Stuck on this S1 question

Discussion in 'UMAT Question Discussion' started by tod, Jul 20, 2017.

  1. tod

    tod New Member

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    In Jane's house one room has 7 lights in a row. Each light can either be switched ON or OFF. Jane makes the following two statements:
    1. More than 2 Lights are OFF
    2. Any OFF Light must not have exactly one adjacent ON light.

    Given that 1 is False and 2 is TRUE, how many different unique arrangements are possible?

    1. 5
    2. 9
    3. 12
    4. 16
     
  2. tod

    tod New Member

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    My answer was 5
     
  3. Sherlock

    Sherlock Breaker of order Happy Valentine's Day

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    Perhaps a good place to start would be to provide your method and how you got the answer 5.
     
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  4. A1

    A1 Admissions Speculator Moderator

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    Statement 1 false means we can have 0 or 1 or 2 lights off = 7 or 6 or 5 lights on. I worked out 9 patterns (o on x off)

    7 on = ooooooo
    6 on combos noting no 1s allowed
    0-6 = xoooooo
    2-4 = ooxoooo
    3-3 = oooxooo
    5 on combos
    0-0-5 = xxooooo
    0-2-3 = xooxooo
    0-3-2 = xoooxoo
    0-5-0 = xooooox
    2-0-3 = ooxxooo

    This assumes mirror images (eg. 4-2 vs 2-4, 3-0-2 vs 2-0-3) don't count. If included they add up to 15 but is not in the answers.
     
  5. sammy04

    sammy04 New Member

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    its 16
    because the combinations go upto more than 12 lol.

    but i wouldnt waste ime on that question if its just 1 pointer, :)
     
  6. LMG!

    LMG! Moderator (UTAS MBBS) Most Helpful Member and Staff Member of the Year 2017-2018

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    I don't necessarily agree that statement one being FALSE means there are 0, 1, or 2 lights on. I'd interpret it as meaning more than two lights are on... how do you come to your conclusion, @A1? What am I interpreting incorrectly?
     
  7. LMG!

    LMG! Moderator (UTAS MBBS) Most Helpful Member and Staff Member of the Year 2017-2018

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    All questions are worth more than one point. Just to clarify, do you mean if there weren't multiple questions based on the same stimulus?
     
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  8. sammy04

    sammy04 New Member

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    yeah, !!! I mean if its just one question based on a big excerpt....
     
  9. A1

    A1 Admissions Speculator Moderator

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    More than 2 lights on can make but does not guarantee the statement false. Like 4 on + 3 off, the latter makes it true.
     
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  10. tod

    tod New Member

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    So this is the way I saw it.

    From Rule 1, we say that 0,1 or 2 lights are off.
    'Any OFF Light must not have exactly one adjacent ON light.'
    If we take this as xxxxxxx, we can already exclude the off lights from being on either end of the sequence as this would mean there is only 1 adjacent ON light if the on light was 2nd in the sequence.

    So you would be left with five possible places for the off lights.
    xxxxxxx
    x0xxxxx
    x0x0xxx
    x0xx0xx
    x0xxx0x
    xx0x0xx
    xx0xx0x
    xxx0x0x
    xx0xxxx
    xxx0xxx
    xxxx0xx
    xxxxx0x
    ....actually i just realised its a hec more than five :)

    Guess the answer was 12 C
     
    Last edited: Jul 21, 2017
  11. A1

    A1 Admissions Speculator Moderator

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    Oops my interpretation above of the rule 2 was incorrect. Indeed it means an Off light must be between two On lights, and 12 is the answer.
     
  12. Kat92

    Kat92 (Student BDemCare|BBeSt|BNurs) Hopeful for JMP

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    I am thinking it could be 16 as I end up with 15 doing it both ways.
     
  13. Mana

    Mana Registrar Administrar

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    Here's my attempt at it.

    So if you have two statements:

    1. More than 2 Lights are OFF - False (essentially, you have the statement that there are 0, 1, or 2 lights off)
    2. Any OFF Light must not have exactly one adjacent ON light

    for 0 lights off, there is 1 combination (all on)
    for 1 light off, there are exactly five combinations (the single off light being anywhere except either end)
    for 2 lights off, as per Rule 2, the off lights cannot be on either end and the off lights cannot be adjacent to each other. Basically this means in the middle 5 lights, you have (where 1 is an OFF light)
    lights 4 apart x 1 10001
    lights 3 apart x 2 10010 and 01001
    lights 2 apart x 3 10100 01010 00101

    This gives the requisite 12.
     
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