A train covers 360 km at a uniform speed. If the speed had been 10 km/h more. It would have taken 3 hours less for the same journey. What is the speed of the train (in km/h)?

Options

A. 40
B. 25
C. 30
D. 50

Answer

To solve the problem, we can use the relationship between distance, speed, and time. The formula we will use is:

\[ \text{Distance} = \text{Speed} \times \text{Time} \]

Let the speed of the train be \( x \) km/h.

### Step 1: Write the equation for the time taken at the original speed.
The time taken to cover 360 km at speed \( x \) is given by:

\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{360}{x} \]

### Step 2: Write the equation for the time taken at the increased speed.
If the speed is increased by 10 km/h, the new speed becomes \( x + 10 \) km/h. The time taken to cover the same distance at this new speed is:

\[ \text{Time} = \frac{360}{x + 10} \]

### Step 3: Set up the equation based on the information given.
According to the problem, the time taken at the original speed is 3 hours more than the time taken at the increased speed. Thus, we can set up the equation:

\[ \frac{360}{x} = \frac{360}{x + 10} + 3 \]

### Step 4: Solve the equation.
To eliminate the fractions, we can multiply through by \( x(x + 10) \):

\[ 360(x + 10) = 360x + 3x(x + 10) \]

Expanding both sides gives:

\[ 360x + 3600 = 360x + 3x^2 + 30x \]

Now, simplify the equation:

\[ 3600 = 3x^2 + 30x \]

Rearranging gives:

\[ 3x^2 + 30x - 3600 = 0 \]

### Step 5: Simplify the quadratic equation.
Divide the entire equation by 3:

\[ x^2 + 10x - 1200 = 0 \]

### Step 6: Factor the quadratic equation.
We need to find two numbers that multiply to -1200 and add to 10. The numbers are 40 and -30. Thus, we can factor the equation as:

\[ (x + 40)(x - 30) = 0 \]

### Step 7: Solve for \( x \).
Setting each factor to zero gives:

\[ x + 40 = 0 \quad \Rightarrow \quad x = -40 \quad (\text{not valid since speed cannot be negative}) \]
\[ x - 30 = 0 \quad \Rightarrow \quad x = 30 \]

### Conclusion:
The speed of the train is \( 30 \) km/h.

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