A refrigerated truck averages 42 km/h over a whole day. In the morning it drove for 4 hours at an average of 50 km/h; in the afternoon it averaged 37 km/h. How many hours did it drive in the afternoon?

Step-by-step solution

1. 1

   Do not average the two speeds. The tempting move is $(50+37)/2 = 43.5$, or to reason that $42$ sits between $50$ and $37$ and stop there. Average speed is never the average of the speeds unless the times are equal. The only safe definition is

   $\bar v = \frac{\text{total distance}}{\text{total time}}$

2. 2

   Name the unknown and write both totals. Let $t$ be the afternoon driving time in hours. Morning distance is $4 \times 50 = 200$ km; afternoon distance is $37t$ km. Total time is $4 + t$ hours. So

   $\frac{200 + 37t}{4 + t} = 42$

3. 3

   Clear the fraction. Multiply both sides by $(4+t)$, which is safe because a driving time makes $4+t>0$:

   $200 + 37t = 42(4 + t) = 168 + 42t$

4. 4

   Collect the $t$ terms. Subtract $168$ and $37t$ from both sides:

   $200 - 168 = 42t - 37t \;\Longrightarrow\; 32 = 5t \;\Longrightarrow\; t = \frac{32}{5} = 6.4$

5. 5

   Interpret and verify. The truck drove $6.4$ hours (6 hours 24 minutes) in the afternoon. Checking: afternoon distance $= 37 \times 6.4 = 236.8$ km, total distance $= 200 + 236.8 = 436.8$ km, total time $= 4 + 6.4 = 10.4$ h, and $436.8 \div 10.4 = 42$ km/h exactly. The answer also passes a sanity test: because $42$ is closer to $37$ than to $50$, the slower afternoon leg must take the longer time, and $6.4 > 4$.