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There are 6 refineries, 7 depots, 9 districts. The refineries are R1, R2, R3, R4, R5, R6. The depots are D1, D2, D3, D4, D5, D6 and D7 and the districts are DR1, DR2, DR3, DR4, DR5, DR6, DR7, DR8, DR9.

Table A shows the cost of transporting one unit from refinery to depot.

Table B shows the cost of transporting one unit from depot to districts.

Table - A

R1

R2

R3

R4

R5

R6

D1

928.2

537.2

567.8

589.9

800.1

323.4

D2

311.8

595.7

885.7

759.9

793.1

420.1

D3

451.1

0

320.1

720.1

1000.1

404.5

D4

371.1

50.1

350.1

650.4

980.1

525.3

D5

1137.3

314.5

0

1157.7

406.3

617.5

D6

617.1

516.8

756.5

1065.9

623.9

509.4

D7

644.3

299.2

537.2

1093.1

725.8

827.4

Table - B

D1

D2

D3

D4

D5

D6

D7

DR1

571.1

205

352

159

434.5

178

337

DR2

200

337.5

291

201

0

980.7

434

DR3

100

0

275

277

850

770.5

835

DR4

0

415.7

350

760

300

560

444.7

DR5

223.5

300

440

1033

880

325

526.5

DR6

577.5

725

443.5

560

1035.3

570

530

DR7

340

410.6

886.7

0

800.7

680.5

800

DR8

627

556.5

1023

1024

759

1025.7

300

DR9

439

738

980

1031.7

1024

900

757

The minimum cost of transportation of one unit from refinery BD to any district is

Option 1 -

Rs. 125

Option 2 -

Rs. 0

Option 3 -

Rs. 375

Option 4 -

None of the above

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

R4 → D5 → DR2

The least cost to reach to AAB is for AE. And that is BD to AE is zero.

0 Upvotes 0 Downvotes

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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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According to the information provided following arrangement will be obtained:

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Order according to height								Student	Belong
1	2	3	4	5	6	7	8		K	G	D
×	×	×	×	×	×	×	√	Arun	×	×	√
×	×	√	×	×	×	×	×	Ajit	√	×	×
√	×	×	×	×	×	×	×	Rohan	√	×	×
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×	√	×	×	×	×	×	×	Ronit	×	×	√
×	×	×	×	√	×	×	×	Shubham	√	×	×
×	×	×	×	×	×	√	×	Sweety	×	√	×
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Looking at the table given in the first question of the set, we get that Talwar lives on the sixth floor.

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The data available is not enough to determine where Rubi belongs to. The given data is insufficient to answer the above question.

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Table - A
	R1	R2	R3	R4	R5	R6
D1	928.2	537.2	567.8	589.9	800.1	323.4
D2	311.8	595.7	885.7	759.9	793.1	420.1
D3	451.1	0	320.1	720.1	1000.1	404.5
D4	371.1	50.1	350.1	650.4	980.1	525.3
D5	1137.3	314.5	0	1157.7	406.3	617.5
D6	617.1	516.8	756.5	1065.9	623.9	509.4
D7	644.3	299.2	537.2	1093.1	725.8	827.4

Table - B
	D1	D2	D3	D4	D5	D6	D7
DR1	571.1	205	352	159	434.5	178	337
DR2	200	337.5	291	201	0	980.7	434
DR3	100	0	275	277	850	770.5	835
DR4	0	415.7	350	760	300	560	444.7
DR5	223.5	300	440	1033	880	325	526.5
DR6	577.5	725	443.5	560	1035.3	570	530
DR7	340	410.6	886.7	0	800.7	680.5	800
DR8	627	556.5	1023	1024	759	1025.7	300
DR9	439	738	980	1031.7	1024	900	757