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Running Rates, Line Segments, and Basic Algebra

Date: 6/26/96 at 17:2:57
From: Anonymous
Subject: Running Rates, Line Segments, and Basic Algebra

Problem No. 1:

Adam ran the same race at an average speed of 6 miles per hour, how 
many minutes longer than Steve did Adam take to complete the race?

Problem No. 2:


Figure not drawn to scale.
If AB > CD, which of the following must be true?
  I. AB > BC
 II. AC > BD

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

Problem No. 3


Date: 6/27/96 at 9:13:17
From: Doctor Patrick
Subject: Re: Running Rates, Line Segments, and Basic Algebra

Hi!  Let's take the problems in the order you sent them:

Problem 1:

First let's figure out how long it took Steven to run the 12 miles.  
Do you know the formula Distance = rate*time?  It means that the 
distance you travel is equal to the speed you are moving * how long 
you move at that speed.  Does that make sense to you?

We know that Steven ran 12 miles (distance) and that he ran at 8 mph 
(rate).  All we have to do is to solve for the time, which turns out 
to be 12(miles)/8(mph) = time when we rewrite the formula from before.  
12/8=1.5, or 1 hour, 30 minutes. Do you understand up to this part?

Now we need to find out how long it took Adam to run the same 12 miles 
at 6mph so that we can compare the two times and find the difference.  
Why don't I let you try and figure out how long it took Adam using the 
same method as above - if you need more help with this one just write 
me back.

Problem 2:

Ok.  We know that AB > CD and we know how the lines are set up in 
relation to each other.  From this we can figure out which of the 
statements have to be true.

Let's look at them one at a time.  The first (I) says AB > BC.  Since 
the only information we have about BC is where it falls in the line, 
and NOT anything about how long it is, there is no way to know if this 
is true.  Therefore any answer with I in it will be wrong.

Now on to the second statement(II) - AC > BD.  Looking at the picture 
I see that AC = AB+BC and that BD = CB+CD.  We know that AB > CD, and 
that BC, although unknown, is the same in both segments.  Since we are 
adding equal parts to AB and CD the relationship stays the same - if 
AB > BC,  AB+BC > CD+BC, right? Then since AB+BC = AC and BC+CD = BD
AC > BD.  So II is true.

Problem 3:

"3 more then x" is the same as saying x+3, and "2 more then y" is the 
same as y+2. So x+3 = y+2.  To find x in terms of y you just need to 
get the x alone on one side of the equation by subtracting the 3 from 
both sides - x+3 - 3 = y+2 - 3. When we add the like terms we get 
x = y-1, which is what the problem wanted.  

The point of this kind of problem is usually to see if you can 
understand the equation when it is written as a sentence and turn it 
into an algebraic expression.  Some key terms to keep in mind:

"is" is the same as =
terms like "more then" and "greater then" will usually mean addition
and "less then" or "fewer then" are signs of subtraction 

I hope all this helps.  Please write back if you need more help,

-Doctor Patrick,  The Math Forum

Date: 6/27/96 at 23:49:7
From: Anonymous

Thanks man.
Associated Topics:
Middle School Algebra
Middle School Word Problems

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