I: A sum of money is to be divided into two parts. One part is invested at 10% p.a. simple interest for 3 years, and the other part at 8% p.a. simple interest for 3 years. The total simple interest earned from both parts together in 3 years is ₹10,800. II: If the entire sum had been invested at 10% p.a. simple interest for 3 years, the total simple interest would have been ₹12,000. What was the amount invested at 8% p.a.?

Solution

ATQ, Let total principal = P. From Condition II: If full P at 10% for 3 years: Interest = P × 10 × 3 / 100 = 30P / 100 = 3P/10 This is given as ₹12,000: 3P/10 = 12,000 ⇒ 3P = 1,20,000 ⇒ P = 40,000 Let amount invested at 10% be x. Then amount at 8% is P − x = 40,000 − x. From Condition I: Interest from 10% part (3 years) = x × 10 × 3 / 100 = 3x/10 Interest from 8% part (3 years) = (40,000 − x) × 8 × 3 / 100 = (40,000 − x) × 24 / 100 = 24(40,000 − x)/100 = (24 × 40,000)/100 − 24x/100 = 9,600 − 6x/25 Total interest = 10,800: 3x/10 + 9,600 − 6x/25 = 10,800 Take LCM of 10 and 25 → 50: (15x/50) + 9,600 − (12x/50) = 10,800 (3x/50) + 9,600 = 10,800 3x/50 = 1,200 3x = 60,000 x = 20,000 So amount at 10% = ₹20,000 Amount at 8% = 40,000 − 20,000 = ₹20,000 Required: amount invested at 8% p.a. = ₹20,000.