The frequency table below shows the number of books read by a group of students in a month.

1. Compare the frequency of 3 books with the other frequencies

The frequencies are $9, 4, x, 8, 6, 5$.

For the mode to be $3$, the frequency for $3$ books, which is $x$, must be the greatest frequency.

The greatest known frequency is $9$, so the least possible value is
$x = 10.$

1. Locate where the values 3 and 4 occur in the ordered data

Cumulative frequency up to 2 books:
$9+4=13$
So the students who read 3 books occupy positions
$14 \text{ to } 13+x.$
The students who read 4 books occupy positions
$14+x \text{ to } 21+x.$

2. Find the largest $x$ such that the median position is a 4

Total number of students:
$9+4+x+8+6+5=32+x.$
Testing the largest possible boundary, if $x=5$,
$32+x=37,$
so the median is the $\frac{37+1}{2}=19\text{th}$ value.

For $x=5$, the students who read 3 books occupy positions
$14 \text{ to } 18,$
and the students who read 4 books occupy positions
$19 \text{ to } 26.$
Hence the median is $4$.

If $x=6$, the 19th value would be $3$, so the median would no longer be $4$. Therefore the largest possible value is
$x=5.$