Directions for questions: Answer the questions on the basis of the information given below. Given below is a network of oil pipelines and nine hubs– A through I. The requirement of oil at the hubs is exactly met by the flow in the pipelines. The flow in some of the pipelines and the requirement at some of the hubs is also mentioned. A is the only hub which is a source hub, i.e. the hub A does not have any requirement of its own but it supplies the requirement of all the other hubs.
It is also known that
I. The maximum capacity of any pipeline is 1000.
II. The requirement at any hub except A and I is equal to the flow in exactly one of the pipelines directly connected to it.
III. No two pipelines directly connected to the same hub carry the same amount of oil.
IV. The flow in none of the pipelines shown above is zero.
As not two pipelines connecting same place are carrying same amount, and at least one of the pipelines should carry same as that of the place, either H-I or G-H must carry 300. If G-H carries 300, then H-I must carry 200, which is not possible.
H-I must carry 300 and G-H must carry 400. D-G should carry 400 more than the requirement at G.
G's requirement must equal that of G-H
Requirement at G = 400 D-G = 800
The flow in El must be equal to the requirement at E, and each of them must be equal to 100. Therefore the requirement at I = 100 + 300 = 400.
Considering F, the flow in FB must be equal to the requirement at F, i.e., 100, and the flo
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As not two pipelines connecting same place are carrying same amount, and at least one of the pipelines should carry same as that of the place, either H-I or G-H must carry 300. If G-H carries 300, then H-I must carry 200, which is not possible.
H-I must carry 300 and G-H must carry 400. D-G should carry 400 more than the requirement at G.
G's requirement must equal that of G-H
Requirement at G = 400 D-G = 800
The flow in El must be equal to the requirement at E, and each of them must be equal to 100. Therefore the requirement at I = 100 + 300 = 400.
Considering F, the flow in FB must be equal to the requirement at F, i.e., 100, and the flow in CF must be 100 + 100 + 200 = 400 (balancing the flow at F). Now, balancing the flows at B, BC must be equal to the requirement at B, i.e., 300.
Now, to meet the conditions at hub C, the flow in exactly one of AC and DC must be equal to the requirement at C, i.e., 200 and the other must be 100. Now, considering D the maximum capacity of A-D is 100.
D cannot be more than 100, as then A-D will be more than 1000.
D must be 100 and D-C must be 100. A-D is 1000. Hence the final diagram will be as below:
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<p>As not two pipelines connecting same place are carrying same amount, and at least one of the pipelines should carry same as that of the place, either H-I or G-H must carry 300. If G-H carries 300, then H-I must carry 200, which is not possible.</p><p>H-I must carry 300 and G-H must carry 400. D-G should carry 400 more than the requirement at G.</p><p>G's requirement must equal that of G-H</p><p>Requirement at G = 400 D-G = 800</p><p>The flow in El must be equal to the requirement at E, and each of them must be equal to 100. Therefore the requirement at I = 100 + 300 = 400.</p><p>Considering F, the flow in FB must be equal to the requirement at F, i.e., 100, and the flow in CF must be 100 + 100 + 200 = 400 (balancing the flow at F). Now, balancing the flows at B, BC must be equal to the requirement at B, i.e., 300.</p><p>Now, to meet the conditions at hub C, the flow in exactly one of AC and DC must be equal to the requirement at C, i.e., 200 and the other must be 100. Now, considering D the maximum capacity of A-D is 100.</p><p>D cannot be more than 100, as then A-D will be more than 1000.</p><p>D must be 100 and D-C must be 100. A-D is 1000. Hence the final diagram will be as below:</p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1754362748phpoYz34f.jpeg" alt="" width="444" height="306" loading="lazy"></picture></div></div>
From the 1st statement: B2 is now as old as B3 was in the past. Hence B2 is younger to B3 or B2 < B3. Also sometime in the past B1 was twice as old as B4. So B1 is elder to B4 or B1 > B4. B3 will be as old as B5 in future, hence B3 < B5. The second statement suggests: B1 > B6. B1 was as old as B7 in the past. Hence B1 > B7. B4 will be as old as B6 in future. Hence B6 > B4. B6 will be as old as B7 now in future. Hence B7 > B6. B7 was as old as B2, when B1 was as old as B7. Hence B1 = B2. Combining both the results, we get: and B5 > B3 > B2 = B1 > B7 > B6 > B4 (Note by B1 = B2, it is meant that they are of similar age group, not necessarily the same).
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