Comments on: Extremely trick problem ( at least for me) about interest gained?

By: John V

John V — Fri, 01 Aug 2008 08:13:13 +0000

to keep this sane, we cacluclate the monthly interest i compounded to produce 13.5% per year, so
1.135 = (1+i)^12. Some log work gives i = .010553 (rounded).

A) Now 2000 * [(1+i)^126 - 1]/i = 2000 * 260.92 = $521,846.96

B) 1500000 = 2000 * [(1+i)^n - 1]/i, or
1+750i = (1+i)^n;
7.9145 = (1.010553)^n, and after log work,
n = 197 months = 16.4 years to accumulate $1.5 million.

By: DC . . . ドン

DC . . . ドン — Wed, 30 Jul 2008 23:17:07 +0000

let r = (1.135)^(1/12) … this is the interest earned after one month. We will now be computing the results monthly based on r.

A. 10.5 years is also 126 months.
Amount: 2000 [ r + r^2 + r^3 + ... + r^126]
= 2000 r*{r^126 -1} / {r - 1} = $529,604.77
this last formula is from the geometric sequence.

B. 2000 [1+r+...rª] = 1500000 … we need ‘a’ (in months yet)
(r^(a+1)-1)/(r-1) = 750
r^(a+1) = 750 *(r-1)
(a+1) lnr = ln(750*(r-1)) … take the natural logarithm of both sides
a+1 = 196.5
a = 195.5 months = 16.29 years
Answer: After 17 years.

By: c j

c j — Tue, 29 Jul 2008 05:59:44 +0000

how often is the interest compounded?

if it is compounded monthly, then its as complicated as you made it… if it is compounded yearly, it is fairly simple.