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Math Help - Two word problems due tomorrow! (LCM, ratios..)

  1. #1
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    Unhappy Two word problems due tomorrow! (LCM, ratios..)

    1) Two friends live out of town and come back to town periodically for business. The judge comes for one day every 15 days, while the accountant comes for one day every 66 days. They were together in town this year on January 11. On what day will they next be together in town?

    -Can't seem to find a solution to this one but I'm pretty sure it involves finding the lowest common multiple of 15 and 66 which is 165..but then how do you find the day that they will be together?

    2) Two turtles run a race. The first turtle runs at 2 centimetres per second while the second turtle runs at 80 metres per hour. Evaluate their speed in kilometres per day. Which is faster?

    -not sure how to solve this one, could it be solved by setting up ratio's? Not sure how to convert the speed measurements.

    Thanks!
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  2. #2
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    Quote Originally Posted by wickedcharchar View Post
    1) Two friends live out of town and come back to town periodically for business. The judge comes for one day every 15 days, while the accountant comes for one day every 66 days. They were together in town this year on January 11. On what day will they next be together in town?

    ...

    2) Two turtles run a race. The first turtle runs at 2 centimetres per second while the second turtle runs at 80 metres per hour. Evaluate their speed in kilometres per day. Which is faster?

    -not sure how to solve this one, could it be solved by setting up ratio's? Not sure how to convert the speed measurements.

    Thanks!
    To #1:

    All your considerations are OK. You only have to add 165 days to the date of 11-01-year:

    \begin{array}{cccccc}Jan&Feb&Mar&Apr&Mai&Jun \\ 20&28&31&30&31&25\end{array} . So the date when they will meet again is the 25-06-year.

    to #2:

    You are supposed to know that

    1 km = 1000 m = 100,000 cm
    1 h = 60 min = 3600 s
    v_{T_1} = 2 \frac{cm}{s} = 2 \frac{\frac1{100,000} km}{\frac1{3600} h} = 2 \frac{3600}{100,000}\frac{km}{h} = \frac9{125} \frac{km}{h} = 0.072 \frac{km}{h}

    Since 1 m = 0.001 km the speed of the 2nd turtle is:

    v_{T_2} = 0.080 \frac{km}{h}
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