# Bus Route Question

phgao
Posers and Puzzles 01 May '05 05:52
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?

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. Daemon Sin
I'm A Mighty Pirate™
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 08:141 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. 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 10:302 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 11:101 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