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- Question
- How many different starting cities are possible such that the above restriction is satisfied?
Options- A. One
- B. Zero
- C. Three
- D. Two
- Correct Answer
- Two
ExplanationOn the basis of above given figure , we can see that
City A is connected by 2 roads, B by 4 roads, C by 3 roads, D by 3 roads and E by 2 roads.
For a city to be starting city for such a route, it has to be connected by odd number of roads.
Hence the required answer is 2, i.e. C and D. - 1. For a route that satisfies the above restrictions, which of the following statements is true?
Options- A. There is no route that satisfies the above restriction.
- B. A route can either start at C or end at C, but not both.
- C. D can be only an intermediate city in the route.
- D. The route has to necessarily end at E. Discuss
- 2. If toll tax at all of the toll booths is same and is devised such that the ratio of maximum cost to minimum cost of travel from X to Y is 28 : 19 then which one of the following is the cost of travel from X to Y through toll Both T3?
Options- A. 320 or 510
- B. 330 or 560
- C. 560 or 510
- D. None of these Discuss
- 3. 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
- 4. 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
- 5. 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
- 6. What is the distance between A and B?
Options- A. 5 km
- B. 6 km
- C. 7 km
- D. 8 km Discuss
- 7. What is the length of route - 2 ?
Options- A. 11 km
- B. 12 km
- C. 13 km
- D. Insufficient information Discuss
- 8. What is the number of routes from P to Q?
Options- A. 5
- B. 6
- C. 9
- D. 12 Discuss
- 9. If 1st October is Sunday, then 1st November will be
Options- A. Monday
- B. Wednesday
- C. Thursday
- D. Tuesday Discuss
- 10. How many 3 are there in the following list in which followed by 9 and preceded 2,3 or 4?
1 1 2 3 9 4 4 5 5 5 5 3 3 9 8 7 7 7 8 8 8 5 5 4 3 9 6 6 6 6
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
Routes and Networks problems
Search Results
Correct Answer: A route can either start at C or end at C, but not both.
Explanation:
According to question ,
Here ,we will eliminate the options one by one. Option (A) cannot be true as there are many routes that satisfy the given condition. Option (C) is also not true as we can have a route starting from D (e.g. DEBDCBAC).
The route need not necessarily end at E, which is apparent. Option B satisfies all the conditions.
Hence , A route can either start at C or end at C, but not both.
Correct Answer: 560 or 510
Explanation:
If toll tax at all of the toll booths is same and let it be k then expense in each path is as follows:
Path 1: 240 + 3k
Path 2: 260 + 5k
Path 3: 260 + 6k
Path 4: 260 + 5k
Path 5: 230 + 3k
So from the given condition On solving we will get k = 50
Tool booth 3 will come in Path 3 and Path 4.
So cost incurred is:
Path 3: 260 + 6k = 260 + 300 = 560
Path 4: 260 + 5k = 260 + 250 = 510
Therefore , 560 or 510 is the cost of travel from X to Y through toll Both T3 .
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 .
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: 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: 8 km
Explanation:
As per the above result ,
Route -1 :- 5 + AB = 13 ? AB = 13 - 5 = 8 km
Hence , required distance between A and B is 8 km .
Correct Answer: 11 km
Explanation:
From the above result length of route -2 = 11 km
Therefore , the length of route -2 is 11 km .
Correct Answer: 6
Explanation:
As per the given above figure , we can see that
The number of routes from P to Q = 2 + 4 = 6
Hence , the number of routes from P to Q is 6 .
Correct Answer: Wednesday
Explanation:
Clearly 1st, 8th, 15th, 22nd and 29th October are Sundays.
So, 31st October is Tuesday.
? 1st November will be Wednesday.
Correct Answer: 3
Explanation:
Three numbers in which followed by 9 and preceded 2, 3 or 4
11 [2 3 9] 4 4 5 5 5 5 [3 3 9] 8 7 7 7 8 8 8 5 5 [4 3 9] 6 6 6 6
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