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6. A is 60 miles from B. An automobile at A starts for B at the rate of 20 miles an hour atthe same time that an automobile at B starts for A at the rate of 25 miles an hour. Howlong will it be before the automobiles meet?RateTimeDistanceRyanCastel

Sagot :

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

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