Transcript Solving Verbal Problems, VGA - LSC

Solving Verbal Problems
Kitty Jay
© 2002 Tomball College LAC
Directions
• Elements on each page are animated automatically.
– Wait for items to appear on the page.
– A right arrow button will automatically appear when
it is time to move to the next page.
• Do not right click on a page to return to the previous
page.
– Use the buttons on each page to return to the menu,
application type, etc.
• If a link takes you to an Internet page, do not use the
back arrow on the web menubar.
– Close the web page which will expose the current
PowerPoint slide.
Verbal Problems, Your Worst
Nightmare
• Do you avoid
homework
assignments that
involve verbal
problems?
• Are you confused
by all the words?
• Do you have
trouble knowing
where to start?
Solving verbal problems is typically one of
the more challenging math topics that
students encounter.
• This presentation has
some of the typical
types of verbal
problems worked out in
detail.
• After viewing this
presentation you
should be able to
identify each type of
verbal problem and an
appropriate approach
for solving it.
I know the answer is
here someplace.
Table of
Table
ofContents
Contents
Click on a button to go to the page.
Strategies
Coins
List of steps to follow for solving word problems
Solving problems involving money
Distance
Solving uniform motion problems, sound clip included
Geometry
Solving problems involving geometric formulas
Number
Solving consecutive integer number problems
Mixture
Solving mixture problems
Practice
Additional problems, answers included
GENERAL STRATEGY STEPS
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
IDENTIFY
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
FORMULA
IDENTIFY
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
DIAGRAM
FORMULA
IDENTIFY
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
EQUATION
DIAGRAM
FORMULA
IDENTIFY
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
SOLVE
EQUATION
DIAGRAM
FORMULA
IDENTIFY
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
EQUATION
DIAGRAM
FORMULA
IDENTIFY
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
EQUATION
DIAGRAM
FORMULA
IDENTIFY
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
CHECK
SOLVE
EQUATION
DIAGRAM
FORMULA
IDENTIFY
Click on each button
to read a description.
READ
Contents
GENERAL
STRATEGY
STEPS
QUESTION
CHECK
SOLVE
EQUATION
DIAGRAM
FORMULA
IDENTIFY
Click on each button
to read a description.
READ
Contents
Read the problem
carefully, as many
times as is
necessary to
understand what the
problem is saying
and what it is
asking.
Strategies
Clearly identify the
unknown quantity
(or quantities) in
the problem, and
label it (them) using
one variable.
Strategies
Is there some
underlying
relationship or
formula you need to
know? If not, then
the words of the
problem themselves
give the required
relationship.
Strategies
When
appropriate,
use
use diagrams,
diagrams,
tables, or
charts to
organize
information.
Strategies
Translate the
information in
the problem
into an
equation or
inequality.
Strategies
Solve the
equation or
inequality.
Strategies
Check the
answer(s) in the
original words of
the problem to
make sure you
have met all of the
conditions stated
in the problem.
Strategies
Make sure you
have answer
answered
the original
question.
Contents
Solving Verbal Problems - Coins
• In a collection of
nickels, dimes, and
quarters, there are
twice as many dimes
as nickels, and 3
fewer quarters than
dimes. If the total
value of the coins is
$4.50, how many of
each type of coin are
there?
Contents
• Read the problem
carefully, as many
times as is
necessary to
understand what the
problem is saying
and what it is
asking.
Clearly identify the unknown quantity
(or quantities) in the problem, and
label it (them) using one variable.
• there are twice as many dimes as nickels
if n represents the number of nickels
then 2n will represent the number of dimes
• 3 fewer quarters than dimes
if 2n represents the number of dimes
then 2n - 3 will represent the number of
quarters
Contents
Coins
Use diagrams or a table whenever you think
it will make the given information clearer.
Nickels
Number of coins
Value of coins
Contents
Coins
Dimes
Quarters
To fill in the value of each amount of
coins, remember:
• each nickel is worth 5 cents
n nickels will be worth 5n
• each dime is worth 10 cents
2n dimes will be worth 10(2n) (twice the # of nickels)
• each quarter is worth 25 cents
2n - 3 quarters will be worth 25(2n - 3) (3 fewer quarters
than dimes)
Example:
Nickels
Number of
Value of
Contents
4
5(4)=20 ¢
Coins
Dimes
Quarters
2(4)=8
2(4)-3=5
10(8)=80 ¢
25(5)=125 ¢
In a collection of nickels, dimes, and quarters,
there are twice as many dimes as nickels, and 3
fewer quarters than dimes. If the total value of
the coins is $4.50, how many of each type of
coin are there?
Change the total money to cents also.
Fill in the table:
Nickels
Dimes
Number of
n
2n
Value of
5n
Contents
Coins
10(2n)
Quarters
Total
2n-3
25(2n-3)
450
Using the information in the “value of coins”
row of the table, write an equation that can be
used to find the number of each type of coin.
value of nickels + value of dimes + value of quarters = $4.50
5n
+
10(2n)
Contents
25(2n-3)
Nickels
Dimes
n
2n
Number of
Value of
+
5n
Coins
10(2n)
=
450
Quarters
Total
2n-3
25(2n-3)
450
Solve the equation.
5n + 10(2n) + 25(2n-3) = 450
75n - 75 = 450
Distribute and collect like terms.
75n = 525
Use the Addition Property
n=7
Use the Multiplication Property
Contents
Coins
Make sure you have answered the
question that was asked.
If there are 7 nickels then
there are twice as many
dimes or 14 dimes and three
fewer quarters or 11 quarters.
Contents
Coins
Check the answer(s) in the original
words of the problem.
• In a collection of nickels, dimes, and
quarters, there are twice as many dimes
as nickels, and 3 fewer quarters than
dimes. If the total value of the coins is
$4.50, how many of each type of coin are
there?
5(7) + 10(14) +25(11) = 450
35 + 140 + 275 = 450
450 = 450
Contents
Coins
Distance Problems
A bike race consists of two
segments whose total length
is 90 kilometers.
The first segment is covered
at 10 kph and takes 2 hours
longer to complete than the
second segment, which is
covered at 25 kph.
How long is each segment?
Contents
Read the problem
carefully to
understand what
is being asked.
Identify the Unknowns
How long is each segment?
The length of the second segment of the
race is equal to the total distance minus
the length of the other segment of the
race.
Contents
Distance
Draw a picture
Finish
90 km
later
90 - d km @ 25 kph
d km @ 10 kph
Start
Contents
Distance
Audio Clip from “Bicycle” by Queen
Use a Formula
Since the problem gives information
about the time involved, use the
formula:
t = d/r (time equals distance divided by the rate)
to fill in the table below.
d
r
t
d/10
First
segment
d
10 kph
Second
segment
90-d
25 kph
Contents
Distance
(90-d)/25
Write the Equation
d
r
First
segment
d
10 kph
Second
segment
90-d
25 kph
t
d/10
(90-d)/25
It takes two hours longer to cover the first segment of the race.
To make the two times equal, add two hours to the time it takes
to cover the second segment
d/10 = (90-d)/25 + 2
Contents
Distance
For example, if it takes 4 hours to
cover the first segment, it will take 2
hours to cover the second segment.
To make the two times equal add 2
hours to the shorter time.
Solve the Equation
d/10 = (90-d)/25 + 2
5d = 2(90-d) + 100
Multiply by 50 to clear the fractions.
5d = 180 - 2d + 100
Use the distributive property.
7d = 280
Combine like terms.
d = 40
Use the multiplication property
Contents
Distance
Answer the Question Asked
A bike race consists of two segments whose
total length is 90 kilometers.
The first segment is covered at 10 kph and
takes 2 hours longer to complete than the
second segment, which is covered at 25 kph.
How long is each segment?
The first segment is 40 kilometers long so
the second segment is 90 - 40 or 50
kilometers long.
Contents
Distance
Check the answer(s) in the
original words of the problem.
A bike race consists of two segments whose
total length is 90 kilometers.
The first segment is covered at 10 kph and takes
2 hours longer to complete than the second
segment, which is covered at 25 kph.
How long is each segment?
40 km + 50 km = 90 km
•
Contents
Distance
Geometric Problems
Find the length
of a rectangle
whose width is
4 feet and
whose area is
22 square feet.
Contents
Read the
problem carefully
to understand
what is being
asked.
Identify the Unknown
Find the length of a rectangle whose
width is 4 feet and whose area is 22
square feet.
x = the length of the rectangle
Contents Geometry
Draw a Picture
Area = 24 square feet
Contents Geometry
Width = 4 ft
Use the Formula
Find the length of a rectangle whose width is 4 feet
and whose area is 22 square feet.
22
area = length
x times width
4
Contents Geometry
Solve the Equation
22  4 x
22
 x divide on both sides by 4
4
11
 x reduce the fraction
2
Contents Geometry
Solve the Equation
22  4 x
22
 x divide on both sides by 4
4
11
 x reduce the fraction
2
Contents
Geometry
Solve the Equation
22  4 x
22
 x divide on both sides by 4
4
11
 x reduce the fraction
2
Contents
Geometry
Make sure you have answered
the question that was asked.
Find the length of a rectangle whose
width is 4 feet and whose area is 22
square feet.
The length of the rectangle is
11/2 feet or 5.5 feet.
Contents Geometry
Check the answer in the original
words of the problem.
Find the length of a rectangle whose
width is 4 feet and whose area is 22
square feet.
area = length times width
length = 5.5 feet
22 = ( 5.5 )( 4 )
22 = 22
Contents Geometry
Consecutive Integer Problems
Use the hotlink, then
click on c in the web
page for a definition.
Close the page to
return to this lesson.
The sum of four
consecutive integers
is 14 less than 5
times the smallest
integer. Find the
four integers.
Contents
Read the
problem carefully
to understand
what is being
asked.
Identify the Unknown
The sum of four consecutive integers
is 14 less than 5 times the smallest
integer. Find the four integers.
x = the first integer
x+1= the second integer
x+2= the third integer
x+3= the fourth integer
Contents
Number
Write the Equation
The sum of four consecutive integers
is 14 less than 5 times the smallest
integer. Find the four integers.
x = the smallest integer
x+1 = the second integer
x+2 = the third integer
x+3 = the fourth integer
x + x + 1 + x + 2 + x + 3 = 5x
Contents
Number
- 14
Solve the Equation
x + x + 1 + x + 2 + x + 3 = 5x – 14
4x + 6 = 5x – 14 Collect like terms
20 = x
Contents
Number
Addition Property
Make sure you have answered
the question that was asked.
The sum of four consecutive integers
is 14 less than 5 times the smallest
integer. Find the four integers.
20 = x
20 is the smallest integer
21 is the second
22 is the third
23 is the fourth
Contents
Number
Check the answer in the
original words of the problem.
The sum of four consecutive integers is
14 less than 5 times the smallest integer.
Find the four integers.
20 + 21 + 22 + 23 = 5(20) – 14
86 = 100 –14
86 = 86
Contents
Mixture Problems
How many liters of pure
acid should be added to 22
liters of a 30% acid
solution to obtain a 45%
acid solution?
Contents
Read the problem
carefully to
understand what
is being asked.
Identify the unknown quantity and
label it using one variable. Draw a
picture.
22 liters of
30% acid
x liters of
pure acid
(100% acid)
x + 22 liters
of 45% acid
solution
Contents
Mixture
Use a Table to Organize
•Label the rows and columns.
•Fill in the cells with the given information.
Liters of
solution
% of Acid
Liters of
pure acid
Pure
acid
X
100%
X
30%
solution
22
30%
0.3(22)
45%
solution
X + 22
45%
0.45(X + 22)
Contents
Mixture
Write the Equation
Liters of
solution
% of Acid
Liters of
pure acid
Pure
acid
X
100%
X
30%
solution
22
30%
0.3(22)
45%
solution
X + 22
45%
0.45(X + 22)
X + 0.3(22) = 0.45(x + 22)
Contents
Mixture
Solve the Equation
x + 0.3(22) = 0.45(x + 22)
x + 6.6 = 0.45x + 9.9
0.55x = 3.3
x=6
6 liters of pure acid should be added
Contents
Mixture
Check the answer in the original
words of the problem.
How many liters of pure acid should be added to
22 liters of a 30% acid solution to obtain a 45%
acid solution?
6 + 0.3(22) = 0.45(6 + 22)
6 + 0.66 = 0.45(28)
12.6 = 12.6
Contents
Mixture
Practice Problems
This lesson on
solving application
problems is over.
Return to the
Contents page for
more practice
problems.
Contents