[tex]\huge{\fcolorbox{maroon}{s} {\large{\fcolorbox{white}{mistyrose}{{\fcolorbox{gold} {White}{Solution}}}}}}[/tex]

Let the principle amount be P and rate of interest be r.

From the beginning of 2001 and beginning of 2004, there will be accumulated interest of 3 yrs.

So, we can write,

⇒ P (1 + r/100)3 = P + 10000 ---- (1)

Similarly, from the beginning of 2001 and beginning of 2007, there will be accumulated interest of 6 years.

We can write,

[tex]⇒ P ( \frac{1 + r}{100} )6 = P + 25000 ----- (2) \\ [/tex]

[tex]Let, ( \frac{1 + r}{100} )3 = A \\ [/tex]

We can write, equation (1) as,

⇒ P ⨯ A = P + 10000

⇒ P (A – 1) = 10000 ---- (3)

Similarly, we can write equation (2) as,

⇒ P × A2 = P + 25000

⇒ P (A2 – 1) = 25000

⇒ P (A – 1) (A + 1) = 25000 ----- (4)

Dividing equation (4) by equation (3), we get,

⇒ A + 1 = 5/2

⇒ A + 1 = 2.5

⇒ A = 1.5

Putting A = 1.5 in eq (3), we get,

⇒ P × (1.5 - 1) = 10000

∴ P = 20000

Therefore, ₹20000 was the principal amount invested.

Explanation:

Let the principle amount be P and rate of interest be r.

From the beginning of 2001 and beginning of 2004, there will be accumulated interest of 3 yrs.

So, we can write,

⇒ P (1 + r/100)3 = P + 10000 ---- (1)

Similarly, from the beginning of 2001 and beginning of 2007, there will be accumulated interest of 6 years.

We can write,

⇒ P (1 + r/100)6 = P + 25000 ----- (2)

Let, (1 + r/100)3 = A

We can write, equation (1) as,

⇒ P ⨯ A = P + 10000

⇒ P (A – 1) = 10000 ---- (3)

Similarly, we can write equation (2) as,

⇒ P × A2 = P + 25000

⇒ P (A2 – 1) = 25000

⇒ P (A – 1) (A + 1) = 25000 ----- (4)

Dividing equation (4) by equation (3), we get,

⇒ A + 1 = 5/2

⇒ A + 1 = 2.5

⇒ A = 1.5

Putting A = 1.5 in eq (3), we get,

⇒ P × (1.5 - 1) = 10000

∴ P = 20000