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Solving Verbal Problems Kitty Jay © 2002 Tomball College LAC.

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Presentation on theme: "Solving Verbal Problems Kitty Jay © 2002 Tomball College LAC."— Presentation transcript:

1 Solving Verbal Problems Kitty Jay © 2002 Tomball College LAC

2 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.

3 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?

4 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 is here someplace.

5 Table of Contents Click 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

6 READ Click on each button to read a description. Contents GENERAL STRATEGY STEPS

7 READ Click on each button to read a description. IDENTIFY Contents GENERAL STRATEGY STEPS

8 READ IDENTIFY FORMULA Click on each button to read a description. Contents GENERAL STRATEGY STEPS

9 READ IDENTIFY FORMULA Click on each button to read a description. DIAGRAM Contents GENERAL STRATEGY STEPS

10 READ IDENTIFY FORMULA Click on each button to read a description. DIAGRAM EQUATION Contents GENERAL STRATEGY STEPS

11 READ IDENTIFY FORMULA Click on each button to read a description. DIAGRAM EQUATION SOLVE Contents GENERAL STRATEGY STEPS

12 READ IDENTIFY FORMULA Click on each button to read a description. DIAGRAM EQUATION Contents GENERAL STRATEGY STEPS

13 READ IDENTIFY FORMULA Click on each button to read a description. DIAGRAM EQUATION Contents GENERAL STRATEGY STEPS

14 READ IDENTIFY FORMULA Click on each button to read a description. DIAGRAM EQUATION SOLVE CHECK Contents GENERAL STRATEGY STEPS

15 READ IDENTIFY FORMULA Click on each button to read a description. DIAGRAM EQUATION SOLVE CHECK QUESTION Contents GENERAL STRATEGY STEPS

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

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

18 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

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

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

21 Solve the equation or inequality. Strategies Solve the equation

22 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)

23 Make sure you have answered the original question. Contents answer

24 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

25 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 CoinsContents

26 Use diagrams or a table whenever you think it will make the given information clearer. CoinsContents NickelsDimesQuarters Number of coins Value of coins

27 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) CoinsContents 4 5(4)=20 ¢ 2(4)=8 10(8)=80 ¢ 2(4)-3=5 25(5)=125 ¢ NickelsDimesQuarters Number of Value of Example:

28 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 NickelsDimesQuarters Number of Value of Total 2n2n-3 5n5n10(2n)25(2n-3)450 Fill in the table:

29 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 5n10(2n)25(2n-3)450 Coins ++= Contents n NickelsDimesQuarters Number of Value of Total 2n2n-3 10(2n)25(2n-3)450 5n5n

30 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

31 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

32 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 CoinsContents

33 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

34 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. Distance Contents

35 d 10 kph 90 - d 25 kph Finish 90 km later Start Audio Clip from “Bicycle” by Queen Distance Contents Draw a picture

36 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-d25 kph d/10 (90-d)/25 Distance Contents Use a Formula

37 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)/ 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. Distance Contents First segment Second segment d d r t 10 kph 90-d25 kph d/10 (90-d)/25 Write the Equation

38 d/10 = (90-d)/ d = 2(90-d) Multiply by 50 to clear the fractions. 5d = d Use the distributive property. 7d = 280 Combine like terms. d = 40 Use the multiplication property Distance Contents Solve the Equation

39 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 or 50 kilometers long. Distance Contents Answer the Question Asked

40 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? Distance Contents Check the answer(s) in the original words of the problem.

41 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

42 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

43 Width = 4 ft Area = 24 square feet GeometryContents Draw a Picture

44 area = length times width Find the length of a rectangle whose width is 4 feet and whose area is 22 square feet. 22x4 GeometryContents Use the Formula

45 GeometryContents Solve the Equation

46 GeometryContents

47 Solve the Equation GeometryContents

48 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.

49 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 GeometryContents Check the answer in the original words of the problem.

50 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. consecutive integers Read the problem carefully to understand what is being asked. Contents Consecutive Integer Problems

51 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

52 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 NumberContents Write the Equation

53 x + x x x + 3 = 5x – 14 4x + 6 = 5x – 14 Collect like terms 20 = x Addition Property NumberContents Solve the Equation

54 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.

55 The sum of four consecutive integers is 14 less than 5 times the smallest integer. Find the four integers = 5(20) – = 100 –14 86 = 86 Contents Check the answer in the original words of the problem.

56 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

57 x + 22 liters of 45% acid solution ContentsMixture 22 liters of 30% acid x liters of pure acid (100% acid) Identify the unknown quantity and label it using one variable. Draw a picture.

58 Liters of solution % of AcidLiters of pure acid Pure acid 30% solution 45% solution X 22 X % 30% 45% X 0.3(22) 0.45(X + 22) Label the rows and columns. Fill in the cells with the given information. ContentsMixture Use a Table to Organize

59 Liters of solution % of AcidLiters of pure acid Pure acid 30% solution 45% solution X 22 X % 30% 45% X 0.3(22) 0.45(X + 22) X +0.3(22)= 0.45(x + 22) ContentsMixture Write the Equation

60 Contents x +0.3(22)= 0.45(x + 22) x = 0.45x x = 3.3 x = 6 6 liters of pure acid should be added Mixture Solve the Equation

61 How many liters of pure acid should be added to 22 liters of a 30% acid solution to obtain a 45% acid solution? (22)= 0.45(6 + 22) = 0.45(28) 12.6 = 12.6 ContentsMixture Check the answer in the original words of the problem.

62 This lesson on solving application problems is over. Return to the Contents page for more practice problems. Contents Practice Problems


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