On a typical morning Aiden gets out of bed, goes through his morning preparation, rides the bus, and walks from the bus stop to work arriving at work 120 minutes after getting out of bed. One morning Aiden got out of bed late, so he rushed through his morning preparation getting onto the bus in half the usual time, the bus ride took 25 percent longer than usual, and he ran from the bus stop to work in half the usual time it takes him to walk arriving at work 96 minutes after he got out of bed. The next morning Aiden got out of bed extra early, leisurely went through his morning preparation taking 25 percent longer than usual to get onto the bus, his bus ride took 25 percent less time than usual, and he walked slowly from the bus stop to work taking 25 percent longer than usual. How many minutes after Aiden got out of bed did he arrive at work that day?

1. Let \( a \) be the time it takes Aiden for his morning preparation.
2. Let \( b \) be the time it takes Aiden to ride the bus.
3. Let \( c \) be the time it takes Aiden to walk from the bus stop to work.

From the problem, we have the following information:
- On a typical morning, Aiden arrives at work 120 minutes after getting out of bed:
  \[
  a + b + c = 120
  \]

- One morning, Aiden rushed through his morning preparation, taking half the usual time, the bus ride took 25% longer, and he ran from the bus stop to work in half the usual time, arriving at work 96 minutes after getting out of bed:
  \[
  0.5a + 1.25b + 0.5c = 96
  \]

  Doubling both sides of the second equation to eliminate the fractions:
  \[
  a + 2.5b + c = 192
  \]

- Subtracting the first equation from the modified second equation:
  \[
  (a + 2.5b + c) - (a + b + c) = 192 - 120
  \]
  \[
  1.5b = 72
  \]
  \[
  b = \frac{72}{1.5} = 48
  \]

- Substituting \( b = 48 \) back into the first equation:
  \[
  a + 48 + c = 120
  \]
  \[
  a + c = 72
  \]

- Substituting \( b = 48 \) into the second equation:
  \[
  0.5a + 1.25 \cdot 48 + 0.5c = 96
  \]
  \[
  0.5a + 60 + 0.5c = 96
  \]
  \[
  0.5a + 0.5c = 36
  \]
  \[
  a + c = 72
  \]

- The next morning, Aiden got out of bed extra early, taking 25% longer for his morning preparation, the bus ride took 25% less time, and he walked slowly from the bus stop to work taking 25% longer:
  \[
  1.25a + 0.75b + 1.25c
  \]

- Substituting \( a + c = 72 \) and \( b = 48 \):
  \[
  1.25a + 0.75 \cdot 48 + 1.25c
  \]
  \[
  1.25a + 36 + 1.25c
  \]
  \[
  1.25(a + c) + 36
  \]
  \[
  1.25 \cdot 72 + 36
  \]
  \[
  90 + 36 = 126
  \]

Thus, Aiden arrived at work 126 minutes after getting out of bed that day.

The final answer is \(\boxed{126}\)