CHAPTER 6. FINANCE 6.6
- I will need R 450 for some university textbooks in 1,5 years time. I currently have R 400. What
interest rate do I need toearn to meet this goal?
Each time that you seesomething different fromwhat you have seen before, start off with the basic
equation that you should recognise very well:
A = P .(1 + i)nIf this were an algebra problem, and you were told to “solve for i”, you should be able toshow that:
A
P
= (1 + i)nn�
A
P
= 1 + in�
A
P
− 1 = i∴ i = n�
A
P
− 1
You do not need to memorise this equation, it iseasy to derive any timeyou need it!
So let us look at the twoexamples mentioned above.
- Check that you agreethat P =R2 500, A =R3 000, n = 128 =^23. This means that:
i =(^23)
�
3000
2500
− 1
= 0, 314534 ...
= 31,45%
Ouch! That is not a verygenerous neighbour youhave.- Check that P =R 400 , A =R 450 , n = 1, 5
i =^1 ,^5�
450
400
− 1
= 0, 0816871 ...
= 8,17%
This means that as longas you can find a bank which pays more than 8 ,17% interest, you should
have the money you need!Note that in both examples, we expressed n as a number of years ( 128 years, not 8 because that is the
number of months) which means i is the annual interest rate. Always keep this in mind — keep years
with years to avoid making silly mistakes.
Exercise 6 - 4
- A machine costs R45 000 and has a scrap value of R9 000 after 10 years. Determine the annual
rate of depreciation if itis calculated on the reducing balance method. - After 5 years an investment doubled in value. At what annual rate was interest compounded?
More practice video solutions or help at http://www.everythingmaths.co.za(1.) 017t (2.) 017u