MHB Calculate Compound Interest: Easy Step-by-Step Guide | Calculator Tips

AI Thread Summary
The discussion centers on calculating compound interest using the formula I = P((1 + R)^n - 1). A user is confused about their calculator results for a loan of £500 at a 12% annual interest rate compounded quarterly over two years. They mistakenly calculated the interest using the wrong interpretation of the formula, leading to an incorrect figure of £614.9 instead of the correct £133.38. Clarification was provided on the importance of correctly applying the formula and using brackets appropriately. The user expressed gratitude for the guidance and acknowledged the need to improve their understanding of the calculations.
logicandtruth
Messages
14
Reaction score
1
Hi new to the forum and would like to improve my level of maths. I am working through a text but need some help with a compound interest question.

the formula to find compound interest is I = P (1 + R)n–1.

P= principal sum
R= interest rate
n= number of periods for which interest is calculated

John borrows £500 over 2 years from a building society at a rate of 12% per annum compounded
quarterly. How much interest will Shifty have to pay at the end of the 2-year loan?

If £500 is loaned for 2 years at a rate of 12% per annum, compounded quarterly, the
calculations need to be made on a quarterly basis. So the value of n will be 4 (quarters) × 2 (years)
= 8, and the value of r will be 12⁄4 = 3% (per quarter).
According to the question the answer in book is I = 500(1.03)8–1 = £133.38.

Now my issue is when i try to do this with my calculator i get the figure 614.9

I am not sure what I am doing wrong. There are other practice questions, but I want to be sure I am following the correct stages on the calculator before I attempt these. I am using this calculator model View attachment 6280

Any advice would be much appreciated
 

Attachments

  • Calculator.jpg
    Calculator.jpg
    36.8 KB · Views: 129
Mathematics news on Phys.org
You are misunderstanding the formula, "I = P (1 + R)n–1".
To get 614.9 you must have interpreted it as I= P(1+ R)^{n-1}:
500(1+ .03)^7= 500(1.03)^7= 500(1.29)= 614.9

But it is I= P((1+ R)^n- 1):
500((1+ .03)^8- 1)= 500(1.03^8- 1)= 500(1.2667- 1)= 500(0.267)= 133.38.

P(1+ R)^n is the amount, both initial amount and interest, that must be repaid. The -1, which, after multiplying by P is -P subtracts off the initial amount to leave interest only.
 
HallsofIvy said:
You are misunderstanding the formula, "I = P (1 + R)n–1".
To get 614.9 you must have interpreted it as I= P(1+ R)^{n-1}:
500(1+ .03)^7= 500(1.03)^7= 500(1.29)= 614.9

But it is I= P((1+ R)^n- 1):
500((1+ .03)^8- 1)= 500(1.03^8- 1)= 500(1.2667- 1)= 500(0.267)= 133.38.

P(1+ R)^n is the amount, both initial amount and interest, that must be repaid. The -1, which, after multiplying by P is -P subtracts off the initial amount to leave interest only.

Thank you so much HallsofIvy for your prompt reply I suspected it was something to do with my use of brackets. Its just something I need to improve on. Apologies for late response and thanks again.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
6
Views
1K
Replies
9
Views
2K
Replies
2
Views
1K
Replies
1
Views
3K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
8
Views
2K
Replies
2
Views
2K
Back
Top