Anil invests an amount for 2 years at the rate of 15% per annum at simple…

Concept: For any sum invested for exactly 2 years at a rate of r% per annum, the compound interest exceeds the simple interest by precisely the extra interest that the first year's own interest earns during the second year. This gives the identity Difference = P × (r/100)2, where P is the principal and r is the annual rate — it falls out of expanding the compound-interest amount and subtracting the simple interest, since the linear terms cancel and only the rate-squared term survives.

Application:

1. Let the principal be P. At r = 15% per annum for 2 years, the simple interest is SI = P × 15 × 2 / 100.

2. The compound interest for 2 years is CI = P × [(1 + 15/100)2 − 1].

3. Subtracting, CI − SI algebraically reduces to just P × (15/100)2 — the linear (2 × r/100) parts of CI and SI cancel exactly, leaving only the rate-squared term.

4. Substitute the given gap of Rs. 450: P × (15/100)2 = 450, i.e. P × 0.0225 = 450.

5. Solve for P: P = 450 / 0.0225 = Rs. 20,000.

Cross-check: Compute both interests directly for P = Rs. 20,000. SI = 20000 × 15 × 2 / 100 = Rs. 6,000. CI = 20000 × [(1.15)2 − 1] = 20000 × 0.3225 = Rs. 6,450. The difference, Rs. 6,450 − Rs. 6,000 = Rs. 450, exactly matches the given gap — confirming the principal.

Hence, the principal Anil invested is Rs. 20,000.