C3 Jan 11 Edexcel - Solutions and Paper in the first post (Now On) + TIPS

well done I got that too, I did it as a mock in college. I lost a mark because i didn't know how to convert the second to last question answer in to hours minutes and second :P I still dont think i know it to be honest. Do you know how you find it??

Ohh I've got that one in front of me right now :P
Basically, you end up with t (hours)=2.098 OR t (hours) = 6.715...
If you multiply each by 60 you get the number of minutes.
Hence t = 125.88...minutes OR 402.691...minutes
The question asks to the nearest minute, so t = 126 OR 403 minutes.
I just worked out the rest "manually" - so we know 126 minutes = 2 hours something (t in hours was 2.098) just work out the remainder -> 126 minutes = 2 hours, (126 - (60x2)) minutes = 2 hours, 6 minutes.
Likewise, 403 minutes = 6 hours, (403 - (60x6)) minutes = 6 hours 43 minutes.

As t is the time after noon, the final result is 14:06 or 18:43.

Hope that helps , it's alright once you get it!

Ohh I've got that one in front of me right now :P
Basically, you end up with t (hours)=2.098 OR t (hours) = 6.715...
If you multiply each by 60 you get the number of minutes.
Hence t = 125.88...minutes OR 402.691...minutes
The question asks to the nearest minute, so t = 126 OR 403 minutes.
I just worked out the rest "manually" - so we know 126 minutes = 2 hours something (t in hours was 2.098) just work out the remainder -> 126 minutes = 2 hours, (126 - (60x2)) minutes = 2 hours, 6 minutes.
Likewise, 403 minutes = 6 hours, (403 - (60x6)) minutes = 6 hours 43 minutes.

As t is the time after noon, the final result is 14:06 or 18:43.

Hope that helps , it's alright once you get it!

That last question from the June 2010 paper was very hard! How you guys did it just...

Well part a, what you needed to do was factorise each expression and then parts cancel out

For b, when you do the log rules, you see that you get the same simplified expression when you put the fraction in one (LogA-Log B=Log (A/B) ) Then solve for x

Tip:
Easy way to convert minutes to hours on the calculator. Round to the nearest minute, divide by 60, and press the button that looks like this: º' ''
E.g. Take 126minutes. Do the calculation 126÷60, hit the º' '' button.

Which solomon paper is the hardest? Which solomon paper is the closest to the real thing and hard? I only have time to do a few.

Thanks.

Impressive, but my friend got
C1-100
C2-100
C3-98
C4-96
S1-100
M1-100
I know, she's mad

I got 98 in C1, 82 in C2 (been too lazy), 43 in S1 (haha, re-took for probably over 85) and I need high A for C3... I hope this Jan will be different - that it will be easier than usual, cuz my teacher said we won't get a chance to retake it in summer :/

I just did the june 2010 paper, I got 74/75 But lost one mark for forgetting to put down a solution

Quick question guys.

You know the question where you have to find the equation of a line e.g. a tangent, in the mark scheme they don't use this method, but are we allowed to use y = mx + c and plug in a known coordinate to get the line? If you get what I mean.

As long as you ger the right equation for the tangent in the form their asking, then you can use whatever method you like, unless it specifically said "show by calculus"

OK.
So "Show by calculus" wouldn't fit with my method?
The method I use:
So basically what I would do is find the gradient and we know a point on the line.

Example: (5,1) and m = 5

y = mx + c
y = 5x + c
1 = 5(5) + c
1 = 25 + c
c = - 24

equation of line:

y = 5x - 24

How do you find the gradient in the first place? Because thats the part where you would use calculus/differention such as dy/dx of the curve -> put in x=5 -> find out m=5 then you can use method above for finding out the tangent like you did But if you found out m=5 by other method (most of the time it's probably not even possible) eg. using perpendicular lines etc. then of course that wouldn't count as using calculus/differention.

But don't worry really, they hardly ever specifiy a technique and I don't want to confuse you xD If you got a method, and it works you'll be good

I think he means doesn't use the formula y-y1=m(x-x1). He just uses y=mx+c using a point on the line (or the point itself).

OK.
So "Show by calculus" wouldn't fit with my method?
The method I use:
So basically what I would do is find the gradient and we know a point on the line.

Example: (5,1) and m = 5

y = mx + c
y = 5x + c
1 = 5(5) + c
1 = 25 + c
c = - 24

equation of line:

y = 5x - 24

Just got a bucketload of papers from my teachers. Have 16 to do by monday, as well as revising for M2, physics and chemistry! xD . Though... I want some more.

The edexcel site is missing some papers and markschemes by the looks of it. Does anyone have the following (+markschemes)?:

Jan/June 03
Jan/June 04
Jan 05
June 07

Thanks!

Mine only start from june 05... but here's the june 07 one