M a t h 152A
Chapter 2,4; More Application Problems
Objectives:
• Finding distance, rate, and time for going
in the same direction
• Finding distance, rate, and time for going
in towards each other
• Finding distance rate, and time for going
in opposite directions
• Finding distance, rate, and time for going
slow, then going faster
• Solve mixture problems
C h a p t e r 2.4
Steps for Solving Application Problems:
1. Read, throw out nonsense numbers
2. Assign a variable (What is it asking
for?)
3. Write an equation
4. Solve the equation
5. Answer the question!
6. Check, does it make sense?
Distance = Rate x
Time
Finding Distance, Rate, and Time For Going in the Same Direction
Ex: Two people start walking at the same time in the same direction. One person walks 4 mph
and the other person walks 6 mph. How long until they are 0.5 miles apart.
«.fjijV^^(j-^
Let -
P^^y^
= t^r>|f S v W ^ : \ \ t \ j c ^
Time
Rate
Slower
t
Faster
t
Distance
^
Equation:
HiTaster's distance
slower's distance
total distance apart
^ " ^ ' ^
Answer: 0.25 hours = time it takes to be 0.5 miles apart
Finding Distance, Rate, and Time For Going in the Same Direction and Catching Up
Ex: Say you and your friend are hanging out. Your friend takes off going 55 mph down the
freeway. Five minutes later, you decide to catch up with her and take off going 60 mph down the
same freeway. How long does it take you to catch up to her, and how long has she been driving?
Let
X- =
WccVt'^ 6sK\iK^c^
aa,o;
Rate
Time
Distance
Your friend
X
You
Answer: 55 minutes = time it takes took you to catch up
1 hour = time she has been driving
Finding Distance, Rate, and Time For Going Towards Each Other
Ex: The distance between Gilroy and WatsonviUe is 24 miles. If you left Gilroy going 40 mph at
the same time as your friend leaves WatsonviUe, going 45 mph, how long will it take to meet up?
Let J -
_
Time
Rate
you
t
friend
t
Distance
Equation: ^ ^ - ^ ^ ^ ^ ^ ^
your distance
friend's distance
total distance traveled
Answer: 0.28 hours (about 17 minutes)« time it takes for both to travel 24 miles
You try:
1. Two cars are traveling toward each other from two cities 270 miles apart. One car is going 5
mph slower than the other. They meet up after 2 hours. How fast are they each going?
Let_
Then
*
Rate
Faster
Slower
X *•
?•
Time
'
Distance
/A
»
—
^
^ — —
Equation:
2x .Jlx-^J^= 2iCL
raster's distance
slower's distance
H>c-\0- vvC
Answer: 70 mph = faster car's rate
65 mph = slower car's rate
total distance traveled
Finding Distance, Rate, and Time For Going In Opposite Directions
Ex: Two cars are going in opposite directions and one car is going 10 mph slower than the other.
After 1.5 hours they are 150 miles apart. How fast are they each going?
„
X
Let
Then X " \ C .
- irvi\if cars -^^rK
Faster
(c<'i':^f6.
^MCH
Rate
Time
.'^"^^^ (^~\o)'ci.^
Distance
A
Equation:
J / 2 ) c + i £ ^ = JSC
Slower
raster's distance
Answer: 55 mph = faster car's rate
45 mph = slower car's rate
X - \
^
slower's distance
' 6 5 - \
^
total distance traveled
'
1
j
a
a
You try:
1. A freight train and a passenger train start from the same place and travel in opposite directions.
The passenger train travels 15 mph faster than the freight train. After 2 hours, they are 278
miles apart. How fast are they each going?
^
C^¥fS)i>^pi^,
Let
= "iA'iYAr-^
Then
V r f Ar\>V W ^ n
>
Xi-lO
Rate
Time
Freight
z
Passenger
z
Distance
Equation:
fasler's distance
slower's distance
-•3C
1±
Answer: 62 mph = freight train's rate
77 mph = passenger train's rate
total distance traveled
Solving Mixture Problems
Ex: Three pounds of type A sunflower seed, which sells for $3.60 per pound, are mixed with 2
pounds of type B sunflower seeds, which sells for $4.80 per pound. What should the mixture cost
per pound?
Let X = (m c] \iMm p r pcmi
Equation:
$3.60
$4.80
+
3 pounds
$x
=
2 pounds
5 pounds
3(3M) .aim") =
3 pounds for $3.60each
2 pounds for S4.80 each
'5
5 pounds for Sxeach
^
Answer: $4.08 per pound = cost of mixture
You try:
1. A nurse mixes 3.3 quarts of a 12% saline solution and 6.7 quarts of a 9% sahne solution. What
percent of saline solution will the mixture be?
Let
A
=
j.kinnt (j
Equation:
/4
3i, 5
" ^ i ^
quarts
quartsof//%solution
Sciifu
+
SDlufen
=
quarts
quarts of '^1% solution
/ ( _ quarts
iQ
quartsofX %solu
Answers: 10% = percent of saline solution in the 10 quart mixture
fti
viM^
Ex: How many ounces of 16% and 30% iodine solution should be mixed together to get 20
ounces of an iodine solution that is 25%?
X
Let
Then
Equation:
=
Lm(t:^
-
f IMV^S i%
16%
scltdicn
^
+
jc ounces
X ounces of 16% solution
Answer:
^itytta)
30%
=
20-Jc ounces
25%
20 ounces
20-x ounces of 30% solution
20 ounces of 25% solution
^2 »fo(c?CXli^J* "^M
7.14 ounces = number ofounces of 16% iodine soTuuon
12.86 ounces = number of ounces of 30% iodine solution
'.I^;
~~
' ~
You try:
Grass seed A sells for $2.65 per pound and grass seed B sells for $2.80 per pound. How many
pounds of each type should be used to get a 12-pound mixture that sells for $2.70 per pound?
Let
Then
17-x = fimh
iff
cycm ^mX
ft
Equation:
+
=
X pounds IJ~K pounds
pounds for'^ tl&'Veach fJ
pounds for-
* each
/2- pounds
'
pounds for ^ '••
each
Answer: 8 pounds = the number of pounds of grass seed A, that sells for $2,65/lb
4 pounds = the number of pounds of grass seed B, that sells for $2,80/lb
Ex: How many liter of an 8% sulfuric acid solution must be mixed with 6 liters of a 15 % solution
to get a solution that is 12% sulfuric acid?
Let
X
- lV\f\ C>f
Then
KHC
=
sduJiUci
Equation:
8%
15%
+
X liters
V
^
=
(x+6) liters
6 liters
^
\
X liters of 8% solution
\
12%
^
6 liters of l5%solution
x+6 liters of 12% solution
OH
Answer:
4.5 = number of liters o f S^Tsulfuric acid solution
10.5 = number of liters of 12% sulfuric acid solution
~ —
You try:
1. How many liter of a 12% sulfuric acid solution must be mixed with 2 liters of a 9% solution to
get a solution that is 11% sulfuric acid?
Let
X
-
mjriM:n
Then
Xt/>
=
Equation:
=
+
X liters
X
liters o£?2-% solution
<,/2x
liters
liters of
117
) liters
solution
iX* Z liters of M % solution
Jfc^ =ftUy'f.22-
-J(x -yS ^ ; c - J &
Answer: 4 liters = number of liters of 12% sulfuric acid solutions
6 liters = number of liters of 11% sulfuric acid solution