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

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Math Expert
Joined: 02 Sep 2009
Posts: 51073
The number of years it would take for the value of an investment to do  [#permalink]

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30 Jul 2018, 23:28
00:00

Difficulty:

45% (medium)

Question Stats:

54% (01:32) correct 46% (01:26) wrong based on 46 sessions

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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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31 Jul 2018, 00:53
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!

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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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31 Jul 2018, 06:46
1
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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Abhishek....

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

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01 Aug 2018, 17:38
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

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.

The number of years it would take for the value of an investment to do &nbs [#permalink] 01 Aug 2018, 17:38
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