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Which one is usually harder- Non calculator paper or calculator paper?


scstdnt

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I was wondering which of the HL maths paper are generally harder- the non calculator paper or the calculator paper?

Thank you.

PS. Sorry Mahuta, please forgive me, I haven't read all the forum rules (only read the rules on the EE forum).

My apologies.

Edited by scstdnt
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PS. Sorry Mahuta, please forgive me, I haven't read all the forum rules (only read the rules on the EE forum).

My apologies.

No problem! See now it looks less confusing, lol.

Anyways, now I can give my opinion.

They're both hard for me in different ways.

Paper 1 is sometimes hard in that you dont have a calculator to check your answers and you're screwed if you end up forgetting how to do a certain thing manually. Also, because I know as a fact that there's nothing I should need the GDC for, which makes me think of it as something simple and easy.

Paper 2 can be hard because having a GDC is sometimes confusing in terms of it sometimes giving you weird answers which makes you panic and lose concentration.

However, I tend to like Paper 1 more.

Like Joel said, Paper 2 was a disaster this year in TZ2 as well. BUT, I screwed up paper 1 as well, so yeah not a person to follow. :P

The thing is, if you are confident and know what you're doing, you will not get confused nor panic, this is what happened to me in the mocks which gave me a 7, but in the real ones..it was different for me.

Having concentration in the exam will get you a 7, in both papers.

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Paper 1 is sometimes hard in that you dont have a calculator to check your answers and you're screwed if you end up forgetting how to do a certain thing manually.

We had a question once on a test where we were supposed to find the sum of the first 8 terms of the series (1/4)(1/4)n, which requires you to find (1/4)8 and I simply don't see how that question only was one mark. That's stuff you never ever claculate manually.

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I tend to find paper 1 (non-calculator) easier than paper 2 (calculator). That was definitely the case in my mock, however in the actual IB exams (Nov 2008) I found paper 2 to be easier. Once I got my component grades, I scored higher on paper 1, but that's because my calculator died on the last question of paper 2 and I forgot to take spare batteries, so I lost half of the points on the last question because my time ran out.

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We had a question once on a test where we were supposed to find the sum of the first 8 terms of the series (1/4)(1/4)n, which requires you to find (1/4)8 and I simply don't see how that question only was one mark. That's stuff you never ever claculate manually.

(1/4) (1/4)n = (1/4)n+1 , right? It can be done manually, why couldn't it? Although you're probably right; it will not be the first 8 terms but certainly the first few terms.

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I say it depends on who you are. Personally I find the calculator paper waaaay easier. I can't add up or do any mental maths very fast (or accurately) and the functions the calculator makes available (ahh my beloved graphs!) really do improve your ability to solve questions :)

I always do better with a calculator to help me work out the answer. In paper 2, if you understand the principles, it's just a question of typing it in, whereas with paper 1 it's understanding the principles, taking the time to work it all out, not making any errors and knowing how to do it all manually :( Murder!

Different people find different things easier/harder. Personally I think loads of it is to do with how well you were taught mental maths as a kid. Definitely accquaint yourself with the calculator though, it's so handy! :P

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We had a question once on a test where we were supposed to find the sum of the first 8 terms of the series (1/4)(1/4)n, which requires you to find (1/4)8 and I simply don't see how that question only was one mark. That's stuff you never ever claculate manually.

(1/4) (1/4)n = (1/4)n+1 , right? It can be done manually, why couldn't it? Although you're probably right; it will not be the first 8 terms but certainly the first few terms.

It was a 2 point question, so I suppose there is some easy way of doing it, but I don't understand it. Appearently is the answer 1/3, if that helps?

Feel free to explain it so me.

Edited by Tilia
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We had a question once on a test where we were supposed to find the sum of the first 8 terms of the series (1/4)(1/4)n, which requires you to find (1/4)8 and I simply don't see how that question only was one mark. That's stuff you never ever claculate manually.

(1/4) (1/4)n = (1/4)n+1 , right? It can be done manually, why couldn't it? Although you're probably right; it will not be the first 8 terms but certainly the first few terms.

It was a 2 point question, so I suppose there is some easy way of doing it, but I don't understand it. Appearently is the answer 1/3, if that helps?

Feel free to explain it so me.

The first 8 terms would give you a geometric progression:

1/4 + (1/4)(1/4) + (1/4)(1/4)^2 + (1/4)(1/4)^3 + (1/4)(1/4)^4 + ... + (1/4)(1/4)^7

The common ratio, r, is 1/4. The first term, a, is 1/4. And the number of terms, n, is 8, so we can apply the formula for the sum of a gemoetric sequence.

Sum of first eight terms = a(1-r^n) / (1-r)

Sum = (1/4)(1 - (1/4)^8) / (1-1/4) = (1/4)(1 - (1/4)^8) / (3/4) = (1 - (1/4)^8) / 3

Now we're left with the task of calculating (1/4)^8 = (1/2)^16

We know that 2^16 (or 4^8, whichever you prefer. I just find it easier to imagine how small or large numbers are with 2 as the base) is very large, therefore we know that (1/2)^16 (or (1/4)^8 ) will be VERY small.

So we can say that (1/2)^16 (or (1/4)^8) is approximately equal to 0.

So, Sum = (1-0) / 3 = 1/3 (APPROXIMATELY, because we approximated (1/4)^8 to 0)

I hope that helps =D

Edited by Redstar
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Anyway, getting back to the question "which paper is harder?"

Paper 1 requires you to know various numerical and trig values by heart since you have no calculator. Paper 1 this year has a vector question stating that cos x = 1/2 and you have to find the value of sin x.

You can do it by using sin^2 x + cos^2 x = 1 , or you can do it by knowing from the information that angle x=60 and that sin 60 = (root 3)/2

The trouble with paper 1 is any arithmetical errors will lead to stuff like factorising being impossible or stuff like simple arithmetic you should be able to do in your head getting too hard without a calculator.

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