The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 4?
A. a
B. b
C. c
D. d
E. e

(C) Think of the polygon as you would a wheel being rotated at its center. A wheel sweeps a 360º angle everytime it returns to its initial positon. Since this is a regular pentagon, we can increment our rotations by dividing 360/5 = 72.

So everytime the center spins 72º clockwise, the polygon moves by one vertex. So, when n = 4 the pentagon rotates by 4 vertices and the initial b position is replaced by c or answer choice (C).

An alternative way to consider the problem is to realize that everytime n = 5, it returns to its original configuration. So, when n = 4, the position is occupied by the vertex that is one vertex away in the clockwise direction because it lags by one 72º turn.
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"Since this is a regular pentagon, we can increment our rotations by
dividing 360/5 = 72. So everytime the center spins 72º clockwise, the polygon moves by one vertex."
Doesn't pentagon has 540 degrees?

The sum of internal pentagon angles is 540 degrees indeed, but 360 degrees - is "full rotation angle" when a point on a circle returns to its own position.

So in the statement
"Since this is a regular pentagon, we can increment our rotations by dividing 360/5 = 72. So everytime the center spins 72º clockwise, the polygon moves by one vertex."
we note and use the statement that it is a REGULAR pentagon. I'd like to bring your attention to the term "REGULAR", because regular polygons can be inscribed in a circle and have vertexes "evenly distributed" on this circle.



Following the same logic it will be true for any REGULAR polygon of n sides that by turning (360/n) degrees it rotates by one vertex.

The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 19?
A. a
B. b
C. c
D. d
E. e

Right Answer is (C).
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Since the rotation is clockwise and the position b is 72 degrees short of returning to its original configuration, shouldn't its position be at a? (c would replace b at 21n)

Imagine that we rotated the pentagon 20 × 72º clockwise. That makes it look the same:



If we rotate it (1 × 72) more degrees clockwise then it will look:



We see that in this case (n = 21) point a stands at original position of point b.

If we rotate (1 × 72) less degrees clockwise than (20 × 72º) clockwise turn then in this case it will look:



For better understanding, please, take a look at this animation:

The operation above rotates the regular pentagon clockwise by (72n) degrees from its center. What replaces the initial b position when n = 4?
A. a
B. b
C. c
D. d
E. e
The right answer is choice (C). (See explanation above in this topic).
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This answer is incorrect, if the polygon turns "clockwise" it should be turning from "b" to "c" in the first 72 degrees. Another 72 degrees 3 times would then leave the original point in upper most angle. The correct answer should be "a".

You're right about what place the point b itself will be at when n = 4, but the question asks about what point will be at original b-point position when n = 4.

So when point b is at original a-point position then point c is at original b-point position. The answer is choice (C).

When n = 4 the situation is following: