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If a sum of money grows to 144/121 times when invested for two years

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If a sum of money grows to 144/121 times when invested for two years  [#permalink]

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New post 29 Jan 2019, 08:09
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Question Stats:

47% (02:27) correct 53% (02:19) wrong based on 22 sessions

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If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to treble if invested at the same rate of interest in a scheme where interest is computed using simple interest method?

(A) 9 years
(B) 18 years
(C) 22 years
(D) 31 years
(E) 33 years

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Re: If a sum of money grows to 144/121 times when invested for two years  [#permalink]

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New post 29 Jan 2019, 10:12
Bunuel wrote:
If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to trebleif invested at the same rate of interest in a scheme where interest is computed using simple interest method?

(A) 9 years
(B) 18 years
(C) 22 years
(D) 31 years
(E) 33 years


Bunuel please check highlighted part ; seems to be a typo error
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Re: If a sum of money grows to 144/121 times when invested for two years  [#permalink]

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New post 30 Jan 2019, 07:45
Bunuel wrote:
If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to trebleif invested at the same rate of interest in a scheme where interest is computed using simple interest method?

(A) 9 years
(B) 18 years
(C) 22 years
(D) 31 years
(E) 33 years


Bunuel I think it should be 'triple'

Also, I think the OA is wrong.

Compound annually: Total = P (1+r/n)^nt...where p is original amount & r is interest rate & t is number of years & n: number of compounding annually.

Total = \(\frac{144}{121}\)P

\(\frac{144}{121}\)P = P (1+r)^2.......\(\frac{12}{11}\)= (1+r).........r = \(\frac{1}{11}\)

Apply in simple interest formula

Total = P * r * #of years

3P = P * \(\frac{1}{11}\) * Y.............Y = 33

Answer must be E

Dear GMATPrepNow Brent,

Can you help and shed lught on the solution above?

Thanks in advance
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Re: If a sum of money grows to 144/121 times when invested for two years  [#permalink]

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New post 30 Jan 2019, 08:01
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Mo2men wrote:
Bunuel wrote:
If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to trebleif invested at the same rate of interest in a scheme where interest is computed using simple interest method?

(A) 9 years
(B) 18 years
(C) 22 years
(D) 31 years
(E) 33 years


Bunuel I think it should be 'triple'

Also, I think the OA is wrong.

Compound annually: Total = P (1+r/n)^nt...where p is original amount & r is interest rate & t is number of years & n: number of compounding annually.

Total = \(\frac{144}{121}\)P

\(\frac{144}{121}\)P = P (1+r)^2.......\(\frac{12}{11}\)= (1+r).........r = \(\frac{1}{11}\)

Apply in simple interest formula

Total = P * r * #of years

3P = P * \(\frac{1}{11}\) * Y.............Y = 33

Answer must be E

Dear GMATPrepNow Brent,

Can you help and shed lught on the solution above?

Thanks in advance


Be careful, the formula you stated "Total = P * r * #of years" is a formula for finding the INTEREST (not the total value of the investment)
There's also a problem with the 3P part of your equation.

Notice that, if we start with a $10 investment, and its value triples to $30, this means the investment received $20 in interest.

We want the investment to go from P to 3P
Well P + 2P = 3P

So, we want the TOTAL INTEREST to equal 2P
We get: 2P = (P)(interest rate)(#of years)
Or: 2P = (P)(1/11)(x)

Solve to get x = 22

Cheers,
Brent
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Re: If a sum of money grows to 144/121 times when invested for two years  [#permalink]

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New post 30 Jan 2019, 08:05
GMATPrepNow wrote:
Mo2men wrote:
Bunuel wrote:
If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to trebleif invested at the same rate of interest in a scheme where interest is computed using simple interest method?

(A) 9 years
(B) 18 years
(C) 22 years
(D) 31 years
(E) 33 years


Bunuel I think it should be 'triple'

Also, I think the OA is wrong.

Compound annually: Total = P (1+r/n)^nt...where p is original amount & r is interest rate & t is number of years & n: number of compounding annually.

Total = \(\frac{144}{121}\)P

\(\frac{144}{121}\)P = P (1+r)^2.......\(\frac{12}{11}\)= (1+r).........r = \(\frac{1}{11}\)

Apply in simple interest formula

Total = P * r * #of years

3P = P * \(\frac{1}{11}\) * Y.............Y = 33

Answer must be E

Dear GMATPrepNow Brent,

Can you help and shed lught on the solution above?

Thanks in advance


Be careful, the formula you stated "Total = P * r * #of years" is a formula for finding the INTEREST (not the total value of the investment)
There's also a problem with the 3P part of your equation.

Notice that, if we start with a $10 investment, and its value triples to $30, this means the investment received $20 in interest.

We want the investment to go from P to 3P
Well P + 2P = 3P

So, we want the TOTAL INTEREST to equal 2P
We get: 2P = (P)(interest rate)(#of years)
Or: 2P = (P)(1/11)(x)

Solve to get x = 22

Cheers,
Brent


Thanks GMATPrepNow Brent

Nicely explained as always :thumbup: Thanks a lot.

I treated the simple interest equation as the compound equation above it.
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Re: If a sum of money grows to 144/121 times when invested for two years  [#permalink]

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New post 31 Jan 2019, 18:39
Bunuel wrote:
If a sum of money grows to 144/121 times when invested for two years in a scheme where interest is compounded annually, how long will the same sum of money take to treble if invested at the same rate of interest in a scheme where interest is computed using simple interest method?

(A) 9 years
(B) 18 years
(C) 22 years
(D) 31 years
(E) 33 years


We can let P = the principal and r = the interest rate and create the equation:

P(1 + r)^2= (144/121)P

(1 + r)^2 = 144/121

1 + r = √(144/121)

1 + r = 12/11

r = 1/11

Therefore, the annual interest rate is 1/11. If we let t = the number of years for the principal to triple using simple interest method with this rate, we can create the equation:

P + P(1/11)t = 3P

P(t/11) = 2P

t/11 = 2

t = 22

Answer: C
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Re: If a sum of money grows to 144/121 times when invested for two years   [#permalink] 31 Jan 2019, 18:39
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