Percent: Successive Changes (Compound Growth)

Transcription

1 The Mathematics 11 Competency Test Percent: Successive Changes (Compound Growth) This short note is intended to emphasize that when the language of percent is used to describe a sequence of changes in some quantity, the percentage values each refer to different (and successive values as) bases. Example 1: Hank is currently paid $11.40 per hour. He signs a contract which will give him 2.5%, 3.2%, and 3.0% pay increases at one year intervals. Compute his hourly pay rate at the end of this contract, and calculate the net percent increase in his hourly rate of pay over the course of the contract, compared to his current rate of pay. solution: The easiest way to solve this problem is to calculate Hank s new rate of pay after each increase. So (i) Before the first pay increase, his rate of pay is $/hr, which is the base value to which the first pay increase is applied. This base value is increased by 2.5% in the first year of the contract. Thus, his new rate of pay is: new rate of pay = new value rounded to the nearest cent. = (11.40 $/hr)( ) = (11.40 $/hr)(1.025) = $/hr (ii) For the second year of the contract, a pay increase of 3.2% is given. This 3.2% increase is applied to the base pay rate of $/hr at the beginning of that second year. So rate of pay in year 2 = new value = (11.69 $/hr)( ) = (11.69 $/hr)(1.032) = $/hr (iii) Finally, in the third year, he receives a pay raise of 3.0% applied to the hourly pay rate of $12.06 at the beginning of that year. This means that for the third year, his rate of pay is rate of pay in year 3 = new value = (12.06 $/hr)( ) David W. Sabo (2003) Percent: Successive Changes (Compound Growth) Page 1 of 5

2 = (12.05 $/hr)(1.030) = $/hr So, at the end of this three-year contract, Hank s rate of pay will be $12.42 per hour. Now, compared with the beginning of the contract, Hanks rate of pay has gone up by $/hr $/hr = 1.02 $/hr To calculate the overall percent increase in his rate of pay for this three-year period, we need to use the formula rate = Amount base where Amount = the actual increase in pay over the three years, in $/hr base = Hank s initial rate of pay at the beginning of the contract Then 1.02 $ / hr rate = $ / hr rounded to four decimal places, or 8.95%. Now, the sum of the percent increases for the three years is 2.5% + 3.2% + 3.0% = 8.7%, which appears to be less than what we just claimed is the actual percent increase over three years. However, the value 8.7% is really quite meaningless here because it is a sum of percent increases which all refer to different base values. Even though on the surface it may appear that his employer is giving him an increase of a total of only 8.7% in rate of pay over the three years, in fact, Hank s net gain is somewhat higher because the percent increase in the second and third years of the contract are applied to the somewhat higher rates of pay at the end of the first and second years of the contract, respectively. Thus, they result in a slightly higher actual Amount of increase in pay than would occur if they were applied to Hank s original rate of pay at the beginning of the contract. When percent increases are to be applied in stages in this way, you will get an incorrect answer if you simply add the percents and they apply their sum to the original value before the first stage. The successive percent increases are always considered to be applied to the successive base values. Example 2: Hank buys a house for $225,000. Hoping to make some easy money in a hot real estate market, he instructs his agent to put the house up for sale at a price 40% higher than what he paid for it. After some months, a buyer contacts Hank, asking Hank whether he d consider reducing the price of the house by 30%. Hank agrees, thinking that since he had originally increased the price by 40%, there s still profit to be made here. When the cheque arrives from the sale of his house, how much profit does Hank find he s actually earned? (Ignore all issues of taxes, legal fees, commissions, etc.) David W. Sabo (2003) Percent: Successive Changes (Compound Growth) Page 2 of 5

3 solution: We might jump to the simple conclusion that Hank should have gained a 10% profit or about $22,500, because the difference between a 40% increase and a 30% decrease might prompt one to think is still a 10% increase. However, we are cautious about jumping to superficial conclusions in percent problems. It is prudent to carry out these calculations in detail before arriving at a conclusion. The facts of the situation are these: (i) Hank initially pays $225,000 for the house. (ii) Hank then puts the house on the market at a price 40% higher than what he paid for it. This 40% increase is applied to the base price of $225,000 to get the initial selling price. So initial selling price = new value = ($225,000)( ) = ($225,000)(1.40) = $315,000 Thus, at Hank s initial instruction, the house is advertised for sale at a price of $315,000. This is the price that potential buyers will see. (iii) Now, when the potential buyer contacts Hank and offers to buy the house at a 30% discount, the price that the buyer is working with is the $315,000 in the advertisement. So, when Hank agrees to allow the buyer to discount the price by 30%, this 30% really refers to the base price of $315,000. The actual selling price is then the new value obtained by applying this percent decrease to the base of $315,000. This means that for the house, Hank will receive actual selling price = new value = ($315,000) (1 + [-0.30]) = ($315,000)(-.70) = $220,500 Since a 30% decrease is involved here, we use rate = in the calculation. So, Hank ends up selling the house for $220,500. (iv) So, in summary, Hank bought the house for $225,000 and he apparently sold it for $220,500. He didn t gain a profit at all in fct he lost $4,500 on the transaction. In this sad story, the 30% discount was applied to a much higher base price than was the original 40% increase. Thus, the 30% discount (which might at first seem like it should be smaller than a 40% increase) actually corresponds to a larger Amount of money than does the initial 40% increase. This example illustrates that it is very, very important to sort out percents and their bases correctly. The percents by themselves really don t mean very much unless you know what the base value is with which they are associated. David W. Sabo (2003) Percent: Successive Changes (Compound Growth) Page 3 of 5

4 Example 3: Iodine-131 is a form of the element iodine which is used to treat certain medical disorders. It decays radioactively, decreasing at a rate of 54.94% each week. (This means that at the end of each week, 54.94% of the iodine-131 present at the beginning of the week will have disappeared.) If a patient is given a treatment containing 35, determine how many milligrams of iodine-131 will remain in her body three weeks later. solution: At first you might think that this is a bit of a silly question. If more than half of the iodine-131 disappears in a single week, surely after even two weeks, all of it is gone, and so the obvious answer here is that there are zero milligrams of iodine-131 left in this patient s body after three weeks have elapsed. But, this is not quite the situation that the problem describes. Actually, there s two twists here. First, if we are calculating how much iodine-131 is left after a number of weeks, we need to recognize that since 54.94% of the iodine-131 disappears in one week, this means that the rate at which the iodine-131 remains in 100% % = 45.06% Secondly, and less obviously, we need to realize that the percents here (either the 54.94% of iodine-131 that disappears during the week, or the 45.06% of iodine-131 that remains at the end of the week) are based on the amount of iodine-131 present at the beginning of that week and not on the amount of iodine-131 present at some arbitrary time in the past (such as when this patient first received the treatment). It might be easiest to track what happens here by constructing a table: week present at the beginning of the week (= base ) disappearing by the end of the week (= x base) that remain at the end of the week (= x base) According to this sequence of calculations, it appears that the patient still has mg of iodine-131 in her body at the end of three weeks. We can also see why the loss of 54.94% of the iodine-131 in a one week period does not mean that all of the iodine-131 is gone before two weeks have elapsed. In the first week, 54.94% of the original treatment of 35 disappears an actual amount of mg, leaving mg of the iodine-131 in this patient s body. During the second week, 54.94% of this amount of iodine-131 disappears. But this corresponds to only because it is 54.94% of the present at the beginning of that week. Each week, 54.94% of the iodine-131 disappears, but that is a successively smaller actual amount of iodine-131, because it is 54.94% of the successively decreasing amounts of iodine-131 present at the beginning of each week. (You can also see that it would appear that technically, the amount of iodine-131 will never completely disappear from the patient s body. In practice, though, you can verify that after 5 weeks, less than 1 milligram remains, so in practical terms, eventually all of the iodine-131 will have disappeared.) David W. Sabo (2003) Percent: Successive Changes (Compound Growth) Page 4 of 5

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