Let s suppose that the manufacturer of a popular washing powder announced a change in how it packages its product.

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1 Show Me: Rate of Change M8049 Let s suppose that the manufacturer of a popular washing powder announced a change in how it packages its product. The original amount of washing powder in a pack was eighty ounces. The amount of washing powder in the new pack is sixty-four ounces. Now, with the new pack the customer gets less washing powder but let s say the manufacturer also reduces the price of the new pack. So, how do I tell if I m getting the same value for the money with this new pack? This is where being able to calculate percentage change comes in handy. If I can calculate by what percent the amount of washing powder in the pack decreased, then I can check to see if the price has been decreased by a similar percent. Being able to analyze a change like this helps make us smart shoppers. In this lesson, we will analyze change in terms of percent increase and decrease, as well as change expressed as rates. Let s start by defining a percent decrease using the washing powder example. We know that the amount of washing powder decreased from eighty ounces in the original pack to sixty-four ounces in the new pack. To analyze the effect of this change, let s calculate the percent decrease between the two amounts. The size of the change can be expressed as the ratio of the amount of change over the initial value of a quantity.

2 The amount of change, or decrease, between the two quantities is the difference between the initial value and the new value. Then, because we re looking for the percent decrease, we can set this ratio, equal to the ratio P over one hundred, and solve the resulting proportion for P the percent change. Let s do it. Taking the proportion, first substitute the initial value, eighty ounces, and the new value, sixty-four ounces. Now, the difference between these two quantities is sixteen ounces. Sixteen divides into eighty, five times. So, simplifying the fraction sixteen over eighty, and dividing out the like units above and below, we find that one over five, equals P over one hundred. Multiplying on both sides by one hundred, and dividing out the one hundred above and below, we get one hundred over five equals P So P is equal to twenty. Therefore, we can say that the amount of washing powder in the pack decreased by twenty percent. A percent decrease, as this example shows, means that a quantity has been reduced. Let s look at another example involving change. A pack of baseball cards had twenty-four cards. The manufacturer then decreased the number of cards in the pack to eighteen cards. Well, we can let P represent the percent decrease in the number of cards in the pack and set up a proportion that can be solved to find P. The percent decrease proportion that we can use is made up of two equal ratios: the difference between the initial value and the new value, all over the initial value, which equals P over one hundred.

3 So, for this problem, which are the correct values for the ratio on the left of the proportion? Now we have enough information to solve this problem for the percent decrease, P Subtracting eighteen from twenty-four, we can simplify the proportion to six over twenty-four, equals P over one hundred This can be further simplified to one over four, equals P over one hundred. Multiplying both sides by one hundred, and then dividing out one hundred above and below on the right, gives us one hundred over four equals P. So P equals twenty-five. Therefore, changing the number of cards in a pack from twenty-four to eighteen decreased the number of cards in the pack by twenty-five percent. We talked about percent decreases earlier, but you know, a change doesn t always involve a decrease it can involve an increase too. We can calculate a percent increase using a proportion similar to that for a decrease, except this time the values that we subtract to find the amount of change are swapped. We do this because for an increase, the new value will be greater than the initial value, and we want to have a positive value when we subtract these two quantities. So, the proportion to find a percent increase is: the new value minus the initial value, all over the initial value, equals P over one hundred, where P represents the percent increase. Let s look at an example. A group of children set up a lemonade stand and sold fifty cups of lemonade one summer day. They sold eighty cups the next day. What was the percent increase in sales of lemonade from the first day to the second day? In this problem, the new value is eighty cups, and the initial value is fifty cups. Substituting these values into the proportion, we have eighty cups minus fifty cups, all over fifty cups, equals P over one hundred.

4 Now we can solve for P. Eighty cups minus fifty cups equals thirty cups. Next, we can divide out the units. The equation can be further simplified as three over five, equals P over one hundred. Multiplying on both sides by one hundred and then dividing out one hundred above and below on the right, gives us three hundred over five, equals P. Simplifying, we find that P equals sixty. So, selling thirty more cups of lemonade on the second day than were sold on the first day equals a sixty-percent increase in lemonade sales. Now, let s say the children sold fifty cups of lemonade on the first day and ninety cups on the second day. Which proportion can be used to calculate the percent increase in the number of lemonade cups sold? Many problems in the real world involve percent increases or decreases. A common example involves the purchase of a new car. Let s say Alicia paid twenty-two thousand dollars for a brand new car. As soon as she drove the car out of the dealer s lot, the value of the car dropped to eighteen thousand seven hundred dollars. What was the percent decrease in the value of the car from the moment it was sold to when it was driven off the lot? We can calculate the percent decrease, P, using the percent decrease proportion. In this problem, the initial value of the car was twenty-two thousand dollars, and the new value was reported as eighteen thousand seven hundred dollars. Substituting these values into the proportion, we get twenty-two thousand dollars minus eighteen thousand seven hundred dollars, all over twenty-two thousand dollars equals P over one hundred. Now, let s find the value of P.

5 Type in the percent decrease in the value of the car. The area of the back yard at Terry s old house was five hundred square feet. The area of the back yard at Terry s new house is seven hundred-fifty square feet. If P represents the percent increase in the area of Terry s backyard, what is P? Click Solution when you think you know the answer. Solution This problem involves a percent increase. To find P the percent increase in Terry s back yard, we can set up this proportion where we subtract the smaller initial value from the greater new value. The new value is seven hundred fifty square feet, and the initial value is five hundred square feet. Now, we just need to substitute these values into the proportion, and solve for P. Substituting gives us seven hundred-fifty square feet minus five hundred square feet, all over five hundred square feet equals P over one hundred. Subtracting on the left gives us two hundred-fifty square feet above the fraction bar. Now, we can divide out like units, and simplify the fraction on the left. Then, multiply both sides by one hundred to isolate P Finally, divide out the one hundred above and below on the right, and simplify to get P equals fifty. So, the area of Terry s new back yard represents a fifty percent increase in size compared to the area of his old back yard. The area of Terry s room at the old house was ninety-six square feet. The area of Terry s room at the new house is one hundred eight square feet.

6 What is the percent change, P, in the size of Terry s room to the nearest tenth? So far, we have looked at problems about percent increase and decrease in which only one quantity changes. However, in other situations we use ratios that compare the changes in two quantities. A common example is the ratio of a change in distance to a corresponding change in time, which we call a rate. We often write this ratio using the variables R for rate, D for distance, and T for time, giving the equation R equals D over T. The variable R is a constant rate measured in units such as miles per hour or feet per second. The variable D represents a distance traveled measured in units such as feet, miles, or kilometers. The variable T represents the time taken to travel the distance D and it is measured in units such as seconds, minutes, or hours. We can use this relationship between R, D, and T to solve problems about rates of change. Let s suppose that the driver of a car travels for four hours at a constant rate and covers two hundred miles. What is the constant rate, R at which the driver travels? To find out, we can substitute two hundred miles for the distance, D and four hours for the time, T to get two hundred miles over four hours. This ratio simplifies to fifty miles over one hour, which we can write as fifty miles per hour. So the car travels at a constant rate of fifty miles per hour. Here s an example for you to try.

7 Troy cycled fifty-seven miles in three hours. If he cycled at a constant rate, what was his rate in miles per hour? We can use the ratio R equals D over T to solve for either D or T if we know the value of the other two variables. For example, suppose we know the rate, R and the distance, D. Then we can rearrange the formula to solve for the time, T. Doing this, we get T equals D over R. In other words, the time of travel is equal to the ratio of the distance traveled to the constant rate. Let s see how it works. Suppose that Ashley drove one hundred fourteen miles at a constant rate of thirty-eight miles per hour. How many hours did she drive? Using the formula for T and substituting the values for D and R, we get T equals one hundred fourteen miles over thirty-eight miles per hour. To simplify this, we can divide out the units of miles on the right, and then dividing one hundred fourteen by thirty-eight, we get T equals three hours. So, it took Ashley three hours to travel one hundred fourteen miles at a constant rate of thirty-eight miles per hour. It s your turn again. A cyclist traveled forty-four miles at a constant rate of eleven miles per hour. How many hours did he cycle? We can also rearrange the original equation to solve for the distance D if we know the values of R and T. Doing this, we can see that the distance D is equal to the rate R, times the time T.

8 For example, suppose that a tourist hiked at a constant rate of eighteen miles per day for seven days. How many miles did the tourist hike during that time? Well, we can use the formula for distance D and substitute eighteen miles per day for the rate R and seven days for the time T. Solving this equation, we find that D is equal to one hundred twenty-six miles. So, the tourist hiked one hundred twenty-six miles in seven days at a constant rate of eighteen miles per day. Now it s your turn. If a biking tour group traveled at a constant rate of thirty-four miles per day for five days, how many miles did they cycle? In this lesson, we explored two types of problems involving change. Change in one quantity can be expressed in terms of a percent decrease or a percent increase. A percent decrease between an initial amount and a new amount is calculated using the following proportion: the initial value minus the new value, all over the initial value, equals P over one hundred where P represents the percent decrease. A percent increase between an initial amount and a new amount is calculated using the following proportion: the new value minus the initial value, all over the initial value, equals P over one hundred where P represents the percent increase. A ratio can be used to compare the change in two different quantities. For example, the rate, R at which a car travels is equal to the distance it travels D, divided by the time it travels, T.

9 If any two of the variables D, R, or T are known, then the equation can be used to solve for the third variable. Rearranging the equation, we can see that the time T is equal to the distance D divided by the constant rate, R. We can also rearrange the equation to show that the distance D equals the rate R times the time, T. If you d like to review this activity again, click Review. If you re ready to exit, click Done.