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.​

Answer 1

Given info:

Compound Interest = Rs.1290

Rate of Interest = 15% p.a

Time = 2 years

Formula we have to know:-

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

Where

C.I = Compound Interest

P = Principle

R = Rate of Interest

N = Time

$\textsf{ \underline{Solution-}}$

Here

C.I = Rs.1290

R = 15%

N = 2 years

Principle = ?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence,

The sum (Principle) is Rs.4000.