# investment

• Sep 15th 2012, 05:54 AM
rcs
investment
if an investment of \$ 25,000 accumulates to P27,500 in 18 months, find the single interest rate?

this is my attempt: 27,000 - 25,000 = 2000 / 18 then the quotient is divided by 25,000. is this possible?
• Sep 15th 2012, 08:25 AM
DeMath
Re: investment
I think it is

$\frac{27500-25000}{25000}=\frac{1}{10}=0.1\quad (\text{or}~10\%)$
• Sep 15th 2012, 08:57 AM
rcs
Re: investment
[QUOTE=DeMath;736289]I think it is

$\frac{27500-25000}{25000}=\frac{1}{10}=0.1\quad (\text{or}~10\%)$[/QUOTE

thanks but... the key answer from the book said that it is 6 2/3 % hmmmmmmmm... im confused... im still solving it here huuhuhuuhuh :(
• Sep 15th 2012, 09:10 AM
DeMath
Re: investment
[QUOTE=rcs;736298]
Quote:

Originally Posted by DeMath
I think it is

$\frac{27500-25000}{25000}=\frac{1}{10}=0.1\quad (\text{or}~10\%)$[/QUOTE

thanks but... the key answer from the book said that it is 6 2/3 % hmmmmmmmm... im confused... im still solving it here huuhuhuuhuh :(

Need to find the annual single interest rate

$\frac{1}{10}\cdot \frac{12}{18}=\frac{1}{10}\cdot \frac{2}{3}= \frac{1}{15}$, i.e. $\left(\frac{1}{15}\cdot 100\right)\!\%= \frac{20}{3}\%= 6\tfrac{2}{3}\%$ (annual rate).
• Sep 17th 2012, 07:34 AM
Wilmer
Re: investment
Very unclear question.

25000(1 + i)^1.5 = 27500
(1 + i)^1.5 = 27500/25000
1 + i = (27500/25000)^(1/1.5) : remember that if a^p = b, then a = b^(1/p)
i = (27500/25000)^(1/1.5) - 1
i = .0656... : so annual rate = ~6.56%

Your 6 2/3% result assumes SIMPLE interest; if so, then SHOULD be specified in the problem statement.