A boy invests a sum of $4000 on his 10th birthday in the year 1996, with an annual interest rate of 7%, compounded every six months. During some stage of the investment, the interest rate changes to an annual rate of 6%, still compounded every six months. On the boy's 18th birthday, the boy checks his account to see a sum of $6770. At what point in time did the interest rate change from 7% to 6%?

(In reply to I love to excel by Leming)

The interest was 3.5% or 3% for each 6-month compounding period.

(1.035)^n * (1.03)^(16-n) = 677/400

n log 1.035 + (16-n) log 1.03 = log (677/400)

(log 1.035 - log 1.03) n = log(677/400) - 16 log 1.03

n = (log(677/400) - 16 log 1.03) / (log 1.035 - log 1.03) = 10.99938

So the interest posted midway between his 15th and 16th birthdays was the last at the old rate and the one on his 16th birthday was the first at the new rate.  The rate changed somewhere between those dates, though perhaps the rate changed just prior to his last posting at the old rate, but the bank was contractually bound for that payment at the old rate since he had already left the money in the bank at the beginning of the half-year.

As banks only post to the penny (and also that non-integral n above), the closest I come is the following balances after each compounding period:

4140
4284.9
4434.87
4590.09
4750.74
4917.02
5089.12
5267.24
5451.59
5642.4
5839.88
6015.08
6195.53
6381.4
6572.84
6770.03

This is with rounding to the nearest penny at each compounding period. (If it weren't rounded, but carried out to many places, the final amount would be off by about 2 cents from 6770.)

DEFDBL A-Z
amt = 4000
FOR i = 1 TO 11
 amt = INT(amt * 103.5 + .5) / 100
 PRINT amt
NEXT
FOR i = 1 TO 5
 amt = INT(amt * 103 + .5) / 100
 PRINT amt
NEXT

Edited on October 4, 2006, 11:11 am

Posted by Charlie on 2006-10-04 11:08:24