Question
A sum of money at compound interest amounts to thrice itself in 3 years. In how many years will it be 9 times itself ?
Answer: Option C
Answer: (c)A = P$(1 + R/100)^T$Let P = Rs.1, then A = Rs.33 = 1$(1 + R/100)^3$On squaring both sides,9 = 1$(1 + R/100)^6$Time = 6 yearsUsing Rule 11,Here, $x = 3, n_1 = 3, y = 9, n_2$ = ?Using, $x^{1/n_1} = y^{1/n_2}$$(3)^{1/3} = (9)^{1/n_2}$$3^{1/3} = (3^2)^{1/n_2}$$3^{1/3} = 3^{2/n_2}$$1/3 = 2/n_2 ⇒ n_2$ = 6 years
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Answer: (c)A = P$(1 + R/100)^T$Let P = Rs.1, then A = Rs.33 = 1$(1 + R/100)^3$On squaring both sides,9 = 1$(1 + R/100)^6$Time = 6 yearsUsing Rule 11,Here, $x = 3, n_1 = 3, y = 9, n_2$ = ?Using, $x^{1/n_1} = y^{1/n_2}$$(3)^{1/3} = (9)^{1/n_2}$$3^{1/3} = (3^2)^{1/n_2}$$3^{1/3} = 3^{2/n_2}$$1/3 = 2/n_2 ⇒ n_2$ = 6 years
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