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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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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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Please explain this bit :

"I divided the principal", i.e, how does division of principal yield interest ?
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I rounded to 33.3333% or 1/3

Since the formula is:

P1 = P0 (1 + r)^n

3 = 1(1+.333333333)^n

3 = 1(4/3)^n

3= (4^n / 3^n)

From here, start with n = 4 and see if the ratio is greater or less than 3

4^4 = 16^2 = 256
3^4 = 9^2 = 81

81*3 = 243...256>243 so 4 is sufficient if the rate were 33.333%. Since the rate is larger than 33.33% it is definitely sufficient
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subhashghosh
Please explain this bit :

"I divided the principal", i.e, how does division of principal yield interest ?

We can quickly get the percentage if we divide the number for which the percentage is found:
such as,
20% of 40 = 40/5 = 8
10% of 80 = 80/10 = 8
50% of 35 = 35/2 = 17.5
33.33% of 100 = 100/3 = 33 approx

Since 34.1% is close to 33.33%; we can divide the number(or principal by 3 and get the interest); Investment: $100

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.
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subhashghosh
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?
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I liked this approach. This question is not difficult, but with this approach it can be done more quickly as well.

fluke
drifting
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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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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bhandariavi
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.

Answer: A
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Quote:
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

Let the principal amount be p and time period of investment be t.
So, 3p=p\((1+\frac{34.1}{100})^t\)
=> 3=\((\frac{134.1}{100})^t\)
=>3=\((1.341)^t\)
For t=3, \((1.341)^t\)=2.41 and for t=4, \((1.341)^t\)=3.23
Thus, smallest investment period is 4 years.

Therefore, the correct answer is option B.
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Imo. B

Initial investment = p, final amount (triple) = 3p, r = 34.1 % and n=?

3p = p (1+34.1/100)n
3 = (1.341)n, as p cancels out from the equation
Let's workout from answer choices to get the answer in quickest way,
Let's try, n=3, 1.3(41)*1.3 = 1.69, 1.69*1.3 = 1.6 * 1.3 = 2.xy, So, A is not the answer
Let's try, n=4, 1.34*1.34 = ~ 1.8, 1.8*1.8 = 3.xy, So, B can be answer.
Rest options are greater than 4, hence, it can satisfy the equation. But, we need smallest investment period to triple the investment. Hence, n = 4.

A. 3
B. 4
C. 6
D. 9
E. 12
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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

Here Rule of 72 can be applied. The rule states that number '72' if divided by number of interest rate(34.1) would give a number of period in which the investment amount doubles. So to triple the investment amount we simply multiple the resultant number by 3/2. Hence

\(\frac{72}{34.1} * \frac{3}{2}\) = 3.16

Since our investment period is annual, the smallest investment period is 4.

Alternatively,

Let investment = 100
Rate r = 34.1%
Initial amount = 100
Amount after 1st year = 134.1
Amount after 2nd year = 179.8
Amount after 3rd year = 241.1
Amount after 4th year = 323.3 (more than triple)

Answer B.

P.S.: In finance rule of 72 is not accurate so other measures are taken to make it as accurate as possible
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2 times: 134.1*1.341 approximately 180
3 times: 180*1.341 is approximately 240
Therefore 4 times to get more than 300%

B
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For P and I questions like these I think a really useful trick to learn to help reduce time is this

for an investment to triple through compound interest it will take (114/interest rate) years for this to occur. for it to double it will take (72/interest rate) years. this question took me <20 seconds to figure out: 114/34.1 = 3.xxx so minimum of 4 years
bhandariavi
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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