2007 Mathematics paper 8a)ii) (1 Viewer)

[lyounamu]

A lot of people have been arguing over whether the answer for this november or october.

I am definitely sure that it is 2010 november but I want to see what others think.

http://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/pdf_doc/maths-07.pdf

Thanks in advance for your contribution.

 

[Azreil]

Well.

I got 34.xxxxxxxxxx months after January 2008.

Which I interpret to by 2010 November.

 

[alez]

i wasnt sure when i did it either
but worked it out to be november
bit iffy though

 

[Arithela]

got november as well

 

[lyounamu]

Can anyone get an answer from the past papers book like Success One?

 

[Absolutezero]

I've just looked it up in success one. It says the answer is November 2010.

 

[Baggygreen408]

its november, cause u get the year and then in terms of months, something like 10.23, whihc means the 11th month, november.

 

[henry08]

its november, cause u get the year and then in terms of months, something like 10.23, whihc means the 11th month, november.

Seconded.

 

[wozzame]

It's definitely November 2010. If you go through all your log laws, etc., you should get that t=2.8872... which is in years. So that's 2 years + 0.8872... years.

Multiply the 0.8872... to get it into months, and you get 10.6473... months. 2 years and 10 months from Jan 2008 is Nov 2010. Therefore, the amount of mobile phones will first exceed 4000 million in November 2010.

 

[stormz89]

N = Ae^kt
where N is the estimate for the number of mobile phones in use in mil and t is the time in years after 1 January 2008

if you read the question carefully it says which month and year will the number of mobile phones in use first exceed 4000mil

It cannot be october because that would be 1st of oct and its actually the 18th of oct, although it doesn't ask for the day so the first month where it exceeds 4000mil is november. Makes sense?