Question
A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself ?
Answer: Option C
Answer: (c)A = P$(1 + R/100)^T$Let P be Rs.1, then A = Rs.22 = 1$(1 + R/100)^4$$2^2 = (1 + R/100)^8$Time = 8 yearsUsing Rule 11,Here, $x = 2, n_1 = 4, y = 4, n_2$ = ?Using $x^{1/n_1} = y^{1/n_2}$$(2)^{1/4} = (4)^{1/n_2}$$(2)^{1/4} = (2^2)^{1/n_2}$$2^{1/4} = 2^{1/n_2}$$1/4 = 2/n_2$$n_2$ = 8 years
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Answer: (c)A = P$(1 + R/100)^T$Let P be Rs.1, then A = Rs.22 = 1$(1 + R/100)^4$$2^2 = (1 + R/100)^8$Time = 8 yearsUsing Rule 11,Here, $x = 2, n_1 = 4, y = 4, n_2$ = ?Using $x^{1/n_1} = y^{1/n_2}$$(2)^{1/4} = (4)^{1/n_2}$$(2)^{1/4} = (2^2)^{1/n_2}$$2^{1/4} = 2^{1/n_2}$$1/4 = 2/n_2$$n_2$ = 8 years
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