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# An investment compounds annually at an interest rate of 34.1% What is

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An investment compounds annually at an interest rate of 34.1% What is [#permalink]

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14 Feb 2011, 20:25
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65% (hard)

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59% (01:17) correct 41% (01:47) wrong based on 189 sessions

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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. 4
B. 5
C. 6
D. 7
E. 8

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Re: An investment compounds annually at an interest rate of 34.1% What is [#permalink]

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15 Feb 2011, 07:39
1
By the Compound Interest formula :

$$A = P(1 + r/100)^nt$$

It comes to (1.341)^n = 3 (since A = 3P, and t = 1)

On multiplying 1.341 further with itself, it can be seen that n = 4, so answer is A.
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Re: An investment compounds annually at an interest rate of 34.1% What is [#permalink]

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15 Feb 2011, 08:24
Please correct me if I am wrong but it seems you need to workout the multiplication of 1.341 at least 3 times very quickly which seems rather daunting given the tight time constraints of the GMAT..

Is there a faster way of doing this?
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Re: An investment compounds annually at an interest rate of 34.1% What is [#permalink]

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15 Feb 2011, 10:29
1
drifting wrote:
Please correct me if I am wrong but it seems you need to workout the multiplication of 1.341 at least 3 times very quickly which seems rather daunting given the tight time constraints of the GMAT..

Is there a faster way of doing this?

I agree; I divided the principal by 3 by considering approx 33.33% and rounded the value to greater nearest integer;

1st year: 34.1% interest is gotten back on 100: 100/3 = 33.33 approx $34: New principal becomes: Principal+Interest: 100+34=134 2nd year: 34.1% interest on 134; 134/3 = 45: Total: 134+45 = 179(new principal) 3rd year: 179/3 = 60: Total: 179+60 = 239 4th year: 239/3 = 80: Total: 239+80 = 319 > 3(100); end of 4 years, the return was already more than triple. _________________ Director Status: Up again. Joined: 31 Oct 2010 Posts: 514 Concentration: Strategy, Operations GMAT 1: 710 Q48 V40 GMAT 2: 740 Q49 V42 Re: An investment compounds annually at an interest rate of 34.1% What is [#permalink] ### Show Tags 17 Feb 2011, 05:38 subhashghosh wrote: By the Compound Interest formula : $$A = P(1 + r/100)^nt$$ It comes to (1.341)^n = 3 (since A = 3P, and t = 1) On multiplying 1.341 further with itself, it can be seen that n = 4, so answer is A. Thanks for the explanation! But what if this equation consists of a number larger than 3, say 300! Another question, Is the T in your formula really necessary? _________________ My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html Manager Joined: 17 Feb 2011 Posts: 156 Concentration: Real Estate, Finance Schools: MIT (Sloan) - Class of 2014 GMAT 1: 760 Q50 V44 Re: An investment compounds annually at an interest rate of 34.1% What is [#permalink] ### Show Tags 17 Feb 2011, 09:55 I liked this approach. This question is not difficult, but with this approach it can be done more quickly as well. fluke wrote: drifting wrote: Please correct me if I am wrong but it seems you need to workout the multiplication of 1.341 at least 3 times very quickly which seems rather daunting given the tight time constraints of the GMAT.. Is there a faster way of doing this? I agree; I divided the principal by 3 by considering approx 33.33% and rounded the value to greater nearest integer; 1 year: 100/3 = 33.33 approx$34: Total: 134
2nd year: 134/3 = 45: Total: 134+45 = 179
3rd year: 179/3 = 60: Total: 179+60 = 239
4th year: 239/3 = 80: Total: 239+80 = 319 > 3(100)

;4 years;

Perhaps there is a better way...
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Re: An investment compounds annually at an interest rate of 34.1% What is [#permalink]

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01 Dec 2016, 19:25
100(1+0.341)^t = 301
(1.341)^t=3.01
t ln(1.341) = ln(3.01)
t = ~3.76

4 is closest to this choice.

A.
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Re: An investment compounds annually at an interest rate of 34.1% What is [#permalink]

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04 Dec 2016, 06:39
I used the approximation to 1/3 and I started with a capital of 10

1)10--13
2)13--17
3)17--23
4)23--30+

So answer is A. I reckon in these cases approximation and easy numbers are the only way
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Re: An investment compounds annually at an interest rate of 34.1% What is [#permalink]

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13 Jan 2018, 10:48
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Top Contributor
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bhandariavi 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. 4
B. 5
C. 6
D. 7
E. 8

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.

Cheers,
Brent
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An investment compounds annually at an interest rate of 34.1% What is [#permalink]

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18 May 2018, 11:27
bhandariavi 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. 4
B. 5
C. 6
D. 7
E. 8

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.

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An investment compounds annually at an interest rate of 34.1% What is   [#permalink] 18 May 2018, 11:27
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