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An investment compounds annually at an interest rate of 34.1

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by guerrero » Sun Jun 23, 2013 8:13 am
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

Can this be done without using std. formula ?

Source : Grockit

OA B
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by Brent@GMATPrepNow » Sun Jun 23, 2013 8:45 am
guerrero 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

Can this be done without using std. formula ?

Source : Grockit

OA B
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 Amrabdelnaby » Thu Dec 10, 2015 4:47 am
Brent is it possible to solve it using the formula as well?
Brent@GMATPrepNow wrote:
guerrero 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

Can this be done without using std. formula ?

Source : Grockit

OA B
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

Cheers,
Brent
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by Brent@GMATPrepNow » Thu Dec 10, 2015 6:51 am
Amrabdelnaby wrote:Brent is it possible to solve it using the formula as well?
You bet.
Give it a try.
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by [email protected] » Thu Dec 10, 2015 9:53 am
Hi Amrabdelnaby,

I assume you're talking about the Compound Interest Formula. You could certainly use that Formula here, but it likely won't be any faster than the approach that Brent used.

(Principal)(1 + R)^T = Total

X = original investment

(X)(1 + .341)^T > 3X

If we estimate 1.341 to be 1.3333 to be 4/3, then we have...

(X)(4/3)^T > 3X
(4/3)^T > 3

From here, you could either TEST THE ANSWERS or continue algebraically....

(4^T)/(3^T) > 3
4^T > (3)(3^T)
4^T > 3^(T+1)

Although here you'd likely end up TESTing THE ANSWERS regardless.

Final Answer: B

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by vs224 » Fri Dec 09, 2016 2:40 am
'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) '

can someone explain concept behind it, or show forward me where i can read more about it.
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by [email protected] » Fri Dec 09, 2016 11:33 am
Hi vs224,

Questions that involve 'exponential growth' are not that common on the GMAT. You will likely see one 'interest rate' question on the Exam, and you might see a story problem that involves exponential growth or exponential decay - but that's about it. The 'math' behind all of these questions is essentially just multiplication (and in many cases, you do NOT actually have to do lots of complex math to get to the solution). This is all meant to say that it's not a high-value subject and the math behind it is rarely all that complicated.

Here's a similar example to the one that you described... If you start off with a $100 investment, the investment increases by 50% of its value each year and you do not remove any of the money from the investment, you would have....

Start = $100
Year 1 = $100(1.5) = $150
Year 2 = $150(1.5) = $225
Year 3 = $225(1.5) = $337.50
Etc

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by DavidG@VeritasPrep » Fri Dec 09, 2016 1:24 pm
vs224 wrote:'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) '

can someone explain concept behind it, or show forward me where i can read more about it.
If x increases by 1/3 of itself, it increases by (1/3)(x). The new value would be x + (1/3)(x), which simplifies to (4/3)x. So when a value increases by 1/3 it's the same as multiplying that initial value by 4/3.
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by Brent@GMATPrepNow » Sat Jan 13, 2018 9:46 am
guerrero 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

Cheers,
Brent
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