Given info:
Compound Interest = Rs.1290
Rate of Interest = 15% p.a
Time = 2 years
[tex] \dag{\underline{\boxed{\sf{C.I = P \bigg(1 + \dfrac{R}{100} \bigg)^{n} - 1}}}}[/tex]
Where
C.I = Compound Interest
P = Principle
R = Rate of Interest
N = Time
[tex]\textsf{ \underline{Solution-}}\\[/tex]
Here
C.I = Rs.1290
R = 15%
N = 2 years
Principle = ?
Now,Calculating the sum (Principle) borrowed by Rachna
[tex] \quad{: \implies{\sf{C.I = \Bigg[P \bigg(1 + \dfrac{R}{100} \bigg)^{n} - 1 \Bigg]}}}[/tex]
Substituting the given values
[tex]\quad{: \implies{\sf{1290 = \Bigg[P \bigg(1 + \dfrac{15}{100} \bigg)^{2} - 1 \Bigg]}}}[/tex]
[tex]\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{(1 \times 100) + 15}{100} \bigg)^{2} - 1 \Bigg]}}}[/tex]
[tex]\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{115}{100} \bigg)^{2} - 1 \Bigg]}}}[/tex]
[tex]\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{115}{100} \times \dfrac{115}{100} \bigg) - 1 \Bigg]}}}[/tex]
[tex]\quad{: \implies{\sf{1290 = \Bigg[P \bigg(\dfrac{13225}{10000} \bigg) - 1 \Bigg]}}}[/tex]
[tex]\quad{: \implies{\sf{1290 = \Bigg[P \bigg( \cancel{\dfrac{13225}{10000}} \bigg) - 1 \Bigg]}}}[/tex]
[tex]\quad{: \implies{\sf{1290 = \Bigg[P \bigg({1.3225 - 1} \bigg) \Bigg]}}}[/tex]
[tex]\quad{: \implies{\sf{1290 = P \times {0.3225}}}}[/tex]
[tex]\quad{: \implies{\sf{\dfrac{1290}{0.3225} = P}}}[/tex]
[tex]\quad{: \implies{\sf{\dfrac{1290 \times 1000}{0.3225 \times 1000} = P}}}[/tex]
[tex]\quad{: \implies{\sf{\dfrac{12900000}{3225} = P}}}[/tex]
[tex]\quad{: \implies{\sf{\cancel{\dfrac{12900000}{3225}} = P}}}[/tex]
[tex]\quad{: \implies{\sf{Rs.4000 = P}}}[/tex]
[tex]\quad{\dag{\underline{\boxed{\tt{\blue{Principle} = \purple{Rs.4000 }}}}}}[/tex]
[tex]\begin{gathered}\end{gathered}[/tex]
Hence,
The sum (Principle) is Rs.4000.
