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- Question
- 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.
- Correct Answer
- Statement 1 is true and so is statement 2.
ExplanationAs 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 - 1. 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
- 2. 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 Discuss
- 3. 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
- 4. 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
- 5. What is the minimum distance between H to A through D?
Options- A. 1800
- B. 2000
- C. 2200
- D. None of these Discuss
- 6. 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
- 7. 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
- 8. 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
- 9. What is the difference between maximum amount and minimum amount that Mr. Ricky can spend on a day if he wants to travel from his home (X) to office (Y) if toll tax at each toll booth is Rs 50?
Options- A. 180
- B. 210
- C. 240
- D. None of these Discuss
- 10. If toll tax is devised in such a way that total cost of travel remains same irrespective of the path then what is the minimum cost of travel from X to Y?
Options- A. 240
- B. 260
- C. 310
- D. None of these Discuss
Routes and Networks problems
Search Results
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: 90
Explanation:
Map 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.
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: 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 .
Correct Answer: 180
Explanation:
On the basis of above given diagram , we can see that
Maximum expenditure is in path 3 and that is 260 + 50 x 6 = 260 + 300 = 560
Minimum expenditure is in path 5 and that is 230 + 50 x 3 = 230 + 150 = 380
? Required difference = Maximum expenditure is in path 3 - Minimum expenditure is in path 5 = 560 ? 380 = 180
Hence , Required difference is 180.
Correct Answer: 260
Explanation:
From the given condition ,
cost incurred in path 2 is equal to that in path 3 then T3 = 0
Similarly ,equating cost of path 3 and path 4 T7 = 0
Equating cost of path 4 and path 5 we will get
260 + T6 + T8 = 230 + T5
? T5 = T6 + T8 + 30
Equating cost of path 1 and path 2 we will get ,
240 +T1 = 260 + T2+ T4
or T1 = T2 + T4 + 20
For minimum total cost T1 = 20 and T5 = 30 then cost incurred is Rs 260.
Therefore , the minimum cost of travel from X to Y is Rs 260 .
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