Rajat invested Rs. (P-100) on compound interest at the rate of ‘R’ % per annum compounded annually. Ravi invested Rs. (P+300) on simple interest at the rate of (R-2) % per annum. Rehan 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. If at the end of two years, the interest obtained by Rajat is 368 more than the interest obtained by Ravi, then find out the value of ‘R’.

Solution

Rehan 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

Rajat invested Rs. (P-100) on compound interest at the rate of ‘R’ % per annum compounded annually. Ravi 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 Rajat is 368 more than the interest obtained by Ravi.

(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.

64R²-800R-9600 = 0

2R²-25R-300 = 0

2R²-(40-15)R-300 = 0

2R²-40R+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 .