Question

Thank you

Answer 1

S=Samantha's time
L=Logan's time
T=total time
$$
T=45=S+L\\
S=\cfrac{L}{3}
$$
Perhaps the tricky part is that Samantha is 3 times faster than Logan means she can do it in 1/3 the time that Logan can.

These are two equations and two unknowns. Substitute second equation into the first and solve for L, then plug back into second to solve for S.

Answer 2

What does x mean to you?

Answer 3

yes

Answer 4

The average rate is 1 route in 45 mins, so using the harmonic mean.
Let s= Samantha's rate per route
and l=Logan's rate per route
$$
\cfrac{1}{T}=\cfrac{\cfrac{1}{s}\times\cfrac{1}{l}}{\cfrac{1}{s}+\cfrac{1}{l}}(\star)\\
=\cfrac{sl}{s+l}=\cfrac{S(3s)}{s+3s}=\cfrac{3s^2}{4s}=\cfrac{3s}{4}\\
s=\cfrac{4}{T\times3}=\cfrac{4}{45\times3}
$$
Plug S back in to \(\star\) to solve for Logan's rate.

Answer 5

Two equations, two unknowns:
We had,
$$
T=45=S+L\\
S=\cfrac{L}{3}\\
$$
The number of equations you need to solve any problem depends on the number of unknowns, which in this case is S and L. So, we need (ahd have) two equations.
There are various methods to solve, one way is to just substitute the second into the first:
$$
T=45=\cfrac{L}{3}+L=\cfrac{4L}{3}\\
\implies L=\cfrac{3T}{4}=\cfrac{3\times 45}{4}
$$

Answer 6

Plug this back into the second equation:
$$
S=\cfrac{3\times 45}{3\times4}=\cfrac{45}{4}
$$

Answer 7

The rates (rather than using my harmonic mean approach above) is simply the reciprocal of each:
$$
s=\cfrac{1}{S}\\
l=\cfrac{1}{L}\\
$$

Answer 8

*Correction to harmonic mean approach above:
$$
\large{
\cfrac{1}{T}=\cfrac{\cfrac{1}{s}\times\cfrac{1}{l}}{\cfrac{1}{s}+\cfrac{1}{l}}(\star)\\
=\cfrac{sl}{s+l}=\cfrac{3l(l)}{3l+l}=\cfrac{3l^2}{4l}=\cfrac{3l}{4}\\
l=\cfrac{4}{T\times3}=\cfrac{4}{45\times3}
}
$$

The reason you need to use harmonic mean to find the means for each is because we are dealing with rates: http://en.wikipedia.org/wiki/Mean#Harmonic_mean_.28HM.29

Answer 9

I think that for the work done, you know that S did 3/4 of the work in the 45 minutes and L only did 1/4 of the work. So S did 3/4 of the route and L did 1/4 of the route. S did 3 times more work than L.