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?

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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?

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Answer 1 · 2017-09-05T14:50:49

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, $\frac{x}{55} - \frac{x}{60} = \frac{1}{3}$ $x/55-x/60 = 1/3$

$\Rightarrow \frac{12 x - 11 x}{5.11 .12} = \frac{1}{3}$ $rArr (12x-11x)/(5.11.12) = 1/3$

$\Rightarrow \frac{x}{5.11 .12} = \frac{1}{3}$ $rArr x/(5.11.12) = 1/3$

$\Rightarrow x = \frac{1}{3} . 5.11 .12$ $rArr x = 1/3. 5.11.12$

$\Rightarrow x = 220$ $rArr x = 220$

Answer 2 · 2017-09-05T14:55:35

See a solution process below:

Explanation:

The formula for finding distance traveled is:

$d = s \times t$ $d = s xx t$

Where:

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

$s$ $s$ is the average speed traveled:

$60 mph$ $60"mph"$ on the way there
$55 mph$ $55"mph"$ on the way back

$t$ $t$ is the time travel.

We can write an equation for the trip out as:

$d = \frac{60 mi}{hr} \times t$ $d = (60"mi")/"hr" xx t$

We can write an equation for the trip back as:

$d = \frac{55 mi}{hr} \times (t + \frac{1}{3} hr)$ $d = (55"mi")/"hr" xx (t + 1/3"hr")$

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

$\frac{60 mi}{hr} \times t = \frac{55 mi}{hr} \times (t + \frac{1}{3} hr)$ $(60"mi")/"hr" xx t = (55"mi")/"hr" xx (t + 1/3"hr")$

$\frac{60 t mi}{hr} = (\frac{55 mi}{hr} \times t) + (\frac{55 mi}{hr} \times \frac{1}{3} hr)$ $(60t"mi")/"hr" = ((55"mi")/"hr" xx t) + ((55"mi")/"hr" xx 1/3"hr")$

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

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

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

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

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

$color(red)("hr")/color(blue)(5"mi") xx (5t"mi")/"hr" = color(red)("hr")/color(blue)(5"mi") xx (55"mi")/3$

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

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

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

$t = 11/3"hr"$

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 · 2017-09-05T16:21:35

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

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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?

3 Answers

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