'A' invested Rs. 20,000 on compound interest at 'p%' p.a. (compounded annually) for two years. 'B' invested Rs. 9,600 on simple interest at 'q%' p.a. for five years. Amount received by 'A' is twice of that received by 'B'. 'q' is ten less than 'p'. Find the amount received on investing Rs. (p²q + pq) on compound interest at (4q -
p)% p.a. (compounded annually) for two years. ('p' is a two-digit number)

Solution

Amount = Principal + Principal × rate × time ÷ 100 So, 20000 × [1 + (p/100)]² = 2 × (9600 + 9600 × q × 5 ÷ 100) Or, 10000 × [1 + (p²/10000) + (p/50)] = 9600 + 480q Or, 10000 + p² + 200p = 9600 + 480q Or, p² + 200p - 480q + 400 = 0 'q' = p – 10 So, p² + 200p - 480 × (p - 10) + 400 = 0 Or, p² + 200p - 480p + 4800 + 400 = 0 Or, p² - 280p + 5200 = 0 Or, p² - 260p - 20p + 5200 = 0 Or, p(p - 260) - 20(p - 260) = 0 Or, (p - 260)(p - 20) = 0 So, 'p' = 260 or 'p' = 20 But 'p' is a two-digit number. So, 'p' = 20 And 'q' = 20 - 10 = 10 p²q + pq = 20² × 10 + 20 × 10 = 4000 + 200 = Rs. 4,200 4q - p = 4 × 10 - 20 = 40 - 20 = 20 Required amount = 4200 × 1.20² = Rs. 6,048