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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 .
Find the wrong number in given number series.
2206, 2230, 2278, 2394, 2566, 2950
128 384 48 144 20 54
1728, 1762, 1804, 1854, 1912, 1978
213, 215, 225, 257, 313, 393
Find the wrong number in the given number series.
4, 8, 24, 84, 480, 2880
Find the wrong number in given number series.
2835, 3460, 3335, 3360, 3354, 3356
Identify the term in the sequence that does not fit, and then determine the 9th term of the corrected series. The answer choices are formatte...
112, 119, 133, 154, 180, 217
2824 2314 1973 1759 1634 1574
1990,1877, 1794, 1711, 1638