Dinesh invested Rs. (P-500) at the rate of 27% per annum on simple interest and at the end of three years, he got Rs. 4860 as an interest. Pankaj invested Rs. (P+300) on simple interest at the rate of (R-2) % per annum. Bhanu invested Rs. (P-100) on compound interest at the rate of ‘R’ % per annum compounded annually. If at the end of two years, the interest obtained by Bhanu is 368 more than the interest obtained by Pankaj, then find out the value of ‘R’.

Dinesh invested Rs. (P-500) at the rate of 27% per annum on simple interest and at the end of three years, he got Rs. 4860 as an interest. (P-500)x27%x3 = 4860 (P-500)x81% = 4860 (P-500) = 486000/81 (P-500) = 6000 P = 6000+500 P = 6500 Bhanu invested Rs. (P-100) on compound interest at the rate of ‘R’ % per annum compounded annually. Pankaj invested Rs. (P+300) on simple interest at the rate of (R-2) % per annum. If at the end of two years, the interest obtained by Bhanu is 368 more than the interest obtained by Pankaj. (P-100)[(1+(R/100)) ² - 1] = [(P+300)x(R-2)x2]/100 + 368 Put the value of ‘P’ in the above equation. (6500-100)[(1+(R/100)) ² -1] = [(6500+300)x(R-2)x2]/100 + 368 6400[(1+(R/100)) ² -1] = [6800x(R-2)x2]/100 + 368 After solving the above equation, we will get a quadratic equation which is given below. 64R ² -800R-9600 = 0 2R ² -25R-300 = 0 2R ² -(40-15)R-300 = 0 2R ² - 40R+15R-300 = 0 2R(R-20)+15(R-20) = 0 (R-20) (2R+15) = 0 R = 20, -(15/2) As we know that the negative value of ‘R’ is not possible. So the value of ‘R’ is 20 .