Calculating the stoppage time of a bus 
Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
A. 9   B.  10
  C.  12  D.  20

Solution is given as- Due to stoppages, it covers 9 km less.
Time taken to cover 9 km =      $\frac{9*60}{54}$    min = 10 min.
I am not able to imagine this.
I feel that distance would be same in both cases. From 1st, we get 54km. In 2nd speed is given. So, time is $\frac{54*60}{45}=72$mins. But it has taken 60 mins, that is either the speed was more or the distance was less. But there is no hint that distance is less. And speed is explicitly given less. I am stuck!
 A: When bus isn't stopping his average speed is $54 km/h$. Note that in the second trip, the trip with stoppages, while he's travelling he'll move with speed of $54 km/h$, but he also stops so the average speed will decrease. From the problem we can conclude that:
$$\frac{54 km/h \cdot t + 0 km/h \cdot (1h - t)}{1h} = 45 km/h$$
From this we can conclude that $t = \frac 56$ hours. So the bus stops for $\frac 16$ hours per hour, or in minutes that's $10$ minutes per hour.  
A: In an actual elapsed hour, the bus travels 45km.
Now you exclude the periods of time the bus is at rest during that hour. In some fraction of an hour, it covers that distance, averaging 54kmh - from here you work out the length of time it was in motion, and hence the desired answer.
A: Speed without stoppages: $54$km/h
Speed with stoppages: $45$km/h
Therefore, if the bus stops, it goes $9$km less every hour.
We know initially, $54$km is travelled in $1$ hour. That means $1$km is travelled in $\frac{1}{54}$hours. Therefore $9$km is travelled in $\frac{9}{54}$hours = $\frac{9}{54}\cdot 60$mins$=10$mins.
Therefore bus stops for $10$ mins.
Why use complex formulae??
