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

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If a sum of money grows to 144/121 times when invested for [#permalink] New post 28 Nov 2004, 12:49
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? (1) 9 years (2) 22 years (3) 18 years (4) 33 years

Correct Answer - (2)
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 [#permalink] New post 28 Nov 2004, 18:17
P(1+r)^2 = 144/121 * P
r = 1/11

P+Prt = 3P
P(1+rt) = 3P; rt = 2;
t = 2/(1/11) = 22
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 [#permalink] New post 29 Nov 2004, 08:55
can you please explain in more details
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 [#permalink] New post 29 Nov 2004, 09:07
the formula for compound interest is p(1+r)^n
where n=time period
r= rate of interest
p= principal amount

first half of the problem states that if compounded it becomes 141/121 times P

=> p(1 + r/100)^2=144/121P
=>1+r/100= 12/11
therefore, r= 100/11

the second half of the problem states asks for the time need for tripling the amount which would be principal + 2 times the principal.

2P must be from simple interest ( i assume.. correct me if I am wrong)

2P= pnr/100

where r=100/11(from earlier deduction)

therefore, n= 22 years.
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 [#permalink] New post 29 Nov 2004, 17:17
I missed the key word "treble" means 3p!

thanks for the explanation
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Re: money [#permalink] New post 30 Nov 2004, 10:39
revital 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? (1) 9 years (2) 22 years (3) 18 years (4) 33 years

Correct Answer - (2)


I got 22, also. If the interest squared times the original is 144/121, then the interest is 12/11, which means the actual interest is 1/11. so you want to triple the principle by figuring out how many times you should add the product of itself and 1/11 to itself. Meaning, the final equation I got would be p + x([1/11]p) = 3p, and solving for x --> x=22
Re: money   [#permalink] 30 Nov 2004, 10:39
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