Routes and Networks
Direction: The figure (not drawn on scale) given below shows the aerial distance (in km) between 8 cities, e.g. distance between H and F is 600 km etc.
- Find the ratio of maximum to minimum distance between A and H if condition is that traveling a city more than once is not allowed.
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From the above given diagram , we can see that
Maximum distance is when path taken is-
A – D – C – B – G – E – F – H = 1300 + 500 + 200 + 400 + 400 + 300 + 600 = 3700 kmCorrect Option: A
From the above given diagram , we can see that
Maximum distance is when path taken is-
A – D – C – B – G – E – F – H = 1300 + 500 + 200 + 400 + 400 + 300 + 600 = 3700 km
Minimum distance is when path taken is-
A – C – G – H = 600 + 400 + 400 = 1400 km
Hence , Required ratio is 37 : 14.
- Find the total number of paths from city H to city A if condition is that traveling a city more than once is not allowed.
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Number of paths is as follows:
Starting from HGB: HGBA, HGBCA, HGBCDA, HGBCEDA, HGBCEFDA (Total 5 paths)
Starting from HGC: HGCA, HGCBA, HGCDA, HGCEDA, HGCEFDA (Total 5 paths)
Starting from HGE: HGECA, HGECBA, HGECDA, HGEDCA, HGEDCBA, HGEDA, HGEFDCBA, HGEFDCA, HGEFDA (Total 9 paths)
Starting from HEG: HEGBA, HEGCBA, HEGCA, HEGCDA (Total 4 paths)
Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths) Starting from HED: HEDCBGA, HEDCBA, HEDCA, HEDA (Total 4 paths)
Starting from HEF: HEFDCGBA, HEFDCBA, HEFDCBA, HEFDCA, HEFDA (Total 5 paths)
Starting from HFE: HFEGBA, HFECGBA, HFECBA, HFECA, HFEDCGBA, HFEDCBA, HFEDCA, HFEDA, (Total 8 paths)
Starting from HFD: HFDEGBA, HFDEGCA, HFDEGBCA, HFDECGBA, HFDECBA, HFDECA, HFDCEBGA, HFDCGBA, HFDCBA, HFDCA, HFDA (Total 11 paths)Correct Option: B
Number of paths is as follows:
Starting from HGB: HGBA, HGBCA, HGBCDA, HGBCEDA, HGBCEFDA (Total 5 paths)
Starting from HGC: HGCA, HGCBA, HGCDA, HGCEDA, HGCEFDA (Total 5 paths)
Starting from HGE: HGECA, HGECBA, HGECDA, HGEDCA, HGEDCBA, HGEDA, HGEFDCBA, HGEFDCA, HGEFDA (Total 9 paths)
Starting from HEG: HEGBA, HEGCBA, HEGCA, HEGCDA (Total 4 paths)
Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths) Starting from HED: HEDCBGA, HEDCBA, HEDCA, HEDA (Total 4 paths)
Starting from HEF: HEFDCGBA, HEFDCBA, HEFDCBA, HEFDCA, HEFDA (Total 5 paths)
Starting from HFE: HFEGBA, HFECGBA, HFECBA, HFECA, HFEDCGBA, HFEDCBA, HFEDCA, HFEDA, (Total 8 paths)
Starting from HFD: HFDEGBA, HFDEGCA, HFDEGBCA, HFDECGBA, HFDECBA, HFDECA, HFDCEBGA, HFDCGBA, HFDCBA, HFDCA, HFDA (Total 11 paths)
Total number of paths is: 5 + 5 + 9 + 4 + 4 + 4 + 5 + 8 + 11 = 55
Hence , the total number of paths from city H to city A is 55 .
- Find the minimum distance between H to A without visiting city E.
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As per the given above diagram , we can see that
For minimum distance path is HGCA = 400 + 400 + 600 = 1400Correct Option: C
As per the given above diagram , we can see that
For minimum distance path is HGCA = 400 + 400 + 600 = 1400
Therefore , the minimum distance between H to A without visiting city E is 1400 km .
- What is the minimum distance between H to A through D?
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On the basis of above given diagram , we can say that
The minimum distance between A and H through D is HEDCA and distance is 500 + 500 + 500 + 600 = 2100.Correct Option: D
On the basis of above given diagram , we can say that
The minimum distance between A and H through D is HEDCA and distance is 500 + 500 + 500 + 600 = 2100.
Hence , required answer will be 2100 km.
- If path H E C is taken and no city is visited twice then find the total number of such paths.
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From solution of question number 26 ,
Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths)Correct Option: B
From solution of question number 26 ,
Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths)
Hence , the total number of such paths is 4 .
