The speed of A is 20 km/h more than the speed of B. The time that A takes to travel a distance of 300 km is 2 hours less than the time that B takes to travel a distance of 240 km. What is the speed (in km/h) of A?

Options

A. 30
B. 40
C. 45
D. 50

Answer

To solve the problem step-by-step, we will define the variables and set up the equations based on the information given.

### Step 1: Define the Variables
Let the speed of B be \( V_B \) km/h. Then the speed of A, which is 20 km/h more than B, can be expressed as:
\[ V_A = V_B + 20 \]

### Step 2: Set Up the Time Equations
The time taken by A to travel 300 km is given by the formula:
\[ \text{Time}_A = \frac{300}{V_A} \]
The time taken by B to travel 240 km is:
\[ \text{Time}_B = \frac{240}{V_B} \]
According to the problem, the time A takes is 2 hours less than the time B takes:
\[ \frac{300}{V_A} = \frac{240}{V_B} - 2 \]

### Step 3: Substitute \( V_A \) in the Equation
Substituting \( V_A \) from Step 1 into the time equation:
\[ \frac{300}{V_B + 20} = \frac{240}{V_B} - 2 \]

### Step 4: Clear the Denominators
To eliminate the fractions, we can multiply through by \( V_B(V_B + 20) \):
\[ 300V_B = 240(V_B + 20) - 2V_B(V_B + 20) \]

### Step 5: Expand and Rearrange the Equation
Expanding the right side:
\[ 300V_B = 240V_B + 4800 - 2V_B^2 - 40V_B \]
Combining like terms gives:
\[ 300V_B = 200V_B + 4800 - 2V_B^2 \]
Rearranging terms leads to:
\[ 2V_B^2 + 100V_B - 4800 = 0 \]

### Step 6: Simplify the Quadratic Equation
Dividing the entire equation by 2 simplifies it:
\[ V_B^2 + 50V_B - 2400 = 0 \]

### Step 7: Factor the Quadratic Equation
To factor the quadratic equation, we look for two numbers that multiply to -2400 and add to 50. The numbers are 60 and -40:
\[ (V_B + 60)(V_B - 40) = 0 \]

### Step 8: Solve for \( V_B \)
Setting each factor to zero gives:
\[ V_B + 60 = 0 \quad \text{or} \quad V_B - 40 = 0 \]
Thus,
\[ V_B = -60 \quad \text{(not possible since speed cannot be negative)} \quad \text{or} \quad V_B = 40 \]

### Step 9: Find \( V_A \)
Now substituting \( V_B = 40 \) back to find \( V_A \):
\[ V_A = V_B + 20 = 40 + 20 = 60 \text{ km/h} \]

### Final Answer
The speed of A is:
\[ \boxed{60} \text{ km/h} \]