A cricket team scored runs in each of the first six overs (the 'powerplay') of its innings. The following information is available about the runs scored in these six overs:
- The team scored a different number of runs in each of the six overs, and a total of 54 runs in the six overs.
- The team's run rate (average runs per over) at the end of the 3rd over was exactly 8.
- The fewest runs in any of the six overs were scored in the 2nd over, and the most runs — 15 — in the 5th over.
- The runs scored in the 3rd over were three times the runs scored in the 2nd over.
- The team scored exactly 3 more runs in the 6th over than in the 4th over.
- The team scored a different number of runs in each of the six overs, and a total of 54 runs in the six overs.
- The team's run rate (average runs per over) at the end of the 3rd over was exactly 8.
- The fewest runs in any of the six overs were scored in the 2nd over, and the most runs — 15 — in the 5th over.
- The runs scored in the 3rd over were three times the runs scored in the 2nd over.
- The team scored exactly 3 more runs in the 6th over than in the 4th over.
Question 1
Single correct option
How many runs were scored in the 1st over?
- A. 8
- B. 6
- C. 12
- D. 9
Show answer and solution
Answer: (A) 8
Deduction: Overs 1–3 total 24 (run rate 8). Overs 4–6 therefore total 54 − 24 = 30; with over 5 = 15, overs 4 + 6 = 15 and over 6 = over 4 + 3 ⇒ over 4 = 6, over 6 = 9.
Let over 2 = x; then over 3 = 3x and over 1 = 24 − 4x. Over 2 must be the smallest of all, so x < 6, and over 1 > x ⇒ x < 4.8, so x ≤ 4. x = 4 gives 8, 4, 12 ✓. x = 3 gives 12, 3, 9 — but 9 repeats over 6 ✗. x = 2 gives 16 > 15 (the maximum) ✗; x = 1 gives 20 ✗. Hence the overs read 8, 4, 12, 6, 15, 9.
| Over | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Runs | 8 | 4 | 12 | 6 | 15 | 9 |
| Cumulative | 8 | 12 | 24 | 30 | 45 | 54 |
| Run rate | 8 | 6 | 8 | 7.5 | 9 | 9 |
Let over 2 = x; then over 3 = 3x and over 1 = 24 − 4x. Over 2 must be the smallest of all, so x < 6, and over 1 > x ⇒ x < 4.8, so x ≤ 4. x = 4 gives 8, 4, 12 ✓. x = 3 gives 12, 3, 9 — but 9 repeats over 6 ✗. x = 2 gives 16 > 15 (the maximum) ✗; x = 1 gives 20 ✗. Hence the overs read 8, 4, 12, 6, 15, 9.
| Over | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Runs | 8 | 4 | 12 | 6 | 15 | 9 |
| Cumulative | 8 | 12 | 24 | 30 | 45 | 54 |
| Run rate | 8 | 6 | 8 | 7.5 | 9 | 9 |
Question 2
Numerical answer
How many runs were scored in the 4th over?
Show answer and solution
Answer: 6
Deduction: Overs 1–3 total 24 (run rate 8). Overs 4–6 therefore total 54 − 24 = 30; with over 5 = 15, overs 4 + 6 = 15 and over 6 = over 4 + 3 ⇒ over 4 = 6, over 6 = 9.
Let over 2 = x; then over 3 = 3x and over 1 = 24 − 4x. Over 2 must be the smallest of all, so x < 6, and over 1 > x ⇒ x < 4.8, so x ≤ 4. x = 4 gives 8, 4, 12 ✓. x = 3 gives 12, 3, 9 — but 9 repeats over 6 ✗. x = 2 gives 16 > 15 (the maximum) ✗; x = 1 gives 20 ✗. Hence the overs read 8, 4, 12, 6, 15, 9.
| Over | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Runs | 8 | 4 | 12 | 6 | 15 | 9 |
| Cumulative | 8 | 12 | 24 | 30 | 45 | 54 |
| Run rate | 8 | 6 | 8 | 7.5 | 9 | 9 |
The 4th over yielded 6 runs.
Let over 2 = x; then over 3 = 3x and over 1 = 24 − 4x. Over 2 must be the smallest of all, so x < 6, and over 1 > x ⇒ x < 4.8, so x ≤ 4. x = 4 gives 8, 4, 12 ✓. x = 3 gives 12, 3, 9 — but 9 repeats over 6 ✗. x = 2 gives 16 > 15 (the maximum) ✗; x = 1 gives 20 ✗. Hence the overs read 8, 4, 12, 6, 15, 9.
| Over | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Runs | 8 | 4 | 12 | 6 | 15 | 9 |
| Cumulative | 8 | 12 | 24 | 30 | 45 | 54 |
| Run rate | 8 | 6 | 8 | 7.5 | 9 | 9 |
The 4th over yielded 6 runs.
Question 3
Numerical answer
The team's run rate at the end of an over is its total so far divided by the number of overs bowled. In how many of the overs 2, 3, 4, 5 and 6 was the run rate at the end of that over strictly higher than at the end of the previous over?
Show answer and solution
Answer: 2
Deduction: Overs 1–3 total 24 (run rate 8). Overs 4–6 therefore total 54 − 24 = 30; with over 5 = 15, overs 4 + 6 = 15 and over 6 = over 4 + 3 ⇒ over 4 = 6, over 6 = 9.
Let over 2 = x; then over 3 = 3x and over 1 = 24 − 4x. Over 2 must be the smallest of all, so x < 6, and over 1 > x ⇒ x < 4.8, so x ≤ 4. x = 4 gives 8, 4, 12 ✓. x = 3 gives 12, 3, 9 — but 9 repeats over 6 ✗. x = 2 gives 16 > 15 (the maximum) ✗; x = 1 gives 20 ✗. Hence the overs read 8, 4, 12, 6, 15, 9.
| Over | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Runs | 8 | 4 | 12 | 6 | 15 | 9 |
| Cumulative | 8 | 12 | 24 | 30 | 45 | 54 |
| Run rate | 8 | 6 | 8 | 7.5 | 9 | 9 |
Run rate increases after over 3 (6 → 8) and after over 5 (7.5 → 9). After over 6 it stays at 9 (not strictly higher). Answer: 2.
Tip: the run rate rises exactly when that over's runs exceed the run rate before it: 12 > 6 and 15 > 7.5, while 9 = 9.
Let over 2 = x; then over 3 = 3x and over 1 = 24 − 4x. Over 2 must be the smallest of all, so x < 6, and over 1 > x ⇒ x < 4.8, so x ≤ 4. x = 4 gives 8, 4, 12 ✓. x = 3 gives 12, 3, 9 — but 9 repeats over 6 ✗. x = 2 gives 16 > 15 (the maximum) ✗; x = 1 gives 20 ✗. Hence the overs read 8, 4, 12, 6, 15, 9.
| Over | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Runs | 8 | 4 | 12 | 6 | 15 | 9 |
| Cumulative | 8 | 12 | 24 | 30 | 45 | 54 |
| Run rate | 8 | 6 | 8 | 7.5 | 9 | 9 |
Run rate increases after over 3 (6 → 8) and after over 5 (7.5 → 9). After over 6 it stays at 9 (not strictly higher). Answer: 2.
Tip: the run rate rises exactly when that over's runs exceed the run rate before it: 12 > 6 and 15 > 7.5, while 9 = 9.
Question 4
Single correct option
In which over was the third-highest number of runs scored?
- A. 3rd
- B. 4th
- C. 6th
- D. 1st
Show answer and solution
Answer: (C) 6th
Deduction: Overs 1–3 total 24 (run rate 8). Overs 4–6 therefore total 54 − 24 = 30; with over 5 = 15, overs 4 + 6 = 15 and over 6 = over 4 + 3 ⇒ over 4 = 6, over 6 = 9.
Let over 2 = x; then over 3 = 3x and over 1 = 24 − 4x. Over 2 must be the smallest of all, so x < 6, and over 1 > x ⇒ x < 4.8, so x ≤ 4. x = 4 gives 8, 4, 12 ✓. x = 3 gives 12, 3, 9 — but 9 repeats over 6 ✗. x = 2 gives 16 > 15 (the maximum) ✗; x = 1 gives 20 ✗. Hence the overs read 8, 4, 12, 6, 15, 9.
| Over | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Runs | 8 | 4 | 12 | 6 | 15 | 9 |
| Cumulative | 8 | 12 | 24 | 30 | 45 | 54 |
| Run rate | 8 | 6 | 8 | 7.5 | 9 | 9 |
In descending order: 15 (5th), 12 (3rd), 9 (6th), 8 (1st), 6 (4th), 4 (2nd). The third-highest is the 6th over.
Let over 2 = x; then over 3 = 3x and over 1 = 24 − 4x. Over 2 must be the smallest of all, so x < 6, and over 1 > x ⇒ x < 4.8, so x ≤ 4. x = 4 gives 8, 4, 12 ✓. x = 3 gives 12, 3, 9 — but 9 repeats over 6 ✗. x = 2 gives 16 > 15 (the maximum) ✗; x = 1 gives 20 ✗. Hence the overs read 8, 4, 12, 6, 15, 9.
| Over | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Runs | 8 | 4 | 12 | 6 | 15 | 9 |
| Cumulative | 8 | 12 | 24 | 30 | 45 | 54 |
| Run rate | 8 | 6 | 8 | 7.5 | 9 | 9 |
In descending order: 15 (5th), 12 (3rd), 9 (6th), 8 (1st), 6 (4th), 4 (2nd). The third-highest is the 6th over.
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