Question

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. What is the original speed?

• A. 40 km/h
• B. 45 km/h
• C. 50 km/h
• D. 60 km/h

Hint:

Time difference problems usually lead to rational equations.

Answer 1

The Correct Option is B

To solve the problem, we need to find the original speed of the train under the given conditions.

1. Let the original speed of the train be \(x\) km/h. The time taken to travel 360 km is: \[ t = \frac{360}{x} \text{ hours} \]
2. If the speed had been increased by 5 km/h, the speed would be \(x+5\) km/h and the time taken would be: \[ \frac{360}{x+5} \text{ hours} \]
3. The increased speed results in 1 hour less travel time: \[ \frac{360}{x} - \frac{360}{x+5} = 1 \]
4. Now we solve the equation to find the value of \(x\).

Solve: \[ \frac{360}{x} - \frac{360}{x+5} = 1 \] Take common denominator \(x(x+5)\):

Multiply both sides by \(x(x+5)\): \[ 360(x+5) - 360x = x(x+5) \]

Expand and simplify:

\[ 360x + 1800 - 360x = x^2 + 5x \] \[ 1800 = x^2 + 5x \]

Rearrange into quadratic form:

\[ x^2 + 5x - 1800 = 0 \]

Apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = 5\), and \(c = -1800\).

\[ x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-1800)}}{2} \] \[ x = \frac{-5 \pm \sqrt{25 + 7200}}{2} \] \[ x = \frac{-5 \pm \sqrt{7225}}{2} \] \[ x = \frac{-5 \pm 85}{2} \]

So the solutions are:

\[ x = \frac{80}{2} = 40 \quad \text{or} \quad x = \frac{-90}{2} = -45 \]

Since speed cannot be negative, the valid solution is: \[ x = 40 \text{ km/h} \]

Therefore, the original speed of the train is 40 km/h.