Answer :
6. A is 60 miles from B. An automobile at A starts for B at the rate of 20 miles an hour at
the same time that an automobile at B starts for A at the rate of 25 miles an hour. How
long will it be before the automobiles meet?
Rate
Time
Distance
Ryan
Castel
Step 1
Assume they are were they meet
Let
Ryan goes from A to B
rate = 20 miles per hour
time = unknow (t), this time is the same for the two automobile
distance1=unknown( x)
Castel goes from B to A
rate2= 25 miles per hour
time= unknown(t)
distance2 =unknown(y)
we also know that distance from A to b is 60 miles, so
[tex]\begin{gathered} \text{distance}1+\text{distance}2=60\text{ miles} \\ x+y=600 \\ y=600-x \end{gathered}[/tex]Step 2
make the equations
for Ryan
[tex]\begin{gathered} \text{rate1}=\frac{dis\tan ce}{\text{time}} \\ 20=\frac{x}{t} \\ t=\frac{x}{20} \end{gathered}[/tex]For Castel
[tex]\begin{gathered} \text{rate}=\text{ }\frac{dis\tan ce}{\text{time}} \\ \text{25}=\frac{y}{t}=\frac{600-x}{t} \\ 25\cdot t=600-x \\ t=\frac{600-x}{25} \end{gathered}[/tex]Now, the time is the same
[tex]\begin{gathered} t=t \\ \frac{x}{20}=\frac{600-x}{25} \end{gathered}[/tex]solve for x
[tex]\begin{gathered} 25x=20(600-x) \\ 25x=12000-20x \\ 25x+20x=12000 \\ 45x=12000 \\ x=\frac{12000}{45} \\ x=266.66\text{ miles} \end{gathered}[/tex]now, with the value of x, replace it to find t
[tex]\begin{gathered} t=\frac{x}{20} \\ t=\frac{266.66}{20} \\ t=13.33\text{ hours} \end{gathered}[/tex]finally, replace the value of x to find y