A sum of money is divided into two parts and invested at

Solution

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.