Consider the kangaroo's starting position as the origin of coordinate axes, with East and North being the positive x and y directions, respectively, and one metre being one unit along the axes.
We begin by considering the first quadrant. If the kangaroo's end point has coordinates $(a,b)$, then a and b must be integers. Also, after $10$ jumps, it must be that $a+b \leq 10$. Hence his end points are bounded by the right-angled triangle with vertices at $(10,0)$, $(0,10)$ and $(0,0)$.
He can finish at any point on the hypotenuse of the triangle since all these points satisfy $a+b=10$ and so can be reached by a jumps East and b jumps North. But he can only end up at a point $(a,b)$ on the other two edges or inside the triangle if $a+b$ is even. (He can certainly reach all such points in $a+b\leq 10$ jumps and if $a+b$ is even, with $a+b< 10$, he can jump away and back again using up $2$ jumps, and can repeat this until he has made $10$ jumps, and so end up at $(a,b)$.)
By symmetry we see that the possible end points form a square of side $11$, and so there are $121$ of them, as shown in the diagram.
This problem is taken from the UKMT Mathematical Challenges.