DILR Prep Tips for MBA
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New answer posted
8 months agoContributor-Level 10
In the 1st round in one group number of matches is (7 * 8/2) = 28
Let's consider bottom 3 they will have 3 matches between them, so remaining 28-3 = 25 matches have involvement of top 5 team, if they won equal number of matches, i.e. 5 each then decision will be taken based on tie breaker rule, hence a team may be eliminated even after winning 5 matches.
New answer posted
8 months agoContributor-Level 10
Number of matches in stage 1 is 2 (8C2) = 2 (7 * 8/2) = 56
Stage 2 will be a knock out tournament with 8 teams so in this stage number of matches will be 7
Total 56 + 7 = 63
New answer posted
8 months agoContributor-Level 10
The maximum quantity of natural gas that can be transported that can be transported from M to Q is 6
New answer posted
8 months agoContributor-Level 10
From the figure and the conditions R can receive only 6 units (5 + 1) of natural gas if utilization is 100 Percent
New answer posted
8 months agoContributor-Level 10
From the figure the maximum quantity of natural gas that S can receive = 6 + 2 + 9 = 17 units.
New answer posted
8 months agoContributor-Level 10
| Mohit | Manohar | Prashant | Dinesh |
Round – 4 | 3200 | 3200 | 3200 | 3200 |
Round – 3 | 1600 | 1600 | 1600 | 1600 |
Round – 2 | 800 | 800 | 7200 | 4000 |
Round – 1 | 400 | 6800 | 3600 | 2000 |
Initially | 6600 | 3400 | 1800 | 1000 |
New answer posted
8 months agoContributor-Level 10
| Mohit | Manohar | Prashant | Dinesh |
Round – 4 | 3200 | 3200 | 3200 | 3200 |
Round – 3 | 1600 | 1600 | 1600 | 1600 |
Round – 2 | 800 | 800 | 7200 | 4000 |
Round – 1 | 400 | 6800 | 3600 | 2000 |
Initially | 6600 | 3400 | 1800 | 1000 |
New answer posted
8 months agoContributor-Level 10
| Mohit | Manohar | Prashant | Dinesh |
Round – 4 | 3200 | 3200 | 3200 | 3200 |
Round – 3 | 1600 | 1600 | 1600 | 1600 |
Round – 2 | 800 | 800 | 7200 | 4000 |
Round – 1 | 400 | 6800 | 3600 | 2000 |
Initially | 6600 | 3400 | 1800 | 1000 |
New answer posted
8 months agoContributor-Level 10
| Mohit | Manohar | Prashant | Dinesh |
Round – 4 | 3200 | 3200 | 3200 | 3200 |
Round – 3 | 1600 | 1600 | 1600 | 1600 |
Round – 2 | 800 | 800 | 7200 | 4000 |
Round – 1 | 400 | 6800 | 3600 | 2000 |
Initially | 6600 | 3400 | 1800 | 1000 |
New answer posted
8 months agoContributor-Level 10
From the given condition g = 10,
Number of students who like both Physics and Maths = 10 + f
Number of students who like both chemistry and Maths = 10 + e
From given condition 10 + f = 10 + e or e = f
Similarly from 3rd condition ¾ (10 + e) = 10 + d
Or 30 + 3e = 40 + 4d
Or
d = 3e - 10/4
Since 15 students like only chemistry hence b = 15
Number of students who like only chemistry is 40
Hence e + g + d + b = 40
or e + 10 + (3e – 10)/4 + 15 = 40
or (7e – 10)/4 = 15 or 7e = 70 or e = 10
Hence a = 5, c = 20, f = e = 10, d = 5
Hence the distribution is as follows
As per the observation of class teacher actual numbers of students who lik
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