Transcript Advanced Bar Modeling

Demystify Challenging
Problems with Bar
Modeling
Gregg Velatini
Dianna Spence
2012 Georgia Mathematics Conference
Simple Ratios and Proportions
 The lengths of three rods are in the ratio of 1:3:4. If the total
length is 72 inches find the length of the longest rod.
Rod 1
9
Rod 2
9
9
9
Rod 3
9
9
9
72 inches
9
9 x 4 = 36 inches
The length of the longest rod is 36 inches
72 / 8 = 9 inches
Ratios and Proportions
 A garbage man had 3 times as much money as a teacher. After
the teacher earned an extra $200 moonlighting, the garbage
man only had twice as much money. How much money did the
teacher have at first?
Garbage
3 Parts
Teacher
1 Part
Ratios and Proportions
 A garbage man had 3 times as much money as a teacher. After
the teacher earned an extra $200 moonlighting, the garbage
man only had twice as much money. How much money did the
teacher have at first?
Garbage
2 Parts
200
Teacher
1 Part
The teacher had $400 at first.
Ratios - Practice
 Karen’s cat condo boards cute calicos for
companionless curmudgeons. In September,
the condo boarded cats and the ratio of
female to male cats was 3:2. In October, she
boarded several more cats, 35 of which were
female. After adding the new cats, the ratio
of female to male cats was reduced to 1:1 If
she wound up with 250 cats, how many of the
original cats were female?
Karen’s cat condo boards cute calicos for companionless curmudgeons. In September,
the condo boarded cats and the ratio of female to male cats was 3:2. In October, she
boarded several more cats, 35 of which were female. After adding the new cats, the ratio
of female to male cats was reduced to 1:1 If she wound up with 250 cats, how many of
the original cats were female?
Before
Female Cats
30
30
30
Male Cats
After
Female Cats
30
35
250 Cats
35
Male Cats
90 of the original cats were
female.
Solving Fraction Equations
 Mom bought 1 carton of eggs. She used 1/6 of the eggs to make
cookies and 1/4 of the eggs to bake a cake. How many eggs did mom
have left?
12 Eggs
1
Cookies
7 eggs
Cake
Solving Fraction Equations-- Practice
 Brad spent 1/3 of his money on Beanie
Babies and 1/2 of it on Nascar collectables.
 What fraction of his money did he spend
altogether?
 What fraction did he have remaining?
Solving Simple Fraction Problems
 Brad spent 1/3 of his money on Beanie Babies and 1/2 of it on
Nascar collectables.
 What fraction of his money did he spend altogether?
 What fraction did he have remaining?
Brad’s Money
1/3
1/2
Beanies
Nascar
1/6
 Brad spent 5/6 of his money.
 Brad had 1/6 of his money remaining.
Solving an Algebraic Equation
 Six less than three times a number is fifteen.
What is the number ?
21/3=7
7
21
6
15
The number is 7
Solving an Algebraic Equation--Practice
 Three more than twice a number is eleven.
What is the number ?
Solving a Simple Algebraic Equation
 Three more than twice a number is eleven.
What is the number ?
2x + 3 = 11
8
2x = 8
4
1 1 1
x = 8/2
x=4
11
The number is 4
 The combined weight of Brad, John and Gregg is 409
lbs. Gregg is 32 lbs heavier than Brad and Brad is 17 lbs
lighter than John. Find John’s weight.
17 lbs
John
409 lbs
Brad
Gregg
32 lbs
17 lbs
John
409 - 17 - 32 lbs
= 360 lbs
Brad
Gregg
John
32 lbs
120 lbs
137 lbs
17 lbs
360 lbs / 3 = 120 lbs
John weighs 120 + 17 = 137 lbs
Solving an Algebraic Equation - Practice
 The combined IQ’s of Mitt, Gary, and Barack is 397.
Barack’s IQ is 7 points higher than Mitt’s, which is 15
points less than Gary’s. Find Barack’s IQ.
 The combined IQ’s of Mitt, Gary,and Barack is 397.
Barack’s IQ is 7 points higher than Mitt’s, which is 15
points less than Gary’s. Find Barack’s IQ.
Barack
7
397
Mitt
Gary
15
Barack
7
397 - 7 - 15 =
375
Mitt
Gary
Barack
15
125
132
7
375 lbs / 3 = 125
Barack’s IQ is 125 + 7 = 132
Mixture Problems
 A “recipe” requires mixing 1 oz of 20% alcohol with 2 oz of 80% alcohol
and 5 oz of orange juice. What is the resulting alcohol concentration?
1 oz
20 %
2 oz
+
80 %
5 oz
+
0%
18/80 = 22 1/2 %
The final concentration is 22 1/2 % alcohol
8 oz
=
?%
Mixture Problems -- Practice
 2 liters of 30% acid are mixed with 1 liter of
60% acid. What is the resulting acid
concentration?
2 liters of 30% acid are mixed with 1 liter of 60% acid. What
is the resulting acid concentration?
2 liters
30 %
1 liter
+
60 %
3 liters
=
?%
The final concentration is 40% acid
Mixture Problems
 What amount and concentration of acid solution must be added
to 1 gal of 60% acid solution in order to get 3 gal of 80% acid
solution?
2
? gal
1 gal
60 %
+
?%
3 gal -1 gal = 2 gal
3 gal
=
80 %
There are 24 shaded
units here. 6 come
from the first bucket.
18 must come from the
second bucket.
Shading each gallon equally
to get 18 total shaded units
results in each gallon with 9
of 10 shaded units
2 gal of 90% acid solution must be
added to 1 gal of 60 % acid
solution to yield 3 gal of 80% acid
solution.
Mixture Problems -- Practice
 What amount and concentration of acid solution must
be added to 2 gal of 30% acid solution in order to get
5 gal of 60% acid solution?
What amount and concentration of acid solution must be
added to 2 gal of 30% acid solution in order to get 5 gal of
60% acid solution?
2 gallons
30 %
3 gallons
+
?%
5 gallons
=
3 gallons of 80% acid must be added.
60 %
Mixture Problems
 How much $1.20 per pound chocolate must be added to 4 pounds of $0.90 per
pound chocolate to get chocolate that averages $1.00 per pound?
? pounds
$1.20 /lb
? pounds
4 pounds
+
$0.90 /lb
=
$1.00 /lb
Each
segment
represents
$0.10
$0.90
2 pounds of $1.20 per pound chocolate must be added to 4 pounds of $0.90 per
pound chocolate to get 6 pounds of chocolate that averages $1.00 per pound
Mixture Problem - Practice
 How much $1.20 per pound chocolate must
be added to 4 pounds of $0.90 per pound
chocolate to get chocolate that averages
$1.10 per pound?
How much $1.20 per pound chocolate must be
added to 4 pounds of $0.90 per pound chocolate to
get chocolate that averages $1.10 per pound?
? pounds
$1.20 /lb
? pounds
4 pounds
+
$0.90 /lb
=
8 pounds of $1.20 per pound chocolate must
be added to 4 pounds of $0.90 per pound
chocolate to get 12 pounds of chocolate that
averages $1.10 per pound
$1.10 /lb
Half Life – Radioactive Decay
A 32 pound of radioactive material decays to 4 lbs in
3000 years
Half - Life in years
Half - Life
Half - Life
Amount
Remaining
4 lbs …
32 lbs
16 lbs
8 lbs
3000 years
Definition: The amount of
time it takes for a material to
decay to ½ of it’s original
amount is called the half-life
Half life =3000/3 =1000 years
Half Life - Practice
 If it takes 2 hrs for a sample to decay from 96
pounds to 12 lbs, how long will it take to
decay to 3 lbs?
half life
If it takes 2 hrs for a sample to decay from 96 pounds to 12
lbs, how long will it take to decay to 3 lbs?
96 lbs
48 lbs
6 lbs
3 lbs
Amount
Remaining
24 lbs 12 lbs
2 hrs
The half life is 120/3
min. = 40 min
It will take two more “half
lifes” to get from 12 pounds
to 3 pounds.
It will take 5 x 40 min. = 200 minutes
to decay from 96 pounds to 3 pounds.
Decibels
Fact: A 3 dB increase is equivalent to a doubling in sound volume.*
Intensity of original
sound = V
Volume
V
2V
4V
8V
16V
0 dB
3 dB
6 dB
9 dB
12 dB
* L  10logIntensityRatio
L  10log2  3.0103dB  3dB
Decibels
Fact: A 3 dB increase is equivalent to a doubling in sound volume.*
A sound engineer finds that adjusting the volume on his
console results in an increase of 15 decibels. By what factor
has the volume increased?
V
2V
4V
8V
16V
0 dB
3 dB
6 dB
9 dB
12 dB
15 dB
32V
15/3=5, so the volume will be
doubled 5 times.
25  32
A 15 dB increase results in the
volume increasing by a factor of
32
Decibels
Axel hears his favorite song on his fancy stereo which and he
turns up the volume such that the volume is increased by a
factor of 64. How many decibels did the sound level increase?
 Axel hears his favorite song on his fancy stereo which and he turns up
the volume such that the volume is increased by a factor of 128. How
many decibels did the sound level increase?
V
0 dB
2V
3 dB
6 dB
4V
9 dB
8V
16V
12 dB
15 dB
32V
64V
18 dB
128V
21 dB
256V
24 dB
 The sound level increased by 21 dB.
System of Equations
y
Solve
y  x 1
x
2 x  3 y  13
x
x
y
y
y=x+1
1
y
13
x
x
x
x
1
1
x
1
 Remove the three “1’s”
y=x+1
x
x
x
10
x
x
2
x=2,y=3
2
2
10
2
2
Systems of Equations - Practice
 A local bake sale sells brownies for $2 each
and cakes for $6 each. At the end of the day
60 more cakes were sold than brownies and
the total revenues were $600. How many
brownies and cakes were sold?
A local bake sale sells brownies for $2 each and cakes for $6
each. At the end of the day 60 more cakes were sold than
brownies and the total revenues were $600. How many
brownies and cakes were sold?
2 B  6C  600
C  B  60
B
B
C
C
C
C
C
C
600
B
B
B
B
B
B
60
B
B
B
240
B
60
B
B
60
B
B
B
60
60
B
60
30
There were 30 Brownies and 90 cakes sold.
Geometry –
 A path up the side of a 500 foot tall hill is 1000 ft. long
A hiker travels 800 feet up the path. What was his
change in elevation?
500 ft
Geometry –
 A path up the side of a 500 foot tall hill is 1000 ft. long
A hiker travels of 800 feet up the path. What was his
change in elevation?

200
The ratio of the line
segments on both
sides must be the
same.
y
500 ft
x
800
y
500 ft
x
100
100
100
100
His change in elevation was 400 feet.
500/5 = 100
Geometry – Practice
 The triangles shown are similar. Find z.
z
9
Geometry –
 The triangle ABC has angles such that angle B is 3
times the measure of angle C and ½ the measure of
angle A. Find the measures of angles A,B, and C.
C
18
54
B
108
A
C
18
B
18
18
18
A
18
18
18
180
degrees
18
18
18
180/10 = 18
Geometry – Practice
 Angles A and B are complementary. Angle A is 2/3
the measure of angle B. Find the measure of angles
A and B
A
B
Rate of Work Problems
Sue can paint a mailbox in 2 hours. It takes Bill 3 hours to
paint the same mailbox. How long will it take them to paint
three of the mailboxes working together?
1/2 Mailbox per hour
Bar represents
one mailbox
Sue
Bill
1/3 Mailbox per hour
Both
5/6 Mailbox per hour
Sue and Bill can
paint 5/6 of a
mailbox in one
hour if they work
together.
Rate of Work Problems
Sue can paint a mailbox in 2 hours. It takes Bill 3 hours to
paint the same mailbox. How long will it take them to paint
three of the mailboxes working together?
1 hour
Both
12 Min
5/6 Mailbox per hour
Sue and Bill can
paint 5/6 of a
mailbox in one
hour if they work
together.
1 mailbox
12
First Hour
Second Hour
Third Hour
36 min
Rate of Work Problems -- Practice
A pro cyclist can complete a race in 2 hours. A
teacher takes 4 hours to complete the same race.
If they share a tandem bike, how long will it take
them to complete the race pedaling together?
Rate of Work Problems
A pro cyclist can complete a race in 2 hours. A teacher takes
4 hours to complete the same race. If they share a tandem
bike, how long will it take them to complete the race pedaling
together?
1/2 race per hour
Bar represents
one race
Pro
Teacher
1/4 race per hour
Both
3/4 race per hour
They can complete
3/4 of the race in
one hour if they
work together.
Rate of Work Problems -- Practice
A pro cyclist can complete a race in 2 hours. A teacher takes 4
hours to complete the same race. If they share a tandem bike,
how long will it take them to complete the race pedaling together?
1 hour
Both
20 Min
They can complete
3/4 of the race in
one hour if they
work together.
¾ race per hour
1 race
20
One hour
It will take them 1
hour and 20
minutes working
together.