This is a distance/rate/time problem so we remember that $\text{rate} \cdot \text{time} = \text{distance}$. We might assign a variable to the total distance driven, and it is solvable that way, but I'm going to assign a variable, say $t$, for the amount of minutes he will drive at $n$ km/hr. This is also equal to how long he will take driving if he arrives on time. We have rates in hours and time in minutes. You might convert minutes to hours, but i don't like big fractions like $\frac{1}{20}$ (even though it's easy to work with), so I'm going to convert hours to minutes. We have $60 \text{km/hr} = 1 \text{km/min}$ and $90 \text{km/hr} = \frac{3}{2} \text{km/min}$. The total distance is always equal, so we will equate the two distances. This is not for a physics contest, so I'm not going to bother to type all the units. We have: $1 \cdot (t+5) = \frac{3}{2} \cdot (t-5)$. Expanding the RHS gives us $t + 5 = \frac{3}{2}t - \frac{15}{2}$. Rearranging the terms gives us $\frac{1}{2}t = \frac{25}{2}$. Finally, multiplying both sides by $2$ gives us $t = 25$. We will now use this to find out the total distance traveled. We will use the first fact (the second fact is also valid). Since he will arrive 5 minutes late by driving at $1$ km/min, he will drive for $t + 5 = 30$ minutes, for a total of $30$km. Now we can find $n$. However, before we do so, we have $t = 25 \text{min} = \frac{5}{12} \text{hr}$. Rate is distance divided by time, so we have $n = \frac{30}{\frac{5}{12}} = 30 \cdot \frac{12}{5} = \boxed{72}$.

Average out the earliness and lateness, so it is just $v_{\text{avg}}=\frac2{\frac1{60}+\frac1{90}}=72$.

Say he will arrive 1/12 of an hour late if he travels at 60 kph and 1/12 of an hour early if he travels at 90 kph. Let x be the distance to David's House- x/60=n+1/12 x/90=n-1/12 Add the two equations- x/36=2n Multiply by 36- x=72n 72 kilometer per hour!

Average out the earliness and lateness, so it is just $v_{\text{avg}}=\frac2{\frac1{60}+\frac1{90}}=72$.