Question
If a sum of money compounded annually becomes 1.44 times of itself in 2 years, then the rate of interest per annum is
Answer: Option A
Answer: (a)A = P$(1 + R/100)^T$1.44P = P$(1 + R/100)^2$$(1.2)^2 = (1 + R/100)^2$$1 + R/100$ = 1.2R = 0.2 × 100 = 20%Using Rule 8,Here, n = 1.44, t = 2 yearsR% = $(n^{1/6} - 1) × 100%$= $[(1.44)^{1/2} - 1] × 100%$= [(1.2) - 1] × 100%= 0.2 × 100% ⇒ R% = 20%
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Answer: (a)A = P$(1 + R/100)^T$1.44P = P$(1 + R/100)^2$$(1.2)^2 = (1 + R/100)^2$$1 + R/100$ = 1.2R = 0.2 × 100 = 20%Using Rule 8,Here, n = 1.44, t = 2 yearsR% = $(n^{1/6} - 1) × 100%$= $[(1.44)^{1/2} - 1] × 100%$= [(1.2) - 1] × 100%= 0.2 × 100% ⇒ R% = 20%
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