# Algebra: Word Problems

## Video Lectures

Displaying all 62 video lectures.

Lecture 1Play Video |
Mixture Problems (1 of 5)Everyone at one point in their lives dreaded the infamous word problems in their algebra class, but fear not, once you learn to do them you'll find them easy and maybe even a little fun! (OK maybe not fun, but definitely easy) In this lecture we're going to cover the method used to obtain our desired concentration of an acid solution from two solutions of different concentrations. It's much easier than it sounds, so stick with it and I'll show you how its done! |

Lecture 2Play Video |
Mixture Problems (2 of 5)Everyone at one point in their lives dreaded the infamous word problems in their algebra class, but fear not, once you learn to do them you'll find them easy and maybe even a little fun! (OK maybe not fun, but definitely easy) In this lecture I'll cover a different type of mixture problem. Practice makes perfect right? So stay tuned and get ready to learn! |

Lecture 3Play Video |
Mixture Problems (3 of 5)Everyone at one point in their lives dreaded the infamous word problems in their algebra class, but fear not, once you learn to do them you'll find them easy and maybe even a little fun! (OK maybe not fun, but definitely easy) This time, we're going to tackle the mixture problem from a business angle! I hope you're catching on now and don't hesitate to leave me a comment if you have a question or a specific problem you would like to cover! |

Lecture 4Play Video |
Mixture Problems (4 of 5)This lecture series will cover the methods of solving mixture problems. Due to the popularity of mixture problems, I decided to expand this section. |

Lecture 5Play Video |
Mixture Problems (5 of 5)This lecture series will cover the methods of solving mixture problems. Due to the popularity of mixture problems, I decided to expand this section. |

Lecture 6Play Video |
The Sum of Consecutive Integers (1 of 3)Everyone at one point in their lives dreaded the infamous word problems in their algebra class, but fear not, once you learn to do them you'll find them easy and maybe even a little fun! (OK maybe not fun, but definitely easy) OK, enough mixtures! Now we're gonna tackle number problems. In this lecture we're going to find the value of four consecutive integers! It's as easy as it sounds! |

Lecture 7Play Video |
The Sum of Consecutive Integers (2 of 3)Everyone at one point in their lives dreaded the infamous word problems in their algebra class, but fear not, once you learn to do them you'll find them easy and maybe even a little fun! (OK maybe not fun, but definitely easy) Here's another example of finding consecutive numbers. |

Lecture 8Play Video |
The Sum of Consecutive Integers (3 of 3)Everyone at one point in their lives dreaded the infamous word problems in their algebra class, but fear not, once you learn to do them you'll find them easy and maybe even a little fun! (OK maybe not fun, but definitely easy) Let's tie these number problems into real life. Say you have 78 problems due in 4 days and you want to do 5 less problems each day. How would you space these problems out across each day? Lets find out! |

Lecture 9Play Video |
Age Word Problems (1/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try one that deals with age: Problem Text: A man is 21 years older than his son. Five years ago he was 4 times as old as his son. What are their ages now? |

Lecture 10Play Video |
Age Word Problems (2/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try a problem that deals with age: Problem Text: A man is 6 years older than 3 times his son's age. Five years from now the sum of their ages will be 56. How old are they now? |

Lecture 11Play Video |
Age Word Problems (3/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try one that deals with age: Problem Text: A man is 6 times as old as his son. In 3 years he will be 4 times as old. How old are the man and his son? |

Lecture 12Play Video |
Number Word Problems (1/4)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try a problem that deals with generic numbers: Problem Text: The larger of 2 numbers is 10 more than the smaller if the larger number is 6 times the smaller number. What are the two numbers? |

Lecture 13Play Video |
Number Word Problems (2/4)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try a problem that deals with football teams and their scores: Problem Text: Two football teams played a game. One team scored four less than twice the other. Together they scored 50 points. How many points did each team score? |

Lecture 14Play Video |
Number Word Problems (3/4)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try a problem that deals with more generic numbers. Problem Text: One number is 4 times another, and their difference is 69. What are the numbers? |

Lecture 15Play Video |
Number Word Problems (4/4)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try a problem that deals with more generic numbers. Problem Text: One number is 4 less than another number. The sum of the two numbers is -22. What are the numbers? |

Lecture 16Play Video |
Rate and Distance (1/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. In this problem we'll find the speed of a driver by comparing their driving time, distance, and speed of another driver. Problem Text: Sammy leaves town drive 40 miles per hour. 2 hours later, Carol leaves from the same place traveling the same road. She catches Sammy after 5 hours and 20 minutes. How fast was Carol driving? |

Lecture 17Play Video |
Rate and Distance (2/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. In this problem we'll find the speed of a driver by comparing their driving time, distance, and speed of another driver. Problem Text: It takes 1 1/4 hours more time for Jim, traveling at 12 miles per hour, to ride two miles farther than Tom, who travels at 16 miles per hour. How long (in time) did each ride? |

Lecture 18Play Video |
Rate and Distance (3/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. In this problem we'll find the speed of a driver by comparing their driving time, distance, and speed of another driver, but this time the drivers are heading towards each other! Problem Text: A police car and a sports car are 15 miles apart and traveling towards one another. The sports car is traveling 10 miles per hour faster than the police car. They meet in 6 minutes. How fast are they traveling? |

Lecture 19Play Video |
Percentages (1/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals with percentages. Problem Text: A retailer has some sweaters that cost $38 each. She wants to sell them at a profit of 20% of her cost. What price should she charge? |

Lecture 20Play Video |
Percentages (2/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals with percentages and price. Problem Text: A toy store is selling bags of marbles at 20% off. The sale price is $2.24. What was the original price? |

Lecture 21Play Video |
Percentages (3/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals with percentages and price. Problem Text: Mary is paid $325 dollars per week, plus a commission of 8% on sales. How much does she need to sell to make $500 per week. |

Lecture 22Play Video |
Investment (1/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals interest rate percentages. Problem Text: Mary invested a certain amount of money at a 10% interest rate and $2000 more than that at a 12% interest rate. Her total yearly interest earned is $1340. How much money did she invest at each rate? |

Lecture 23Play Video |
Investment (2/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals interest rate percentages. Problem Text: A total of $6000 was invested, a portion at 6% and the remainder at 8%. The total amount of interest earned was $450. How much was invested at each rate? |

Lecture 24Play Video |
Investment (3/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals interest rate percentages. Problem Text: A sum of $4000 is to be invested in 2 portions, on at 4%, the other at 6%. If the interest earned with the portion at 4% exceeds the interest earned by the portion at 6% by $40, how much is invested at each rate? |

Lecture 25Play Video |
Ratios (1/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals with ratios. Problem Text: A board, 18 feet long, is cut into 2 pieces such that one piece is 1/5 the length of the other piece. Find the length of each piece. |

Lecture 26Play Video |
Ratios (2/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals with ratios. Problem Text: Find a number such that one half of the number is 3 feet less than two thirds the number. |

Lecture 27Play Video |
Ratios (3/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This problem deals with ratios. This problem is a little more complicated than the first two, but it's bark is worse than its bite. Problem Text: One third of a number plus five fourths of the number is 5 more than four thirds of the number. |

Lecture 28Play Video |
Rates of Performing Work (1/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This series introduces the rates of performing work problems. A lot of people have trouble with these kinds of problems, but keep watching and you'll see it's not nearly as hard as you think! Problem Text: Mary can paint a garage by herself in 6 hours. John can paint the same garage by himself in 9 hours. How fast can they paint it together. |

Lecture 29Play Video |
Rates of Performing Work (2/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This series introduces the rates of performing work problems. A lot of people have trouble with these kinds of problems, but keep watching and you'll see it's not nearly as hard as you think! Problem Text: It takes 12 hours to fill a water tank. It takes 16 hours to drain the same water tank. How long will it take to fill the tank if the drain is left open? |

Lecture 30Play Video |
Rates of Performing Work (3/3)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. This series introduces the rates of performing work problems. A lot of people have trouble with these kinds of problems, but keep watching and you'll see it's not nearly as hard as you think! Problem Text: Jim can dig a hole by himself in 12 hours. John can do it in 8 hours. Jack can do it in 6. How long will it take them if they all work together? |

Lecture 31Play Video |
Find the equation from 2 points (1 of 2)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. These problems will deal with linear equations. Problem Text: Find the equations given these two points: (-1,2) (2,5). |

Lecture 32Play Video |
Find the equation from 2 points (2 of 2)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. These problems will deal with linear equations using y=mx+b. Problem Text: Find the equations given these two points: (-2, 5) (3, -3). |

Lecture 33Play Video |
Finding the slopeWord problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. Remember: Slope = rise / run. Problem Text: Find the slope given these two points: (-2, 4) (3, -2). |

Lecture 34Play Video |
Finding the slope (when given an equation).Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. Remember: Slope = rise / run. Problem Text: Find the slope given this equation: 4x + y = 7. |

Lecture 35Play Video |
Find the X and Y InterceptsWord problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. Remember: Slope = rise / run. Problem Text: Find the equation of the line: x - 2y = 6 |

Lecture 36Play Video |
Finding the equation of a line from the slope and a pointWord problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem. Remember: y = mx + b. Problem Text: Find the equation of a line given the slop and one point on the line: m = -2 ( -1,6). |

Lecture 37Play Video |
Problems with Two Variables (1 of 5)This five part lecture series will cover methods of solving word problems in algebra that have two variables. |

Lecture 38Play Video |
Problems with Two Variables (2 of 5)This five part lecture series will cover methods of solving word problems in algebra that have two variables. |

Lecture 39Play Video |
Problems with Two Variables (3 of 5)This five part lecture series will cover methods of solving word problems in algebra that have two variables. |

Lecture 40Play Video |
Problems with Two Variables (4 of 5)This five part lecture series will cover methods of solving word problems in algebra that have two variables. |

Lecture 41Play Video |
Problems with Two Variables (5 of 5)This five part lecture series will cover methods of solving word problems in algebra that have two variables. |

Lecture 42Play Video |
Growth and Decay Word Problems (1 of 7) Population GrowthIn this first of seven part lecture series I will show you how to calculate what the population growth of a town of 100,000 will be in 10 years. |

Lecture 43Play Video |
Growth and Decay Word Problems (2 of 7) Population GrowthIn this video I will show you how to calculate the original bacteria culture if the culture was found to have a population of 8000 after 5 hours and the growth constant is k=0.25. |

Lecture 44Play Video |
Growth and Decay Word Problems (3 of 7) Population GrowthIn this video I will show you how to calculate how many years it would take for the population of a town to double if the town grows at a rate of 1% per year. |

Lecture 45Play Video |
Growth and Decay Word Problems (4 of 7) Air PressureIn this video I will show you how to calculate the air pressure of Denver and Denver's air pressure as a % of the pressure at sea level. |

Lecture 46Play Video |
Growth and Decay Word Problems (5 of 7) Air Pressure on Mt. EverestIn this video I will show you how to calculate at what elevation is the air pressure 1/2 of what it is at sea level; and what fraction of atmospheric pressure does a climber experience at Mt. Everest. |

Lecture 47Play Video |
Growth and Decay Word Problems (6 of 7) Population GrowthIn this video I will show you how to calculate the population of a town in 25 years if the town experiences a population growth of 5% every 10 years. |

Lecture 48Play Video |
Growth and Decay Word Problems (7 of 7) Radioactive DecayIn this video I will show you how to calculate the age of an artifact that has 12% of its original carbon-14 using radioactive carbon-14 dating. |

Lecture 49Play Video |
Numbers (1 of 2)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try a problem that deals with more generic numbers. Problem Text: One number is 2 less than 4 times the other. The sum of the two is 33. Find the numbers. |

Lecture 50Play Video |
Numbers (2 of 2)Word problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try a problem that deals with more generic numbers. Problem Text: One number is 5 more than 4 times another. The larger number plus six times the smaller number is 55. Find the numbers. |

Lecture 51Play Video |
Consecutive Even IntegersWord problems are challenging, but rewarding once you learn how they're done! Let's take a look at another word problem, this time we'll try a problem that deals with more generic numbers. Problem Text: Find three consecutive even integers. Five times the difference between the third and the first number is ten less than five times the second number. |

Lecture 52Play Video |
Geometric Shapes: Rectangles (1 of 2)Word problems are challenging, but rewarding once you learn how they're done! Problem Text: The length of a rectangle is three less than four times the width. The perimeter is 34. Fine the length and width. |

Lecture 53Play Video |
Geometric Shape: Rectangles 2 of 2Word problems are challenging, but rewarding once you learn how they're done! Problem Text: The length of a rectangle is two more than twice the width. If the length is increased by 8 and the width by 5 the perimeter will be 60. What are the dimensions of the original rectangle? |

Lecture 54Play Video |
Geometric Shapes: TrianglesWord problems are challenging, but rewarding once you learn how they're done! Problem Text: The perimeter of a triangle is 35". The second side is 1" more than twice the first, the third side is 1" less than four times the first. Find the length of the three sides. |

Lecture 55Play Video |
MixturesWord problems are challenging, but rewarding once you learn how they're done! Problem Text: Two grades of hamburger are mixed. The mixed hamburger totals 300lbs. Grade A sells for $2.50/lb. Grade B hamburger sell for $1.80. If the total mixture sold for $638.00, how much of each type was used? |

Lecture 56Play Video |
Word Problems With Movie TicketsWord problems are challenging, but rewarding once you learn how they're done! Problem Text: For a theater showing 202 tickets were sold. A child's ticket costs $6.00 and an adult's ticket costs $10.00. How many tickets of each were sold if the receipts totaled $1708.00? |

Lecture 57Play Video |
Word Problems With MoneyWord problems are challenging, but rewarding once you learn how they're done! Problem Text: A bank tell has 180 bills in the drawer consisting of $20.00 bills and $10.00 bills. If the total cash value is $3050.00. How many of each type of bills are in the drawer? |

Lecture 58Play Video |
Mixture Problems 1 of 2Word problems are challenging, but rewarding once you learn how they're done! Problem Text: 260cm3 of a 40% acid solution is added to an 100% acid solution to form a 60% solution. How much of the 100% acid solution is needed? |

Lecture 59Play Video |
Mixture Problems 2 of 2Word problems are challenging, but rewarding once you learn how they're done! Problem Text: A 10% alcohol solution and a 40% alcohol solution are mixed to form 80ml of a 35% solution. How much of each mixture was needed? |

Lecture 60Play Video |
Investments 1 of 3Word problems are challenging, but rewarding once you learn how they're done! Problem Text: Jim invested some money at 6% and $4800 more at 8%. Total income from the investments was $1084. How much did he invest at each rate? |

Lecture 61Play Video |
Investments 2 of 3Word problems are challenging, but rewarding once you learn how they're done! Problem Text: Sally invested a portion of $12,000 at 7% and the remainder at 10%. How much was invested at each rate if the total interested earned was $1032? |

Lecture 62Play Video |
Investments 3 of 3Word problems are challenging, but rewarding once you learn how they're done! Problem Text: A total of $8000 was invested, part at 6% and part at 9%. How much was invested at each rate if the total interest earned was $645? |