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/ Maths Query

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Yanis Nayu - on 14 May 2018

Trying to help my daughter with her GCSE revision...

Is there an elegant way of calculating the annual percentage increase if a business increases in value from £250k to £325k over 6 years? I can only do it iteratively  

Thanks in advance!

 

wintertree - on 14 May 2018
In reply to Yanis Nayu:

(325/250) to the power of (1/6)

 

edit: to explain 

If is the annual rate, 325 = 250 * r^6.  

You solve for r by first dividing both sides by 250, then taking the 6th root of both sides, equivalent to raising to the power of 1/6.  That gives the rate as an annual fractional increase, which is about 1.044, so that’s 4.4%.

 

Post edited at 18:55
MG - on 14 May 2018
In reply to Yanis Nayu:

Sixth root of 325/250, minus 1, times 100?

Yanis Nayu - on 14 May 2018
In reply to wintertree:

Cheers!

Yanis Nayu - on 14 May 2018
In reply to MG:

Cheers!

Philip on 14 May 2018
In reply to Yanis Nayu:

I'm intrigued. The question is simple but the maths seems A-level. It's like the questions used to explain the use of logarithms.

The opposite, what is the sales value of a 250 company that grows 4.4% annually for 6 years, is easier. And part of explaining geometric series.

I think you need to find out what element of maths she is trying to learn with this exercise. Simply showing how to solve won't help her, but this is the maths you need to reverse compound interest - which I think is an important skill.

Dave B on 14 May 2018
In reply to wintertree:

I intuitively do this with logs, but your easy is faster and more elegant.

 

Yanis Nayu - on 14 May 2018
In reply to Philip:

I’ve spent nearly 16 years thinking pretty much my only use to her would be to help her with maths, but I’m useless. GCSEs seem to have gone from piss-easy to nearly A-level this year, combined with me having done my O-levels over 30 years ago...

Wil Treasure - on 14 May 2018
In reply to Philip:

>The question is simple but the maths seems A-level.

This is something which has always been on the GCSE syllabus. The new spec can ask questions in more complex scenarios, but this is something which would have appeared on the old one too.

I don't know why anyone would try to solve this with logarithms, it's just a rearranged equation which would appear on a calculator paper. You'd only need logarithms to solve for the power (which is an A-Level topic).

irt OP:

As Philip says, just showing her the solution isn't in itself helpful. The key thing here is that she understands how compound interest works (i.e. where the power comes from) I train my students to set up the equation as if they were calculating the interest (as wintertree did) and then to rearrange. Setting up an equation is a really important skill on the new specification.

wintertree - on 14 May 2018
In reply to Wil Treasure:

> I don't know why anyone would try to solve this with logarithms

If you had log and anti-log tabled to hand but no calculator it would make sense and would be the easiest way for sure.  I’ve worked with someone who still had rote memorised log tables in their head from 40 years previously.

If I didn’t have a calculator, and didn’t have the necessary tables, I would partially solve it in my head to get r^6 = 1.3.  Then as this isn’t much more than 1, I would take an estimate of r = 1 + (1.3 - 1) / 6 = 1.05 as my upper bound and half that amount above 1, r = 1.025, as my lower bound and do a few rounds of bisection to get the answer, with a lot of pen and paper multiplication.

Or perhaps I’d calculate the r^6 for 4 values between and including my bounds, plot them on a paper graph and draw a smooth curve, before reading off the answer.

Log tables definitely win in the absence of a calculator!

There’s presumably a pen and paper algorithm for computing fractional powers more directly, but I don’t know it.  Edit:  okay, I googled it - computers either use log/anti-log, bisection or Newton’s method.  Two of which are embodied in my all time favourite WTF piece of code - https://en.m.wikipedia.org/wiki/Fast_inverse_square_root

 

Post edited at 21:29
Wil Treasure - on 14 May 2018
In reply to wintertree:

> I’ve worked with someone who still had rote memorised log tables in their head from 40 years previously...

This makes sense. It also explains why it wouldn't have occurred to me; I'm not old enough ;-)

 

captain paranoia - on 14 May 2018
In reply to Wil Treasure:

> I don't know why anyone would try to solve this with logarithms,

Because, without access to a calculator, but with access to log/antilog tables, you can compute a numerical result.

And, when we were being taught maths, those were the exam conditions.

That sixth root computation isn't something I can do in my head. Or on paper...


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