Two friends

Two friends on a bicycle rode against each other from two places 49 km apart. The first set off at 8.00 at a speed of 20km/h, the second 12 minutes later at a speed of 25km/h. What time do they meet? How many km will each of them travel since then?

Result

t = 9:12
a =  24 km
b =  25 km

Step-by-step explanation:

$$\begin{gathered} s=49\text{km} \\ {v}_1=20\text{km/h} \\ {v}_2=25\text{km/h} \\ {t}_1=8.00\text{h} \\ {t}_2=8\frac{12}{60}=8+\frac{12}{60}=\frac{8\cdot 60+12}{60}=\frac{492}{60}=\frac{41}{5}=8\frac{1}{5}=8.2\text{h} \\ a+b=s \\ a=v_1\cdot (t-t_1) \\ b=v_2\cdot (t-t_2) \\ a+b=49 \\ a=20\cdot (t-8) \\ b=25\cdot (t-8.2) \\ a+b=49 \\ a-20t=-160 \\ b-25t=-205 \\ Row2-Row1\to Row2 \\ a+b=49 \\ -b-20t=-209 \\ b-25t=-205 \\ Row3+Row2\to Row3 \\ a+b=49 \\ -b-20t=-209 \\ -45t=-414 \\ t=\frac{-414}{-45}=9.2 \\ b=\frac{-209+20t}{-1}=\frac{-209+20\cdot 9.2}{-1}=25 \\ a=49-b=49-25=24 \\ a=24 \\ b=25 \\ t=\frac{46}{5}=9.2=9:12 \end{gathered}$$

$a=24=24\text{km}$

$b=25=25\text{km}$

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