Bus speed distance

A bus travels between places A and B. If it increases its average speed by 5 km/h, the travel time will be reduced by 20 minutes. If he reduces his original speed by 4 km/h, the driving time is increased by 20 minutes. What is the average speed of the bus? What is the running time of the bus? What is the length of its route?

Final Answer:

v =  40 km/h
t =  3 h
s =  120 km

Step-by-step explanation:

$$\begin{gathered} t_1=20\text{min}\to \text{h}=20:60\text{h}=0.33333\text{h} \\ \Delta v_1=5\text{km/h} \\ \Delta v_2=4\text{km/h} \\ s=v\cdot t \\ s=(v+\Delta v_1)\cdot (t-t_1) \\ s=(v-\Delta v_2)\cdot (t+t_1) \\ (v+5)\cdot (t-1/3)=(v-4)\cdot (t+1/3)=vt \\ 27t=2v+1 \\ 12t+4=v \\ 27\cdot t=2\cdot v+1 \\ 12\cdot t+4=v \\ 27t-2v=1 \\ 12t-v=-4 \\ Row2-\frac{12}{27}\cdot Row1\to Row2 \\ -0.11v=-4.44 \\ v=\frac{-4.44444444}{-0.11111111}=40 \\ t=\frac{1+2v}{27}=\frac{1+2\cdot 40}{27}=3 \\ t=3 \\ v=40=40\text{km/h} \end{gathered}$$

$t=3=3\text{h}$

$$\begin{gathered} s=v\cdot t=40\cdot 3=120=120\text{km} \\ \text{ Verifying Solution: } \\ {s}_1=(v+\Delta v_1)\cdot (t-t_1)=(40+5)\cdot (3-0.3333)=120\text{km} \\ {s}_2=(v-\Delta v_2)\cdot (t+t_1)=(40-4)\cdot (3+0.3333)=120\text{km} \end{gathered}$$

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