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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 ishere someplace.
Table of ContentsClick on a button to go to the page.Strategies
Coins
Distance
Geometry
Number
List of steps to follow for solving word problems
Solving problems involving money
Solving uniform motion problems, sound clip included
Solving problems involving geometric formulas
Solving consecutive integer number problems
Mixture Solving mixture problems
Practice Additional problems, answers included
Table of Contents
READ
Click on each button to read a description.
Contents
GENERAL STRATEGY STEPS
READ
Click on each button to read a description.
IDENTIFY
Contents
GENERAL STRATEGY STEPS
READ
IDENTIFY
FORMULA
Click on each button to read a description.
Contents
GENERAL STRATEGY STEPS
READ
IDENTIFY
FORMULA
Click on each button to read a description.
DIAGRAM
Contents
GENERAL STRATEGY
STEPS
READ
IDENTIFY
FORMULA
Click on each button to read a description.
DIAGRAM
EQUATION
Contents
GENERAL STRATEGY STEPS
READ
IDENTIFY
FORMULA
Click on each button to read a description.
DIAGRAM
EQUATION
SOLVE
Contents
GENERAL STRATEGY
STEPS
READ
IDENTIFY
FORMULA
Click on each button to read a description.
DIAGRAM
EQUATION
Contents
GENERAL STRATEGY STEPS
READ
IDENTIFY
FORMULA
Click on each button to read a description.
DIAGRAM
EQUATION
Contents
GENERAL STRATEGY STEPS
READ
IDENTIFY
FORMULA
Click on each button to read a description.
DIAGRAM
EQUATION
SOLVE
CHECK
Contents
GENERAL STRATEGY STEPS
READ
IDENTIFY
FORMULA
Click on each button to read a description.
DIAGRAM
EQUATION
SOLVE
CHECK
QUESTION
Contents
GENERAL STRATEGY
STEPS
carefully, as many times as is necessary to understand what the problem is saying and what it is asking.
Strategies
Read the problem
Clearly identify the unknown quantity (or quantities) in the problem, and label it (them) using one variable.Strategies
Clearly identify
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
underlying relationship
When appropriate, use diagrams, tables, or charts to organize information.Strategies
use diagrams,
Translate the information in the problem into an equation or inequality.
Strategies
Translate the information
Solve the equation or inequality.
Strategies
Solve the equation
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
Check the answer(s)
Make sure you have answered the original question.
Contents
answer
Solving Verbal Problems - Coins
• Read the problem carefully, as many times as is necessary to understand what the problem is saying and what it is asking.
• 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
Clearly identify the unknown quantity (or quantities) in the problem, and label it (them) using one variable. • there are twice as many dimes as
nickelsif n represents the number of nickels
then 2n will represent the number of dimes
• 3 fewer quarters than dimesif 2n represents the number of dimes
then 2n - 3 will represent the number of quarters
CoinsContents
Use diagrams or a table whenever you think it will make the given information clearer.
CoinsContents
Nickels Dimes Quarters
Number of coins
Value of coins
To fill in the value of each amount of coins, remember:
• each nickel is worth 5 centsn nickels will be worth 5n
• each dime is worth 10 cents2n dimes will be worth 10(2n) (twice the # of nickels)
• each quarter is worth 25 cents2n - 3 quarters will be worth 25(2n - 3) (3 fewer quarters than dimes)
CoinsContents
4
5(4)=20 ¢
2(4)=8
10(8)=80 ¢
2(4)-3=5
25(5)=125 ¢
Nickels Dimes Quarters
Number of
Value of
Example:
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.
CoinsContents
n
Nickels Dimes QuartersNumber
ofValue of
Total
2n 2n-3
5n 10(2n)
25(2n-3) 450
Fill in the table:
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) 25(2n-3) 450
Coins
+ + =
Contents
n
Nickels Dimes QuartersNumber
ofValue of
Total
2n 2n-3
10(2n)
25(2n-3) 4505n
Solve the equation.
75n - 75 = 450 Distribute and collect like terms.
75n = 525 Use the Addition Property
n = 7 Use the Multiplication Property
5n + 10(2n) + 25(2n-3) =
450
CoinsContents
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.
CoinsContents
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
CoinsContents
Distance ProblemsA 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?
Read the problem carefully to understand what is being asked.
Contents
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.
DistanceContents
d km @ 10 kph
90 - d km @ 25 kph
Finish 90 km later
Start
Audio Clip from “Bicycle” by Queen
DistanceContents
Draw a picture
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.
First segment
Second segment
d
d
r t
10 kph
90-d 25 kph
d/10
(90-d)/25
DistanceContents
Use a Formula
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
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.DistanceContents
First segment
Second segment
d
d
r t
10 kph
90-d 25 kph
d/10
(90-d)/25
Write 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
DistanceContents
Solve the Equation
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.
DistanceContents
Answer the Question Asked
•40 km + 50 km = 90 km
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?
DistanceContents
Check the answer(s) in the original words of the
problem.
Read the problem carefully to understand what is being asked.
Find the length of a rectangle whose width is 4 feet and whose area is 22 square feet.
Contents
Geometric Problems
Find the length of a rectangle whose width is 4 feet and whose area is 22 square feet.
x = the length of the rectangle
GeometryContents
Identify the Unknown
Width = 4 ftArea = 24 square
feet
GeometryContents
Draw a Picture
area = length times width
Find the length of a rectangle whose width is 4 feet
and whose area is 22 square feet.
22 x 4
GeometryContents
Use the Formula
fraction thereduce 2
11
4by sidesboth on divide 4
22
422
x
x
x
GeometryContents
Solve the Equation
Solve the Equation
fraction thereduce 2
11
4by sidesboth on divide 4
22
422
x
x
x
GeometryContents
Solve the Equation
fraction thereduce 2
11
4by sidesboth on divide 4
22
422
x
x
x
GeometryContents
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.
GeometryContents
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.
area = length times width
length = 5.5 feet
22 = ( 5.5 )( 4 )
22 = 22GeometryContents
Check the answer in the original words of the
problem.
Use the hotlink, then click on c in the web page for
a definition. Close the page to return
to this lesson.
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.
Read the problem carefully to understand what is being asked.
Contents
Consecutive Integer Problems
Number
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
Identify the Unknown
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 integerx + x + 1+ x + 2 + x + 3= 5x - 14
NumberContents
Write the Equation
x + x + 1 + x + 2 + x + 3 = 5x – 14
4x + 6 = 5x – 14 Collect like terms
20 = x Addition Property
NumberContents
Solve the Equation
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
NumberContents
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 + 21 + 22 + 23 = 5(20) – 14
86 = 100 –14
86 = 86Contents
Check the answer in the original words of the
problem.
Contents
How many liters of pure acid should be added to 22 liters of a 30% acid solution to obtain a 45% acid solution?
Read the problem carefully to understand what is being asked.
Mixture Problems
x + 22 liters
of 45% acid
solutionContents Mixture
22 liters of 30%
acidx liters of pure acid
(100% acid)
Identify the unknown quantity and label it using one variable. Draw a picture.
Liters of solution
% of Acid Liters of pure acid
Pure acid
30% solutio
n
45% solutio
n
X
22
X + 22
100%
30%
45%
X
0.3(22)
0.45(X + 22)
•Label the rows and columns.
•Fill in the cells with the given information.
Contents Mixture
Use a Table to Organize
Liters of solution
% of Acid Liters of pure acid
Pure acid
30% solutio
n
45% solutio
n
X
22
X + 22
100%
30%
45%
X
0.3(22)
0.45(X + 22)
X + 0.3(22) = 0.45(x + 22)
Contents Mixture
Write the Equation
Contents
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
Mixture
Solve the Equation
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
Check the answer in the original words of the
problem.
This lesson on solving
application problems is over.
Return to the Contents page for
more practice problems.
Contents
Practice Problems