Question

You ride your bike to campus a distance of 6 miles and return home on the same route. Going to​ campus, you ride mostly downhill and average 5 miles per hour faster than on your return trip home. If the round trip takesnbsp two hours and nbsp10 minuteslong dashthat ​is, StartFraction 13 Over 6 EndFraction hourslong dashwhat is your average rate on the return​ trip?

Answer 1

Answer:

1.98 miles per hour.

Step-by-step explanation:

Let x miles per hour be the average rate on the return​ trip,

∵ The speed on round trip is 5 miles per hour faster than on return trip,

So, the speed in round trip = (x+5) miles per hour,

Now, distance in one sided trip = 6 miles,

Since,

$Speed = \frac{Distance}{Time}\implies Time = \frac{Distance}{Speed}$

According to the question,

Time taken in return trip - Time taken in round trip = $\frac{13}{6}$ hours,

$\implies \frac{6}{x}-\frac{6}{x+5}=\frac{13}{6}$

$\frac{6x+30-6x}{x(x+5)}=\frac{13}{6}$

$\frac{30}{x^2 + 5x}=\frac{13}{6}$

$180 = 13x^2 + 65x$

$13x^2 + 65x - 180 = 0$

By quadratic formula,

$x=\frac{-65\pm \sqrt{65^2 - 4\times 13\times -180}}{26}$

$x=\frac{-65\pm \sqrt{4225 + 9360}}{26}$

$x=\frac{-65\pm \sqrt{13585}}{26}$

$x\approx 1.98\text{ or }x=-6.98$

∵ speed can not be negative,

Hence, average rate on the return​ trip would be 1.98 miles per hour.

Answer 2

Answer:v=4 mph

Step-by-step explanation:

Given

Distance between school and home is 6 miles

let v be the speed during return trip in miles/hr

total time taken $=2 hr 10 min \approx \frac{13}{6} hr$

During arrival $t_1=\frac{6}{v+5}$

During Return $t_2=\frac{6}{v}$

Total time $t_1+t_2=\frac{13}{6}$

$\frac{6}{v+5}+\frac{6}{v}=\frac{13}{6}$

$6(\frac{1}{v+5}+\frac{1}{v})=\frac{13}{6}$

$6\cdot \frac{2v+5}{v(v+5)}=\frac{13}{6}$

$72v+180=13v^2+65v$

$13v^2-7v-180=0$

$v=\frac{7\pm \sqrt{7^2+4\times 13\times 180}}{2\times 13}$

$v=4 mph$

Thus velocity during return trip $v=4 mph$