Note: The question states the time to go upstream is a number of times (not explicitly written) the time to go downstream. We'll assume a general number N
Answer:
[tex]\displaystyle v_b=\frac{N+1}{N-1}(4\ mph)[/tex]
Explanation:
Relative Speed
If a boat is going upstream against the water current, the true speed of motion is [tex]v_b-v_w[/tex], being [tex]v_b[/tex] the speed of the boat and [tex]v_w[/tex] the speed of the water. If the boat is going downstream, the true speed becomes [tex]v_b+v_w[/tex].
The question states the time to go upstream is a number of times N (not explicitly written) the time to go downstream. The speed of an object is computed as
[tex]\displaystyle v=\frac{x}{t}[/tex]
Where x is the distance traveled and t the time taken for that. The time can be computed by
[tex]\displaystyle t=\frac{x}{v}[/tex]
If [tex]t_u[/tex] is the time for the upstream travel and [tex]t_d[/tex] is the time for the downstream travel, then
[tex]t_u=Nt_d[/tex]
Siince the same distance x= 10 miles is traveled in both cases:
[tex]\displaystyle \frac{10}{v_b-v_w}=N\frac{10}{v_b+v_w}[/tex]
Simplifying and rearrangling
[tex]v_b+v_w=N(v_b-v_w)[/tex]
Operating
[tex]v_b+v_w=Nv_b-Nv_w[/tex]
Solving for [tex]v_b[/tex]
[tex]\displaystyle v_b=\frac{N+1}{N-1}v_w[/tex]
[tex]If\ N=2,\ v_w=4\ mph[/tex]
[tex]\displaystyle v_b=\frac{3}{1}(4)=12\ mph[/tex]
If N=3
[tex]\displaystyle v_b=\frac{4}{2}v_w=2(4)=8\ mph[/tex]
We can use the required value of N to compute the speed of the boat as explained
