091 Motion and Mixture Problems – Section 3.4
Summary of Distance Problems: a problem in which an object is moving at a specific rate for a specific
period of time. First, specify the rate and time for each person or object. Note that the unit of rate
determines the unit of time. Next, use the formula Rate Time = Distance to determine the distance for
each person/object. Finally, add, subtract, or set the distance equal to one another.
When to add distances, subtract distances, or set distances equal:
Addition:

1. l
l
2. 
starting at same place, traveling in opposite directions

starting apart, traveling toward each other
3. 
starts at one rate, rate changes, but continue in same direction
(unless specifically states difference such as #21, pg 222)
Subtraction:
1. A
l A is x miles ahead of B, or A and B are x miles apart
B
l (Both start at the same place and time, but one travels faster
x miles
so gets farther in a given amount of time.)
Set Equal:
1. 



Race (both start at point A and end at point B)
l
“There and Back” (ex. boat out to an island and back)
2. l
3. 
l

l
4.


“Catch up” (1 person leaves later than another, but travels faster:
when do they catch up?)

walk
Change transportation (ex. walks to work; bikes to work)

bike
Summary of Mixture Problems : interest problems or any problem in which two or more quantities are combined to
produce a different quantity. Note that “beakers” or “buckets” are used to group factors for each quantity.
Remember that the concentration of water is 0% and the concentration of pure solution is 100%.
In mixture problems in this class, three different things can happen:
1. Two or more mixtures are added together to form a new mixture (for example with chemical mixtures)
amt 1 +
rate 1
amt 2 =
rate 2
tot amt
rate 3
Equation is:
rate 1 (amount 1) + rate 2 (amount 2) = rate 3 (total amount)
2. Two or more mixtures are added together to equal a blended dollar amount (ex. the interest earned on
two or more accounts with different interest rates is added together to equal the total interest earned.)
amt 1 +
rate 1
amt 2 =
rate 2
total dollar
amount
Equation is:
rate 1 (amount 1) + rate 2(amount 2) = given dollar amount
3. Two mixtures are set equal to each other (for example when two investments having different
interest rates both earn the same dollar amount of interest.)
amt 1
rate 1
= amt 2
rate 2
Equation is:
amount 1 (rate 1) = amount 2 (rate 2)
******Note that if the rates are percents, you must convert them to decimals for use in your equation!!
Motion Problem Example:
Two trains going in opposite directions leave from the same station on parallel tracks. One train travels at
50 miles per hour and the other at 60 miles per hour. How long will it take for the trains to be 440 miles
apart?
1.
Draw a picture and determine whether the distances will be added, subtracted, or set equal.
Train 1 distance


In this case, train 1’s distance needs to be
Train 2 distance
added to train 2’s distance to get the total
distance of 440 miles.
Total distance is 440 miles
2.
Define variables.
In this problem, the question being asked is, “How long will it take for the trains to be 440 miles
apart?” Thus the variable is time.
x = number of hours that the trains will travel.
Note that in this problem, both trains travel for the same length of time, and that the unit of time
is hours, since the rate is in miles per hour.
3. Set up a chart.
Train 1
Train 2
Total
Rate
50 mph
60 mph

Time
x hours
x hours
=
Distance
4.
Use the distance formula to calculate the distance for Train 1 and Train 2. (Multiply across the
rows.) Note that the total distance can also be entered.
 Time
Rate
=
Distance
Train 1
50 mph
x hours
50x miles
Train 2
60 mph
x hours
60x miles
Total
440 miles
5.
Since it was already determined that the distances for train 1 and train 2 would be added, the
equation can now be set up from the distance column.
50x + 60x = 440
6.
Now solve the equation
50x + 60x = 440
110x = 440
110 110
x = 4
7.
Thus, to answer the question, after 4 hours, train 1 and train 2 will be 440 miles apart.
8. To check the answer, use the rate time = distance formula to calculate the distance that each
train travelled.
Train 1: 50 mph 4 hours = 200 miles
Train 2: 60 mph 4 hours = 240 miles
The total distance covered by both trains is 200 + 240 = 440 miles, which matches the given
information.
Mixture Problem Example 1: Interest
Brad entered a building design contest and won the grand prize of $5,000. He invested the money in two
accounts, one earning 3.5% interest, and the other earning 5% interest. If he earned a total of $227.50 in
interest in one year, how much did he invest in each account?
In this problem, a dollar amount of interest will be earned in the account earning 3.5% and in the account
earning 5%. The total amount of interest earned in the two accounts is given as $227.50. To figure out
this problem you will be using the fact that principal rate time = interest. (Note that in this class it
will be assumed that time always equals 1 year, and thus time won’t be factored into the equation.)
1. x = amount of money invested at 3.5%
5000  x = amount of money invested at 5% The total amount of money available to
invest was $5000. So if $x was invested at 3.5%,
then that leaves $5000  x to invest at 5%
principal
$ x + $5000  x = $ 227.50
rate
3.5%
5%
(assume time of 1 year)
2.
.035x + .05(5000  x) = 227.50
.035x + 250  .05x = 227.50
 0.015x + 250 = 227.50
 250  250
=  22.50
 0.015x
 0.015
 0.015
x
= 1500
3.
So, Brad invested $1500 at 3.5% and
$5000  $1500 = $3500 at 5%
**Note that when the equation was written, the percents
were converted into decimals.
4. To check this answer, you could calculate the dollar amount of interest earned in each account.
In the account earning 3.5% interest, the dollar amount of interest earned would be: .035(1500) =
$52.50
In the account earning 5% interest, the dollar amount of interest earned would be : .05(3500) =
$175
So, the total interest earned would be $52.50 + 175.00 = $227.50, which matches the amount
given in the problem.
Mixture Problem Example 2: Chemical Solution
A. Margaret has 6 ounces of 60% cleaning solution. She is worried that the solution will be too harsh on
her antique claw tub, so she wants to dilute it with water. If she wants to make a 45% solution, how
much water should she add?
1.
x
= ounces of water (0%)
x + 6 = ounces of 45% solution (The total amount of solution will be the sum of the amount of
60% solution, 6 oz, plus the amount of water, x oz.)
amount 6 oz
strength 60%
2.
.6(6) +
3.6 +
+
0(x)
0
3.6
 2.7
.9
.45
2
x oz
0%
=
(x + 6) oz
45%
= .45 (x+6)
= .45x + 2.7
= .45x + 2.7
 2.7
= .45x
.45
= x
** Note that when the equation was written, the percents
were converted into decimals.
3. Margaret needs to add 2 ounces of water to the 6 ounces of 60% cleaning solution to make a
45% solution.
B. In the mixture problem above, the amount of one solution was given, the amount of the other
solution was designated as “x”, and the total amount of the new 45% mixture was calculated by
adding the amounts of each of the individual solutions.
I could adjust the problem as follows:
Margaret has a 60% cleaning solution. She is worried that the solution will be too harsh on her antique
claw tub, so she wants to dilute it with water. If she wants to make 10 ounces of a 45% solution, how
much water and how much of the 60% cleaning solution should she use?
In this case, no amount is given for the 60% solution or for the water. However, the total amount of
the new 45% mixture was given. This problem would be set up as follows:
x = ounces of 60% solution
10  x = ounces of water (0%)
x oz + (10  x) oz =
10 oz
60%
0%
45%
Thus, the equation can be set up as:
Notice now that the amount of one solution (in this case the 60%
solution) has been defined as “x” and the amount of the other
solution, the water, has been defined as “10  x” the amount
of the total 45% solution, 10, minus the amount of the 60%
solution, x.
.6x + 0(10  x) = .45 (10)
.6x + 0
= 4.5
.6x
= 4.5
.6
.6
X
= 7.5 So, Margaret should mix 7.5 oz of 60% solution with (10  7.5) = 2.5 oz of water.