Cricket Run Rates Analysis

XAT 2024 · Quantitative Ability & Data Interpretation · Hard · Data Interpretation

Passage / data set

A T20 cricket match consists of two teams playing twenty overs each, numbered 1 to 20. The runs scored in any over is a non-negative integer. The run rate at the end of any over is the average runs scored up to and including that over, i.e., the run rate at the end of the kk-th over is the average number of runs scored in overs numbered 1,2,…,k1, 2, \dots, k, where 1≤k≤201 \le k \le 20, kk a positive integer.

The following table indicates the run rate of a team at the end of some of the overs during a T20 cricket match (correct up to 2 decimal places), where 1≤N−2<N+6≤201 \le N - 2 < N + 6 \le 20, NN a positive integer. It is also known that the team did not score less than 6 runs and more than 15 runs in any over.

Over NumberRun Rate
N−2N - 28.00
NN7.43
N+2N + 28.11
N+4N + 48.45
N+6N + 68.08

Question 1 of 3

What is the value of NN?

  1. A.

    7

  2. B.

    13

  3. C.

    14

  4. D.

    9

  5. E.

    12

Answer

A

Explanation

The total score after NN overs is N×7.43N \times 7.43, which must be close to an integer. Testing values of N∈[3,14]N \in [3, 14]:

  • For N=7N = 7, total score =7×7.43=52.01≈52= 7 \times 7.43 = 52.01 \approx 52. Check consistency:
  • N−2=5N - 2 = 5 overs: score =5×8.00=40= 5 \times 8.00 = 40
  • N=7N = 7 overs: score =52= 52
  • N+2=9N + 2 = 9 overs: score =9×8.11=72.99≈73= 9 \times 8.11 = 72.99 \approx 73
  • N+4=11N + 4 = 11 overs: score =11×8.45=92.95≈93= 11 \times 8.45 = 92.95 \approx 93
  • N+6=13N + 6 = 13 overs: score =13×8.08=105.04≈105= 13 \times 8.08 = 105.04 \approx 105 All over-by-over run totals fall within the valid range of 66 to 1515 runs per over. Thus N=7N = 7.

Question 2 of 3

In which of these pairs of over numbers, the team could have scored 22 runs in total?

  1. A.

    6 and 7

  2. B.

    7 and 8

  3. C.

    8 and 9

  4. D.

    9 and 10

  5. E.

    10 and 11

Answer

D

Explanation

Scores after overs:

  • Over 5: 40
  • Over 7: 52   ⟹  \implies Runs in overs 6 & 7 =12= 12 (must be 6 each since minimum runs per over is 6).
  • Over 9: 73   ⟹  \implies Runs in overs 8 & 9 =21= 21. Since Over 7 =6= 6, max in Over 8 is 15   ⟹  \implies Runs in overs 7 & 8 max =21= 21.
  • Over 11: 93   ⟹  \implies Runs in overs 10 & 11 =20= 20.
  • Runs in overs 9 & 10: Since runs in (8&9) =21= 21 and runs in (10&11) =20= 20, depending on distribution, runs in 9 and 10 can be up to 15+10=2515 + 10 = 25, so 22 runs is possible.

Question 3 of 3

In which of the following over numbers, the team MUST have scored the least number of runs?

  1. A.

    7

  2. B.

    8

  3. C.

    9

  4. D.

    10

  5. E.

    11

Answer

A

Explanation

Total runs in overs 6 and 7 =52−40=12= 52 - 40 = 12. Since minimum runs per over is 6, runs scored in over 6 =6= 6 and runs scored in over 7 =6= 6. 6 is the minimum allowed runs per over, so over 7 must have scored the least possible number of runs.

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