# On a trip from Detroit to Columbus, Ohio, Mrs. Smith drove at an average speed of 60 MPH. Returning, her average speed was 55MPH. If it took her ⅓ hour longer on the return trip, how far is it from Detroit to Columbus?

Sep 5, 2017

220 miles

#### Explanation:

Let the distance be x Miles

From Detroit to Columbus,Ohio, she took x/60 hrs

And while returning she took x/55 hours.

Now as per question, $\frac{x}{55} - \frac{x}{60} = \frac{1}{3}$

$\Rightarrow \frac{12 x - 11 x}{5.11 .12} = \frac{1}{3}$

$\Rightarrow \frac{x}{5.11 .12} = \frac{1}{3}$

$\Rightarrow x = \frac{1}{3.} 5.11 .12$

$\Rightarrow x = 220$

Sep 5, 2017

See a solution process below:

#### Explanation:

The formula for finding distance traveled is:

$d = s \times t$

Where:

$d$ is the distance traveled, what we are solving for.

$s$ is the average speed traveled:

• $60 \text{mph}$ on the way there
• $55 \text{mph}$ on the way back

$t$ is the time travel.

We can write an equation for the trip out as:

d = (60"mi")/"hr" xx t

We can write an equation for the trip back as:

$d = \left(55 \text{mi")/"hr" xx (t + 1/3"hr}\right)$

Because the distance both ways was the same we can now equate the right side of each equation and solve for $t$:

$\left(60 \text{mi")/"hr" xx t = (55"mi")/"hr" xx (t + 1/3"hr}\right)$

$\left(60 t \text{mi")/"hr" = ((55"mi")/"hr" xx t) + ((55"mi")/"hr" xx 1/3"hr}\right)$

(60t"mi")/"hr" = ((55"mi")/"hr" xx t) + ((55"mi")/color(red)(cancel(color(black)("hr"))) xx 1/3color(red)(cancel(color(black)("hr"))))

$\frac{60 t \text{mi")/"hr" = (55t"mi")/"hr" + (55"mi}}{3}$

$\frac{60 t \text{mi")/"hr" - color(red)((55t"mi")/"hr") = (55t"mi")/"hr" - color(red)((55t"mi")/"hr") + (55"mi}}{3}$

$\left(60 - 55\right) \frac{t \text{mi")/"hr" = 0 + (55"mi}}{3}$

$\frac{5 t \text{mi")/"hr" = (55"mi}}{3}$

$\frac{\textcolor{red}{\text{hr")/color(blue)(5"mi") xx (5t"mi")/"hr" = color(red)("hr")/color(blue)(5"mi") xx (55"mi}}}{3}$

cancel(color(red)("hr"))/color(blue)(color(black)(cancel(color(blue)(5)))color(black)(cancel(color(blue)("mi")))) xx (color(blue)(cancel(color(black)(5)))tcolor(blue)(cancel(color(black)("mi"))))/color(red)(cancel(color(black)("hr"))) = color(red)("hr")/color(blue)(5color(black)(cancel(color(blue)("mi")))) xx (55color(blue)(cancel(color(black)("mi"))))/3

$t = \frac{55 \textcolor{red}{\text{hr}}}{\textcolor{b l u e}{5} \times 3}$

$t = \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{55}}} 11 \textcolor{red}{\text{hr}}}{\cancel{\textcolor{b l u e}{5}} \times 3}$

$t = \frac{11}{3} \text{hr}$

Now, substitute $\frac{11}{3} \text{hr}$ for $t$ in the first equation and calculate the distance traveled:

d = (60"mi")/"hr" xx t becomes:

d = (60"mi")/"hr" xx 11/3"hr"

d = (color(blue)(cancel(color(black)(60)))20"mi")/color(red)(cancel(color(black)("hr"))) xx 11/color(blue)(cancel(color(black)(3)))color(red)(cancel(color(black)("hr")))

$d = 20 \text{mi" xx 11}$

$d = 220 \text{mi}$

Sep 5, 2017

242 miles

#### Explanation:

Distance is speed x time

The journey out is the same distance as the journey back

Set the distance as $d$ miles

Set the time out as $t$ hours

So journey out we have $d = t \times 60 m p h \text{ } \ldots \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

So journey back we have $d = \left(t + \frac{1}{3}\right) \times 55 m p h \text{ } E q u a t i o n \left(2\right)$

Equating $E q n \left(1\right) \text{ to "Eqn(2) " through } d$

$60 t = d = \left(t + \frac{1}{3}\right) 55$

$60 t = 55 t + \frac{55}{3}$

Subtract $55 t$ from both sides

$5 t = \frac{55}{3}$

Divide both sides by 5

$t = \frac{55}{15} \text{ hours }$

$t = \frac{55 \div 5}{15 \div 5} = \frac{11}{3} \text{ hours} \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(3\right)$

Using $E q n \left(3\right)$ substitute for $t$ in $E q n \left(1\right)$

$d = \frac{11}{3} \times 66$

$d = 11 \times 22$

$d = 242$ miles