Two cities are 120 km apart. A car drives from the first city to the second and back, taking 4 hours 24 minutes in total. The return speed is 20% greater than the outbound speed. Find the outbound speed.

Step-by-step solution

1. 1

   Convert the mixed time into a single unit. Speeds are in km/h, so the time must be in hours:

   $4\text{ h }24\text{ min}=4+\frac{24}{60}=4.4\text{ h}=\frac{22}{5}\text{ h}$

   Leaving the $24$ minutes as minutes is the single most common way to wreck this problem.

2. 2

   Name the unknown and express the second speed in terms of it. Let $v$ be the outbound speed in km/h, with $v>0$. A $20\%$ increase multiplies by $1.2$, so the return speed is

   $1.2v=\frac{6v}{5}\ \text{km/h}$

   Choosing the smaller speed as the unknown keeps the fractions simple.

3. 3

   Write each leg's time as distance divided by speed. Both legs cover $120$ km:

   $t_{\text{out}}=\frac{120}{v},\qquad t_{\text{back}}=\frac{120}{1.2v}$

   The times, not the speeds, are what add up — average speed on a round trip is not the average of the two speeds, so never average $v$ and $1.2v$.

4. 4

   Simplify the second fraction before combining. Dividing $120$ by $1.2$ is exact:

   $\frac{120}{1.2v}=\frac{100}{v}$

   so the total-time equation becomes

   $\frac{120}{v}+\frac{100}{v}=4.4\qquad\Longrightarrow\qquad \frac{220}{v}=4.4$

5. 5

   Solve and interpret. Multiplying both sides by $v$ and dividing by $4.4$:

   $v=\frac{220}{4.4}=50$

   $\boxed{v=50\ \text{km/h}}$

   The return speed is $1.2\times 50=60$ km/h.

6. 6

   Check against the original wording. Outbound: $120/50=2.4$ h. Return: $120/60=2$ h. Total $2.4+2=4.4$ h $=4$ h $24$ min, and $60$ is indeed $20\%$ more than $50$. Note the round-trip average speed is $240/4.4\approx 54.5$ km/h, not $55$ — another reminder that speeds do not average.