Sam's daily morning routine involves getting breakfast before going to work. He travels by car.
On a particular day, Sam leaves home at 8:00 am and arrives at the breakfast place at 8:15 am.
He leaves the breakfast place at 8:30 am. He arrives at his office at 9:00 am. Sam's average speed for the whole journey is 42 km/h.
His average speed from home to the breakfast place is 10 km/h faster than his average speed from the breakfast place to the office.
Find the distance between the breakfast place and Sam's office. Answer ...km

1. Convert the journey times into hours and find the total distance for the whole journey.

From home to breakfast place: $15$ minutes $=\dfrac{1}{4}$ h.

From breakfast place to office: $30$ minutes $=\dfrac{1}{2}$ h.

From 8:00 am to 9:00 am, total time $=1$ h.

Average speed for the whole journey $=42$ km/h, so
$\text{total distance}=42\times 1=42\text{ km}.$

2. Form an equation using the two parts of the journey.

Let the average speed from the breakfast place to the office be $x$ km/h.

Then the average speed from home to the breakfast place is $(x+10)$ km/h.

Distance from home to breakfast place:
$\dfrac{1}{4}(x+10)$

Distance from breakfast place to office:
$\dfrac{1}{2}x$

Since the total distance is $42$ km,
$\dfrac{1}{4}(x+10)+\dfrac{1}{2}x=42.$

3. Solve the linear equation for the speed from the breakfast place to the office.

$\dfrac{1}{4}(x+10)+\dfrac{1}{2}x=42$

Multiply by $4$:
$x+10+2x=168$

$3x=158$

$x=\dfrac{158}{3}.$

4. Find the required distance from the breakfast place to the office.

Distance from breakfast place to office:
$\dfrac{1}{2}\times \dfrac{158}{3}=\dfrac{79}{3}=26\dfrac{1}{3}.$