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- Question
- Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is
Options- A. 60
- B. 75
- C. 45
- D. 90
- Correct Answer
- 90
ExplanationMap of the town is given as below:
It is given that Neelam took the shortest path hence she has to pass through path EF.
Number of ways to reach E from A is 2+2C2 = 4C2 = 6.
Number of ways to reach B from F is 4+2C2 = 6C2 = 15.
Hence total number of ways to reach B from A is 6 x 15 = 90. - 1. Find the maximum distance between A and H if condition is that traveling a city more than once is not allowed.
Options- A. 3800
- B. 3700
- C. 3200
- D. None of these Discuss
- 2. If path H E C is taken and no city is visited twice then find the total number of such paths.
Options- A. 3
- B. 4
- C. 5
- D. None of these Discuss
- 3. What is the minimum distance between H to A through D?
Options- A. 1800
- B. 2000
- C. 2200
- D. None of these Discuss
- 4. Find the minimum distance between H to A without visiting city E.
Options- A. 1200
- B. 1300
- C. 1400
- D. None of these Discuss
- 5. 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.
Options- A. 44
- B. 55
- C. 59
- D. None of these Discuss
- 6. Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is
Options- A. 1170
- B. 630
- C. 792
- D. 1200 Discuss
- 7. Which of the following is true?
Options- A. Statement 1 is true but statement 2 is false.
- B. Statement 1 is false but statement 2 is true.
- C. Statement 1 is true and so is statement 2.
- D. Statements 1 and 2 are false. Discuss
- 8. Which nodes should be selected for installation of surveillance cameras?
Options- A. 3, 6 and 7
- B. 3, 1 and 4
- C. 2, 5 ,1 and 4
- D. None of these Discuss
- 9. If a new path is constructed from T7 to Y then find the total number of paths from X to Y?
Options- A. 7
- B. 6
- C. 8
- D. None of these Discuss
- 10. Which one of the following could be the value of toll taxes at T1, T2, T3?............... T8 such that total cost of travel is same irrespective of route?
Options- A. 50, 10, 0, 10, 60, 20, 0, 10
- B. 50, 20, 0, 10, 60, 20, 0, 10
- C. 50, 20, 0, 10, 40, 20, 0, 10
- D. None of these Discuss
Routes and Networks problems
Search Results
Correct Answer: 3700
Explanation:
From the solution of question number 25 ,
Maximum distance is when path taken is:
A ? D ? C ? B ? G ? E ? F ? H = 1300 + 500 + 200 + 400 + 400 + 300 + 600 = 3700 km.
Therefore , the maximum distance between A and H is 3700 km.
Correct Answer: 4
Explanation:
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 .
Correct Answer: None of these
Explanation:
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.
Correct Answer: 1400
Explanation:
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 .
Correct Answer: 55
Explanation:
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 .
Correct Answer: 1170
Explanation:
From the answer of previous question,
Total number of ways from A to B is 90.
Now we need to calculate total number of ways from B to C that is 13 (on the same logic)
Hence, total number of paths from A to C is 90 x 13 = 1170.
Correct Answer: Statement 1 is true and so is statement 2.
Explanation:
As per the given conditions,
we require surveillance cameras that would cover all roads converging at an intersection. The cost for covering all the roads has to be minimum.
If we put a camera on '1', roads from '7' and '6' to 1 will be covered.
If we put a camera on intersection '2', roads converging from '3' to '2' will be covered.
If we put a camera on '3', roads from '3' to '5', '6' to '5' and '7' to '5' will be covered.
Now we are left with roads from 7 to 4 and from 6 to 4 both of them converge from 4 hence these will be covered by putting a camera on '4'. And then all the roads will be covered and the minimum cost is 2 + 5 + 1 + 4 = 12 lakhs
Correct Answer: 2, 5 ,1 and 4
Explanation:
On the basis of above given question ,we can say that
Similar to the above answer cameras are put on 2, 5, 1 and 4 to minimize the cost.
Correct Answer: 7
Explanation:
As per the given above diagram , we can see that
Total number of paths from X to Y is 7.
Correct Answer: 50, 20, 0, 10, 60, 20, 0, 10
Explanation:
From the solution of previous question ,
we have following results:
T3 = T7 = 0,
T1 = 20 + T2 + T4, T5 = 30 + T6 + T8.
From the given options option B satisfies the condition.
Hence , required answer will be 50, 20, 0, 10, 60, 20, 0, 10 .
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