Let's analyze the problem step by step.

Step 1: Define the initial values

• Let the initial speed be $S$.
• Let the initial time be $T$.

Step 2: Calculate the initial distance

• The initial distance $D_1$ is given by: $D_1 = S \times T$

Step 3: Determine the new speed and time

• The speed is increased by $\frac{100}{3} \%$, so the new speed $S'$ is: $S' = S \times \left(1 + \frac{100}{3} \times \frac{1}{100}\right) = S \times \left(1 + \frac{1}{3}\right) = \frac{4S}{3}$

• The time is decreased by 25%, so the new time $T'$ is: $T' = T \times \left(1 - \frac{25}{100}\right) = T \times 0.75 = \frac{3T}{4}$

Step 4: Calculate the new distance

• The new distance $D_2$ is given by: $D_2 = S' \times T' = \frac{4S}{3} \times \frac{3T}{4} = S \times T = D_1$

Step 5: Determine the percentage change in distance

Since $D_2 = D_1$, there is no change in the distance.

Therefore, the correct answer is:

• (d) None of these

Would you like more details or have any questions? Here are 5 related questions you might find interesting:

1. How would the distance change if the speed increased by 50% and the time decreased by 50%?
2. What is the formula for calculating percentage change in general?
3. If the speed and time both increased by 25%, what would be the percentage change in distance?
4. How does the distance traveled relate to speed and time graphically?
5. Can you explain how to solve similar problems using proportional reasoning?

Tip: When analyzing problems involving percentage changes, always ensure to apply the percentage change to the correct base quantity to avoid calculation errors.