The document discusses various mathematical concepts and how to use bar modeling to represent and solve related word problems. It provides examples of using bar modeling to solve problems involving ratios and proportions, fractions, algebraic equations, mixtures, radioactive decay, decibels, systems of equations, geometry, and rate of work. For each type of problem, it presents a sample word problem, models the problem using bars or diagrams, and shows the steps to arrive at the solution. The overall document demonstrates how bar modeling can help demystify and build understanding of challenging math concepts.

1. Demystify Challenging
Problems with Bar
Modeling
Gregg Velatini
Dianna Spence
2012 Georgia Mathematics Conference

2. 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 2
9 x 4 = 36 inches
Rod 1
72 inches
72 / 8 = 9 inches
Rod 3
9
9 9 9 9
The length of the longest rod is 36 inches
9 9 9

3. 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
Teacher
3 Parts
1 Part

4. 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
Teacher
2 Parts
1 Part
200
The teacher had $400 at first.

5. 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?

6. Female Cats
After
Before
90 of the original cats were
female.
Male Cats
Female Cats 35
Male Cats 35
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?
250 Cats
30
30
30 30

7. 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
Cookies
Cake
1
7 eggs

8. 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?

9. 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?
1/6
1/3
 Brad spent 5/6 of his money.
Brad’s Money
1/2
Beanies Nascar
 Brad had 1/6 of his money remaining.

10. Solving an Algebraic Equation
 Six less than three times a number is fifteen.
What is the number?
15
21
The number is 7
6
7 21/3=7

11. Solving an Algebraic Equation--Practice
 Three more than twice a number is eleven.
What is the number?

12. Solving a Simple Algebraic Equation
 Three more than twice a number is eleven.
What is the number?
11
8
2x + 3 = 11
2x = 8
x = 8/2
x = 4
4 1 1 1
The number is 4

13.  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.
409 lbs
32 lbs
17 lbs
Brad
John
Gregg
409 - 17 - 32 lbs
= 360 lbs
32 lbs
17 lbs
Brad
John
Gregg 360 lbs / 3 = 120 lbs
John 17 lbs
120 lbs John weighs 120 + 17 = 137 lbs
137 lbs

14. 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.

15.  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.
397
15
7
Mitt
Barack
Gary
397 - 7 - 15 =
375
15
7
Mitt
Barack
Gary 375 lbs / 3 = 125
Barack 7
125 Barack’s IQ is 125 + 7 = 132
132

16. 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 2 oz 8 oz
20 % 80 % ? %
18/80 = 22 1/2 %
The final concentration is 22 1/2 % alcohol
+
5 oz
0 %

17. Mixture Problems -- Practice
 2 liters of 30% acid are mixed with 1 liter of
60% acid. What is the resulting acid
concentration?

18. 2 liters of 30% acid are mixed with 1 liter of 60% acid. What
is the resulting acid concentration?
=
2 liters 1 liter 3 liters
+
30 % 60 % ? %
The final concentration is 40% acid

19. 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?
+ =
1 gal ? gal 3 gal
60 % ? % 80 %
3 gal -1 gal = 2 gal
2 gal
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.

20. 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?

21. 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 3 gallons 5 gallons
+
30 % ? % 60 %
3 gallons of 80% acid must be added.

22. 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
? pounds
$1.20 /lb $1.00 /lb
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
4 pounds
$0.90 /lb
Each
segment
represents
$0.10
$0.90

23. 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?

24. 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?
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
+ =
? pounds 4 pounds ? pounds
$1.20 /lb $1.10 /lb
$0.90 /lb

25. Half Life – Radioactive Decay
A 32 pound of radioactive material decays to 4 lbs in
3000 years
Half life =3000/3 =1000 years
3000 years
Amount
Remaining
Half - Life in 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
32 lbs 16 lbs 8 lbs
4 lbs …
Half - Life Half - Life

26. 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?

27. 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?
2 hrs
Amount
Remaining
96 lbs 48 lbs
24 lbs 12 lbs
3 lbs
6 lbs
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.

28. Decibels
Volume
 
atio
IntensityR
L log
10
* 
  dB
dB
L 3
0103
.
3
2
log
10 


Fact: A 3 dB increase is equivalent to a doubling in sound volume.*
Intensity of original
sound = V
V
2V
4V
8V
16V
0 dB
3 dB
6 dB
9 dB
12 dB

29. Decibels
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?
Fact: A 3 dB increase is equivalent to a doubling in sound volume.*
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.
32
25

A 15 dB increase results in the
volume increasing by a factor of
32

30. 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?

31.  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
2V
4V
8V
16V
0 dB
3 dB
6 dB
9 dB
12 dB
15 dB
32V
18 dB
21 dB
24 dB
64V
128V
256V
 The sound level increased by 21 dB.

32. System of Equations
Solve
13
3
2
1




y
x
x
y
13
x x y y y
y=x+1
x x 1
1 x x
x 1
x x x x
x
10
2 2 2 2
2
10
x=2,y=3
y
x 1
y=x+1
 Remove the three “1’s”

33. 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?

34. 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?
600
B
60
600
6
2




B
C
C
B
B C C C
240 There were 30 Brownies and 90 cakes sold.
C C C
B B B B B B B B
60 60 60 60 60 60
B B B B B B B B 30

35. 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

36. Geometry –
500 ft
x
y
 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?
 The ratio of the line
segments on both
sides must be the
same.
800
200
x
y
500 ft 500/5 = 100
100
100
His change in elevation was 400 feet.
100
100

37. Geometry – Practice
 The triangles shown are similar. Find z.
z 9

38. 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.
A
B
C
C
B
A
180
degrees
180/10 = 18
18
18
18
18
18
18
18 18 18
18
18
54 108

39. 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

40. 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?
Sue
Bill
Both
1/3 Mailbox per hour
1/2 Mailbox per hour
Bar represents
one mailbox
5/6 Mailbox per hour
Sue and Bill can
paint 5/6 of a
mailbox in one
hour if they work
together.

41. 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?
Both
5/6 Mailbox per hour
Sue and Bill can
paint 5/6 of a
mailbox in one
hour if they work
together.
12 Min
1 hour
12
1 mailbox
First Hour Second Hour Third Hour 36 min

42. 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?

43. 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?
Pro
Both
1/4 race per hour
1/2 race per hour
Bar represents
one race
3/4 race per hour
They can complete
3/4 of the race in
one hour if they
work together.
Teacher

44. 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?
Both
¾ race per hour
They can complete
3/4 of the race in
one hour if they
work together.
20 Min
1 hour
1 race
One hour
It will take them 1
hour and 20
minutes working
together.
20