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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.

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Solving verbal problems l.jpg

Solving Verbal Problems

Kitty Jay

© 2002 Tomball College LAC


Directions l.jpg

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 l.jpg

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?


Slide4 l.jpg

Solving verbal problems is typically one of the more challenging math topics that students encounter.

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.


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Table of Contents

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


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GENERAL STRATEGY STEPS

Click on each button

to read a description.

READ

Contents


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GENERAL STRATEGY STEPS

Click on each button

to read a description.

IDENTIFY

READ

Contents


General strategy steps8 l.jpg

GENERAL STRATEGY STEPS

FORMULA

Click on each button

to read a description.

IDENTIFY

READ

Contents


General strategy steps9 l.jpg

GENERAL STRATEGY STEPS

DIAGRAM

FORMULA

Click on each button

to read a description.

IDENTIFY

READ

Contents


General strategy steps10 l.jpg

GENERAL STRATEGY STEPS

EQUATION

DIAGRAM

FORMULA

Click on each button

to read a description.

IDENTIFY

READ

Contents


General strategy steps11 l.jpg

GENERAL STRATEGY STEPS

SOLVE

EQUATION

DIAGRAM

FORMULA

Click on each button

to read a description.

IDENTIFY

READ

Contents


General strategy steps12 l.jpg

GENERAL STRATEGY STEPS

EQUATION

DIAGRAM

FORMULA

Click on each button

to read a description.

IDENTIFY

READ

Contents


General strategy steps13 l.jpg

GENERAL STRATEGY STEPS

EQUATION

DIAGRAM

FORMULA

Click on each button

to read a description.

IDENTIFY

READ

Contents


General strategy steps14 l.jpg

GENERAL STRATEGY STEPS

CHECK

SOLVE

EQUATION

DIAGRAM

FORMULA

Click on each button

to read a description.

IDENTIFY

READ

Contents


General strategy steps15 l.jpg

GENERAL STRATEGY STEPS

QUESTION

CHECK

SOLVE

EQUATION

DIAGRAM

FORMULA

Click on each button

to read a description.

IDENTIFY

READ

Contents


Read the problem l.jpg

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 l.jpg

Clearly identify

Clearly identify the unknown quantity (or quantities) in the problem, and label it (them) using one variable.

Strategies


Underlying relationship l.jpg

underlying relationship

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


Use diagrams l.jpg

use diagrams,

When appropriate, use diagrams, tables, or charts to organize information.

Strategies


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Translate the information

Translate the information in the problem into an equation or inequality.

Strategies


Solve the equation l.jpg

Solve the equation

Solve the equation or inequality.

Strategies


Check the answer s l.jpg

Check the answer(s)

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


Answer l.jpg

answer

Make sure you have answered the original question.

Contents


Solving verbal problems coins l.jpg

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?

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

Contents


Slide25 l.jpg

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 l.jpg

Nickels

Dimes

Quarters

Number of coins

Value of coins

Use diagrams or a table whenever you think it will make the given information clearer.

Contents

Coins


To fill in the value of each amount of coins remember l.jpg

Nickels

Dimes

Quarters

Example:

Number of

Value of

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)

2(4)-3=5

4

2(4)=8

5(4)=20 ¢

10(8)=80 ¢

25(5)=125 ¢

Contents

Coins


Slide28 l.jpg

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


Slide29 l.jpg

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

Nickels

Dimes

Quarters

Total

Number of

n

2n

2n-3

5n

10(2n)

25(2n-3)

450

Value of

Contents

Coins


Solve the equation30 l.jpg

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

Contents

Coins


Make sure you have answered the question that was asked l.jpg

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 l.jpg

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 l.jpg

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?

Read the problem carefully to understand what is being asked.

Contents


Identify the unknowns l.jpg

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 l.jpg

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 l.jpg

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.

t

r

d

First segment

d

d/10

10 kph

Second segment

90-d

25 kph

(90-d)/25

Contents

Distance


Write the equation l.jpg

Write the Equation

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


Solve the equation38 l.jpg

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 l.jpg

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 problem40 l.jpg

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 l.jpg

Geometric Problems

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


Identify the unknown l.jpg

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 picture43 l.jpg

Draw a Picture

Width = 4 ft

Area = 24 square feet

Contents

Geometry


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Use the Formula

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


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Solve the Equation

Contents

Geometry


Slide46 l.jpg

Solve the Equation

Contents

Geometry


Slide47 l.jpg

Solve the Equation

Contents

Geometry


Make sure you have answered the question that was asked48 l.jpg

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 l.jpg

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 l.jpg

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.

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


Identify the unknown51 l.jpg

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 equation52 l.jpg

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

- 14

Contents

Number


Solve the equation53 l.jpg

Solve the Equation

x + x + 1 + x + 2 + x + 3 = 5x – 14

4x + 6 = 5x – 14 Collect like terms

20 = x Addition Property

Contents

Number


Make sure you have answered the question that was asked54 l.jpg

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 problem55 l.jpg

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 l.jpg

Mixture Problems

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


Identify the unknown quantity and label it using one variable draw a picture l.jpg

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 l.jpg

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 equation59 l.jpg

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 equation60 l.jpg

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 problem61 l.jpg

Check the answer in the original words of the problem.

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


Practice problems l.jpg

Practice Problems

This lesson on solving application problems is over. Return to the Contentspage for more practice problems.

Contents


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