Comprehension
The batting average (BA) of a Test batsman is computed from runs scored and innings played — completed innings and incomplete innings (not out) in the following manner:
\[ r_1 = \text{Number of runs scored in completed innings} \] \[ n_1 = \text{Number of completed innings} \] \[ r_2 = \text{Number of runs scored in incomplete innings} \] \[ n_2 = \text{Number of incomplete innings} \]
\[ BA = \frac{r_1 + r_2}{n_1} \]
To better assess a batsman’s accomplishments, the ICC is considering two other measures MBA1 and MBA2, defined as follows:
\[ MBA_1 = \frac{r_1}{n_1} + \frac{n_2}{n_1} \max \left( 0, \left( \frac{r_2}{n_2} - \frac{r_1}{n_1} \right) \right) \]
\[ MBA_2 = \frac{r_1 + r_2}{n_1 + n_2} \]
Question: 1
Based on the above information which of the following is true?
Based on the above information which of the following is true?
Show Hint
Write \(A_1=r_1/n_1\) and \(A_2=r_2/n_2\) and compare the three measures in terms of these averages; test a numerical example against each option.
Updated On: Nov 4, 2026
- \(MBA_1 \le BA \le MBA_2\)
- None of these
- \(MBA_2 \le BA \le MBA_1\)
- \(BA \le MBA_2 \le MBA_1\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Rewrite the measures.
Let \(A_1=\frac{r_1}{n_1}\) and \(A_2=\frac{r_2}{n_2}\) (the average in completed and incomplete innings). Then
\[BA=A_1+\tfrac{n_2}{n_1}A_2,\qquad MBA_1=A_1+\tfrac{n_2}{n_1}\max(0,A_2-A_1),\qquad MBA_2=\frac{n_1A_1+n_2A_2}{n_1+n_2}.\]Step 2: Compare \(MBA_1\) with \(BA\).
\(BA-MBA_1=\frac{n_2}{n_1}\left[A_2-\max(0,A_2-A_1)\right]=\frac{n_2}{n_1}\min(A_1,A_2)\ge 0\). So \(MBA_1\le BA\).
Step 3: Compare \(MBA_2\) with \(MBA_1\).
\(MBA_2=A_1+\frac{n_2}{n_1+n_2}(A_2-A_1)\).
Step 4: Test the options with an example.
Take \(r_1=100,\ n_1=2,\ r_2=100,\ n_2=1\): \(BA=100\), \(MBA_1=50+\frac12\cdot 50=75\), \(MBA_2=\frac{200}{3}\approx 66.7\).
Final Answer:
The correct option is (2): None of these.
Let \(A_1=\frac{r_1}{n_1}\) and \(A_2=\frac{r_2}{n_2}\) (the average in completed and incomplete innings). Then
\[BA=A_1+\tfrac{n_2}{n_1}A_2,\qquad MBA_1=A_1+\tfrac{n_2}{n_1}\max(0,A_2-A_1),\qquad MBA_2=\frac{n_1A_1+n_2A_2}{n_1+n_2}.\]Step 2: Compare \(MBA_1\) with \(BA\).
\(BA-MBA_1=\frac{n_2}{n_1}\left[A_2-\max(0,A_2-A_1)\right]=\frac{n_2}{n_1}\min(A_1,A_2)\ge 0\). So \(MBA_1\le BA\).
Step 3: Compare \(MBA_2\) with \(MBA_1\).
\(MBA_2=A_1+\frac{n_2}{n_1+n_2}(A_2-A_1)\).
- If \(A_2\le A_1\): \(MBA_1=A_1\ge MBA_2\).
- If \(A_2>A_1\): \(MBA_1-MBA_2=(A_2-A_1)\,n_2\left(\frac1{n_1}-\frac1{n_1+n_2}\right)\ge 0\).
Step 4: Test the options with an example.
Take \(r_1=100,\ n_1=2,\ r_2=100,\ n_2=1\): \(BA=100\), \(MBA_1=50+\frac12\cdot 50=75\), \(MBA_2=\frac{200}{3}\approx 66.7\).
- (1) \(75\le 100\le 66.7\): false.
- (4) \(100\le 66.7\le 75\): false.
- (3) \(66.7\le 100\le 75\): false.
Final Answer:
The correct option is (2): None of these.
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Question: 2
An experienced cricketer with no incomplete innings has BA of 50. The next time he bats, the innings is incomplete and he scores 45 runs. It can be inferred that
An experienced cricketer with no incomplete innings has BA of 50. The next time he bats, the innings is incomplete and he scores 45 runs. It can be inferred that
Show Hint
Before the innings \(n_2=0\), so all three measures equal 50. Recompute each measure after adding one incomplete innings of 45 runs.
Updated On: Nov 4, 2026
- \(BA\) and \(MBA_1\) will both increase
- \(BA\) will increase and not enough data is available to assess change in \(MBA_1\) and \(MBA_2\)
- \(BA\) will increase and \(MBA_2\) will decrease
- None of these
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Before the innings.
No incomplete innings, so \(n_2=0\), \(r_2=0\), and \(r_1=50\,n_1\). Then \(BA=\frac{r_1}{n_1}=50\), \(MBA_1=\frac{r_1}{n_1}=50\) and \(MBA_2=\frac{r_1}{n_1}=50\).
Step 2: After the incomplete innings (\(r_2=45,\ n_2=1\)).
(1) is false because \(MBA_1\) does not increase. (2) is false because the data suffices (the result holds for every \(n_1\)). (3) states exactly what we found: BA increases and \(MBA_2\) decreases.
Final Answer:
The correct option is (3): BA will increase and MBA\(_2\) will decrease.
No incomplete innings, so \(n_2=0\), \(r_2=0\), and \(r_1=50\,n_1\). Then \(BA=\frac{r_1}{n_1}=50\), \(MBA_1=\frac{r_1}{n_1}=50\) and \(MBA_2=\frac{r_1}{n_1}=50\).
Step 2: After the incomplete innings (\(r_2=45,\ n_2=1\)).
- \(BA=\frac{50n_1+45}{n_1}=50+\frac{45}{n_1}>50\). BA increases (the denominator \(n_1\) does not change since the innings is incomplete).
- \(MBA_1=50+\frac{1}{n_1}\max(0,\,45-50)=50+0=50\). MBA\(_1\) is unchanged.
- \(MBA_2=\frac{50n_1+45}{n_1+1}=50-\frac{5}{n_1+1}<50\). MBA\(_2\) decreases.
(1) is false because \(MBA_1\) does not increase. (2) is false because the data suffices (the result holds for every \(n_1\)). (3) states exactly what we found: BA increases and \(MBA_2\) decreases.
Final Answer:
The correct option is (3): BA will increase and MBA\(_2\) will decrease.
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