DILR Preparation Tips for MBA
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New answer posted
2 months agoContributor-Level 10
R4 → D5 → DR2
The least cost to reach to AAB is for AE. And that is BD to AE is zero.
New answer posted
2 months agoContributor-Level 10
If you look for large figures you would find them in both tables in D5.
the largest cost = 1157.7 + 1035.3
= 2193.00
New answer posted
2 months agoContributor-Level 10
Number of refineries = 6
Number of depots = 7
Number of districts = 9
Therefore, number of possible ways to send petrol from any refinery to any district is 6 * 7 * 9 = 378.
New question posted
2 months agoNew answer posted
2 months agoContributor-Level 10
Total number of excellent students = 120
Number of female excellent students = 70
Required fraction
New answer posted
2 months agoContributor-Level 10
Number of female good students = 8
Number of males average students = 48
Required ratio = 8 : 48 = 1 : 6
New answer posted
2 months agoContributor-Level 10
From the above table, it is clear that number of female good students = 8
New answer posted
2 months agoContributor-Level 10
From (II), 40% of total number of students = 96
Total number of students = (96/40)*100=240
Number of boys = 240 – 96 = 144
From (I), number of average male students =144/3=48
From (III), number of excellent students = 240/2= 120
Number of excellent female = 120 – 50 = 70
From these calculations, we get the following table.
| Performance | Total | ||
Average | Good | Excellent | ||
Male | 48 | 46 | 50 | 144 |
Female | 18 | 8 | 70 | 96 |
Total | 66 | 54 | 120 | 240 |
From the above table, it is clear that number of male good students = 46
New answer posted
2 months agoContributor-Level 10
The answer can be obtained directly by increasing the answer in question 11 by 50%, as all the Children increase the cost by 50% and hence the total cost will increase by 50% only.
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