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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 .
Find the approximate value of Question mark(?) for given equation.
(71.92% of 149.99) ÷ 12.04 + 107.98 = ?
12.99% of 499.99 ÷ 13.17 = ? ÷ 20.15
(2310.23 ÷ 32.98) + (1008.32 ÷ 23.9) + 1594.11 = ?
?% of 549.83 – 18.05 × 31.96 = 44.94% of 479.84 – 13.98 × 33.13
19.87% of (49.68 × ?) = 19.78% of 1099.87
A can complete a piece of work in 15 days. B is 30% more efficient than A. If they work together, how many days will it take to complete the work?
30.11% of 149.99 + √195.97 ÷ 7.02 = ?
1560.182 ÷ √168 + √143 * √224 – 4649.87 ÷ 30.883= ?
(3/5) of 3025 + (18² + 12²) = ? + 22.22% of 1125