Transcript Lsn_Baum_Feb14_ModellingWPDays3-8Inked_MPM2DI

Grade 10 Academic Math
Chapter 1 – Linear
Systems
Modelling Word Problems
Days 4 through Days 9
Day 4 Agenda
1. Warm-up
2. Types of Modelling Problems
3. Mixture Problems
4. Relative Value Problems
5. Practice
Learning Goal
By the end of the lesson…
… students will be able to read
and interpret a mixture or relative
value word problems and create
a pair of linear relation equations,
resulting in a linear system
Curriculum Expectations
• Solve problems that arise from realistic
situations described in words… by
choosing an appropriate algebraic…
method
• Ontario Catholic School Graduate
Expectations: The graduate is expected
to be… a self-directed life long learner
who CGE4f applies effective… problem
solving… skills
Mathematical Process
Expectations
• Connecting – make connections among
mathematical concepts and procedures;
and relate mathematical ideas to
situations or phenomena drawn from
other contexts
Modelling Types
•
•
•
•
1.
2.
3.
4.
Break-Even Problems
Mixture Problems
Relative Value Problems
Rate Problems
Mixture Problems
• 2 things come together to give a total
number or amount
• 2 things come together to form a total cost,
weight, points, etc.
• Equations are usually in form Ax + By = C
Mixture Problems
• Ex. 1 Henry sharpens figures skates for $3
a pair and hockey skates for $2.50 per
pair. If he earns $240 and sharpens 94
pairs of skates, how many pairs of each
type of skate does he sharpen?
Example 1 Mixture (Cont’d)
Mixture Problems
Let x represent
# of figure skates
Let y represent
# of hockey skates
x + y = 94
(# of skates eq’n)
3x + 2.5y = 240
(earnings eq’n)
Mixture Problems
• Ex. 2 Joe has 38 loonies and toonies
totalling $55. How many of each type of
coin does he have?
Example 2 Mixture (Cont’d)
Mixture Problems
Let l represent
# of loonies
Let t represent
# of toonies
l + t = 38
(# of coins equation)
l + 2t = 55
(value equation)
Mixture Problems
• Ex. 3 (p.44, #11e)
• Benoit invested some money at 8% and
some at 10%. He earned a total of $235 in
interest.
Example 3 Mixture (Cont’d)
Mixture Problems
Let x represent
Amount of $
invested at 8%
Let y represent
Amount of $
invested at 10%
0.08x + 0.1y = 235 (interest equation)
Note: In order to do a $ invested eq’n,
we need the amount invested
Mixture Problems
• Ex. 4, p.51, #4c
• The total value of nickels and dimes is 75¢
Example 4 Mixture (Cont’d)
Mixture Problems
Let n represent
# of nickels
Let d represent
# of dimes
0.05n + 0.10d = 0.75
equation)
(value
Note: In order to do a # of coins
eq’n, we need to know the # coins
Relative Value Problems
• Usually 2 unknown numbers, ages, etc.
• No set form to the equations
• Must follow the directional words such as
more than, less, times, is, twice, sum,
difference, etc.
Relative Value Problems
• Ex. 1, p.51, #7
• The sum of two numbers is 72. Their
difference is 48. Find the numbers.
Example 1 Relative Value
(Cont’d)
Relative Value Problems
Let x represent
the first number
Let y represent
the other number
x + y = 72
x – y = 48
(sum equation)
(difference equation)
Relative Value Problems
• Ex. 2, p.51, #8)
• A number is four times another number.
Six times the smaller number plus half of
the larger number equals 212. Find the
numbers.
Example 2 Relative Value
(Cont’d)
Relative Value Problems
Let x represent
the first number
Let y represent
the other number
x = 4y
(multiplication eq’n)
0.5x + 6y = 212 (difference equation)
Relative Value Problems
• Ex. 3, p.24, #7
• At the December concert, 209 tickets were
sold. There were 23 more student tickets
sold than twice the number of adult tickets.
How many of each were sold?
Example 3 Relative Value
(Cont’d)
Relative Value Problems
Let x represent
# of student tickets
Let y represent
# of adult tickets
x - 23 = 2y
(relative # of tickets)
x + y = 209
(# of tickets)
Relative Value Problems
• Ex. 4, p.24, #8
• A rectangle with a perimeter of 54cm is 3m
longer than it is wide. What are its length
and width?
Relative Value Problems
Let l represent
width of the rect.
Let w represent
length of the rect.
2x + 2y = 54
(perimeter eq’n)
l–3=w
(relative length to
width eq’n)
Humour Break
Break-Even Problems
• Usually look for the point at which two
things cost the same
• Can refer to the point at which cost and
number of things are equal
• Equations usually take the form of
y = mx + b
Break-Even Problems
• Ex. 1. Barney’s Banquet Hall charges
$500 to rent the room, plus $15 for each
meal and Patrick’s Party Palace charges
$400 for the hall plus $18 for each meal.
When will both places cost the same
amount?
Example 1 Break-Even (Cont’d)
Break-Even Problems
Let x represent
# meals
Let y represent
the cost
y = 15x + 500
(Barney’s BH)
y = 18x + 400
(Patrick’s PP)
Break-Even Problems
• Ex. 2. The Millennium Wheelchair Co. has
just started its business. It costs them
$125 to make each wheelchair plus
$15,000 in start-up costs. They plan to sell
the chairs for $500 each. How many chairs
do they have to sell in order to break
even?
Example 2 Break-Even (Cont’d)
Break-Even Problems
Let x represent
# of wheelchairs
Let y represent
cost or revenue
y = 125x + 15000
(Cost eq’n)
y = 500x
(Revenue eq’n)
Break-Even Problems
• Ex. 3. p.44, #11c
• It costs $135 to rent the car, based on $25
per day, plus $0.15/km
Example 3 Break-Even (Cont’d)
Break-Even Problems
Let x represent
# of days
Let y represent
# of km driven
25x + 0.15y = 135 (Cost eq’n)
Note: This is not a usual example.
Usually if you are dealing with car
rental, you have an eq’n like
y = 0.15x + 25
Humour Break
Rate (Speed Distance Time)
Problems (Copy)
• Usually looking for time, speed or distance
• Distance = Speed x Time (from science –
can be rearranged for speed and time
also)
• Easiest to use a chart to help develop the
equations
Rate (Speed Distance Time)
Problems
• But first, we have the
Distance = Speed x Time
Or...
D=SxT
(equation)
Rate (Speed Distance Time)
Problems
• We can also rearrange this eq’n to solve
for speed...
Speed = Distance
-----------Time
Or...
Rate (Speed Distance Time)
Problems
• We can also rearrange this eq’n to solve
for Time...
Time =
Distance
-----------Speed
Rate (Speed Distance Time)
Problems
• Ex. 1 Fred travelled 95 km by car and
train. The car averaged 60 km/h and the
train averaged 90 km/hr. If the trip took 1.5
hours, how long did he travel by car?
• Let’s use a speed distance time chart to
organize our information...
Example 1 Rate (Cont’d)
Rate (Speed Distance Time)
Problems
Let x represent
the time in the car
Let y represent
the time on the
train
Distance
(km)
Car
Train
Total
Speed
(kph)
Time (h)
Rate (Speed Distance Time)
Problems
Distance
(km)
Car
Train
Total
Speed
(kph)
60
Time (h)
Rate (Speed Distance Time)
Problems
Distance
(km)
Speed
(kph)
Car
60
Train
90
Total
Time (h)
Rate (Speed Distance Time)
Problems
Car
Train
Total
Distance
Speed
(km) (recall (kph)
D = S x T...
60x
60
90
Time (h)
x
Rate (Speed Distance Time)
Problems
Distance
(km)
Speed
(kph)
Time (h)
Car
60x
60
x
Train
90y
90
y
Total
Rate (Speed Distance Time)
Problems
Distance
(km)
Speed
(kph)
Time (h)
Car
60x
60
x
Train
90y
90
y
Total
95
Rate (Speed Distance Time)
Problems
Distance
(km)
Speed
(kph)
Time (h)
Car
60x
60
x
Train
90y
90
y
Total
95
1.5
Rate (Speed Distance Time)
Problems
x + y = 1.5
(total travelling time)
60x + 90y = 95 (total distance travelled)
Rate (Speed Distance Time)
Problems
• Ex. 2 (text p.137, #6) A traffic helicopter
pilot finds that with a tailwind, her 120km
trip away from the airport takes 30
minutes. On her return trip to the airport,
into the wind, she finds that her trip is 10
minutes longer. What is the speed of the
helicopter? What is the speed of the wind?
Example 2 Rate (Cont’d)
Rate (Speed Distance Time)
Problems
Let h represent
the speed of the
helicopter
Let w represent
the speed of the wind
Distance
(km)
With tail
wind
With
headwind
Total
Speed
(kph)
Time (h)
Rate (Speed Distance Time)
Problems
Distance
(km)
With
tailwind
With
headwind
Total
120
120
Speed
(kph)
Time (h)
Rate (Speed Distance Time)
Problems
With
tailwind
With
headwind
Total
Distance
(km)
Speed
(kph)
120
h+w
120
Time (h)
Rate (Speed Distance Time)
Problems
With
tailwind
With
headwind
Total
Distance
(km)
Speed
(kph)
Time (h)
120
h+w
½
120
h-w
2/3 (keep
as a
fraction)
Rate (Speed Distance Time)
Problems
• Recall that
Speed = Distance
-----------Time
Rate (Speed Distance Time)
Problems
h + w = 120/0.5
(with tailwind... )
h + w = 240
(simplified)
h – w = 120/(2/3)
(with headwind)
h – w = 120 x (3/2) (flip 3/2 and x’s)
h – w = 180
(simplified)
Humour Break