FP1 is set twice a year in January/February and June. The following exam papers are in PDF format.
June 2015
June 2014 Jan 2014
June 2013 Jan 2013 June 2012 Jan 2012 June 2011
Jan 2011 Jun 2010 Feb 2010 Jun 2009 Jan 2009 Jun 2008 Jan 2008 Jun 2007 Jan 2007
Jun 2006 Jan 2006 Jun 2005 Specimen 2005
The following worked solutions by Bardmeister are in PDF format.
Winter 2011
Summer 2010 Winter 2010 Summer 2009 Winter 2009
Summer 2008 Winter 2008 Summer 2007 Winter 2007
Any worked solutions that appear on this Wiki are the work of the Author and not WJEC. If you find any errors or omissions please leave a comment below.
Comments (7)
lovemath said
at 12:20 pm on Feb 3, 2009
Are there worked solutions/mark schemes for these papers?
Steph Richards said
at 1:10 pm on Feb 3, 2009
Unfortunately I am not allowed to put the mark schemes on the site as they are from the WJEC's secure website. If there is a specific question then I could post a worked solution of my own if that would help. Alternatively (and probably better for you to have immediate access) There is an Author I know (Dr Barry Nix) who publishes worked solutions from 2008 back to 2003. http://www.johnnix.co.uk/css/booklet7.htm gives a link which is a little out of date but I am sure you could get the necessary details from there
Steph Richards said
at 1:20 pm on Feb 3, 2009
I have put a link in the sidebar on the Front Page to the WJEC ON LINE store. This link will take you to the past paper books that you can purchase from the WJEC directly. Hope this helps
Steph Richards said
at 9:28 pm on Jan 11, 2011
Thanks very much for these worked solutions! They should help greatly
DaisyMYH said
at 1:50 pm on Nov 9, 2012
Is there any chance someone could do worked solutions for the other past papers such as both of 2006's papers and Winter/Jan 2011? It would be really, really helpful. Thanks!
Tara said
at 10:03 am on Sep 12, 2014
In the worked solutions for winter 2010, in Q (1a) where does the +(1+2i) go? Why does it become x (1+2i)?
ArthurBaas said
at 2:13 pm on Sep 12, 2014
The 1+2i is taken out as a factor of the first two terms, the step in between would be f(1+2i)=(1+2i)[(1+2i)^2+1]. Hope that helps!
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