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?
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1 edit
Originally posted by phgaoX is the point between A and B; Y between B and C; Z between C and D
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
2 edits
Originally posted by Mephisto2A--------X-----B----Y---------C----------Z----D
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?
Originally posted by phgaoit looks ok, except that you also took over my typo:
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"