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Edexcel C3,C4 June 2013 Thread

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Original post by justinawe
I don't quite understand your point here. You've done all the other past papers at home or in school, this is nothing like sitting for an exam for real. People always think their paper was the hardest, exam pressure makes all the difference here.

Which questions were particularly hard, anyway? Maybe part (a) of question 3 (though honestly, everyone should know that sin[x]=cos[90°-x], this is basic stuff), and the question with the mechanics context may have been a bit confusing for some. Doesn't seem too much different from your ordinary paper. It was more difficult than average, sure, but not significantly so.

I don't really care if you feel insulted or if you think I'm being patronising. It quite simply wasn't "drastically different". If anything, others could say they're insulted by your claim that the papers they sat for were significantly easier.


the paper was very different though! only about 3 questions out of the 8 were the regular questions that come up

it may have been easily done for some people, however the statistics say that it was extremely different. apparently edexcel have had thousands of complaints from students and teachers, this has never happened to this extent before of what I'm aware.

when the majority of people taking the exam were shocked about how hard it was something must be wrong. it's not like every single person was underprepared or freaked out due to the stress in the exam hall


Posted from TSR Mobile
Original post by jadecross
can someone help me with this question please, how do you simplify:
Unparseable latex formula:

-ln|6|-ln|9|+ln|4|[color="#222222"] ? the answer is apparently [tex]ln(| \dfrac{2}{27} |)[/tex]? I'm really confused! [s]frown[/s]



-ln l6l is ln l1/6l. Subtracting lns means dividing so ln l1/6l -ln l9l= ln l1/54l. Then adding lns means multiplying so ln l1/54l + ln l4l= ln l4/54l=ln l2/27l :smile:
Reply 3382
Original post by jadecross
can someone help me with this question please, how do you simplify:ln6ln9+ln4-ln|6|-ln|9|+ln|4| ? the answer is apparently ln(227)ln(| \dfrac{2}{27} |)? I'm really confused! :frown:

My tip is I usually put the positive ln first.

So ln(4) - ln(6) - ln(9)
then you use your log laws so ln(4)-ln(6) would be ln(4/6)

and then ln (4/6) -ln(9) would be ln(4/ 6x9) which is ln(4/54) which simplifies to ln (2/27) :smile:
Original post by samueliscool
-ln l6l is ln l1/6l. Subtracting lns means dividing so ln l1/6l -ln l9l= ln l1/54l. Then adding lns means multiplying so ln l1/54l + ln l4l= ln l4/54l=ln l2/27l :smile:


Original post by TLi1295
My tip is I usually put the positive ln first.

So ln(4) - ln(6) - ln(9)
then you use your log laws so ln(4)-ln(6) would be ln(4/6)

and then ln (4/6) -ln(9) would be ln(4/ 6x9) which is ln(4/54) which simplifies to ln (2/27) :smile:

thank you so much guys! :smile:
Original post by jadecross
can someone help me with this question please, how do you simplify:ln6ln9+ln4-ln|6|-ln|9|+ln|4| ? the answer is apparently ln(227)ln(| \dfrac{2}{27} |)? I'm really confused! :frown:


You should remove the modulus as they're known constants, first of all... |6|=6, |9|=9, etc.

ln6ln9+ln4-\ln 6 - \ln 9 + \ln 4

=ln4ln6ln9= \ln 4 - \ln 6 - \ln 9

Apply your log rules here - logcalogcb=logcab\log_c a - \log_c b = \log_c \dfrac{a}{b}

=ln46ln9=ln23ln9= \ln \dfrac{4}{6} - \ln 9 = \ln \dfrac{2}{3} - \ln 9

=ln239=ln227= \ln \dfrac{\frac{2}{3}}{9} = \ln \dfrac{2}{27}
[video="youtube;0EW8N60kHkU"]http://www.youtube.com/watch?v=0EW8N60kHkU[/video]
LMFAO!!
Reply 3386
Original post by Turnitdown
In what way did they change in January??


The had changed to some extant, the type of questions they normally ask never came up like the max value of the fraction.
Original post by deanna3006
the paper was very different though! only about 3 questions out of the 8 were the regular questions that come up

it may have been easily done for some people, however the statistics say that it was extremely different. apparently edexcel have had thousands of complaints from students and teachers, this has never happened to this extent before of what I'm aware.

when the majority of people taking the exam were shocked about how hard it was something must be wrong. it's not like every single person was underprepared or freaked out due to the stress in the exam hall


Posted from TSR Mobile


It doesn't mean that something is wrong with the exam. It means something is wrong with the way people are learning the content, seeing that everyone loses their mind the moment they have to do something slightly different to what they're used to.

Exams are supposed to be about applying the same knowledge to different situations, not applying the same knowledge to the same old situations over and over again.
Original post by MAD Phil
I haven't seen that anyone's posted any C3R answers yet, so here's my attempt.

Let me know about my mistakes, as usual!


omfg, the international paper or the first link in your post, those questions were sooo eassyy wtfff??????????

was that the international paper right?
Reply 3389
Original post by MAD Phil
I haven't seen that anyone's posted any C3R answers yet, so here's my attempt.

Let me know about my mistakes, as usual!


Is this the original paper? Not the replacement one?
Original post by ACBLISS
The had changed to some extant, the type of questions they normally ask never came up like the max value of the fraction.


Was this just C3 or is it fair to expect the same thing on C4?
Original post by BooAlphie
I know why I struggled. I did understand the concepts before going into the exam very thoroughly, and had I had another hour or so, and had it not been such a shock I probably would have been able to do that paper. The reason it was a shock? Because up until now we've had easier papers in the same format. Is that right? No probably not, but the fact that had they not lost the original paper that was easier, and in line with all the other papers, shows that it's not even like they planned to make it harder because they thought the exams were not differentiating enough.

I agree with what you say about the UMS conversions, but what may have been "bog standard easy papers" for some people were clearly not for others otherwise we would all have been getting 100% and they would have already changed the system.

And what you say about silly mistakes, I can retaliate with silly mistakes. So many people will have made more silly mistakes than usual because of the shock of this paper, as well as losing the marks that they could not do at all. It's therefore in no way an accurate reflection of what people can do.

No, it is not Edexcel's job to prepare me, but when I buy officially endorsed textbooks from them, I expect them to be of some use. When I do past papers of similar format I expect the paper I sit in the exam to be of similar format unless told otherwise, which it would have been if we sat the original paper. Whilst they did not promise the papers to be of similar style and format, they know that schools are preparing students like this.

Essentially what you have said is that hard work is not worth anything because exams such be hard and test natural ability.

Edexcel are at fault because:
1) They lost a bunch of papers, were concerned enough to replace them, and now that some people have accidentally sat the wrong papers, which I agree is not their fault, are going back and saying "we don't have evidence that they've been leaked", because they ceebs to do anything. So now people feel like they have been unfair disadvantaged.
2) They say that this paper was to the same standard as other papers, which it defintely was not.
3) They have misled me, and other pupils by telling allowing us to practice for this exam with normal papers, and suddenly changing them.

You are saying this because you have found the paper ok, which I believe is because you have understood the concepts more thoroughly but you are also a gifted mathematician probably. Not everyone is as lucky. If the questions were slightly different, it could easily have been you in this position, and getting upset.

EDIT- you are completely missing the point that had we recieved the original paper we wouldn't even be having this discussion. It means it's not even like Edexcel decided to make them harder, because they were not challenging enough, as the original paper was easier, and that is what they intended to give us. It just feels like it's something they've thrown together at the last minute.

Not everyone found the old papers easy. You obviously did, but some people worked incredibly hard for their grades in those papers. I'm glad that you liked the paper this year though. As long as you found it challenging and had a ball doing it.

I couldn't have said it better myself. I completely agree with you. I'd rep you if I could
Reply 3392
Original post by justinawe
It doesn't mean that something is wrong with the exam. It means something is wrong with the way people are learning the content, seeing that everyone loses their mind the moment they have to do something slightly different to what they're used to.

Exams are supposed to be about applying the same knowledge to different situations, not applying the same knowledge to the same old situations over and over again.


im guessing you did well
May I ask how you know and why the exam style changed since jan?
Reply 3394
Original post by justinawe
But due to grade boundaries there will still be around the same percentage of people achieving A*, A etc... so how is it unfair? Remember that your grades are a measure of how you did relative to other candidates, not how well you actually did in the paper.

What good does changing the spec do anyway? Everything in this exam was on the spec. The spec isn't what decides the difficultly of the paper in general, it's the exam questions.


Yeah if they significantly lower the grade boundaries then it will work out the same so thats fine. Otherwise, it's not fair. In terms of changing the specification, if edexcel think that their papers are too easy at the moment, they should create a new maths specification with new test papers that are ALL significantly harder, not deciding to have one paper much harder then the rest in our specification. Having said that, if edexcel lower the grade boundaries for this then it will be fine.
Original post by yaboy
im guessing you did well


I did C3 in Jan.
Original post by justinawe
You should remove the modulus as they're known constants, first of all... |6|=6, |9|=9, etc.

ln6ln9+ln4-\ln 6 - \ln 9 + \ln 4

=ln4ln6ln9= \ln 4 - \ln 6 - \ln 9

Apply your log rules here - logcalogcb=logcab\log_c a - \log_c b = \log_c \dfrac{a}{b}

=ln46ln9=ln23ln9= \ln \dfrac{4}{6} - \ln 9 = \ln \dfrac{2}{3} - \ln 9

=ln239=ln227= \ln \dfrac{\frac{2}{3}}{9} = \ln \dfrac{2}{27}


Thanks Justin! :smile:
Original post by EL77
Yeah if they significantly lower the grade boundaries then it will work out the same so thats fine. Otherwise, it's not fair. In terms of changing the specification, if edexcel think that their papers are too easy at the moment, they should create a new maths specification with new test papers that are ALL significantly harder, not deciding to have one paper much harder then the rest in our specification. Having said that, if edexcel lower the grade boundaries for this then it will be fine.


They will significantly lower the grade boundaries if people do significantly worse. There will always be roughly the same percentage of people achieving A*, A etc, so if people do badly they will have to lower the grade boundaries as such.

If they don't significantly lower then, then that would mean others didn't find it as hard as you did, then. Though looking at most of the reactions on here, that probably isn't the case.
Reply 3398
Original post by justinawe
I did C3 in Jan.


yeh I did that paper and accept that it was my fault for flopping that paper, but this paper was ridiculous.
Hey guys, I was wondering if anyone could clarify something for me on this integration question.

I know about taking all the terms to their respective sides, (for example if they were x and y terms) but this is P and a constant k.

My question: Is there a way where I could have taken all of it to the side with respect to P and integrated that accordingly.

Soo.. I would have

1/kp dP = 1 dt

Then I would take the 1/k outside of the integral and carried on like that? Would this have worked? The markscheme says that the k is on the side of dT and it's integrated with respect to T.

However, I'm almost sure I've seen one other question where they took the k to the same side..

Help please and thank you in advance

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