substitute the values.

»» After that we would use the formula of calculating amount where, amount is already given to us that usage of formula would gave us the value of y that is Principal.

»» Further we would again use the given formula of calculating the Simple Interest and apply all the values and the value of y as the calculated Principal and solve it further.

______

Now let's solve it!!

Applying the formula of calculating the Simple Interest and inserting the values. With converting the time into proper time as 2 years 3 months can't be used directly.

  •  \hookrightarrow \:  \bf{1 \:year \:  =  \:12 \: months }
  •  \hookrightarrow \:  \bf{2 \:years \:  =  \:12 \:  \times  \: 2 \: months }
  •  \hookrightarrow \:  \bf{2 \:years \:  =  \:24 \:months }
  • \hookrightarrow \:  \bf{24  \: months\:  +  \: 3} \: months

Converting 27 months into year:

  • \hookrightarrow \:  \bf{ \dfrac{27}{12} }

  • \hookrightarrow \:  \bf{ \dfrac{ \cancel{27}}{ \cancel{12}} }

  •  \hookrightarrow \:  \bf{ \dfrac{9}{4}}

Now applying values in the given formula:

  •  \hookrightarrow \:  \bf{Simple  \: Interest \:  =   \:  \dfrac{Principal \:  \times  \: Rate \:  \times \: Time }{100} }

  •  \hookrightarrow \:  \bf{Simple  \: Interest \:  =   \:  \dfrac{y \:  \times  \: 11 \:  \times \:  9}{100 \times \: 4} }

  • \hookrightarrow \:  \bf{Simple  \: Interest \:  =   \:  \dfrac{99y}{400} }

Now, using the formula of calculating the amount and substituting the required values:

Given,

  • Amount = Rs.4491
  • Principal = y

We know that,

  • Amount = Simple Interest + Principal

Now,

  • \hookrightarrow \:  \bf{Amount \:  =  \: y \:  +  \:  \dfrac{99y}{400} }

  • \hookrightarrow \:  \bf{Amount \:  =  \:  \dfrac{y \:  \times 400}{1 \times 400}  \:  +  \:  \dfrac{99y}{400} }

  • \hookrightarrow \:  \bf{Amount \:  =  \:  \dfrac{499y}{400}}

Applying the value of amount:

  • \hookrightarrow \:  \bf{4491 \:  =  \:  \dfrac{499y}{400}}

Cross multiplying:

  • \hookrightarrow \:  \bf{y \:  =  \:  \dfrac{4491 \:  \times  \: 400}{499} }

  • \hookrightarrow \:  \bf{y \:  =  \:  \dfrac{ \cancel{4491} \:  \times  \: 400}{ \cancel{499}} }

Thus, value of y is 3600. Again using the formula of Simple Interest and formula of calculating the amount. It's the last step then we would get our final answer.

Given,

  • Principal = Rs.3600
  • Time = 3 years (According to the statement)
  • Rate = 11%

Calculations:

\hookrightarrow \:  \bf{Simple  \: Interest \:  =   \:  \dfrac{3600 \: \times 3 \: \times 11}{100} }

\hookrightarrow \:  \bf{Simple  \: Interest \:  =   \:  \dfrac{\cancel{3600} \: \times 3 \: \times 11}{\cancel{100}} }

\hookrightarrow \:  \bf{Simple  \: Interest \:  =   \:  \dfrac{900 \: \times 3 \: \times 11}{1}}

\hookrightarrow \:  \bf{ Simple \: Interest \: = \: 36 × 3 ×11}

\hookrightarrow \:  \bf{ Simple \: Interest \: = \: 36\: \times 33}

\hookrightarrow \:  \bf{ Simple \: Interest \: = \: 1188}

Here again using the given formula of amount to calculate the amount of 3 years:

Given,

  • Principal = Rs.3600
  • Simple Interest = Rs.1188

Formula,

  • Amount = Principal + Simple Interest

Substituting values:

  • \hookrightarrow \:  \bf{3600 + 1188}
  • \hookrightarrow \:  \bf{4788}

 \bold{\underline{\underline{ \Large \mathfrak{Conclusion:-}}}}

  • It will amount to Rs.4788 in 3 years at the same rate of interest.

Answered by Anonymous
388

Answer:

\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}\end{gathered}

  • Amount = Rs.4491
  • Rate of Interest = 11%
  • Time = 2 years 3 months

\begin{gathered}\end{gathered}

\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{To Find :}}}}}}\end{gathered}

  • Amount of 3 years

\begin{gathered}\end{gathered}

\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Using Formula :}}}}}}\end{gathered}

\quad\dag{\underline{\boxed{\sf{S.I = Amount - Principle}}}}

\quad\dag\underline{\boxed{\sf{P =  \dfrac{S.I  \times 100}{R \times T}}}}

\quad\dag{\underline{\boxed{\sf{S.I = \dfrac{ P \times R \times T}{100}}}}}

\quad\dag{\underline{\boxed{\sf{A = P +  S.I}}}}

Where

  • \leadsto P = Principle
  • \leadsto S.I = Simple Interest
  • \leadsto R = Rate of Interest
  • \leadsto T = Time Period
  • \leadsto A = Amount

\begin{gathered}\end{gathered}

\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}\end{gathered}

\bigstar{\underline{\underline{\frak{\red{Here}}}}}

  • Amount = Rs.4491
  • Rate of Interest = 11%
  • Time = 2 years 3 months

\begin{gathered}\end{gathered}

\bigstar\underline{\underline{\frak{\red{Converting  \: the  \: time \:  into  \: years.}}}}

{\bf{\dashrightarrow{1 \:  year = 12 \:  months}}}

  • Converting 2 years 3 months into years

{\sf{\dashrightarrow{Time = 2 \:   year  \: 3 \:  months}}}

{\sf{\dashrightarrow{Time = (2 \times 12) +  \: 3 \:  months}}}

{\sf{\dashrightarrow{Time = 24+  \: 3 \:  months}}}

{\sf{\dashrightarrow{Time = 27\:  months}}}

{\sf{\dashrightarrow{Time = \dfrac{27}{12} \: years}}}

{\sf{\dashrightarrow{Time =  \cancel{\dfrac{27}{12}}\: years}}}

{\sf{\dashrightarrow{Time = \dfrac{9}{4} \: years}}}

{\dag{\underline{\boxed{\sf{\purple{Time = \dfrac{9}{4} \: years}}}}}}

\begin{gathered}\end{gathered}

\bigstar{\underline{\underline{\frak{\red{Calculating \:  the  \: Simple \:  Interest }}}}}

\leadsto{\bf{S.I = A - P}}

  • Substituting the values

\leadsto{\sf{S.I = 4491 - P}}

{\dag{\underline{\boxed{\sf{\purple{Simple \: Interest= 4491 - P}}}}}}

\begin{gathered}\end{gathered}

\bigstar{\underline{\underline{\frak{\red{Now, Finding  \: the \:  Principle }}}}}

: \longmapsto{\bf{P =  \dfrac{S.I  \times 100}{R \times T}}}

  • Substituting the values

: \longmapsto{\sf{P =  \dfrac{(4491 -P ) \times 100}{11 \times\dfrac{9}{4} }}}

: \longmapsto{\sf{P =  \dfrac{(4491 -P ) \times 100 \times 4}{11 \times{9}}}}

: \longmapsto{\sf{P =  \dfrac{(4491 -P ) \times  400}{99}}}

{: \longmapsto{\sf{P =  \dfrac{(4491 \times 400) - (400 \times P )}{99}}}}

{: \longmapsto{\sf{P =  \dfrac{1796400  -  400P }{99}}}}

{: \longmapsto{\sf{99 \times P = {1796400  -  400P }}}}

{: \longmapsto{\sf{99P = {1796400  -  400P }}}}

{: \longmapsto{\sf{99P + 400P= {1796400}}}}

{: \longmapsto{\sf{499P= {1796400}}}}

{: \longmapsto{\sf{P= \dfrac{1796400}{499}}}}

{: \longmapsto{\sf{P=  \cancel{\dfrac{1796400}{499}}}}}

{: \longmapsto{\sf{P= Rs.3600}}}

\dag{\underline{\boxed{\sf{\purple{Principle= Rs.3600}}}}}

  • The Principle is Rs.3600.

\begin{gathered}\end{gathered}

\quad▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

\begin{gathered}\end{gathered}

 \bigstar \underline{\underline{\sf{\red{Now, Calculating \:  the \:  simple \:  Interest  \: on  \: 3 \:  years  \: at \:  rate  \: of \:  11 \%}}}}

{:  \implies{\bf{S.I = \dfrac{ P \times R \times T}{100}}}}

  • Substituting the values

{ :  \implies{\sf{S.I = \dfrac{ 3600 \times 11 \times 3}{100}}}}

{ :  \implies{\sf{S.I = \dfrac{\cancel{3600} \times 11 \times 3}{\cancel{100}}}}}

{ :  \implies{\sf{S.I ={ 36 \times 33}}}}

{ :  \implies{\sf{S.I ={Rs.1188}}}}

 \dag{\underline{\boxed{\sf{\purple{S.I ={Rs.1188}}}}}}

\begin{gathered}\end{gathered}

\bigstar{\underline{\underline{\sf{\red{Finding \:  the \:  Amount \:  for \:  3 \:  years.}}}}}

: \Rightarrow{\bf{A = P +  S.I}}

  • Substituting the values

: \Rightarrow{\sf{A = 3600 +  1188}}

: \Rightarrow{\sf{A =Rs.4788}}

 \dag{\underline{\boxed{\sf{\purple{A =Rs.4788}}}}}

\begin{gathered}\end{gathered}

\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Answer :}}}}}}\end{gathered}

  • The Amount is Rs.4788 in 3 years at the rate of 11%..

\begin{gathered}\end{gathered}

\begin{gathered}{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}\end{gathered}

  • You can solve this question by second method also with is already answered by @poonamtikka87100.
  • Here is the question link : https://brainly.in/question/42759939
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