Data Interpretation
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New answer posted
5 months agoContributor-Level 10
| Day 1 | Day 2 | Day 3 | Day 4 | Day 5 |
Start Price | 100 | 110 | 120 | 130 | 120 |
End Price | 110 | 120 | 130 | 120 | 110 |
Let initial amount with T1 and Michal is .
T1 sold shares on Day 1, Day 2, Day 3, whereas buys shares on Day 4 and Day 5.
Total money with T1 is = Y + 110 * 10 + 120 * 10 + 130 * 10 – 120 * 10 – 110 * 10 = Y + 1300
Total money with T2 = Y + 1200
Total money with T2 = Y + 120 * 10 + 130 * 10 + 120 * 10 = Y + 3700
Total money with T2 & T1 = 2Y + 5000.
Therefore maximum possible increase is 5000.
New answer posted
5 months agoContributor-Level 10
| Day 1 | Day 2 | Day 3 | Day 4 | Day 5 |
Start Price | 100 | 90 | 100 | 110 | 120 |
End Price | 90 | 80 | 110 | 120 | 110 |
Let initial amount with T1 and T2 is Y.
Total money with T1 = Y – 900 + 1000 + 1100 + 1200 – 1100 = Y + 1300
Total money with T2 = Y + 1200
Therefore difference between T1 and T2 is Rs. 100 and Number of shares with T2 and T1 is same.
New answer posted
5 months agoContributor-Level 10
Player 1 | Player 2 | Export of Player 1 to 2 | Export of Player 2 to 1 | Winner |
P1 | P2 | 50 | 35 | P1 |
P1 | P4 | 40 | 17 | P1 |
P1 | P4 | 44 | 70 | P4 |
P1 | P5 | 57 | 98 | P5 |
P1 | P6 | 68 | 35 | P1 |
P2 | P3 | 35 | 9 | P2 |
P2 | P4 | 58 | 32 | P2 |
P2 | P5 | 43 | 64 | P5 |
P2 | P6 | 59 | 25 | P2 |
P3 | P4 | 15 | 28 | P4 |
P3 | P5 | 7 | 53 | P5 |
P3 | P6 | 24 | 57 | P6 |
P3 | P5 | 68 | 45 | P4 |
P3 | P6 | 35 | 62 | P6 |
P5 | P6 | 52 | 35 | P5 |
Rank | Player | Number of times Winner | Points |
1 | P5 | 4 | 8 |
2 | P1 | 3 | 6 |
2 | P2 | 3 | 6 |
2 | P4 | 3 | 6 |
4 | P6 | 2 | 4 |
5 | P3 | 0 | 0 |
We cannot determine the answer as there is a tie between 3 players and we do not know how to break that tie.
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