| Over Number | Run Rate |
|---|---|
| N-2 | 8.00 |
| N | 7.43 |
| N+2 | 8.11 |
| N+4 | 8.45 |
| N+6 | 8.08 |
Step 1: Define and calculate run rate. The run rate after k overs is computed as:
Run Rate at k = $\frac{\text{Total runs scored in overs 1 to } k}{k}$
Based on the provided data:
The total runs scored in overs (<i>N</i> − 1) and N can be determined by the change in total runs:
Total runs in overs (<i>N</i> − 1) and N = (Run Rate at N × N) − (Run Rate at N − 2 × (<i>N</i> − 2)).
Substituting the given values:
Runs in (<i>N</i> − 1) and N = 7.43N − 8(<i>N</i> − 2).
Simplify the expression:
Runs in (<i>N</i> − 1) and N = 7.43N − 8N + 16 = −0.57N + 16.
Step 2: Determine valid range for N. Given that the team scored between 6 and 15 runs per over, the runs in overs (<i>N</i> − 1) and N must fall within this range:
6 ≤ −0.57N + 16 ≤ 15.
Solve the inequalities independently:
The solution to these inequalities indicates that N must be an integer between 7 and 13, inclusive. The value N = 13 satisfies all the given conditions.
Final Answer: 13.
Step 1: Analyze Constraints. Runs per over range from 6 to 15. The sum of runs over two overs must equal 22.
Step 2: Determine Valid Score Combinations. Pairs of scores that sum to 22 and fall within the allowed range per over are: (7, 15), (8, 14), (9, 13), (10, 12), (11, 11).
Step 3: Assign Scores to Overs. Given the valid score pairs are within the specified per-over run range, scores of 8 and 14 can occur in overs 8 and 9 respectively.
Final Answer: Overs 8 and 9.
Step 1: Analyze run rates. The run rate declines at over N (refer to the table in Question 23). This implies that a minimal quantity of runs was accumulated in over N − 1 or N. Step 2: Determine the lowest run accumulation. Given N = 7 (as per Question 23), the team must have scored the fewest runs in over 7, since the run rate decreases at this interval, signifying a minimum increase to the cumulative score. Final Answer: 7.