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

$1$

$2$

$3$

$4$

✅ Correct Option: 2

We need to find the value of n first, then determine the smallest possible value of x.

The batsman played $(n+2)$ innings with an average of 29 runs.

When we know the average and number of innings, we can find the total score using:

Total score = Average × Number of innings

Total score in $(n+2)$ innings = $29 \times (n+2) = 29n + 58$

We know the last two innings were 38 and 15 runs.

Total score in first n innings = Total score in all innings - Last two innings

Total score in first n innings = $(29n + 58) - 38 - 15 = 29n + 5$

We're told the average in the first n innings was 30 runs.

So: Total score in first n innings = $30 \times n = 30n$

Now we have two expressions for the same quantity:

$30n = 29n + 5$

$30n - 29n = 5$

$n = 5$

The total score in the first n innings must be the same whether we calculate it by subtracting the last two scores or by using the given average.

Now we know there were 5 innings in the first part, with:

Total score = $30 \times 5 = 150$ runs

Each score < 38 runs (given condition)

One score = x runs (the minimum we want to find)

To make x as small as possible, we need to make the other 4 scores as large as possible.

Since each score must be less than 38, the maximum any score can be is 37 runs.

Maximum total of the other 4 scores = $37 \times 4 = 148$ runs

Therefore:

$x = \text{Total score} - \text{Maximum sum of other 4 scores}$

$x = 150 - 148 = 2$ runs

The smallest possible value of x is 2 runs.

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