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The number of years it would take for the value of an investment to do

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The number of years it would take for the value of an investment to do  [#permalink]

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New post 31 Jul 2018, 00:28
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The number of years it would take for the value of an investment to double, at 26% interest compounded annually, is approximately which of the following?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

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Re: The number of years it would take for the value of an investment to do  [#permalink]

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New post 31 Jul 2018, 01:53
Bunuel wrote:
The number of years it would take for the value of an investment to double, at 26% interest compounded annually, is approximately which of the following?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6


As we're explicitly asked to estimate, we won't bother with exact calculations.
This is an Alternative approach.

If we start with 100 and increase every year by 25% (which is exactly 1/4 so easier to calculate than 26%)
after 1 year we'd have 100 + 100/4 = 125,
after 2 years we'd have 125 + 125/4 = about 125 + 30 = 155
after 3 years we'd have about 155 + 155/4 = about 155 + 40 = 195.
This is very nearly double what we started!

(B) is our answer.
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Re: The number of years it would take for the value of an investment to do  [#permalink]

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New post 31 Jul 2018, 07:46
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Bunuel wrote:
The number of years it would take for the value of an investment to double, at 26% interest compounded annually, is approximately which of the following?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6


\(A = P(1 + \frac{r}{100})^n\)

Or, \(2p = p(1 + \frac{26}{100})^n\)

Or, \(2 = (1 + \frac{26}{100})^n\)

Or, \(2 = (1 + \frac{26}{100})^n\)

Or, \(2 = (\frac{126}{100})^n\)

Or, \(2 = 1.26^n\)

Now calculate , \(1.26^2 = 1.5876\)

And \(1.26^3 = 2.00\), Thus Answer must be (B) 3
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The number of years it would take for the value of an investment to do  [#permalink]

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New post 01 Aug 2018, 18:38
Bunuel wrote:
The number of years it would take for the value of an investment to double, at 26% interest compounded annually, is approximately which of the following?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

Agreed, estimation is the way to go. A compound interest rate of 26% translates into a multiplier of 1.26. Change to 1.25

\(1.25=1\frac{25}{100}=1\frac{1}{4}=\frac{5}{4}\)

Imagine 1 in an account.

Multiply \(1*\frac{5}{4}*\frac{5}{4}*\frac{5}{4}\) ... until the numerator is about twice the denominator.

Each time the multiplier gets used = 1 year has elapsed
(the multiplier is being used at the end of each year, with a first year base of 1)

\(1*\frac{5}{4}*\frac{5}{4}*\frac{5}{4}=\frac{125}{64}\) => 125 is \(\approx\) double 64

Number of years? \(\frac{5}{4}\) was used three times, i.e., at the end of each of 3 years

Answer B

Same as above with $1 and years shown

Begin: \($1=>\)End Year 1: \(($1*\frac{5}{4})=$\frac{5}{4}\)

Begin: \($\frac{5}{4}=>\)END Year 2: (\($\frac{5}{4}*\frac{5}{4})=$\frac{25}{16}\)

Begin: \($\frac{25}{16}=>\)END Year 3: \(($\frac{25}{16}*\frac{5}{4})=$\frac{125}{64}\)

\($125\) is just about double \($64\)

One more year? Check

Begin: \($\frac{125}{64}=>\)END Year 4:\(($\frac{125}{64}*\frac{5}{4})=$\frac{625}{256}\)

MORE than doubled. (256*2 = 512). Too many years.

End of Year 3 was correct.

Answer B
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