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.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Kitty Jay
© 2002 Tomball College LAC
I know the answer is
here someplace.
Table of Contents
Click on a button to go to the page.
Strategies
List of steps to follow for solving word problems
Coins
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
Click on each button
to read a description.
READ
Contents
Click on each button
to read a description.
IDENTIFY
READ
Contents
FORMULA
Click on each button
to read a description.
IDENTIFY
READ
Contents
DIAGRAM
FORMULA
Click on each button
to read a description.
IDENTIFY
READ
Contents
EQUATION
DIAGRAM
FORMULA
Click on each button
to read a description.
IDENTIFY
READ
Contents
SOLVE
EQUATION
DIAGRAM
FORMULA
Click on each button
to read a description.
IDENTIFY
READ
Contents
EQUATION
DIAGRAM
FORMULA
Click on each button
to read a description.
IDENTIFY
READ
Contents
EQUATION
DIAGRAM
FORMULA
Click on each button
to read a description.
IDENTIFY
READ
Contents
CHECK
SOLVE
EQUATION
DIAGRAM
FORMULA
Click on each button
to read a description.
IDENTIFY
READ
Contents
QUESTION
CHECK
SOLVE
EQUATION
DIAGRAM
FORMULA
Click on each button
to read a description.
IDENTIFY
READ
Contents
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 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 answered the original question.
Contents
Contents
if n represents the number of nickels
then 2n will represent the number of dimes
if 2n represents the number of dimes
then 2n - 3 will represent the number of quarters
Contents
Coins
Nickels
Dimes
Quarters
Number of coins
Value of coins
Contents
Coins
Nickels
Dimes
Quarters
Example:
Number of
Value of
n nickels will be worth 5n
2n dimes will be worth 10(2n)(twice the # of nickels)
2n - 3 quarters will be worth 25(2n - 3)(3 fewer quarters than dimes)
2(4)-3=5
4
2(4)=8
5(4)=20 ¢
10(8)=80 ¢
25(5)=125 ¢
Contents
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?
Change the total money to cents also.
Fill in the table:
Nickels
Dimes
Quarters
Total
Number of
n
2n
2n-3
5n
10(2n)
25(2n-3)
450
Value of
Contents
Coins
value of nickels
+ value of dimes
+ value of quarters
= $4.50
+
+
=
5n
10(2n)
25(2n-3)
450
Nickels
Dimes
Quarters
Total
Number of
n
2n
2n-3
5n
10(2n)
25(2n-3)
450
Value of
Contents
Coins
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
Contents
Coins
If there are 7 nickels then there are twice as many dimes or 14 dimes and three fewer quarters or 11 quarters.
Contents
Coins
5(7) + 10(14) +25(11) = 450
35 + 140 + 275 = 450
450 = 450
Contents
Coins
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?
Read the problem carefully to understand what is being asked.
Contents
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
Finish 90 km later
90 - d km @ 25 kph
d km @ 10 kph
Start
Contents
Distance
Audio Clip from “Bicycle” by Queen
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.
t
r
d
First segment
d
d/10
10 kph
Second segment
90-d
25 kph
(90-d)/25
Contents
Distance
t
r
d
First segment
d/10
d
10 kph
Second segment
(90-d)/25
90-d
25 kph
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
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.
d/10 =
(90-d)/25 + 2
Contents
Distance
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
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
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
Distance
Find the length of a rectangle whose width is 4 feet and whose area is 22 square feet.
Read the problem carefully to understand what is being asked.
Contents
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
Width = 4 ft
Area = 24 square feet
Contents
Geometry
Find the length of a rectangle whose width is 4 feet and whose area is 22 square feet.
22
area = length times width
x
4
Contents
Geometry
Contents
Geometry
Solve the Equation
Contents
Geometry
Solve the Equation
Contents
Geometry
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
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
Use the hotlink, then click on c in the web page for a definition. Close the page to return to this lesson.
Read the problem carefully to understand what is being asked.
The sum of four consecutive integers is 14 less than 5 times the smallest integer. Find the four integers.
Contents
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
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
- 14
Contents
Number
x + x + 1 + x + 2 + x + 3 = 5x – 14
4x + 6 = 5x – 14 Collect like terms
20 = x Addition Property
Contents
Number
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
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
Read the problem carefully to understand what is being asked.
How many liters of pure acid should be added to 22 liters of a 30% acid solution to obtain a 45% acid solution?
Contents
22 liters of 30% acid
x liters of pure acid
(100% acid)
x + 22 liters
of 45% acid solution
Contents
Mixture
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
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
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
6 +
0.3(22)
= 0.45(6 + 22)
How many liters of pure acid should be added to 22 liters of a 30% acid solution to obtain a 45% acid solution?
6 + 0.66 = 0.45(28)
12.6 = 12.6
Contents
Mixture
This lesson on solving application problems is over. Return to the Contentspage for more practice problems.
Contents