Problem 1
Answer: C) 5
This is through point B. We have AB+BC = 2+3 = 5
Another path is AE+EB+BC = 4+1+3 = 8, which is longer than 5 found earlier.
Another path is AE+ED+DC = 4+3+2 = 9, which is longer as well.
Another path is AE+ED+DB+BC = 4+3+5+3 = 19, which is longer as well.
============================================
Problem 2
Answer: A) 6
This route is
AC+CF+FE+ED = 1+2+1+2 = 6
which is the shortest route
A longer route would be
AC+CF+FD = 1+2+4 = 7
Another longer route is
AB+BE+ED = 3+4+2 = 9
There are other longer routes as well.
============================================
Problem 3
Answer: B) 7
Start at point A. There are two paths to take. Take the shortest one AE = 1
Then from there repeat the last step: take the shortest path
Keep this up until you reach point G
AE+ED+DF+FG = 1+2+1+3 = 7
--------
Let's go back to point A. If we go to E, then we travel 1 unit so far. If we go for EH =8 instead of ED = 2, then we have 1+8 = 9 units traveled so far, but that exceeds 7 we found earlier. So we can rule this path out. This is one way you can use process of elimination to find the shortest path.
Similarly, go from A to B to C and we travel 2+5 = 7 units, which is equal to the answer, but we haven't reached G yet. So we can rule out any path that has ABC as the start.
The path ABDFG is not the shortest because AB+BD+DF+FG = 2+3+1+3 = 9, which is not shorter than 7.