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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 .
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