Brothers Santa and Chris walk to school from their house. The former takes 40 minutes while the latter, 30 minutes.One day Santa started 5 minutes earlier than Chris. In how many minutes would Chris overtake Santa?

Correct option is B

Given:

• Santa takes 40 minutes to reach school
• Chris takes 30 minutes to reach school
• Santa starts 5 minutes earlier than Chris
• We need to find when Chris will overtake Santa

Concept Used:
This is a relative speed problem where both start from the same point but at different times and speeds.

Formula Used:
$$\text{Time to overtake} = \frac{\text{Lead time} \times \text{Slower speed}}{\text{Faster speed} - \text{Slower speed}}$$​​

Solution:

Let’s assume the total distance to school is LCM of 40 and 30 = 120 units (to make calculations simple)

Santa's speed = 120 ÷ 40 = 3 units/min
Chris’s speed = 120 ÷ 30 = 4 units/min

Santa gets a 5-minute head start:
Distance covered by Santa in 5 minutes = 5 × 3 = 15 units

Now, Chris starts and starts closing the gap at a relative speed of:
4 – 3 = 1 unit/min

Time to overtake = 15 / 1 = 15 minutes

Correct Option: (B) 15 minutes