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Recent content by Hassan Kazmi

  1. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    Since HCF of N and 500 has 5^2 in it, N must have 5^2 so q=2. Since LCM of N and 500 has 2^3 in it, N must have 2^3 because 500 only has 2^2, so p=3. Since LCM of N and 500 has 7 in it, N must also have a 7 as 500 does not have a 7, so r=1.
  2. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    b(ii) They both threw a six: this question has two events: 1) sarah throws a six and 2) terry throws a six the question asks you the probability that both the events 1 AND 2 happen at the same time. thus, we will multiply. if the question asked you to find the probability that either sarah...
  3. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    Ah, damn it! You're right. If we want only codes that contain both As, your method is fine. Another way of doing it is using the permutation: 2P2 x 5P2 x 3! (just like the blocks in the other question, it will be like [A|A][?][?] where these letters need to randomise). Then divide by 2! for...
  4. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    If you have the questions, post them here and maybe we can help solve them. I'm not sure where to get the mark schemes.
  5. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    Since HCF of N and 500 has 5^2 in it, N must have 5^2 so q=2. Since LCM of N and 500 has 2^3 in it, N must have 2^3 because 500 only has 2^2, so p=3. Since LCM of N and 500 has 7 in it, N must also have a 7 as 500 does not have a 7, so r=1.
  6. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    Yes, you are right for the second question. For the first part the answer is 3! x 6! because there are six blocks of people, like this, where one block contains 3 people: [1][1][1][1][1][3] = 4320 For the second part, again, it will be 3! x 2! x 5! because there are five blocks now...
  7. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    No, wait, I just overlooked the fact that we have to make 4-digit codes! Look at how this is done: All possible 4-letter codes: 7p4 (fair enough? although this contains those repetitive codes) = 840 4-letter codes containing both As: 2p2 x 5p2 (to make sure the 2 As are selected) = 40 (divide...
  8. Hassan Kazmi

    Hope I kept you motivated? :P

    Hope I kept you motivated? :P
  9. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    1) Simply doing 7! gives us all combinations, but, half of them are repetitive due to the 2 As. So, divide by 2! to remove those repetitions. = 7!/2! 2) Do this by first thinking about the possible combinations when the two people sit together, then subtract it from the total. So, when there...
  10. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    Compound Interest is in the CIE O Level Mathematics D (Calculator Version) syllabus. But that is a relatively simple compared to the questions above. For compund interest, you just have to use the I = PRT formula again and again using the modified P (principal) which means adding the interest to...
  11. Hassan Kazmi

    urgent maths paper 4 help needed

    Hm, so I see that 0580/m/j/10: There is no such part as Q7(a)(i).. there is only Q7(a), to which I answer: First convert the radius of 31cm to m, that is 0.31m. Volume of cylinder is pi*r^2*h = 3.142*0.31*0.31*15 = 4.53 m^3.
  12. Hassan Kazmi

    urgent maths paper 4 help needed

    oops, sorry. I thought it was A level math. Let me get back to you on that.
  13. Hassan Kazmi

    urgent maths paper 4 help needed

    For A, integrate the first equation v = A (t - 0.05t^2) with respect to t, with boundary t=0 to t=15. Integrating the velocity function gives us the distance, so, equate the integral you just found to 225, as the distance from t=0 to t=15 is 225m. For B, since v = A (t - 0.05t^2) and v = B/t^2...
  14. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    This is a quadratic graph, right? So the power of the equation has to be 2, thus, n = 2. Considering the possibility that this can be a graph other than a quadratic graph, it can have the power of any positive even number, such as 4, 6, 8 etc. a is basically the coefficient which...
  15. Hassan Kazmi

    Maths, Addmaths and Statistics: Post your doubts here!

    Cheers!
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