Cambridge IGCSE past exam papers are now updated up to June 2025 examination series. Year 2028 Specimen papers are now available.
Important announcement:
Latest updates for year 2025 examination series is out now. Get the November 2025 and March 2026 exam timetables on our new site Skolatis.
Cambridge IGCSE is being offered by CAIE (Cambridge Assessment International Education) board from The University of Cambridge. These Cambridge IGCSE past exam papers are copyrighted by UCLES.
- Updated up to June 2025 series past exam papers. YAY!
- 2028 Specimen papers are now available.
- “Specimen Papers” would be available when there is a change of syllabus for that year. Otherwise, “Specimen Papers” would not be available when the syllabus stays the same.
- DO NOT download the past papers at last minute. We strongly suggest you to download everything you need at least 4 months before the exam.
Click on the below image to be redirected to our new site Skolatis to download your required past exam papers.
* If you are taking Edexcel iGCSE, please visit Edexcel iGCSE past exam papers page.
Any Grade Threshold Speculations, guys? for the feb march 42 0580 Mathematics.
How did you guys find the paper?
Regards,
Vaibhav
the hexagon question pissed me off.
What length you guys got for square, I got 11.4cm.
yeah me too. i agree with you for both. this paper was kinda hard though. i guess it was to make up for the easiness of paper 2.
Hey dude? even I got the same answer.
if so we get 4 marks or no marks!
exactly!! people in my class got the answer 12.68 or something…i think that question ought to be cancelled.
a dude in my class got 12.7
if they do that i’ll change my name to Harambe
Hi Friends!!!
How was you maths paper 4 Feb- March 2017? I think it was easy except for one hexagon question. Don’t forget to leave your thoughts too.
Yeah, indeed. That was an easy paper.But the Hexagon question was really tricky and tough.
the hexagon question just drained the energy out of me! but got that wrong! may lose big marks 🙁
How much marks are you expecting you would be losing? In the 4 mark question I wrote anything and everything I can so that examiner can award me at least 1 or even 2 in some case, probably. Other than that, everything went fine, I suppose.
The prove question before that hexagon question also went wrong! So i suppose i should get above 120 at least! thats a bit shame as it was an easy paper! 🙁
dude in that question,
was one side given as 3X^2??
It was easy except for the hexagon question 🙁
For the hexagon one,
the sides of the square are (20-x) and sqrt(3)x respectively, which are equal since its a square.
So 20-x = sqrt(3)x
square both sides
(20-x)^2=3x^2
x^2-40x+400=3x^2
which simplifes to:
x^2+20x-200=0
Hence, x = -10+10sqrt(3) = 7.32 (((-10-10sqrt(3) can be ruled out)))
Now for the square side length,
20-7.32=12.68
This is the correct answer, even confirmed it with my teacher
yeah same!! except for the hexagon question everything else was easy!!
Hey guys,
How did you find maths paper 4? It was really hard for me..Might loose as many as 20 to 30 marks….
What is the answer you got for pentagon question(second last question) and the sector question(the one with 3 combined sectors with angle 65)?
3.18 cm. I rechecked it and confirmed with my classmates too.
i got it as 6.02
That’s wrong. Put the formulas. the Perimeter was for whole shape. so the equation formed would be 3(2(Pie)Radius) = 20.5 – 2 r.
And I have checked it again by putting values and finding perimeter I got 20.5
3.18 cm for 3 combined factors question. It was hexagon. And in the second last question the answer was already mentioned as it was written ‘show that’. The answer was (20 – x)
3.79 cm
Oh Yes, Correction. I wrote 3.8 cm.
Oh Yes! I wrote 3.8 cm. I just forgot.
indeed friend youre right
same answer here.
you wont get 20.3 with 3.18…i checked it
the answer will be around 17..The answer will only come with 3.79
Yes, true. I agree with you Shruthi.
Yup…even I got that…
I corrected!! I wrote 3.8 but I wrote 3.18 here, somehow! Apologies!!
I amended my comment 🙂
Dear IGCSE Center, On what basis are grades alloted, Percentage or Percentile?
Hi Kanishka,
The grades are being allotted based on grade threshold.
Dear IGCSE Centre,
Can you please let us know on what criteria is Grade threshold set?
Regards
Hi Vaibhav,
It depends on the level of difficulty of the exam questions.
Okay! Can you please speculate it, if possible for the Feb March 2017? According to the opinions listed, I would like to know whether it would rise or fall as compared with the Feb March 2016 which was 184. Also, the opinions summarized are:
1. The Hexagon Question – 10 mark was really hard, and I don’t think most of them get it.
2. The paper was really very easy except for the 10 mark question (not 10 but for sure 4)
So, if you can please help? But that’s fine. Thank you for the reply! 🙂
Regards
Hi Vaibhav,
No one could ?speculate? the threshold, not even CIE. The threshold will only be produced AFTER marking all the exam papers.
But is the great threshold based on percentile or tage.
Hi Raj,
Percentage tells you only that how well you performed in the given test itself. In this case, you are not compared to other students who scored less/more than you. It represents the number of questions you answered correctly.
Percentile gives you only the idea of how good/bad you did as compared to other people.
Grade threshold is based on the level of difficulty of the papers.
How was the feb-march 2017 paper? I found it quite easy, should be able to get 120 up. What about you guys.
I didnt get that square unside a hexagon question.Can any body help??
Same here!
The answer for square was 12.7 cm.
Since all sides of square are equal,
(20 – x ) = 1.732 x
Thus, 2.732 x = 20
x = 7.32 cm.
Side of square = 20 -x = 20 – 7.32 = 12.7 cm
but 20-x was the side of a rectangle and the vertex of rectangle was different from square
that is what I thought the first time viewing the question as well but it remains the same since as the shape changes, the value of x changes, and yet the length of squares remain same as the length and breadth of the rectangle since x is a variable and it remained the same since the hexagon used was the same 🙂
the side was root 3Xsquare and the other side was 20-X, given that it was a square , both side were equal, therefore you had to either diretly equate it as root3Xsquare = 20-x or you could have taken the area as well
It was easy! Except for the Hexagon 4-mark question! That was challenging. I think I would get 120 +. Any Grade Threshold Speculations? I think the threshold may rise!!!