Results 1 to 5 of 5

Math Help - Bicyclist

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    152

    Bicyclist

    A bicylcist cycled from town A to town B and back. He returned into town A in 4 hours.
    His speed on the flat part of the road was 16km/h. His speed on the slope (going up) was 12km/h, and his speed on the slope (going down) was 24km/h.

    What is the distance between town A and B? (there is no info missing - thats all that it is).

    ----------------
    I think I manged to solve it with some logical thinking but I am still having problems doing it properly (with equations and all). I think the answer is 32km. (I won't explain how I got that because I don't want it to affect on your thinking..)
    Last edited by metlx; March 11th 2009 at 08:06 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by metlx View Post
    A bicylcist cycled from town A to town B and back. He returned into town A in 4 hours.
    His speed on the flat part of the road was 16km/h. His speed on the slope (going up) was 12km/h, and his speed on the slope (going down) was 24km/h.

    What is the distance between town A and B? (there is no info missing - thats all that it is).

    ----------------
    I think I manged to solve it with some logical thinking but I am still having problems doing it properly (with equations and all). I think the answer is 32km. (I won't explain how I got that because I don't want it to affect on your thinking..)
    I've got the same result and since we are the majority you are right!

    What I've done:

    1. Calculate the average speed of the trip from A to B:

    v_1 = \dfrac{2\cdot 16 \cdot 12}{16+12} = \dfrac{96}7\ \frac{km}{h}

    2. Calculate the average speed of the trip from B to A:

    v_2 = \dfrac{2\cdot 16 \cdot 24}{16+24} = 19.2\ \frac{km}{h}

    3. Calculate the average speed of the complete trip:

    v= \dfrac{2\cdot \frac{96}7 \cdot 19.2}{\frac{96}7+19.2} = 16\ \frac{km}{h}

    4. The total length of the trip is therefore 4\ h \cdot 16\ \frac{km}h = 64 \ km. And that means the distance between A and B must be 32 km
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    152
    Quote Originally Posted by earboth View Post
    1. Calculate the average speed of the trip from A to B:

    v_1 = \dfrac{2\cdot 16 \cdot 12}{16+12} = \dfrac{96}7\ \frac{km}{h}

    2. Calculate the average speed of the trip from B to A:

    v_2 = \dfrac{2\cdot 16 \cdot 24}{16+24} = 19.2\ \frac{km}{h}
    hmm. why 2 * 16 * 12 and 2 * 16* 24? why the 2 at the beginning?

    btw, i got that he is cycling 2.5h (flat part + uphill) and 1.5 hour (downhill + flat part).
    since the uphill speed : downhill speed = 1:2 -> that means the time he spends going up is twice as big as the time he spends going down.

    so the hill part = 3x
    2 hours up = 2* 12 = 24km
    1 hour down = 1 *24 = 24km
    then i guessed that 3x = 3 hours and that the remaining 1 hour is the time he spent on the flat part of the road (1 hour, both ways -> flat part = 8km in length)
    flat part both ways = 2*8 km = 16km

    24+24+16 = 64
    AB = 32


    but i did a logical jump (or whatever) there :P
    the teacher wanted to know where i got that 3x = 3 hours.. i didnt have an explanation and i still dont have it..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,739
    Thanks
    645
    Herllo, metlx!

    A bicylcist cycled from town A to town B and back in 4 hours.
    His speed on the flat part of the road was 16 km/h.
    His speed going uphill was 12 km/h, and his speed going downhill was 24 km/h.

    What is the distance between towns A and B?
    Code:
                          *
                     x  *    *   y
                      *          *
                    *               *
        A * - * - * - - - - - - - - - - * - * - * - * B
          : - w - :                     : - - z - - :
    Let w,x,y,z be the distances for the four phases of the ride.


    \text{From }A\text{ to }B.

    . . \begin{array}{cc}\text{He rode }w\text{ km at 16 kph.} & \text{This took: }\frac{w}{16}\text{ hours.} \\<br />
\text{He rode }x\text{ km at 12 kph.} & \text{This took: }\frac{x}{12}\text{ hours.} \\<br />
\text{He rode: }y\text{ km at 24 kph.} &\text{This took: }\frac{y}{24}\text{ hours.} \\<br />
\text{He rode: }z\text{ km at 16 kph.} & \text{This took: }\frac{z}{16}\text{ hours.}\end{array}

    His ride from A to B took: . \frac{w}{16} + \frac{x}{12} + \frac{y}{24} + \frac{z}{16}\:=\:\frac{3w+4x+2y+3z}{48} hours.


    \text{From }B\text{ to }A.

    . . \begin{array}{cc}\text{He rode }z\text{ km at 16 kph.} & \text{This took: }\frac{z}{16}\text{ hours.} \\<br />
\text{He rode }y\text{ km at 12 kph.} & \text{This took: }\frac{y}{12}\text{ hours.} \\<br />
\text{He rode }x\text{ km at 24 kph.} & \text{This took: }\frac{x}{24}\text{ hours.} \\<br />
\text{He rode }w\text{ km at 16 kph.} & \text{This took: }\frac{w}{16}\text{ hours.} \end{array}

    His ride from B to A took: . \frac{z}{16} + \frac{y}{12} + \frac{x}{24} + \frac{w}{16} \:=\:\frac{3w + 2x + 4y + 3z}{48} hours.


    The round trip took 4 hours.

    . . \frac{3w + 4x + 2y + 3z}{48} + \frac{3w + 2x + 4y + 3z}{48} \:=\:4 \quad\Rightarrow\quad 6w + 6x + 6y + 6z \:=\:192


    Therefore: . w + x + y + z \:=\:32\text{ km}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    152
    wow..just wow!

    why are there 4 phases?
    aren't there just 3 all together?
    2 in one way: w (flat part) and x (uphill)
    2 in the opposite direction: y (downhill) and w (flat part)
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum