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 .