SOLUTION
$\mathbf{E}$

We note first that if $b$ is the mean of the numbers $a$ and $c$, then $b=\frac{1}{2}(a+c)$. Hence $2 b=a+c$ and therefore $b-a=c-b$. In other words, the numbers $a, b, c$ form an arithmetic sequence. That is, their differences are equal.
We deduce that, because each number in the grid, other than 10 and 25 , is the mean of the two adjacent numbers, the six numbers in the grid form an arithmetic sequence. Suppose that their common difference is $d$. Then the numbers in the grid will form the sequence $10,10+d, 10+2 d, 10+3 d, 10+4 d, 10+5 d$.

It follows that $10+5 d=25$. Hence $d=3$.
The number which is in the square marked $X$ is $10+3 d$. Because $d=3$, this number is 19 .
11. In the grid below each of the blank squares and the square marked $X$ are to be filled by the mean of the two numbers in its adjacent squares.
Which number should go in the square marked $X$ ?

A 15
B 16
C 17
D 18
E 19