Overtake Time Calculation

This problem involves calculating the time it takes for Chris to overtake Santa, considering their different walking speeds and Santa's head start.

Understanding Speeds and Head Start

• Let the distance to school be D units.
• Santa's time = 40 minutes. Santa's speed ($S_{Santa}$) = $\frac{D}{40}$ units/minute.
• Chris's time = 30 minutes. Chris's speed ($S_{Chris}$) = $\frac{D}{30}$ units/minute.
• Santa starts 5 minutes earlier than Chris.

Calculating Overtake Time

Let t be the time in minutes Chris travels until he overtakes Santa.

• In time t, Chris covers a distance: Distance$_{Chris}$ = $S_{Chris} \times t = \frac{D}{30} \times t$.
• Since Santa started 5 minutes earlier, Santa travels for $(t + 5)$ minutes.
• In time $(t + 5)$, Santa covers a distance: Distance$_{Santa}$ = $S_{Santa} \times (t + 5) = \frac{D}{40} \times (t + 5)$.
• Chris overtakes Santa when they have covered the same distance: Distance$_{Chris}$ = Distance$_{Santa}$.
• Set up the equation: $\frac{D}{30} \times t = \frac{D}{40} \times (t + 5)$.
• Divide both sides by D: $\frac{t}{30} = \frac{t + 5}{40}$.
• Multiply both sides by the least common multiple of 30 and 40, which is 120: $120 \times \frac{t}{30} = 120 \times \frac{t + 5}{40}$.
• Simplify: $4t = 3(t + 5)$.
• Expand the right side: $4t = 3t + 15$.
• Solve for t by subtracting 3t from both sides: $4t - 3t = 15$.
• Result: $t = 15$ minutes.

Therefore, Chris would overtake Santa in 15 minutes.