A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is

The correct option is 2.

Let us break down the logical reasoning and mathematical derivations step-by-step to understand why this is the case.

Step 1: Determine the number of innings, $n$
The batsman played $n+2$ innings with an average score of 29 runs.
Therefore, the total runs scored in all the $n+2$ innings is:
$\text{Total Runs}=29\times (n+2)$

We are given that the scores in the last two innings were 38 and 15 runs. Thus, the total runs in the first $n$ innings is the total runs minus the runs scored in these last two innings:
$\text{Runs in first}n\text{innings}=29(n+2)-38-15$
Simplifying the expression:
$\text{Runs in first}n\text{innings}=29n+58-53=29n+5$

We are also given that the average score in these first $n$ innings was 30 runs. Therefore, the total runs in the first $n$ innings is:
$\text{Runs in first}n\text{innings}=30n$

Equating both expressions for the total runs in the first $n$ innings:
$30n=29n+5$
Subtracting $29n$ from both sides gives:
$n=5$

Step 2: Find the total runs in the first 5 innings
Since $n=5$, the total runs scored in these 5 innings is:
$\text{Total runs in 5 innings}=30\times 5=150$

Step 3: Minimize the lowest score $x$
Let the scores in the first 5 innings be $s_1,s_2,s_3,s_4,s_5$ in ascending order, where $s_1=x$ is the lowest score.
We are told that the batsman scored less than 38 runs in each of these first $n$ innings. Because scores in cricket are integers, the maximum possible score in any of these innings is:
$s_i\le 37$

To make the lowest score $x$ as small as possible, we must make the other 4 scores as large as possible. The maximum possible value for each of the remaining 4 innings is 37 runs.
Thus, we set:
$s_2=s_3=s_4=s_5=37$

Now, we write the sum equation:
$x+s_2+s_3+s_4+s_5=150$
Substituting the maximum values:
$x+37+37+37+37=150$
$x+148=150$
$x=150-148=2$

Hence, the smallest possible value of $x$ is 2.