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Math Help - Linear word problems

  1. #1
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    Linear word problems

    Please help me solve this problem by coming up with a linear algebra equation:

    A biathlon event involves running and cycling. Kim can cycle 30km/h faster than she can run If Kim spends 48 minutes and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run?

    Formula: distance= speed * time
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  2. #2
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    Quote Originally Posted by delicate_tears View Post
    Please help me solve this problem by coming up with a linear algebra equation:

    A biathlon event involves running and cycling. Kim can cycle 30km/h faster than she can run If Kim spends 48 minutes and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run?

    Formula: distance= speed * time
    dont know if this is right for sure, let me know x-48 1/3 =30
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  3. #3
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    Quote Originally Posted by Leona_Marie View Post
    dont know if this is right for sure, let me know x-48 1/3 =30
    what i meant was x-48 1/3 /2 = 30
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  4. #4
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    Quote Originally Posted by delicate_tears View Post
    Please help me solve this problem by coming up with a linear algebra equation:

    A biathlon event involves running and cycling. Kim can cycle 30km/h faster than she can run If Kim spends 48 minutes and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run?

    Formula: distance= speed * time
    Hi delicate_tears,

    Let's see if I'm interpreting this problem correctly.

    Let x = speed in km/h running

    Let x + 30 = speed in km/hr cycling

    48 minutes running = 48/60 = 4/5 hours

    one-third as much again cycling seems like 48 + 1/3(48) = 64 minutes = 64/60 = 16/15 hours.

    d = rate X time

    d = 60

    The distance traveled while running would be \frac{4}{5}x

    The distance traveled while cycling would be \frac{15}{16}(x+30)

    The two distances together would equal 60 km.

    The linear equation would then be:

    \frac{4}{5}x+\frac{16}{15}(x+30)=60
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  5. #5
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    Hey Leona_Marie, thanks for replying first to my thread! Unfortunately I don't quite understand the equation you wrote as it contained too many slashes that I got mixed up.

    Quote Originally Posted by masters View Post
    Hi delicate_tears,

    Let's see if I'm interpreting this problem correctly.

    Let x = speed in km/h running

    Let x + 30 = speed in km/hr cycling

    48 minutes running = 48/60 = 4/5 hours

    one-third as much again cycling seems like 48 + 1/3(48) = 64 minutes = 64/60 = 16/15 hours.

    d = rate X time

    d = 60

    The distance traveled while running would be \frac{4}{5}x

    The distance traveled while cycling would be \frac{15}{16}(x+30)

    The two distances together would equal 60 km.

    The linear equation would then be:

    \frac{4}{5}x+\frac{16}{15}(x+30)=60
    Hi masters, my interpretation of the problem is different to yours but your interpretation is actually the correct one.

    Where the word problem says 'as third as much time again cycling' I wrote is as 1/3(48)= 16 minutes= 16/60 hr = 4/15 hr.

    Thankyou so much
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