Solving for Original Car Time After Speed Increase

This problem involves calculating the original time taken by a car to cover a distance when its speed is increased by a certain percentage, resulting in a shorter travel time.

Mathematical Setup

• Let the original speed of the car be $S$.
• Let the original time taken be $T$ (in minutes).
• The distance $D$ is constant. We know that Distance = Speed $\times$ Time, so $D = S \times T$.

Calculating New Speed and Time

• The speed is increased by 30%. The new speed, $S'$, is $S + 0.30S = 1.30S$.
• The time taken is reduced by 24 minutes. The new time, $T'$, is $T - 24$ minutes.
• Since the distance remains the same: $D = S' \times T'$.

Equating Distances and Solving for T

• Using the distance formula for both scenarios: $S \times T = (1.30S) \times (T - 24)$.
• Divide both sides by $S$ (assuming $S \neq 0$): $T = 1.30 \times (T - 24)$.
• Distribute 1.30: $T = 1.30T - (1.30 \times 24)$.
• Calculate $1.30 \times 24$: $1.30 \times 24 = 31.2$.
• The equation becomes: $T = 1.30T - 31.2$.
• Rearrange the terms to solve for $T$: $1.30T - T = 31.2$.
• Simplify: $0.30T = 31.2$.
• Calculate $T$: $T = \frac{31.2}{0.30} = \frac{312}{3} = 104$ minutes.

Converting Time to Hours and Minutes

• The original time taken is 104 minutes.
• To convert this to hours and minutes: $104 \text{ minutes} = 60 \text{ minutes} + 44 \text{ minutes} = 1 \text{ hour} + 44 \text{ minutes}$.

Therefore, the time taken by the car to cover the same distance at its usual speed is 1 hour 44 minutes.