Question

A person reaches his destination 40 minutes late if his speed is 3 km/hr and reaches 30 minutes before time if his speed is 4 km/hr. Then the distance of his destination from his starting point is:

• A. 20 km
• B. 14 km
• C. 24 km
• D. 12 km

Hint:

In time-speed-distance problems, when two speeds result in different times, use the time difference to form an equation and solve for the unknown distance.

Answer 1

The Correct Option is B

Let the distance to the destination be \( D \) km. Step 1: Time difference.
Time taken at 3 km/hr = \( \frac{D}{3} \) hours
Time taken at 4 km/hr = \( \frac{D}{4} \) hours
The difference in time = \( 40 + 30 = 70 \) minutes = \( \frac{7}{6} \) hours. So, \[ \frac{D}{3} - \frac{D}{4} = \frac{7}{6} \] Step 2: Solving the equation.
Take LCM of 3 and 4: \[ \frac{4D - 3D}{12} = \frac{7}{6} \quad \Rightarrow \quad \frac{D}{12} = \frac{7}{6} \] \[ D = \frac{7}{6} \times 12 = 14 \, \text{km} \] Thus, the distance is \( \boxed{14} \, \text{km.}\)