Question

Dave walked to his friend's house at a rate of $4\text{mph}$ and returned back biking at a rate of $10\text{mph}$. If it took him $18\text{minutes}$ longer to walk than to bike, what was the total distance of the round trip?

Answer 1

Find the total distance of the round trip.

Given that, their walking speed Dave to his friend's house is $4\text{mph}$ and his biking speed from his friend's house is $10\text{mph}$.

Time taken for walking is $18\text{minutes}$ longer than for biking.

Let $x$ be time taken for biking, so time taken for walking is $x+18$.

The distance covered by biking is $10x$ and the distance covered by walking is $4\left(x+18\right)$.

From the given information:

$\begin{array}{rcl}10x& =& 4\left(x+18\right)\\ & \Rightarrow & 10x=4x+72\\ & \Rightarrow & 6x=72\\ & \Rightarrow & x=12\end{array}$

Time taken for biking is $12\text{minutes}$, so the time taken for walking is $(x+18)=30\text{minutes}$.

Change the obtained time from minutes to hours.

$$\begin{gathered} 12\text{minutes}=0.2\text{hours} \\ 30\text{minutes}=0.5\text{hours} \end{gathered}$$

Find the total distance covered in the round trip:

$\begin{array}{rcl}\text{Total distance}& =& \text{walking distance}+\text{biking distance}\\ & =& \left(4\times 0.5\right)+\left(10\times 0.2\right)\\ & =& 2+2\\ & =& 4\mathrm{miles}\end{array}$

Therefore, the total distance of the round trip is $4\text{miles}$.