A sum of money is divided into two parts and invested at simple interest for 3 years. One part is invested at 8% p.a. and the other part at 12% p.a. The total simple interest obtained is Rs 864. If the two rates of interest are interchanged on the two parts (keeping the principal amounts and time same), the total interest becomes Rs 960. What is the total amount of money invested?

ATQ, Let the two parts be Rs x and Rs (P − x), where P is total principal. Case 1 (original): Interest₁ = x * 8 * 3 /100 + (P − x) * 12 * 3 /100 = 864 Case 2 (rates interchanged): Interest₂ = x * 12 * 3 /100 + (P − x) * 8 * 3 /100 = 960 Consider the difference: Interest₂ − Interest₁ = 96 Interest₂ − Interest₁ = [x123/100 + (P − x)83/100] − [x83/100 + (P − x)123/100] = (3/100)[(12x + 8P − 8x) − (8x + 12P − 12x)] = (3/100)[(4x + 8P) − (−4x + 12P)] = (3/100)(8x − 4P) So: (3/100)(8x − 4P) = 96 8x − 4P = 96 * 100 / 3 = 3200 Divide by 4: 2x − P = 800 → P = 2x − 800 …(i) Now use Interest₁ = 864: Interest₁ = (3/100)[8x + 12(P − x)] = (3/100)[8x + 12P − 12x] = (3/100)(12P − 4x) = 864 So: 12P − 4x = 864 * 100 / 3 = 28800 Divide by 4: 3P − x = 7200 …(ii) From (i): P = 2x − 800 Substitute into (ii): 3(2x − 800) − x = 7200 6x − 2400 − x = 7200 5x = 9600 → x = 1920 Then P = 2x − 800 = 2*1920 − 800 = 3840 − 800 = 3040 Total principal = Rs 3040.