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Solving Verbal ProblemsPowerPoint Presentation

Solving Verbal Problems

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Solving Verbal Problems

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Kitty Jay

© 2002 Tomball College LAC

- Elements on each page are animated automatically.
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- A right arrow button will automatically appear when it is time to move to the next page.

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- 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.

- Do you avoid homework assignments that involve verbal problems?

- Are you confused by all the words?
- Do you have trouble knowing where to start?

I know the answer is

here someplace.

- 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.

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

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to read a description.

IDENTIFY

READ

Contents

FORMULA

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to read a description.

IDENTIFY

READ

Contents

DIAGRAM

FORMULA

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to read a description.

IDENTIFY

READ

Contents

EQUATION

DIAGRAM

FORMULA

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to read a description.

IDENTIFY

READ

Contents

SOLVE

EQUATION

DIAGRAM

FORMULA

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to read a description.

IDENTIFY

READ

Contents

EQUATION

DIAGRAM

FORMULA

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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

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to read a description.

IDENTIFY

READ

Contents

QUESTION

CHECK

SOLVE

EQUATION

DIAGRAM

FORMULA

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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

- 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?

- Read the problem carefully, as many times as is necessary to understand what the problem is saying and what it is asking.

Contents

- 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

Nickels

Dimes

Quarters

Number of coins

Value of coins

Contents

Coins

Nickels

Dimes

Quarters

Example:

Number of

Value of

- 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)

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

- 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

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?

- 40 km + 50 km = 90 km

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

20 = x

20 is the smallest integer

21 is the second

22 is the third

23 is the fourth

Contents

Number

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

- 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

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