Question

To reduce the usual time taken to cover a distance of 300 km by 1 hour, the speed of a bus is increased by 10 km/h. The usual speed of the bus (in km/h) is:

• A. 62
• B. 60
• C. 55
• D. 50

Hint:

Time difference equation: \(\frac{D}{v} - \frac{D}{v+10} = 1\)

Answer 1

The Correct Option is D

Step 1: Define variables.
Let usual speed = \(v\) km/h.
Usual time = \(\frac{300}{v}\) hours. Step 2: Define new speed and time.
New speed = \(v + 10\) km/h.
New time = \(\frac{300}{v+10}\) hours. Step 3: Set up the time difference equation.
Usual time − New time = 1 hour.
\(\frac{300}{v} - \frac{300}{v+10} = 1\). Step 4: Solve the equation.
Multiply through by \(v(v+10)\):
\(300(v+10) - 300v = v(v+10)\).
\(300v + 3000 - 300v = v^2 + 10v\).
\(3000 = v^2 + 10v\).
\(v^2 + 10v - 3000 = 0\). Step 5: Solve the quadratic.
Discriminant = \(10^2 - 4(1)(-3000) = 100 + 12000 = 12100\).
\(\sqrt{12100} = 110\).
\(v = \frac{-10 \pm 110}{2}\).
Positive root: \(v = \frac{100}{2} = 50\).
Thus, usual speed = 50 km/h.