Q. 4.42 - Intermediate Algebra

Question

Translate to a system of equations and then solve:

Charlie left his mother’s house traveling at an average speed of $36$miles per hour. His sister Sally left $15$minutes $(1/4 hour)$later traveling the same route at an average speed of miles per hour. How long before Sally catches

up to Charlie?

Step-by-Step Solution

Answer

Sally catches Charlie in $2\frac{1}{2}$or $2.5$hours.

Step 1. Given

Charlie and Sally are travelling on the same route after leaving mother's home.
Charlie left home at a speed of $36$mph. Fifteen minutes later, Sally left home at a speed of $42$mph on the same route.

Step 2. Finding the distance.

Assuming the time of Charlie and Sally as $c\&s$, we will find the distance for them.
As we know that ,we will use the same formula.

1. Charlie's rate is $36 mph$and the time is $c$, so the distance comes out to be $36c$.
2. Sally's rate is $42 mph$and the time is $s$, so the distance comes out to be $42s$.

And as we know that Sally left home fifteen minutes later than Charlie, so Charlie's time will be fifteen minutes more than that of Sally. 

Step 3. Substituting the values.

Charlie's time will be fifteen minutes more. This can be written as $c=s+\frac{1}{4}$.

To get a system of equations, we must recognize that both will drive the same distance. So this gives us $36c=42s$.

We have two equations, i.e.,
$$\begin{gathered} 36c=42s \\ c=s+\frac{1}{4} \end{gathered}$$

Substituting value of $c$ in first equation will give us $36(s+\frac{1}{4})=42s$.

Step 4. Finding the values.

We have $36(s+\frac{1}{4})=42s$.

Solving the equation,

$$\begin{gathered} 36s+9=42s \\ 6s=9 \\ s=\frac{3}{2} or 1\frac{1}{2} or 1.5 hours \end{gathered}$$
This shows that Sally will take $1\frac{1}{2}$or $1.5$ hours to catch Charlie.

Substituting this value in $c=s+\frac{1}{4}$will give time travelled by Charlie. So,

$$\begin{gathered} c=s+\frac{1}{4} \\ c=\frac{3}{2}+\frac{1}{4} \\ c=\frac{7}{4} \end{gathered}$$

This can also be written as  $1\frac{3}{4}$hours.

Step 5. Check the solution

We need to check the distance travelled by both Charlie and Sally. If the distance comes out to be the same, our answer is right. 

So,

1. Charlie has travelled $1\frac{3}{4}$hours at speed of $36 mph$, distance $=\frac{7}{4}\times 36$ miles
   $=63$ miles.
2. Sally has travelled $1\frac{1}{2}$ hours at speed of $42$ mph, so distance $=\frac{3}{2}\times 42$ miles
   $=63$ miles.

The distance covered are same, i.e., our answers are right.