Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. 01 May '05 05:52
    A bus route from town A to town D goes along a straight line and passes first through town B and then through town C. Jason has a positioning device that can tell him the distance from his current location to any of the four towns on the route. When he is somewhere between A and B, from the information supplied by the device Jason learns that the sum of the distances to A, to B and to D is greater than the distance to C by 5.9 km. When Jason is between B and C, he finds out that the sum of the distances to all four towns is 16.7 km. When he is between C and D, Jason discovers that the sum of the distances to B and to D is 8.1 km. What is the distance between C and D?

    Show all working for your answer to be deemed correct.
  2. 01 May '05 13:52
    well since the sum to d is greater than that of C by 5.9 km, D is 5.9 km away from C. B to d is 8.1, -5.9 (between C-d), so B to C is 2.2 km long


    wait..... 5.9 km! all the extra between B andC is extra, isn't it?
  3. 01 May '05 13:53
    meh, that's totally wrong...
  4. Standard member Daemon Sin
    I'm A Mighty Pirateā„¢
    01 May '05 14:28
    Errrr... I was never any good at maths but I got that:

    A to B = 8.6
    B to C = 2.2
    C to D = 5.9
  5. 02 May '05 05:53
    Originally posted by Daemon Sin
    Errrr... I was never any good at maths but I got that:

    A to B = 8.6
    B to C = 2.2
    C to D = 5.9
    Can anyone provide a 'algebra' solution?
  6. 02 May '05 08:14 / 1 edit
    Originally posted by phgao
    Can anyone provide a 'algebra' solution?
    X is the point between A and B; Y between B and C; Z between C and D

    at X: AX + XB + XD = XC + 5.9
    or (AX +XB) + (XB + BC + CD) = (XB + BC) + 5.9
    or AB + CD = 5.9

    at Y: AY + BY + YC + YD = 16.7
    or (AB + BY) + BC + (YC + CD) = 16.7
    or AB + 2xBC + CD= 16.7
    or with the first point: 2xBC = 16.7 - 5.9
    or BC = 5.4

    at Z: BZ + ZD = 8.1
    or (BC + CZ) + ZD = 8.1
    or BC + CD = 8.1
    with the above CD= 8.1 - 5.7 = 2.7

    this would make AB= 5.9 - 2.7 = 3.2
  7. Standard member Trains44
    Full speed locomotiv
    02 May '05 21:11
    Originally posted by phgao
    Can anyone provide a 'algebra' solution?
    Yes...E= mc2 Sincerely, Trains44
  8. 03 May '05 12:23
    Originally posted by TRAINS44
    Yes...E= mc2 Sincerely, Trains44
    lol!! I'll get back to you Meph on your answer, really busy now.
  9. 04 May '05 10:30 / 2 edits
    Originally posted by Mephisto2
    X is the point between A and B; Y between B and C; Z between C and D

    at X: AX + XB + XD = XC + 5.9
    or (AX +XB) + (XB + BC + CD) = (XB + BC) + 5.9
    or AB + CD = 5.9

    at Y: AY + BY + YC + YD = 16.7
    or (AB + BY) + BC + (YC + CD) = 16 ...[text shortened]... above CD= 8.1 - 5.7 = 2.7

    this would make AB= 5.9 - 2.7 = 3.2
    A--------X-----B----Y---------C----------Z----D

    When he is somewhere between A and B, from the information supplied by the device Jason learns that the sum of the distances to A, to B and to D is greater than the distance to C by 5.9 km.

    at X: AX + XB + XD = XC + 5.9
    or (AX +XB) + (XB + BC + CD) = (XB + BC) + 5.9
    or AB + CD = 5.9

    Looks right so far. I think you've got the sense of the problem:
    that the distances from X need to be rewritten as distances between
    the points.

    When Jason is between B and C, he finds out that the sum of the distances to all four towns is 16.7 km.

    at Y: AY + BY + YC + YD = 16.7
    AY = AB+BY and BY+YC = BC and YD=YC+CD, and therefore as you said:
    or (AB + BY) + BC + (YC + CD) = 16.7
    Intermediate step that I took a while to see:
    AB + BY + BC + (BC-BY) + CD = 16.7
    or AB + 2xBC + CD= 16.7
    or with the first point: 2xBC = 16.7 - 5.9
    or BC = 5.4

    I agree.

    When he is between C and D, Jason discovers that the sum of the distances to B and to D is 8.1 km.

    at Z: BZ + ZD = 8.1

    True -- or BD = 8.1; CD = BD-BC = 8.1-5.4 = 2.7 km

    or (BC + CZ) + ZD = 8.1
    or BC + CD = 8.1
    with the above CD= 8.1 - 5.4 = 2.7

    What is the distance between C and D?

    this would make AB= 5.9 - 2.7 = 3.2

    It's pretty much your answer, but can you run through that and check?
  10. 04 May '05 11:04
    Originally posted by phgao
    A--------X-----B----Y---------C----------Z----D

    When he is somewhere between A and B, from the information supplied by the device Jason learns that the sum of the distances to A, to B and to D is greater than the distance to C by 5.9 km.

    at X: AX + XB + XD = XC + 5.9
    or (AX +XB) + (XB + BC + CD) = (XB + BC) + 5.9
    or AB + CD = 5.9

    Looks right so f ...[text shortened]... e AB= 5.9 - 2.7 = 3.2

    It's pretty much your answer, but can you run through that and check?
    it looks ok, except that you also took over my typo:
    "with the above CD= 8.1 - 5.7 = 2.7 " should read
    "with the above CD= 8.1 - 5.4 = 2.7"
  11. 04 May '05 11:10 / 1 edit
    Originally posted by Mephisto2
    it looks ok, except that you also took over my typo:
    "with the above CD= 8.1 - 5.7 = 2.7 " should read
    "with the above CD= 8.1 - 5.4 = 2.7"
    oh yeah, but the ans is still good ....clusmy mistake on my part...all fixed now