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Math Help - Don't know how to solve math story problems

  1. #1
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    Don't know how to solve math story problems

    I don't understand how to solve these problems (this is a practice test btw)

    Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 hours. How quickly can all three fill the pool together?


    Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

    If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks?
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Smile ....

    Quote Originally Posted by AmandaRaymond View Post
    I don't understand how to solve these problems (this is a practice test btw)

    Q1Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 hours. How quickly can all three fill the pool together?


    Q2Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?

    Q3If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes, and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them working together to mix the 20 drinks?
    Q1
    Jim can fill 1/2th of bucket in 15 minutes
    Sue can fill 1/3rd in 15 minutes
    Tony can fill 1/6 th in 15 minutes
    together they fill
    1/6+1/3+1/2 = 6/6 = 1 bucket in 15 minutes

    Q2

    Let the time be t (when the two meet)
    so this means the their displacement will be equal
    hence
    6(t+3) = 10t
    thus
    4t = 18
    t=9/2 hours after the first one leaves
    so your answer is (9/2)-3 = (3/2)hours.......................(I have corrected this ,you have to deduct 3 hours as mentioned in question )

    Q3
    Let the time taken =t minutes
    speed at which Steven makes drinks=4 drinks per minute
    speed at which Sue makes drinks= 2 drinks per minute
    speed at which Jack makes drinks=4/3 drinks per minute
    so
    4xt + 2xt + (4/3)xt =20 drinks
    (22/3)t=20
    hence t= 60/22 minutes
    Last edited by ADARSH; November 8th 2008 at 09:31 PM.
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  3. #3
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    thank you for your help!


    but for the last one I don't know how you got 20 drinks can you explain that?

    and for the second one when I did 9/2+3 I got 27/2

    and then i flipped the the fraction to 1/3 and got 3/2
    Last edited by AmandaRaymond; November 7th 2008 at 06:37 AM.
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  4. #4
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    Hello, Amanda!

    I have a different approach . . . Take your pick.


    1) Jim can fill a pool carrying buckets of water in 30 minutes.
    Sue can do the same job in 45 minutes.
    Tony can do the same job in 1 hours.
    How quickly can all three fill the pool together?

    Jim goes the job in 30 minutes.
    In one minute, he does \tfrac{1}{30} of the job.
    In x minutes, he does \tfrac{x}{30} of the job.

    Sue does the job in 45 minutes.
    In one minute, she does \tfrac{1}{45} of the job.
    In x minutes, she does \tfrac{x}{45} of the job.

    Tony does the job in 90 minutes.
    In one minute, he does \tfrac{1}{90} of the job.
    In x minutes, he does \tfrac{x}{90} of the job.

    Working together for x minutes, they do: . \frac{x}{30} + \frac{x}{45} + \frac{x}{90} of the job.

    But in x minutes, we expect them to complete the job (1 job).

    There is our equation . . . . .  \frac{x}{30} + \frac{x}{45} + \frac{x}{90} \;=\;1


    Multiply by 90: . 3x+2x+x \:=\:90 \quad\Rightarrow\quad x \:=\:15 minutes.




    2) Two cyclists, A and B, start biking from a trail's start 3 hours apart.
    B travels at 10 mph and starts 3 hours after A who is traveling at 6 mph.
    How much time will pass before B overtakes A from the time B started biking?

    A travels at 6 mph and has a 3-hour headstart.
    . . He is already 18 miles ahead.

    B travels at 10 mph and must make up the 18 miles.

    Relative to A,\:B travels at 10 - 6 \:=\:4 mph.

    . . . It is as if A is has stopped, and B is approaching him at 4 mph.

    To cover 18 miles, it will take: . \frac{18}{4} \:=\:4\tfrac{1}{2} hours.




    3) If Steven can mix 20 drinks in 5 minutes, Sue can mix 20 drinks in 10 minutes,
    and Jack can mix 20 drinks in 15 minutes, how much time will it take all 3 of them
    working together to mix the 20 drinks?
    Same set-up as #1 . . .

    Steven does the job in 5 minutes.
    In one minute, he can do \tfrac{1}{5} of the job.
    In x minutes, he can do \tfrac{x}{5} of the job.

    Sue does the job in 10 minutes.
    In one minute, she can do \tfrac{1}{10} of the job.
    In x minutes, she can do \tfrac{x}{10} of the job.

    Jack does the job in 15 minutes.
    In one minute, he can do \tfrac{1}{15} of the job.
    In x minutes, he can do \tfrac{x}{15} of the job.

    Working together for x minutes, they do: . \frac{x}{5} + \frac{x}{10} + \frac{x}{15} of the job.

    But in x minutes, we expect them to complete the job (1 job).


    There is our equation . . . . . \frac{x}{5} + \frac{x}{10} + \frac{x}{15} \;=\;1

    Mulitply by 30: . 6x + 3x + 2x \:=\:30 \quad\Rightarrow\quad x \:=\:\frac{30}{11}

    Together, it will take them: . 2\tfrac{8}{11} minutes.

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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    Hi ,Amanda
    Quote Originally Posted by AmandaRaymond View Post
    but for the last one I don't know how you got 20 drinks can you explain that?
    I am sorry for my short steps
    I think this might help
    **
    Steven mixes 20 drinks in 5 minutes
    we can say he mixes  \frac{20}{5} = 4 drinks in 1 minute
    that's what I have called the speed at which he mixes drinks
    **
    Sue mixes 20 drinks in 10 minutes
    so he mixes \frac {20}{10} = 2 drinks in 1 minute
    **
    Jack mixes 20 drinks in 15 minutes
    so he mixes \frac{20}{15} = \frac{4}{3} drinks per minute
    **
    Lets consider the time taken by all three of them be ' t ' minutes
    (Speed of mixing drinks) X (time taken ) = number of drinks made
    all of them take the same time which is t

    And the number of drinks to be made is 20
    hence

     <br />
4*t + 2*t +\frac{4}{3}*t = 20 <br />
     =\frac{12+6+4}{3} * t =20
     hence t = \frac{60}{22} minutes = Answer
    -----------------
    I have corrected the second one in my earlier post
    if still there is trouble feel free to ask
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