Solving Verbal Problems Kitty Jay © 2002 Tomball College LAC.

Solving Verbal Problems

Kitty Jay© 2002 Tomball College LAC

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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

http://wwwtc.nhmccd.edu/bluebonnet/education/resources/mlab/applications.doc

READ

Click on each button to read a description.

Contents

GENERAL STRATEGY STEPS

READ


IDENTIFY

Contents


READ

IDENTIFY

FORMULA


Contents


READ

IDENTIFY

FORMULA


DIAGRAM

Contents

GENERAL STRATEGY

STEPS

READ

IDENTIFY

FORMULA


DIAGRAM

EQUATION

Contents


READ

IDENTIFY

FORMULA


DIAGRAM

EQUATION

SOLVE

Contents

GENERAL STRATEGY

STEPS

READ

IDENTIFY

FORMULA


DIAGRAM

EQUATION

Contents


READ

IDENTIFY

FORMULA


DIAGRAM

EQUATION

Contents


READ

IDENTIFY

FORMULA


DIAGRAM

EQUATION

SOLVE

CHECK

Contents


READ

IDENTIFY

FORMULA


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


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.


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.



DistanceContents

Check the answer(s) 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.

Contents

Geometric Problems


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


11


22

422

x

x

x

GeometryContents

Solve the Equation


11


22

422

x

x

x

GeometryContents


The length of the rectangle is 11/2 feet or 5.5 feet.

GeometryContents

Make sure you have answered the question that

was asked.


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.


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


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.


20 + 21 + 22 + 23 = 5(20) – 14

86 = 100 –14

86 = 86Contents


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?


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


problem.

This lesson on solving

application problems is over.

Return to the Contents page for

more practice problems.

Contents

Practice Problems

Solving Verbal Problems Kitty Jay © 2002 Tomball College LAC.

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