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An investment compounds annually at an interest rate of 34.1% What is the smallest investment period by which time the investment will more than triple in value?
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14 Oct 2019, 23:03
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For simplicity of calculation, I would approximate 34.1% to 34%.
After year 1, The total amount will be 1.34 times the invested amount After year 2, the total amount will be 1.34*1.34 = 1.7956 approximately 1.80 times the invested amount After year 3, the total amount will be 1.34*1.80 = 2.412, approximately 2.4 times the invested amount After year 4, the total amount will be 1.34*2.4 = 3.216, and this is more than 3 times the invested amount.
Therefore the amount has to be invested for at least 4 years at the compound annual interest rate of 34.1% in order to just earn more than three times the invested amount.
The answer is therefore B.
PS. I would be looking forward to shorter alternatives, as this approach is not optimal in my view.
An investment compounds annually at an interest rate of 34.1% What is
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Updated on: 15 Oct 2019, 21:23
CI=p*(1+r/100)'t so R=34.1% P=1 target P =4 will happen IMO B
An investment compounds annually at an interest rate of 34.1% What is the smallest investment period by which time the investment will more than triple in value?
A. 3 B. 4 C. 6 D. 9 E. 12
Originally posted by Archit3110 on 15 Oct 2019, 00:57.
Last edited by Archit3110 on 15 Oct 2019, 21:23, edited 1 time in total.
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15 Oct 2019, 04:48
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Quote:
An investment compounds annually at an interest rate of 34.1% What is the smallest investment period by which time the investment will more than triple in value?
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15 Oct 2019, 06:26
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Quote:
An investment compounds annually at an interest rate of 34.1% What is the smallest investment period by which time the investment will more than triple in value?
A. 3 B. 4 C. 6 D. 9 E. 12
Let the principal amount be p and time period of investment be t. So, 3p=p\((1+\frac{34.1}{100})^t\) => 3=\((\frac{134.1}{100})^t\) =>3=\((1.341)^t\) For t=3, \((1.341)^t\)=2.41 and for t=4, \((1.341)^t\)=3.23 Thus, smallest investment period is 4 years.
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15 Oct 2019, 08:40
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Imo. B
Initial investment = p, final amount (triple) = 3p, r = 34.1 % and n=?
3p = p (1+34.1/100)n 3 = (1.341)n, as p cancels out from the equation Let's workout from answer choices to get the answer in quickest way, Let's try, n=3, 1.3(41)*1.3 = 1.69, 1.69*1.3 = 1.6 * 1.3 = 2.xy, So, A is not the answer Let's try, n=4, 1.34*1.34 = ~ 1.8, 1.8*1.8 = 3.xy, So, B can be answer. Rest options are greater than 4, hence, it can satisfy the equation. But, we need smallest investment period to triple the investment. Hence, n = 4.
A. 3 B. 4 C. 6 D. 9 E. 12
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Re: An investment compounds annually at an interest rate of 34.1% What is
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15 Oct 2019, 11:11
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An investment compounds annually at an interest rate of 34.1% What is the smallest investment period by which time the investment will more than triple in value?
A. 3 B. 4 C. 6 D. 9 E. 12
Here Rule of 72 can be applied. The rule states that number '72' if divided by number of interest rate(34.1) would give a number of period in which the investment amount doubles. So to triple the investment amount we simply multiple the resultant number by 3/2. Hence
\(\frac{72}{34.1} * \frac{3}{2}\) = 3.16
Since our investment period is annual, the smallest investment period is 4.
Alternatively,
Let investment = 100 Rate r = 34.1% Initial amount = 100 Amount after 1st year = 134.1 Amount after 2nd year = 179.8 Amount after 3rd year = 241.1 Amount after 4th year = 323.3 (more than triple)
Answer B.
P.S.: In finance rule of 72 is not accurate so other measures are taken to make it as accurate as possible
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