FNPF: compound interest
davejvh
Registered Posts: 19 Dedicated contributor 🌟 🐵 🌟
Hi,
I came across a question in my financial performance cba yesterday, I'm wondering how everyone else would approach it. I won't repeat the actual question but here's a jist with changed figures:
in year 1 price per kilo was £100
in year 5 price per kilo was £135
question 1: state the year 5 price as an index using year 1 as base (easy enough)
question 2: state the % price increase from year 1 to year 5 (again easy enough)
question 3: this is equivalent to an annual compound interest rate of:
[multiple choice] 5.91%, 7.81%, 8.07%, 35%
I couldn't think of a way to calculate the compound interest rate given just the two prices and time difference in years so I tested each of the multiple choice answers to find the right one.
Is that would everyone else would do too?
I came across a question in my financial performance cba yesterday, I'm wondering how everyone else would approach it. I won't repeat the actual question but here's a jist with changed figures:
in year 1 price per kilo was £100
in year 5 price per kilo was £135
question 1: state the year 5 price as an index using year 1 as base (easy enough)
question 2: state the % price increase from year 1 to year 5 (again easy enough)
question 3: this is equivalent to an annual compound interest rate of:
[multiple choice] 5.91%, 7.81%, 8.07%, 35%
I couldn't think of a way to calculate the compound interest rate given just the two prices and time difference in years so I tested each of the multiple choice answers to find the right one.
Is that would everyone else would do too?
0
Comments

the 4th root of 1.35
on my calculator I would type in 4, 2nd function, y^x, 1.35. The answer on my calc is 1.0779 ie 7.79%, so from the answers you've provided I'd choose 7.81% (slight rounding difference)0 
davejvh,
Nia is correct but she (I am assuming this person is a woman) didn't show you the steps.
From my recollection compound interest formula takes the form
A = P (1 + R/100) ^ n
A is the amount at end of period, P is intial principal invested and R is the percentage rate and n is the number of years.
So in your example we are saying:
135 = 100 (1 + R/100)^4
To find the rate then we make (1 + R/100) the subject of the equation:
(135/100)^(1/4) = (1+ r) here the power of 1/4 is the same as the 4th root of the number i.e 1.35. The little r means the rate divided by 100.
So the 4th root of 1.35 is equal to 1.0779 and so we have
1.0779 = 1 + r
Making r the subject of the equation we are left with
r = 1.0779  1
r = .0779 or 7.79% (multiplying back by 100%)
If we substitute this rate back in the original equation we can prove our answer.
135 = 100 (1 + 7.79/100) ^ 4
It doesnt come exactly to 135 but this is because of the rounding so you would need to round up. Hope this helps.0 
OMG!! this is a bit worrying. I have got this CBA on Monday and ive never done compound interest. Havnt even got those buttons on my calculator!!!! Just checked my book and cant see it even mentioned in the Osbourne Book??0

I agree with Nia, basically you just want to calculate 4th root (as there were 4 movements from original ) of 135/100  using a decent calculator this should be no problem. Seems no need to make something anything more complicated than it actually is0

thank you Crispy! It isn't complicated, so why make it so....
Deborah, don't worry about this for your exam..... and it doesn't matter that you don't have those buttons on your calculator. If you want to pm me, i'll give you some tips on how to deal with it  I promise, it isn't difficult!0 
little bit patronising don't you think Henry, SHE is the cat's mother! and yes, this PERSON is female!
Sorry if I offended you. I certainly wasn't meant to be patronising. Obviously it is easy for you to look at the problem and just find the 4th root of 1.35 but that is because you have a superior knowledge/understanding of compound interest. But if I had no knowledge of how the formula works I would want to know how we arrived at the 4th root of 1.35. This is why I thought I would explain the formula in detail. Didn't mean to "steal your thunder".
PS: I am a GCSE Maths tutor on the side and it is important to me that my student understands the problem fully and this is achievable by going through the steps  so it was the same kind of approach I was taking here with this particular question and not a reflection on you.0 
Nia,
Sorry if I offended you. I certainly wasn't meant to be patronising. Obviously it is easy for you to look at the problem and just find the 4th root of 1.35 but that is because you have a superior knowledge/understanding of compound interest. But if I had no knowledge of how the formula works I would want to know how we arrived at the 4th root of 1.35. This is why I thought I would explain the formula in detail. Didn't mean to "steal your thunder".
PS: I am a GCSE Maths tutor on the side and it is important to me that my student understands the problem fully and this is achievable by going through the steps  so it was the same kind of approach I was taking here with this particular question and not a reflection on you.
"steal your thunder" and "not a reflection on you" .... don't make me laugh, get over yourself!!! and anyway Dave didn't ask for an explanation, just what everyone else would do......so that's what I wrote  what I would do!0 
Nia,
I see you are trying to make this out more than it is and I don't want to get into a quarrel with you. You expressed your answer one way and I explained mine another way. My answer was not meant to undermine yours in anyway.
The truth of the matter is I am a methodical person in the way I think. At first I didn't even get your answer, it was too quick for me. But it is after I worked through it methodically the way I know how that I saw the connection between your answer and mine. So I thought it might be beneficial to the poster to see show you made the leap from the information presented to the answer you gave. That is all.0 
See you're doing it again........... Maybe you should think before you open your mouth then, and not be so patronising. Grow a set! THE END!
One last thing actually. I apologise unreservedly if I came across as being patronising. I promise you it was never my intention.
Over and out.0 
thanks for replying guys...
I have had my calculator inspected in the past and been told scientific calculators are not permitted in the exam, so I only had a 20button basic one with me.
I got the formula down to
a=b*(1+z)^4
a=135
b=100
Find z
on paper but couldn't think of a way to restate the formula (Thanks Henry for laying that out, obvious when I saw it, move ^4 to the other side, ^.025).
even if I had, is it possible to calculate 1.35^0.25 on a basic calculator? forgive me if that's a dumb question!
I think (and my tutor agreed) that the AAT expect candidates to test each of the answers; this shows understanding of the iterations from original to final price and therefore the underlying concept of compound interest, and explains why of the three part question, the first two parts were number entry boxes and the last part multiple choice.
Nia and Henry, you are better than my tutor, she said it was impossible to work it out. I knew it wasn't, it would just take a bit of algebra and/or a better calculator, neither of which I could rustle up in the exam hall.0 
Dave, as far as I am aware you are allowed to use scientific calculators in exams, so long as they are not programmable0

even if I had, is it possible to calculate 1.35^0.25 on a basic calculator? forgive me if that's a dumb question!
not a dumb question. On a basic calculator, with just a square root button, I think I'm right in saying for a fourth root you can square root your number, then square root again....... haven't had time to check this out though0 
Nia,
glad to be wrong about sci calculators
anyway, sqrt button twice does work:
sqrt 1.35=1.1618
sqrt 1.1618=1.0779
forgive the next dumb question...
imagine the time was 6 years:
135=100(1+r)^6
135^(1/6)=1+r
1+r=135^0.1666
1+r=1.0512
r=0.0512
r=5.12%
...how would you reduce that on a normal calculator?
1st root: 1.161895004
2nd root: 1.077912336
3rd root: 1.038225571 [too low]
now, I realize i've strayed into the realms of "does it matter? the exam is done already!".... but I am curious!0 
a third square root would be the eighth root, will have to give the sixth root some thought... straight forward though with a sci calc0
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