# Taxes and Sales

Alignments to Content Standards: A-SSE.B

Judy is working at a retail store over summer break. A customer buys a \$50 shirt that is on sale for 20% off. Judy computes the discount, then adds sales tax of 10%, and tells the customer how much he owes. The customer insists that Judy first add the sales tax and then apply the discount. He is convinced that this way he will save more money because the discount amount will be larger. 1. Is the customer right? 2. Does your answer to part (a) depend on the numbers used or would it work for any percentage discount and any sales tax percentage? Find a convincing argument using algebraic expressions and/or diagrams for this more general scenario. ## IM Commentary This task is not about computing the final price of the shirt but about using the structure in the computation to make a general argument. The key underlying idea is that multiplication is commutative, which we often just take for granted and don't feel needs any explanation. In this case, the context of the problem makes it not obvious at all that we can switch the order of the two computations, but it becomes quite obvious after observing that the application of both the discount and the sales tax are just instances of multiplication. Since the order in which we multiply is irrelevant, the answer must be the same regardless of which we apply first. The solution presents both an algebraic approach to the general result in part (b), and also a diagram that illustrates the same result graphically. This task presents a good opportunity for students to construct a viable argument and critique the reasoning of others (MP3). ## Solution 1. Judy first takes 20% off which gives a new price of$\$50(0.8) = \$40$. She then adds the 10% sales tax for a final price of$\$40(1.1) = \$44$. The customer first adds 10% for a new price of$\$50(1.1) = \$55$. He then takes 20% off for a final price of$\$55(0.8) = \$44$. The customer is right to say that the discount amount will be larger, it is \$11 opposed to \$10 with his method. But the additional \$1 just gets subtracted from the tax amount that was added in the first step. So the final price is the same in both cases.

It does not matter in which order the discount and tax are computed.

2. If we don’t actually perform the computations but just record them we find the following:

Judy: $50(0.8)(1.1) = 44$

customer: $50(1.1)(0.8) = 44$

We see that it is not surprising that both computations get the same answer, since $(0.8) \cdot (1.1) = (1.1) \cdot (0.8)$.

This result will generalize if we replace \$50, 20%, 10% by any other numbers. If we let$P$stand for the original price,$s$for the sales percentage and$t\$ for the tax percentage, we have

$$P \left(1-\tfrac{s}{100} \right)\left( 1 + \tfrac{t}{100} \right) = P \left( 1 + \tfrac{t}{100} \right) \left( 1 - \tfrac{s}{100} \right)$$

We see that changing the order in which the sale and the tax are applied does not matter.

We can also visualize this with the following diagram. Yellow represents the action of subtracting 20% and blue represents the action of adding 10%. We see that both paths result in the same final answer. Even though the diagram uses the numbers from the problem, we can see from the structure in the diagram that both paths will result in the same final price even if the yellow and blue areas are altered.