The Dean’s Column
Don’t be Fooled by Misleading Discounts
Percentage problems have long been the nemesis of most students. Problems get particularly unpleasant when multiple percents need to be processed in the same problem. By showing students that combined percentages don’t always lead to the conclusion that is expected, you turn this one-time nemesis into a delightfully simple arithmetic algorithm that affords lots of useful applications. For example, most people would expect that if an item were increased by 10% and then discounted by 10% (or vise versa) that the price remains unchanged. Well, this is not the case. We will present a not-very-well-known scheme that will simplify this situation and will enchant your students at the same time. We will begin by considering the following problem:
Wanting to buy a coat, Lisa is faced with a dilemma. Two competing stores next to each other carry the same brand coat with the same list price, but with two different discount offers. Store A offers a 10% discount year round on all its goods, but on this particular day offers an additional 20% on top of their already discounted price. Store B simply offers a discount of 30% on that day in order to stay competitive. How many percentage points difference is there between the two options open to Lisa?
At first glance, students will assume there is no difference in price, since 10 + 20 = 30, yielding the same discount in both cases. The clever student will see that this is not correct, since in store A only 10% is calculated on the original list price, with the 20% calculated on the lower price, while at store B, the entire 30% is calculated on the original price. Now, the question to be answered is, what percentage difference is there between the discount in store A and store B?
One expected procedure will have the student assume the cost of the coat to be $100, calculate the 10% discount yielding a $90 price, and an additional 20% of the $90 price (or $18) will bring the price down to $72. In store B, the 30% discount on $100 would bring the price down to $70, giving a discount difference of $2, which in this case is 2%. This procedure, although correct and not too difficult, is a bit cumbersome and does not always allow a full insight into the situation.
An interesting and quite unusual procedure(1*) is provided for entertainment and fresh insight into this problem situation:
Here is a mechanical method for obtaining a single percentage discount (or increase) equivalent to two (or more) successive discounts (or increases).
(1) Change each of the percents involved into decimal form:
.20 and .10
(2) Subtract each of these decimals from 1.00:
.80 and .90 (for an increase, add to 1.00)
(3) Multiply these differences:
(.80)(.90) = .72
(4) Subtract this number (i.e., .72) from 1.00:
1.00 - .72 =.28, which represents the combined discount
(If the result of step 3 is greater than 1.00, subtract 1.00 from it to obtain the percent of increase.)
When we convert .28 back to percent form, we obtain 28%, the equivalent of successive discounts of 20% and 10%.
This combined percentage of 28% differs from 30% by 2%.
This procedure can also be used to combine more than two successive discounts following the same approach. In addition, successive increases, combined or not combined with a discount, can also be accommodated in this procedure by adding the decimal equivalent of the increase to 1.00, where the discount was subtracted from 1.00 and then continue in the same way. If the end result comes out greater than 1.00, then this reflects an overall increase rather than the discount as found in the above problem.
This algorithm not only streamlines a typically cumbersome situation, but also provides some insight into the overall picture. For example, the question “Is it advantageous to the buyer in the above problem to receive a 20% discount and then a 10% discount, or the reverse, 10% discount and then a 20% discount?” The answer to this question is not immediately intuitively obvious. Yet, since the procedure just presented shows that the calculation is merely multiplication, a commutative operation, we find immediately that there is no difference between the two.
So here you have a delightful algorithm for combining successive discounts or increases or combinations of these. Not only is it useful, but it will enchant your students (and probably your colleagues as well). Now try to see what will happen with our opening problem of increasing and decreasing the price of an item by 10%. You will be surprised with the result!
1*. It is provided without justification of its validity so as not to detract from the solution of the problem. However, for further discussion of this procedure, the reader is referred to A. S. Posamentier, B. S. Smith and J. Stepelman, Teaching Secondary School Mathematics: Techniques and Enrichment Units (Columbus, Ohio: Merrill/Prentice Hall, 7th ed. 2006) pp. 271-273.