We are given that two trains start a distance of 992 km apart. If they meet after 4 hours, this means that the sum of the distances of both trains after 4 hours traveling must be 992 km. This is written mathematically as:
[tex]d_a+d_b=992[/tex]
Now, the distance traveled by an object is given by the following relationship:
[tex]\begin{gathered} d_a=r_at \\ d_b=r_bt \end{gathered}[/tex]
Where:
[tex]\begin{gathered} r_a,r_b=\text{ rates of the trains} \\ t=\text{ time} \end{gathered}[/tex]
Now, we substitute the values in the equation for the total distance:
[tex]r_at+r_bt=992[/tex]
Now, the time is 4 hours, therefore, we have:
[tex]r_a(4)+r_b(4)=992[/tex]
Now, we are also given that one train travels 20 km/h slower than the other. If a is the slower train then we have:
[tex]r_a=r_b-20[/tex]
Substituting we get:
[tex]4(r_b-20)+4r_b=992[/tex]
Now, we divide both sides by 4:
[tex]r_b-20+r_b=\frac{992}{4}[/tex]
Solving the operations:
[tex]r_b-20+r_b=248[/tex]
Now, we add like terms:
[tex]2r_b-20=248[/tex]
Now, we add 20 to both sides:
[tex]\begin{gathered} 2r_b=248+20 \\ 2r_b=268 \end{gathered}[/tex]
Now, we divide both sides by 2:
[tex]r_b=\frac{268}{2}=134[/tex]
Therefore the fastest train has a rate of 134 km/h. Now, we determine the rate of the slower train:
[tex]\begin{gathered} r_a=134-20 \\ r_a=114 \end{gathered}[/tex]
Therefore, the slower train travels ar 114 km/h
