Anthony recently completed a trip in 3 hours. If his average speed had been 20 percent faster, how long would the same trip have taken ?

Options

A. `2.2` hours
B. `2.4` hours
C. `2.5` hours
D. `2` hours

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Define the Variables
Let:
- \( S_1 \) = original speed
- \( D \) = distance of the trip
- \( T_1 \) = time taken for the trip = 3 hours

### Step 2: Calculate the Original Speed
Using the formula for speed:
\[ S_1 = \frac{D}{T_1} = \frac{D}{3} \]

### Step 3: Determine the Increased Speed
If Anthony's speed had been 20% faster, his new speed \( S_2 \) can be calculated as:
\[ S_2 = S_1 + 0.2 \times S_1 = 1.2 \times S_1 \]
Substituting the value of \( S_1 \):
\[ S_2 = 1.2 \times \frac{D}{3} = \frac{1.2D}{3} \]

### Step 4: Set Up the Equation for Time with Increased Speed
Using the formula for speed again, we can express the time taken with the new speed \( S_2 \) as:
\[ S_2 = \frac{D}{T_2} \]
Where \( T_2 \) is the new time taken. Setting the two expressions for \( S_2 \) equal gives:
\[ \frac{1.2D}{3} = \frac{D}{T_2} \]

### Step 5: Solve for \( T_2 \)
To find \( T_2 \), we can cross-multiply:
\[ 1.2D \cdot T_2 = 3D \]
Dividing both sides by \( D \) (assuming \( D \neq 0 \)):
\[ 1.2T_2 = 3 \]
Now, divide both sides by 1.2:
\[ T_2 = \frac{3}{1.2} = 2.5 \text{ hours} \]

### Conclusion
Thus, if Anthony's average speed had been 20% faster, the trip would have taken him **2.5 hours**.

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