8. Linda goes to visit a friend who lives in Lompoc. It is 390 miles from Linda's house to her friend's house. When she came back, she didn't stop, so it took her a half an hour less time and her average speed was 5 miles per hour faster. How fast was she going when she went down to Lompoc, and how fast was she going when she drove back?

First we define our unknowns. In this case they are looking for how fast she was driving when she went down to Lompoc and how fast she was driving when she drove back. We could set this up with two unknowns, but since we know that she averaged 5 mi/hr faster coming back we could

let x = how fast she was driving going down

and then we would have that

x + 5 = how fast she drove on the way back.

Some authors suggest we make up a d = rt table to organize our thinking.

distance = rate x time table

Once we have two columns filled in we can fill in the third column by using the appropriate form of the d = rt equation. In this case we need to fill in the column for the time so we use the equation

t = d/r

Completed d = rt table

 

Now the sentence "It took her a half hour less time to get back." can be translated as

 

390/(x+5)=390/x-1/2

This gives us a rational equation. Clear denominators. The smallest common denominator is 2x(x+5)

2x(x+5)(390/(x+5))=2x(x+5)(390/x-1/2)

Multiply the 2x(x+5) by both terms on the right side.

 

2x(x+5)390/x=2x(x+5)390/x-(1/2)2x(x+5)

 

Cancel and assemble the surviving factors.

 

2x(390)=2(x+5)390-x(x+5)

 

Remove parentheses and multiply.

 

780x=780x+3900-x^2-5x

 

After subtracting 780x from both sides, transpose all the terms

to the left.

x^2+5x-3900=0

Factor

(x+65)(x-60)=0

and set the factors = 0.

x+65=0,x=-65;x-60=0,x=60

 

We get two answers. One is 60 mi/hr and the other one is -65 mi/hr. Of the two, 60mi/hr makes more sense. If she was driving to Lompoc at 60 mi/hr then it would take her 6.5 hours .If she came back averaging 5 mi/hr faster then it would take her 6 hours which would be 1/2 hour less time, and the answer checks.

The answer of -65 mi/hr would take a little more imagination to check. It would mean that she was driving to Lompoc at 65 mi/hr backwards. If she were driving backwards then she would have been in Lompoc 6 hours ago. To get her speed coming back, adding 5 mi/hr to -65 mi/hr would give us a speed of -60 mi/hr, and it would then take her -6,5 hrs which is 1/2 less than -6 hours. However, most authors would have us reject such an answer as being meaningless.

 

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