Question

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?

Answer 1

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, `x/55-x/60 = 1/3`

`rArr (12x-11x)/(5.11.12) = 1/3`

`rArr x/(5.11.12) = 1/3`

`rArr x = 1/3. 5.11.12`

`rArr x = 220`

Answer 2

See a solution process below:

Explanation:

The formula for finding distance traveled is:

`d = s xx t`

Where:

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

`s` is the average speed traveled:

• `60"mph"` on the way there
• `55"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 = (55"mi")/"hr" xx (t + 1/3"hr")`

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

`(60"mi")/"hr" xx t = (55"mi")/"hr" xx (t + 1/3"hr")`

`(60t"mi")/"hr" = ((55"mi")/"hr" xx t) + ((55"mi")/"hr" xx 1/3"hr")`

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

`(60t"mi")/"hr" = (55t"mi")/"hr" + (55"mi")/3`

`(60t"mi")/"hr" - color(red)((55t"mi")/"hr") = (55t"mi")/"hr" - color(red)((55t"mi")/"hr") + (55"mi")/3`

`(60 - 55)(t"mi")/"hr" = 0 + (55"mi")/3`

`(5t"mi")/"hr" = (55"mi")/3`

`color(red)("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 = (55color(red)("hr"))/(color(blue)(5) xx 3)`

`t = (color(blue)(cancel(color(black)(55)))11color(red)("hr"))/(cancel(color(blue)(5)) xx 3)`

`t = 11/3"hr"`

Now, substitute `11/3"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"mi" xx 11"`

`d = 220"mi"`

Answer 3

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=txx 60 mph" "..............Equation(1)`

So journey back we have `d=(t+1/3)xx55mph" "Equation(2)`

Equating `Eqn(1)" to "Eqn(2) " through "d`

`60t=d=(t+1/3)55`

`60t=55t+55/3`

Subtract `55t` from both sides

`5t=55/3`

Divide both sides by 5

`t=55/15" hours "`

`t= (55-:5)/(15-:5) = 11/3" hours".................Equation(3)`

Using `Eqn(3)` substitute for `t` in `Eqn(1)`

`d=11/3xx66`

`d=11xx22`

`d=242` miles