Question 1

The quantity moved from Avanti to Vidisha is

Accepted Answer

As per the given above figure , we can see that Since flow from Vaishali to Jyotishmati is 300 where as demand is 400 so the deficient 100 would be met by flow from Vidisha. Again the demand of 700 in Panchal is again to be met by flow from Jyotishmati which can get it from Vidisha. Thus, the quantity moved from Avanti to Vidisha is 200 + 100 + 700 = 1000

Question 2

What is the free capacity available in the Avanti-Vidisha pipeline?

Accepted Answer

On the basis of above given route network diagram , we can see that Free capacity in Avanti-Vidisha is zero. Explanation is similar as in previous answer.

Question 3

The free capacity available at the Avanti-Vaishali pipeline is

Accepted Answer

From above given route diagram , we can see that Free capacity at Avanti-Vaishali pipeline is 300, since capacity of each pipeline is 1000 and demand at Vidisha is 400 and 300 flows to Jyotishmati. Thus, free capacity = {1000 – (400 + 300)} = 300

Question 4

What is the total number of ways to reach A to B in the network given?

Accepted Answer

As per the given above figure , we can see that There are four ways to go from A to the first level of nodes. Each of these 4 nodes in turn leads into two more ways to go to the second level nodes. Each of the second level nodes leads into two more ways to go to the third level nodes. And from here we have only one way each to go to B. Hence by fundamental principal of counting, total number of ways = 4 x 2 x 2 x 1 = 16 ways.

Question 5

4 cities are connected by a road network as shown in the figure. In how many ways can you start from any city and come back to it without travelling on the same road more than once?

Accepted Answer

According to question , It can be seen that every city is connected to all the other cities (i.e. 3 other cities). Step 1: Let starting point is A, there are 3 ways in which we could proceed, viz. AB, AD or AC. Step 2: Once we are at any of these cities (B, D or C), each one of them is connected to 3 other cities. But since we cannot go back to A the originating city, there are only 2 ways in which we could proceed from here. Step 3: let us assume that we are at B, we can only go to D or C by taking BD or BC respectively. From this point we a choice of either directly going back to A (thus skipping 4th city or go to 4th city and come back to A. ) Step 4:- Now if we are at D, we can either take DA or DCA. So there are 2 more ways to go from here. So , total number of ways = 3 x 2 x 2 = 12 ways.

Question 6

What is the maximum quantity of natural gas S can receive?

Accepted Answer

From the above given figure , The maximum quantity of natural gas that S can receive = 6 + 2 + 9 = 17 units. Therefore , required answer will be 17 units.

Question 7

For which 2 cities it can be safely concluded that they have natural gas plants?

Accepted Answer

We know that cities M and P are gas plants.Hence , option A will be correct answer .

Question 8

What is the maximum quantity of natural gas than can be transported from M to R?

Accepted Answer

From the above given figure and the conditions R can receive only 6 units (5 + 1) of natural gas if utilization is 100%. Therefore , required answer will be 6 units.

Question 9

Eight cities A, B, C, D, E, F, G and H are connected with one-way roads R 1, R 2, R 3, R 4, R 5 and R 6 in the following manner: R1 leads from A to C via B; R2 leads from C to D and then via B to F; R3 leads from D to A and then via E to H; R4 leads…

Accepted Answer

On the basis of above given question , we can say that We can block B to D if A to B means R1 and B to F means R2 is blocked. Therefore minimum 2 ways needed to be block.

Question 10

In the adjoining figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

Accepted Answer

Here , number of vertical steps ( v ) = 3 Number of horizontal steps ( h ) = 5 Then in this case total number of ways is given by h+vCh = h+vCv = 8C3 = 6 x 7 x ( 8/6 ) = 7 x 8 = 56. Hence , 56 distinct routes can a car reach point B from point A .

Question 11

If the government wants to ensure that all motorists travelling from S to T pay the same amount ( fuel costs and toll combined ) regardless of the route they choose and the street from B to C is under repairs ( and hence unusable ), then a feasible …

Accepted Answer

As per the given figure , we can see that Let the toll charged at junctions A, B, C and D be a, b, c and d respectively. Then 1st we will list down all the routes and corresponding cost of travel. Since Route BC is under repair hence route S-B-C-T is not in use. Rest all four have the same toll charges hence 14 + a = 9 + a + b ⇒ b = 14 - 9 = 5 Similarly 10 + c + d = 13 + d ⇒ c = 13 - 10 = 3 Hence Options 4 is ruled out, now if we check option rest 3 options we will find out that option 2 and 3 both are correct. Option (2)/(3) Inconsistent options .

Question 12

If the government wants to ensure that all routes from S to T get the same amount of traffic, then a feasible set of toll charged (in rupees) at junctions A, B, C and D respectively to achieve this goal is :

Accepted Answer

As per the given diagram , we can see that Let the toll charged at junctions A, B, C and D be a, b, c and d respectively. Then 1st we will list down all the routes and corresponding cost of travel. Here in this case all 5 routes have the same toll charge hence 14 + a = 7 + b + c = 13 + d = 9 + a + b = 10 + c + d After solving we will get a = 1, b = 5, c = 3 and d = 2

Question 13

The government wants to devise a toll policy such that the total cost to the commuters per trip is minimized. The policy should also ensure that not more than 70 percent of the total traffic passes through junction B. The cost incurred by the commut…

Accepted Answer

On the basis of above given question , we can say that There must be one other route other than those involving B. We must take S - D - C - T as the other route. S - B - C - T, if toll at B = 3, total cost = 10 S - D - C - T, if toll at D and C is 0, total cost is 10. Hence ,$ 10 is the least cost.

Question 14

If the government wants to ensure that the traffic at S gets evenly distributed along streets from S to A, from S to B, and from S to D, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

Accepted Answer

According to question , If all the five routes have the same cost, then there will be an equal flow in all the five routes, i.e. 20% in each route. But then the percentage of traffic in S - A = 20% (Only one route involving S - A) S - B = 40% (As there are two routes involving S - B) S - D = 40% (As there are two routes involving S - D) But here the given condition that traffic in S-A is equal to that in S - B, which in turn is equal to S - D is not satisfied. Of the routes, that can be used the number of routes involving S - A must be the same as S - B, which in turn is same as that as S - D. That is possible only when we block the junction C and that can be done by taking higher toll charge at C to achieve this goal c > 3. Hence , required answer will be option A .

Question 15

If the government wants to ensure that no traffic flows on the street from D to T, while equal amount of traffic flows through junctions A and C, then a feasible set of toll charged (in rupees) at junctions A, B, C and D respectively to achieve this…

Accepted Answer

From the given diagram , Let the toll charged at junctions A, B, C and D be a, b, c and d respectively. Then 1st we will list down all the routes and corresponding cost of travel. Since the cost of travel including toll on routes S-A-T, S-B-C-T, S-B-A-T and S-D-C-T is the same. And D-T has no traffic due to high toll charge at D. From the last solutions we will get b = 5, 14 + a = 7 + b + c = 12 + c, or a + 2 = c 7 + b + c = 10 + c + d = 12 + c or d = 2, hence the result is B = 5, d = 2 and c-a = 2 that is satisfied by option (E).

Question 16

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.

Accepted Answer

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.

Question 17

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.

Accepted Answer

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 .

Question 18

Find the minimum distance between H to A without visiting city E.

Accepted Answer

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 .

Question 19

What is the minimum distance between H to A through D?

Accepted Answer

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.

Question 20

If path H E C is taken and no city is visited twice then find the total number of such paths.

Accepted Answer

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 .

Routes and Networks Questions