Five points on a circle are numbered 1,2,3,4, and 5 in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point 5, after 1995 jumps it will be on point 
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }$
Let us see how the bug moves.
First, we see that if it starts at point 5, it moves to point 1.
At point 1, it moves to point 2.
At point 2, since 2 is even, it moves to point 4.
Then at point 4, it moves to point 1.
We can see that at this point, the bug will cycle between 1, 2, and 4
More specifically, we can see that all numbers congruent to 0 (mod 3) will have the bug on point 4 on that step number.
Thus, we can conclude that the answer is $\fbox{\text{(D)}}$