Page 2 of 4 FirstFirst 1234 LastLast
Results 21 to 40 of 70

Thread: hee hee need help in Pri 4 Maths (again)

  1. #21

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ziploc View Post
    Putting it into equations:

    G = (H + D) % 10, X = INT((H + D) / 10)
    K = (A + E + X) % 10, Y = INT((A + E + X) / 10)
    F = (K + J + Y) % 10, Z = INT((K + J + Y) / 10)
    H = K + K + Z

    As you can see above, I don't think you can't solve it via algebra as there are simply not enough info - nothing that defines D, A, E, J. It can only be done by brute force substitution.
    it's not that you can't.

    can't suggests a lack of ability to do so.

    you can, but brute force does it much more efficiently.. ironically.

  2. #22

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by brapodam View Post
    No they're putting too much stress on pri sch students. What kind of rubbish is this? We learn simple algebra like 3x + 4y = 15, 2x + y = 10, calculate x and y this kind of qn in sec 2, what kind of question is this?

    Next thing you know they'll start teaching complex numbers and hypothesis testing in kindergarten...seriously
    well, no one ever said that examinations should be just about regurgitating methods taught to you in class, with just a few modifications here and there to make them "more challenging" (when they are not).

    they are about differentiating academic ability, is that right?

    if that is the case, if academic ability is about brute force memorisation, then it can encompass brute force trial and error methods too, right?

    i fail to see how setting a goal for primary school students where they could get full marks in mathematics examinations by studying enough will aid them in the long run. it's high time that kids get a taste of real life earlier on, so that the singaporean education system doesn't just produce a bunch of kids waltzing into secondary school, falling flat on their face when they can't get 100 marks for maths, and then start crying in class.
    Last edited by night86mare; 15th July 2010 at 09:43 PM.

  3. #23

    Default Re: hee hee need help in Pri 4 Maths (again)

    Wahlao...I totally stone. Is this math or is this a letter to aunt angie?

    I love A, but A love C while C hate K but C loves K. LOL
    My Flickr iownthislensthatlensthislensthatlens

  4. #24
    Deregistered allenleonhart's Avatar
    Join Date
    Sep 2008
    Posts
    3,656
    Blog Entries
    1

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ziploc View Post
    Putting it into equations:

    G = (H + D) % 10, X = INT((H + D) / 10)
    K = (A + E + X) % 10, Y = INT((A + E + X) / 10)
    F = (K + J + Y) % 10, Z = INT((K + J + Y) / 10)
    H = K + K + Z

    As you can see above, I don't think you can't solve it via algebra as there are simply not enough info - nothing that defines D, A, E, J. It can only be done by brute force substitution.
    btw, i realised ur using programming logic
    cause suddenly i realise the mod and int looks familiar to wad i did for my noi training

  5. #25

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by allenleonhart View Post
    LOL wont lar. brute force is just test numbers. pluck a few and try ur luck. not much to pluck in actually. K is the first one wanna identify, then guess H, (K+K) or (K+K+1)

    which leaves u with 7 more numbers all DIFFERENT to guess with.

    thats why i can do in 5 mins
    hey don't forget, you're a jc1 student, and one who does decently in math. This qn is given to a pri 4 student...can they really solve it? maybe those prc scholars can, but what about locals?

  6. #26
    Deregistered allenleonhart's Avatar
    Join Date
    Sep 2008
    Posts
    3,656
    Blog Entries
    1

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by brapodam View Post
    hey don't forget, you're a jc1 student, and one who does decently in math. This qn is given to a pri 4 student...can they really solve it? maybe those prc scholars can, but what about locals?
    no lar. my jc 1 didnt teach me this

    this one was familiar cause i encountered when i was p4-6 that time

    u can imagine. i used to be a good boy and do 10 pages per assessment book per topic per day one

    encounter this always use logic. what are the stuff i can link and limit first?

    then brute force
    Last edited by allenleonhart; 15th July 2010 at 09:46 PM.

  7. #27

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by brapodam View Post
    hey don't forget, you're a jc1 student, and one who does decently in math. This qn is given to a pri 4 student...can they really solve it? maybe those prc scholars can, but what about locals?
    are you telling me that you need a few more years of education to dial in random numbers and do trial and error?

    essentially, the skills needed here are:

    1) carefulness
    2) determination to get the answer
    3) simple addition skills

    i believe 1 and 2 are part and parcel of examinations, 3 is taught by primary 4. i hope.

  8. #28
    Moderator ziploc's Avatar
    Join Date
    Jan 2002
    Location
    Snoopyland
    Posts
    4,577

    Default Re: hee hee need help in Pri 4 Maths (again)

    Here is another possible answer:

    1172
    1643
    2815

    So as you can see, the answer is not exact, and there are many permutations. The question looks like something that was conjured up by a teacher too creative. And it looks more like IQ test than math.
    Last edited by ziploc; 15th July 2010 at 09:51 PM.

  9. #29

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by night86mare View Post
    are you telling me that you need a few more years of education to dial in random numbers and do trial and error?

    essentially, the skills needed here are:

    1) carefulness
    2) determination to get the answer
    3) simple addition skills

    i believe 1 and 2 are part and parcel of examinations, 3 is taught by primary 4. i hope.
    Yes but I dont think math should be about planting in random numbers and see if they fit. That should perhaps be part of a life skills workshop to develop people's perseverance and patience to solve the problem by testing a random range of numbers in the equation.

    Sure, some logic is involved, but nothing much more than that. I thought they were already trying to discourage the "trial and error" method since the start of our education, why else would they teach us ways to work out "number pattern" style questions?

  10. #30

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ziploc View Post
    Here is another possible answer:

    1172
    1643
    2815

    So as you can see, the answer is not exact, and there are many permutations. The question looks like something that was conjured up by a teacher too creative.
    i am wondering if you can get K = 2.

    i tried for a while, and i couldn't really get anything like that out.

  11. #31

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by brapodam View Post
    Yes but I dont think math should be about planting in random numbers and see if they fit. That should perhaps be part of a life skills workshop to develop people's perseverance and patience to solve the problem by testing a random range of numbers in the equation.

    Sure, some logic is involved, but nothing much more than that. I thought they were already trying to discourage the "trial and error" method since the start of our education, why else would they teach us ways to work out "number pattern" style questions?
    perhaps, but that's your view.

    i think as long as the entire examination doesn't centre around 50 questions of that sort, it should be fine.

    such questions can be a way to differentiate better students from the poorer ones - or perhaps how exam-smart you are. obviously, if i were a pri 4 student and i got this for my final examinations, i would realise quickly that this would be a waste of my time relative to the other "safer" questions, and leave it for last. then if i check my answers and i still have time, then can play with numbers for fun.

  12. #32
    Moderator ziploc's Avatar
    Join Date
    Jan 2002
    Location
    Snoopyland
    Posts
    4,577

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by brapodam View Post
    Yes but I dont think math should be about planting in random numbers and see if they fit. That should perhaps be part of a life skills workshop to develop people's perseverance and patience to solve the problem by testing a random range of numbers in the equation.

    Sure, some logic is involved, but nothing much more than that. I thought they were already trying to discourage the "trial and error" method since the start of our education, why else would they teach us ways to work out "number pattern" style questions?
    Yes agree. This is not math but logic.

  13. #33
    Deregistered allenleonhart's Avatar
    Join Date
    Sep 2008
    Posts
    3,656
    Blog Entries
    1

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ziploc View Post
    Yes agree. This is not math but logic.
    ziploc, try for a permutation for k is not equal to 1.

    so far the ones we did, K=1 only

    qn wants another value of K

    if not K=1 means something else. got hidden logic that limits k=1

    anyway to do it via Cprog? input value of K and guess value.

    i think possible
    Last edited by allenleonhart; 15th July 2010 at 10:00 PM.

  14. #34
    Moderator ziploc's Avatar
    Join Date
    Jan 2002
    Location
    Snoopyland
    Posts
    4,577

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by night86mare View Post
    i am wondering if you can get K = 2.

    i tried for a while, and i couldn't really get anything like that out.
    Yes you can, like this:

    2295
    2731
    5026

  15. #35

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by night86mare View Post
    perhaps, but that's your view.

    i think as long as the entire examination doesn't centre around 50 questions of that sort, it should be fine.

    such questions can be a way to differentiate better students from the poorer ones - or perhaps how exam-smart you are. obviously, if i were a pri 4 student and i got this for my final examinations, i would realise quickly that this would be a waste of my time relative to the other "safer" questions, and leave it for last. then if i check my answers and i still have time, then can play with numbers for fun.
    at that age, i would have probably wasted the whole time trying to figure out how to do that, and will only give up after i realise the exam is about to end lol
    Quote Originally Posted by allenleonhart View Post
    ziploc, try for a permutation for k is not equal to 1.

    so far the ones we did, K=1 only

    qn wants another value of K

    if not K=1 means something else. got hidden logic that limits k=1
    later they change the qn to "find all the possible combinations of values of A, D, E, F, G, H, and K".

    Right, go for it, math geek!

  16. #36

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ziploc View Post
    Yes you can, like this:

    2295
    2731
    5026
    yes, seems correct.

    whoever thought this question up needs to be shot.

    if must give tedious question, at least make sure that you only have one answer, tsk.

  17. #37
    Moderator ziploc's Avatar
    Join Date
    Jan 2002
    Location
    Snoopyland
    Posts
    4,577

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by allenleonhart View Post
    btw, i realised ur using programming logic
    cause suddenly i realise the mod and int looks familiar to wad i did for my noi training
    It is actually quotient and remainder, but since there is no symbol for quotient, I use programming representation. Here is another way of putting it (in words):

    G = remainder((H + D) / 10), X = quotient((H + D) / 10)
    K = remainder((A + E + X) / 10), Y = quotient((A + E + X) / 10)
    F = remainder((K + J + Y) / 10), Z = quotient((K + J + Y) / 10)
    H = K + K + Z
    Last edited by ziploc; 15th July 2010 at 10:11 PM.

  18. #38
    Moderator ziploc's Avatar
    Join Date
    Jan 2002
    Location
    Snoopyland
    Posts
    4,577

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by night86mare View Post
    yes, seems correct.

    whoever thought this question up needs to be shot.

    if must give tedious question, at least make sure that you only have one answer, tsk.
    Well, nowadays they also want to teach creativity to students and not only confine to rules. This question does look like they are teaching trial-and-error.

  19. #39
    Member leowyien's Avatar
    Join Date
    Jan 2010
    Location
    Serangoon
    Posts
    859

    Default Re: hee hee need help in Pri 4 Maths (again)

    LOL WHERE GOT PRIMARY 4!! If really then i really fail......

    Pass PSLE last year, also no such question....
    indie

  20. #40

    Default Re: hee hee need help in Pri 4 Maths (again)

    My answer K = 3

    using logic K + K = H, H <=9 thus K can only be 1-4

    so if K = 1, then H = 2 11A2 + 2JED = 2F1G
    since 1 and 2 is taken, the rest of the letters = 3,4,5,6,7,8,9,0
    using the additional of the second digit 'A' and 'E' A+E = 1 or 11 there for A,E = 5,6/4,7,3/8 (combination to form 11)
    if we choose A, E as 5, 6 then the 1st digit 2 + D must be less than 10 else the 2nd digit is not true, thus if we sub D as 7 then 2 + 7 = G = 9
    now left with 4,3,8,0
    3rd digit K + J = F, since 2nd digit A+E we set as 11 therefore the 3rd digit will have a carry over, become 1+1+J = F === 2+J = F (F = 3/4/8/0) F cannot = 3 cause J is not 1, F cannot = 4 cause J is not 2 (K = 1 and H = 2) leaving 8 and 0 but 2 + J = F cannot have a carry over thus F cannot be 8 as A, E is 5, 6 and 0 is out

    there for K = 1 is not possible, continue with K = 2 and H = 4, bah bah bah


    when u get to K = 3 33A6 + 6JED = 6F3G
    Numbers left = 1,2,4,5,6,7,8,9,0

    same method start with A, E A + E = 13 (combination for A, E = 4,9/5,8/6,7)
    set A, E as 4, 9 then your 3rd digit must be 1+3+J = F (F must be <9 so that there is no carry over) let J = 1 your 3d digit 1+3+1 = F (F=5)
    so far numbers left are ( 2,7,8,0 )
    last set of digits 6 + D = G (G < 10) thus 6 + D(D=2) = G(G=8)
    numbers left ( 7,0 )
    D80/D700 18-135/17-50/14-24/24-70/70-200/70-300/50/60 SB600/900

Page 2 of 4 FirstFirst 1234 LastLast

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •