Question

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Answer 1

Answer:

Given details are, Rate = 15 % per annum

Time = 2 years

CI = Rs 1290

By using the formula,

CI = P [(1 + R/100)n – 1] 1290

P [(1 + 15/100)2 – 1] 1290

P [0.3225]

P = 1290/0.3225 = 4000

Answer 2

Answer:

⚘ Question :-

➳ Rachna borrowed a certain sum at the rate of 15% per annum. if she paid at the end of the two years Rs.1290 as interest Compounded annually, find the sum she borrowed..

$\begin{gathered}\end{gathered}$

⚘ Solution :-

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

• ➳ Compound Interest = Rs.1290
• ➳ Rate of Interest = 15% p.a
• ➳ Time = 2 years

$\begin{gathered}\end{gathered}$

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

• ➳ The Sum (Principle)

$\begin{gathered}\end{gathered}$

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

$\dag{\underline{\boxed{\sf{C.I = P \bigg(1 + \dfrac{R}{100} \bigg)^{n} - 1}}}}$

$\red\bigstar$ Where

• ➽ C.I = Compound Interest
• ➽ P = Principle
• ➽ R = Rate of Interest
• ➽ N = Time

$\begin{gathered}\end{gathered}$

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

$\red\bigstar$ Here

• ➽ C.I = Rs.1290
• ➽ R = 15%
• ➽ N = 2 years
• ➽ Principle = ?

$\begin{gathered}\end{gathered}$

$\red\bigstar$ Now,Calculating the sum (Principle) borrowed by Rachna

$\quad{: \implies{\sf{C.I = \Bigg[P \bigg(1 + \dfrac{R}{100} \bigg)^{n} - 1 \Bigg]}}}$

• Substituting the given values

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(1 + \dfrac{15}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{(1 \times 100) + 15}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{115}{100} \bigg)^{2} - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{115}{100} \times \dfrac{115}{100} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{13225}{10000} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg( \cancel{\dfrac{13225}{10000}} \bigg) - 1 \Bigg]}}}$

$\quad{: \implies{\sf{1290 = \Bigg[P \bigg({1.3225 - 1} \bigg) \Bigg]}}}$

$\quad{: \implies{\sf{1290 = P \times {0.3225}}}}$

$\quad{: \implies{\sf{\dfrac{1290}{0.3225} = P}}}$

$\quad{: \implies{\sf{\dfrac{1290 \times 1000}{0.3225 \times 1000} = P}}}$

$\quad{: \implies{\sf{\dfrac{12900000}{3225} = P}}}$

$\quad{: \implies{\sf{\cancel{\dfrac{12900000}{3225}} = P}}}$

$\quad{: \implies{\sf{Rs.4000 = P}}}$

$\quad{\dag{\underline{\boxed{\tt{\blue{Principle} = \purple{Rs.4000 }}}}}}$

$\begin{gathered}\end{gathered}$

$\red\bigstar$ Hence,

• ➳ The sum (Principle) is Rs.4000.

$\begin{gathered}\end{gathered}$

$\begin{gathered}{\textsf{\textbf{\underline{\underline{Learn More :}}}}}\end{gathered}$

$\quad\circ{\underline{\boxed{\sf{\pink{A ={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\quad\circ{\underline{\boxed{\sf{\pink{Amount = Principle + Interest}}}}}$

$\quad\circ{\underline{\boxed{\sf{\pink{ P=Amount - Interest }}}}}$

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

$\quad\circ{\underline{\boxed{\sf{\pink{P = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\quad\circ{\underline{\boxed{\sf{\pink{P = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$

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