Transcript Mixture Problem - Gloucester Township Public Schools

Over Lesson 2–8
Over Lesson 2–8
Weighted Averages
Lesson 2-9A
Mixture Problems
You translated sentences into equations.
Solve mixture problems.
• weighted average – The sum of the
product of the number of units and the
value per unit divided by the sum of the
number of units, represented by M.
• mixture problems – problems where two
or more parts are combined into a whole.
• In mixture problems, the units are usually
the number of gallons or pounds and the
value is the cost, value, or concentration
per unit.
• It is usually helpful to set up a table for
these problems.
Mixture Problem
PETS Mandisha feeds her cat gourmet cat food that
costs $1.75 per pound. She combines it with cheaper
food that costs $0.50 per pound. How many pounds
of cheaper food should Mandisha buy to go with
5 pounds of gourmet food, if she wants the average
price to be $1.00 per pound?
Let w = the number of pounds of cheaper cat food.
Make a table.
x
=
This is your
equation
Add column Price she wants
Mixture Problem
Write and solve an equation using the information
in the table.
Price of
gourmet
cat food
8.75
plus
price of
cheaper
cat food
equals
+
0.5w
=
price of
mixed
cat food.
1.00(5 + w)
8.75 + 0.5w = 1.00(5 + w)
Original equation
8.75 + 0.5w = 5 + 1w
Distributive
Property
8.75 + 0.5w – 0.5w = 5 + 1w – 0.5w
8.75 = 5 + 0.5w
Subtract 0.5w
from each side.
Simplify.
Mixture Problem
8.75 – 5 = 5 + 0.5w – 5
3.75 = 0.5w
Subtract 5 from
each side.
Simplify.
Divide each side
by 0.5.
7.5 = w
Simplify.
Answer: Mandisha should buy 7.5 pounds of cheaper
cat food to be mixed with the 5 pounds of
gourmet cat food so that the average price is
$1.00 per pound of cat food.
Cheryl bought 3 ounces of glass beads that cost
$1.79 an ounce. The seed beads cost $0.99 an
ounce. How many ounces of seed beads can she
buy if she only wants the beads to be $1.29 an
ounce for her craft project?
Weight of
beads (oz)
x
Price per
ounce ($)
=
Total Price
($/oz)
Glass Beads
3
1.79
5.37
Seed Beads
s
0.99
0.99s
Total Beads
3+s
1.29
1.29(3 + s)
Equation : 5.37 + 0.99s = 1.29(3 + s)
• Solve :
5.37 + 0.99s = 1.29(3 + s)
5.37 + 0.99s = 3.87 + 1. 29s
1.5 + 0.99s = 1.29s
1.5 = 0.3s
5=s
Distributive Property
Subtraction Property of Equality
Subtraction Property of Equality
Division Property of Equality
Cheryl would have to buy 5 ounces of seed beads for
her project to make the total bead cost $1.29 per
ounce .
Percent Mixture Problem
AUTO MAINTENANCE A car’s radiator should
contain a solution of 50% antifreeze. Bob has
2 gallons of a 35% antifreeze. How many gallons of
100% antifreeze should Bob add to his solution to
produce a solution of 50% antifreeze?
Let g = the number of gallons of 100% antifreeze to be
added. Make a table.
What Bob has
What Bob adds
What Bob wants
Percent Mixture Problem
Write and solve an equation using the information
in the table.
Amount of
antifreeze in
35% solution
0.35(2)
plus
+
amount of
antifreeze in
100% solution
1.0(g)
0.35(2) + 1.0(g) = 0.50(2 + g)
0.70 + 1g = 1 + 0.50g
equals
amount of
antifreeze in
50% solution.
=
0.50(2 + g)
Original equation
Distributive
Property
0.70 + 1g – 0.50g = 1 + 0.50g – 0.50g Subtract 0.50g
from each side.
Percent Mixture Problem
0.70 + 0.50g = 1
0.70 + 0.50g – 0.70 = 1 – 0.70
0.50g = 0.30
Simplify.
Subtract 0.70
from each side.
Simplify.
Divide each
side by 0.50.
g = 0.6
Simplify.
Answer: Bae should add 0.6 gallon of 100%
antifreeze to produce a 50% solution.
A recipe calls for mixed nuts with 50% peanuts.
pound of 15% peanuts has already been used.
How many pounds of 75% peanuts needs to be
added to obtain the required 50% mix?
Weight of
Percentage of
x
=
peanuts (lbs.)
peanuts
(decimal form)
Amount of
peanuts
15% peanuts
½
.15
.075
75% peanuts
x
.75
.75x
50% peanuts
Equation:
½+x
.50
0.075 +.75x = .5(½ + x)
.50(½ + x)
Mixture Problem worksheet
Chapter 2 Review due Wednesday
Weighted Averages
Lesson 2-9B
Uniform Motion
Problems
You translated sentences into equations.
Solve uniform motion problems.
• uniform motion problem – problems in
which an object moves at a certain speed or
rate. Also referred to as a rate problem.
• Uniform Motion problems use the formula
d = rt
d represents distance
r represents rate
t represents time
Speed of One Vehicle
AIR TRAVEL Nita took a non-stop flight to visit her
grandmother. The 750-mile trip took three hours and
45 minutes. Because of bad weather, the return trip
took four hours and 45 minutes. What was her
average speed for the round trip?
This is a Round Trip Problem.
Nita traveled 750 mile to her grandmother’s and 750 miles
back so her round trip was1500 miles.
She traveled a 3.75 mi/hr to get there and 4.75 mi/hr on
her return, for a round trip time of 8.5 hours.
Note: you have to change minutes to hours.
Speed of One Vehicle
Use the formula
M = Round Trip Distance ÷ Round Trip Time
Answer: The average speed was about 176 miles per
hour.
Check The solution of 176 miles per hour is between
the going portion rate 200 miles per hour, and
the return rate, 157.9 miles per hour. So, the
answer is reasonable.
In the morning, when traffic is light, it takes
30 minutes to get to work. The trip is 15 miles
through towns. In the afternoon, when traffic is
a little heavier, it takes 45 minutes. What is the
average speed for the round trip?
Convert minutes to hours since we are looking for
speed in mi/hr
The average
speed for the
trip to and
from work is
24 miles per
hour
M = 24 mi/hr
Speeds of Two Vehicles
RESCUE A railroad switching operator has discovered
that two trains are heading toward each other on the
same track. Currently, the trains are 53 miles apart. One
train is traveling at 75 miles per hour and the other
40 miles per hour. The faster train will require 5 miles to
stop safely, and the slower train will require 3 miles to
stop safely. About how many minutes does the operator
have to warn the train engineers to stop their trains?
Step 1
Draw a diagram.
53 miles apart
Takes 5 miles to stop
53 – (5 + 3) = 45 miles
Takes 3 miles to stop
Speeds of Two Vehicles
Step 2 Let m = the number of hours that the operator
has to warn the train engineers to stop their trains safely.
Make a table.
Step 3 Write and solve an equation using the
information in the table.
Distance
traveled by
fast train
75m
plus
distance
traveled by
other train
equals
45 miles.
+
40m
=
45
Speeds of Two Vehicles
75m + 40m = 45
115m = 45
Original equation
Simplify.
Divide each side by 115.
m ≈ 0.39
0.39 × 60 = 23.4
Round to the nearest
hundredth.
Convert to minutes by
multiplying by 60.
Answer: The operator has about 23 minutes to warn
the engineers.
Two students left the school on their bicycles at the same time, one
heading north and the other heading south. The student heading
north travels 15 miles per hour, and the one heading south travels at
17 miles per hour. After about how many minutes will they be 7.5
miles apart?
Let t = time
17t + 15t = 7.5
32t = 7.5
t = .23 hrs
add the distances together to equal 7.5
convert the hours to minutes
.23 x 60 = 14 min
After 14 minutes the students will be 7.5 miles apart.
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Chapter 2 review due tomorrow