An investment compounds annually at an interest rate

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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

The OA is B.

Please, can anyone help me to solve this PS question? Thanks in advance.
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by Brent@GMATPrepNow » Wed May 16, 2018 4:14 pm
BTGmoderatorLU wrote: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
We can use fractions to solve this question.
Each year, the investment increases 34.1%
This is very close to an increase of 1/3 (33.33%)

So, if the investment increases by 1/3 each year, then each year, we can find the value of the investment by multiplying last year's value by 4/3 (this represents a 1/3 increase)

So, let's say the initial investment is $1.
We want to determine how many years it takes the investment to be worth at least $3 (triple)

Year 0: $1
Year 1: ($1)(4/3) = $4/3
Year 2: ($1)(4/3)(4/3) = $16/9 (this is less than $3)
Year 3: ($1)(4/3)(4/3)(4/3) = $64/27 (this is less than $3)
Year 4: ($1)(4/3)(4/3)(4/3)(4/3) = $256/81 (this is more than $3)

So, it takes 4 years for the investment to more than triple in value.

Answer: B

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Brent
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by Jeff@TargetTestPrep » Fri May 18, 2018 10:34 am
BTGmoderatorLU wrote: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

The annually compounded interest formula is A = P(1 + r)^t. Since we want to triple the value of investment, A = 3P. So we have:

3P = P(1 + 0.341)^t

Divide both sides by P, we have:

3 = (1.341)^t

Let's round the base to 1.333 = 4/3. We see that (4/3)^3 = 64/27 < 3 but (4/3)^4 = 256/81 > 3. So t must be 4.

Answer: B

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