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    (Original post by Confused-teen)
    Another question: From CIE Thinking Skills June 2005

    The cog sizes and chain length on a bicycle are designed to even out wear, so that the same hole in the chain does not encounter the same tooth on the cog in every revolution. The chain wheel (front cog) of a particular bicycle has 48 teeth, the rear cog has 18 teeth and the chain has 90 holes.
    How many revolutions must the front cog do before the chain and both cogs are back in exactly the same relative positions (i.e. the same holes are encountering the same teeth on both cogs)?
    A 5
    B 8
    C 15
    D 20
    E 30
    Someone has done this already in this Thread.

    You have to find out the least common multiple of 18, 48 and 90.
    18 is always into 90, so just 48 and 90.

    To do this you must devide the numbers into their prime factors:
    48: 2 * 2 * 2 * 2 * 3
    90: 2 ................* 3 * 3 * 5
    To find the least common multiple you have to mulitply all numbers but those that are double only once:
    .....2 * 2 * 2 * 2 * 3 * 3 * 5 = 720

    The question is about the front wheel, so: 720/48=15
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    (Original post by berlinPPE)
    Jepp, I understand I forgot to add the long... stupid mistakes -.-
    Thanks!!!

    Another one.... can anybody explain the algebra for these age-questions?

    I can´t find them anymore :confused: But it is e.g. the one with 10 years ago, Jane was 2x Marys age, 5 years ago 2/3 (?), Daisy is 3 years older than Mary (or Jane?)...How old is Daisy today?

    I am sorry I was searching for these question, but I really couldn´t find it... I am at the airport now...have to leave now, but wanted to ask before I am going ^^

    Hope someone knows which one I mean:yes:

    Thanks again
    Jane, Mary and Daisy are sisters. Jane is twice as old as Mary currently. In five years time, Mary will be two thirds the age of Jane. Ten years from now, Mary will be three quarters Jane’s age. Daisy is 3 years older than Jane. How old is Daisy currently?

    x = Jane, y = Mary, z = Daisy

    ..I. x = 2y
    .II. 2/3 * (x+5) = y + 5
    III. 3/4 * (z+10) = y + 10
    IV. z = x + 3

    I. in II. 2/3 * (2y+5) = y + 5
    ....<=> 4/3 y + 10/3 = y + 5
    ....<=> 1/3 y = 5/3
    ....<=> y = 5

    in I. x = 10

    in IV. z = 10 + 3 = 13
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    Sorry I didn't see it!

    Thanks alotttttttttt! =D
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    Guys...I just realised I must really be the man for Oxford. While going over my results on the 2008 paper I just noticed that I didn't laugh at the word "tits" in question 17. Not even a little childish giggle! Now how mature is that!? :woo: :woo:

    EDIT: should I perhaps mention that in my interview if I am offered the chance?
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    (Original post by kaderk)
    Yeah that's the one - I didn't realise at the time that logs of fire would have counted as energy seeing as it was separated from electricty and gas
    At first glance I thought: "This question is too easy - I bet lower down there's something that should be included..."

    And I saw the logs for fire and contemplated it and thought, yes I should include these and then...wait is this a trick? Surely logs for fire aren't actual "energy" costs...

    So basically, I overthought it...
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    I don't think so, I didn't laugh as well. But that is perhaps because I'm Polish so words like "tits", "fu*k" or "bicycle" don't move me at all :P
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    (Original post by Confused-teen)
    Another question: From CIE Thinking Skills June 2005

    The cog sizes and chain length on a bicycle are designed to even out wear, so that the same hole in the chain does not encounter the same tooth on the cog in every revolution. The chain wheel (front cog) of a particular bicycle has 48 teeth, the rear cog has 18 teeth and the chain has 90 holes.
    How many revolutions must the front cog do before the chain and both cogs are back in exactly the same relative positions (i.e. the same holes are encountering the same teeth on both cogs)?
    A 5
    B 8
    C 15
    D 20
    E 30
    June 2005? I haven't seen that before!
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    (Original post by Will762)
    Re. question 11: Nope, they are not the same thing re-worded.

    'B' is essentially: "If something (P) doesn't happen then something (Q) will result.

    'E' however, explicitly states that if P does happen then Q will NOT happen.

    But this does not necessarily follow. There could be many other reasons why Q could happen, even if P does happen.

    (In this case -- the business not being saved -- it could be because the business was unpopular with its potential customers, it's too expensive, the Marxist government bans all private enterprise... and so on.)

    This is something pretty crucial for logic / the TSA, so I hope it helps.

    -- From one feverish Oxbridge candidate to another.
    Ah hah, I see. Thanks! It's also the hedging, 'could' versus 'will'.


    I know a lot of you have done the 2007 test, is there a near-unanimous verdict on the correct answers so that I can check to see how I did?
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    again, anyone have the TSA 2006 paper on their computer? Would appreciate it, if they can post it here/email me if they do .

    Also, 2 days away, hows everyone feeling?
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    Ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh hhhh! The Tsa Is Soon!!!!!!! Be Scared!!!!!!!!! Ahhhhhhhhhhhhhhhhhhhhhhhh!!!!!!! !!!!!!!!!!!!!!!
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    I'm on the verge of a mental breakdown, just can't wait until this is all over.

    ALSO- DOES ANYONE HAVE THE CIE 2005 PAPER ?
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    (Original post by j1991)
    again, anyone have the TSA 2006 paper on their computer? Would appreciate it, if they can post it here/email me if they do .

    Also, 2 days away, hows everyone feeling?
    i've already answered this, but the 2006 that everyones been looking at is from here http://www.cie.org.uk/qualifications...def_id=765_804
    and go on paper 1 it is identical to tsa (also it's done by same exam board and is called thinking skills)
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    Please could someone show me the working out to the following:

    Jason left Ranjiv’s house yesterday afternoon, walking home at 5 kilometres per hour. Ranjiv
    quickly discovered that Jason had left his mobile phone behind and, 6 minutes after his
    departure, set off after him on his bicycle at 25 kilometres per hour.
    How far had Jason walked when Ranjiv caught up with him?
    A 375 metres
    B 450 metres
    C 500 metres
    D 600 metres
    E 625 metres

    The answer is E. But i don't understand the method behind it.

    Thanks to anyone who helps in advance
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    (Original post by TopSlacker)
    Ah hah, I see. Thanks! It's also the hedging, 'could' versus 'will'.


    I know a lot of you have done the 2007 test, is there a near-unanimous verdict on the correct answers so that I can check to see how I did?
    i got these in common with 2 others. am pretty sure that they are right.
    in brackets i give you my personal answers for those, where we differ.
    Spoiler:
    Show

    1. B
    2. B
    3. A
    4. C
    5. C
    6. A
    7. B
    8. D
    9. (C)
    10. C
    11. C
    12. D
    13. A
    14. D
    15. B
    16. E
    17. 12/9/7/5
    18. C
    19. A
    20. B
    21. ...
    22. A
    23. B
    24. A, C
    25. (C)
    26. E
    27. E
    28. (B)
    29. B
    30. (B)
    31. (E)
    32. (D)
    33. C
    34. A
    35. A
    36. C
    37. B
    38. E
    39. A
    40. (C)
    41. B
    42. A
    43. (D)
    44. C
    45. C
    46. (?)
    47. D
    48. D
    49. D
    50. B
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    (Original post by amy123123)
    I'm on the verge of a mental breakdown, just can't wait until this is all over.

    ALSO- DOES ANYONE HAVE THE CIE 2005 PAPER ?

    Here you go!

    What college are you applying to?
    Attached Images
  1. File Type: pdf 8436_s05_qp_1.pdf (161.8 KB, 282 views)
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    (Original post by Confused-teen)
    Please could someone show me the working out to the following:

    Jason left Ranjiv’s house yesterday afternoon, walking home at 5 kilometres per hour. Ranjiv
    quickly discovered that Jason had left his mobile phone behind and, 6 minutes after his
    departure, set off after him on his bicycle at 25 kilometres per hour.
    How far had Jason walked when Ranjiv caught up with him?
    A 375 metres
    B 450 metres
    C 500 metres
    D 600 metres
    E 625 metres

    The answer is E. But i don't understand the method behind it.

    Thanks to anyone who helps in advance
    okay, this is the equation ive used-

    x/5 - x/25= 1/10. its 1/10 because, its 6minutes, so 6/60(one hour).
    then- times both sides by 25.
    5x-x=2.5
    4x=2.5
    x=2.5/4
    x=0.625km, so 625m..does that help?
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    (Original post by Confused-teen)
    Here you go!

    What college are you applying to?
    thank you, can i have the mark scheme too??

    Im assuming the only ones we've all got are the 2005,2006,07 and 08?
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    (Original post by Confused-teen)
    Please could someone show me the working out to the following:

    Jason left Ranjiv’s house yesterday afternoon, walking home at 5 kilometres per hour. Ranjiv
    quickly discovered that Jason had left his mobile phone behind and, 6 minutes after his
    departure, set off after him on his bicycle at 25 kilometres per hour.
    How far had Jason walked when Ranjiv caught up with him?
    A 375 metres
    B 450 metres
    C 500 metres
    D 600 metres
    E 625 metres

    The answer is E. But i don't understand the method behind it.

    Thanks to anyone who helps in advance
    v s/t <=> s = v * t = 5 km/h * 0,1h = 0,5 km

    --> Now we see that it can't be A, B or C.

    Now we can look how long Ranjiv needs for the 500m.

    t = s/v = 0,5km / 25 km/h = 0,02 h

    In that time Jason walks s = v * t = 5 km/h * 0,02 h = 0,1 km

    Now we can see that it can't be D either because they are still at different places when Jason reaches the 600m.

    So it must be E.

    Anyhow, there must be a easier way! Hope that helps you though. ;-)

    edit: amy, that looks good. can you explain it a little more?
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    Okay, you see s=d/t. we know Jason's doing 5km/h, and Raj' 25km/h, so its 5=d/t, we want x to be the distance, so its 5=x/t, if we know t it become t=x/5. (i really hope this makes sense, im hopeless at explaining!), so anyway, we do the same for Raj, let it equal the time we know- which is 6minutes, then solve it-
    1. x/5 - x/25= 1/10. its 1/10 because, its 6minutes, so 6/60(one hour).
    2. then- times both sides by 25.
    3. which gives us- 5x-x=2.5
    4x=2.5
    x=2.5/4
    x=0.625km, so 625m
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    (Original post by 09911041)
    v s/t <=> s = v * t = 5 km/h * 0,1h = 0,5 km

    --> Now we see that it can't be A, B or C.

    Now we can look how long Ranjiv needs for the 500m.

    t = s/v = 0,5km / 25 km/h = 0,02 h

    In that time Jason walks s = v * t = 5 km/h * 0,02 h = 0,1 km

    Now we can see that it can't be D either because they are still at different places when Jason reaches the 600m.

    So it must be E.

    Anyhow, there must be a easier way! Hope that helps you though. ;-)

    edit: amy, that looks good. can you explain it a little more?

    Doing TSA hurts your head :|
    I don't get a word of that
 
 
 
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