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Solving Verbal Problems

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

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  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. 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.

  5. 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

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

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

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

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

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

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

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

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

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

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

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

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

  18. 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

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

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

  21. Solve the equation Solve the equation or inequality. Strategies

  22. 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

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

  24. 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

  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 Contents Coins

  26. 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

  27. 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

  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. 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

  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 + + = 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

  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 Contents Coins

  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. Contents Coins

  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 35 + 140 + 275 = 450 450 = 450 Contents Coins

  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. Contents Distance

  35. 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

  36. 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

  37. 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

  38. 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

  39. 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

  40. 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

  41. 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

  42. 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

  43. Draw a Picture Width = 4 ft Area = 24 square feet Contents Geometry

  44. 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

  45. Solve the Equation Contents Geometry

  46. Solve the Equation Contents Geometry

  47. Solve the Equation Contents Geometry

  48. 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

  49. 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

  50. 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

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