Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
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To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.
With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.
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Problem 8)
The answer is choice B) 5
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Work Shown:
abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10
101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10
101 base 2 = (1*4 + 0*2 + 1*1) base 10
101 base 2 = (4 + 0 + 1) base 10
101 base 2 = 5 base 10
