Question
A sum of money becomes double in 3 years at compound interest compounded annually. At the same rate, in how many years will it become four times of itself ?
Answer: Option D
Answer: (d)A = P$(1 + R/100)^T$Let P. Rs., A = Rs.22 = 1$(1 + R/100)^3$On squaring both sides.4 = 1$(1 + R/100)^6$Time = 6 yearsUsing Rule 11,Here, $x = 2, n_1 = 3, y = 4, n_2$ = ?$x^{1/n_1} = y^{1/n_2}$$2^{1/3} = 4^{1/n_2}$$2^{1/3} = (2^2)^{1/n_2}$$2^{1/3} = 2^{2/n_2}$$1/3 = 2/n_2$$n_2$ = 6 Years
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Answer: (d)A = P$(1 + R/100)^T$Let P. Rs., A = Rs.22 = 1$(1 + R/100)^3$On squaring both sides.4 = 1$(1 + R/100)^6$Time = 6 yearsUsing Rule 11,Here, $x = 2, n_1 = 3, y = 4, n_2$ = ?$x^{1/n_1} = y^{1/n_2}$$2^{1/3} = 4^{1/n_2}$$2^{1/3} = (2^2)^{1/n_2}$$2^{1/3} = 2^{2/n_2}$$1/3 = 2/n_2$$n_2$ = 6 Years
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