- 01 May '05 05:52A 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. - 02 May '05 08:14 / 1 edit

X is the point between A and B; Y between B and C; Z between C and D*Originally posted by phgao***Can anyone provide a 'algebra' solution?**

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 ~~Trains44~~Full speed locomotiv- 04 May '05 10:30 / 2 edits

A--------X-----B----Y---------C----------Z----D*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

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? - 04 May '05 11:04

it looks ok, except that you also took over my typo:*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?

"with the above CD= 8.1 - 5.7 = 2.7 " should read

"with the above CD= 8.1 - 5.4 = 2.7"