SAT Math: 1

Short way:

Realize that $time=\frac{distance}{speed}$

So,

$\frac{Xmiles}{5mph}+\frac{Xmiles}{3mph}=1hour$

Thus, solving for the distance ( $X$ ),

$\frac{8Xmiles}{15mph}=1h$ (get the common denominator)

$X=\frac{15}{8}miles$ (multiply both sides by $\frac{15mph}{8miles}$ )

But $X$ is the one-way distance. So for back and forth,

$Distance_{total}=2X=\frac{15}{4}miles=\boxed{3\frac{3}{4}miles}$

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Long way:

Three key ingredients for cooking this problem:

Speed, Distance, and Time.

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From basic mechanics, we know that

$speed=\frac{distance}{time}$ , i.e. $s=\frac{d}{t}$ .

From the problem given, we know that:

Graham's total travel time was 1 hour. On his trip to school, he traveled at an average speed of 3mph.

On his way back, he traveled at an average speed of 5mph.

His total travel time was 1 hour.

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Putting all of this together mathematically, we get the following:

$\begin{cases} t_{ school }+t_{ back }=1_{ hr } \\ t_{school}=\frac{d}{s_{school}} \\ t_{back}=\frac{d}{s_{back}} \end{cases}$

Adding 2nd and 3rd equations together, we get

$t_{ school }+t_{ back }=\frac { d }{ s_{ school } } +\frac { d }{ s_{ back } }$

Combining the right side, we get

$t_{ school }+t_{ back }=\frac { d(s_{school}+s_{back}) }{ s_{ school }s_{back} }$

Now all we have to do is plug in the values we already know:

$t_{ school }+t_{ back }=1hr$ , $s_{school}=3mph$ , $s_{back}=5mph$ .

And so,

$t_{ school }+t_{ back }=\frac { d(s_{school}+s_{back}) }{ s_{ school }s_{back} }$

$\Rightarrow 1hr=\frac { d(3mph+5mph) }{ 3mph5mph}$

$\Rightarrow 1hr=\frac{d(8mph)}{15(mph)^2}$

Note that mph stands for "miles per hour," i.e. $\frac{miles}{hour}=\frac{mi}{hr}$ .

Thus,

$1hr=\frac{d(8\frac{mi}{hr})}{15\frac{mi^2}{hr^2}}$

$\Rightarrow 1hr=\frac{d(8)}{15\frac{mi}{hr}}$

$\Rightarrow 1hr(15\frac{mi}{hr})=8d$

$\Rightarrow 15mi=8d$

$\Rightarrow d=\frac{15mi}{8}$ .

But we're not done yet. Remember from the original system we've set up, the d represented the distance it took Graham to travel from one place to another. Since this distance is the same to and fro, d is the same to school and back. Thus, d denotes HALF the distance.

Hence,

$d=\frac{1}{2}d_{total}=\frac{15mi}{8}$ ,

$d_{total}=\frac{30mi}{8}=\frac{15mi}{4}=\boxed{3\frac{3}{4}mi}$ .

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NOTE: Alternative solutions exist.

total time = t1 + t2 = s/3 + s/5 s= 3.75

let d be the distance of the whole route
d/2 be half the distance of the whole route
s=d/t
t=d/s
(d/2)/3 + (d/2)/5 = 1
d/6 + d/10 = 1
(5d + 3d)/30 = 1
8d/30 = 1
8d=30
d=3.75 or 3 3/4