Q 115 - Intermediate Algebra

Question

Felecia left her home to visit her daughter driving $45$ mph. Her husband waited for the dog sitter to arrive and left home twenty minutes $\left(1/3hour\right)$ later. He drove $55$ mph to catch up to Felecia. How long before he reaches her?

Step-by-Step Solution

Answer

Felecia's husband will take $1.5$ hours to catch up to Felecia.

Step 1. Given information.

Consider the given question,

Driving speed of Felecia is $45$ mph.

Driving speed of her husband is $55$ mph.

Assume $x$ and $y$ represents the lengths of the time of Felecia and her husband. Then,

Distance travelled by Felecia$=45x$

Distance travelled by Felecia's husband$=55y$

Step 2. Form the system of equations.

It can be said that Felecia's husband will have to travel the same distance to catch up to her. Then, $45x=55y$  ....... (i)

As Felecia's husband left $20$ minutes later than Felecia.

So,  $y=x-\frac{1}{3}$

Step 3. Substitute the equations and solve for x .

Substituting the value of $y$ in equation (i),

$$\begin{gathered} 45x=55\left(x-\frac{1}{3}\right) \\ 45x=55x-\frac{55}{3} \\ 55x-45x=\frac{55}{3} \\ x=\frac{55}{3\times 10} \\ x=\frac{11}{6} \end{gathered}$$

Step 4. Substitute the value of x and solve for y .

Substituting the value of $x$,

$$\begin{gathered} y=x-\frac{1}{3} \\ y=\frac{11}{6}-\frac{1}{3} \\ y=\frac{9}{6} \\ y=1.5 \end{gathered}$$

Step 5. Check the answer.

Substitute the value of $x$ in equation (i),

$$\begin{gathered} 45\times \left(\frac{11}{6}\right)=55y \\ \frac{45}{55}\times \left(\frac{11}{6}\right)=y \\ y=1.5 \end{gathered}$$

Therefore, it will take$1.5$ hours for Felecia's husband to catch up to her.