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Math Help - simulataneous equation problem

  1. #1
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    simulataneous equation problem

    hello,

    could someone please help the following

    A climbing centre has 14 fixed staff plus one additional staff member for every six climbers. There are twice as many climbers as staff. Find the number of

    a climbers b staff

    thanks
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Lets say there were 6m climbers

    so number of staff members = 14+m


    But its given that

    2(14+m) = 6m

    28 + 2m = 6m

    4m =28

    Hence m = 7.
    Thus
    Spoiler:

    a = 6m = 42

    b = 14+7 = 21
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  3. #3
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    thanks adarsh

    if we let a be climbers and b staff then

    a=2b
    b=14 + \frac{a}{6}

    ... i know your answer is correct (because the book says ), but im confused as to why statements like 'twice as many climbers than staff' gives

    a=2b

    and not

    b=2a
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  4. #4
    No one in Particular VonNemo19's Avatar
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    if a = climbers and b = staff and there are "twice as many staff as climbers" (keywords of the problem), then if we multiply the number of climbers by 2, then that's how many staff we have. Right? Hence, 2 times the number of climbers = the number of staff. Or better yet a=2b.
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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    Consider these things

    a is twice of b
    .
    .

    Take an example
    6 is twice of 3

    6 = 2 * 3

    similarly

    a = 2 * b

    --this might help
    try to think taking examples
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