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# Algebra 1 PARCC question: shirts and ties

## Part A

At a clothing store, Ted bought 4 shirts and 2 ties for a total price of $95. At the same store, Stephen bought 3 shirts and 3 ties for a total price of$84. Each shirt was the same price, and each tie was the same price. Which system of equations can be used to find s, the cost of each shirt in dollars, and t, the cost of each tie in dollars?

## A

$\begin{cases} 6(s+t) = 95 \\ 3(s+t) = 84 \end{cases}$

## B

$\begin{cases} 4s+2t = 95 \\ 3s+3t = 84 \end{cases}$

## C

$\begin{cases} 7s+5t=179 \\ s+t=12 \end{cases}$

## D

$\begin{cases} 7s+5t=179 \\ 7s+5t=12(s+t) \end{cases}$

## Part B

Linda bought 1 shirt and 2 ties at the same store. What is the total price, in dollars and cents, of Linda’s purchase?

Correct Answers and References

For Part A, the correct answer is B. For Part B, the correct answer is 36.50, but any numerical value that is equivalent will be accepted, such as 36.5.

The question is purportedly aligned to Common Core Math Standard A.REI.6, which says students in algebra should be able to “solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.”

Sample Solution Strategy (there are others)

Examine the equations relative to the scenario.

If the variable t stands for the price of a tie at this store, and the variable s stands for the price of a shirt, then you have to multiply the price each by the number of shirts each customer bought, add that to the product of the price of each tie and the number of ties each customer bought, and get the total price paid for the merchandise.

Consider A: For Ted, the top equation in option A, we multiply 6 times the combined price for one shirt and one tie, s + t. That effectively means we’re going to get a bigger number, since Stephen didn’t buy six of the combined shirt-tie combos. He bought only 4 shirts, for instance, but if we distribute the 6 (4+2) over the sum of the prices, that would be like paying for each item (six total items of clothing) what he would pay for two combined items (a shirt and a tie). That just doesn’t make sense.

The same holds for Stephen’s equation, the bottom row of the system. We can’t distribute the total combined number of items over the sum of the prices and get anything close to the $84 Stephen paid. We eliminate A. Consider now C: The bottom row in the system effectively says that the price of a shirt plus the price of a tie equals$12. If the cost of a shirt and a tie at this store is $12, and Stephen bough six items of clothing, he paid too much.$84 is way more than $12 × 3. Consider finally D: Look at the bottom row in the system: $7s+5d=12(s+t)$ Distribute the 12 on the right-hand side over the sum, s + t. You get something like this, and you can’t solve that, no way, over the real numbers: $7s+5t=12s+12t$ $-5t=5s$ $-t=s$ It is not possible that the price of a tie would be the inverse of the price of a shirt. No store would ever pay people to purchase a tie. As a result of this nonsense, option D doesn’t make any sense. Considering the correct answer, the only one left, we find that both the top and bottom rows explain the scenario described in the text. In Part B, use the solution to Part A and plug in the prices. Solving the top row in option B for s, we find $4s+2t=95$ $4s=95-2t$ $s=\frac{95}{4}-\frac{t}{2}$ Plug s into the bottom equation: $3s+3t=84$ $3(\frac{95}{4}-\frac{t}{2})+3t=84$ $\frac{285}{4} - \frac{3t}{2} + 3t = 84$ $\frac{285}{4} - 84 = -3\frac{t}{2}$ $3\frac{t}{2} = \frac{336}{4} - \frac{285}{4}$ $\frac{3t}{2} = \frac{51}{4}$ $t = \frac{102}{12} = \frac{17}{2} = 8.50$ So each tie costs$8.50, and the shirt, given the solution for s in terms of t right above that, is

$s=\frac{95}{4}-\frac{8.50}{2}$
$s=23.75-4.25 = 19.50$

If Linda buys 1 shirt for $19.50 and 2 ties for$8.50 each ($17 for the two ties), it will cost$36.50.

## Analysis of this question

The question can be delivered online only in Part B, and an appropriate list of incorrect answers for the \$36.50 would have to be developed, probably parallel to the various systems in Part A, although, as shown, some of those are total nonsense. This means a parallel setup with multiple choice options for incorrect answers in Part B, for use with students who need to take the test on paper, would suffer from comparability.

The question is aligned to the Common Core math standard to which it purports to align.

No special accommodation challenges can be identified with this question, other than the possible clue students would get for Part A with a Part B set of options that lack a parallel price for the answer choice made by the student in Part A. Still, I would consider the question fair.

A slight editorial issue arises in the writing of the question. Since math students are often taught to represent variables with a single letter or symbol, it would have been better to use first names that don’t start with the same letter as the articles of clothing. For example, both “Stephen” and “shirt” start with “s,” and both “Ted” and “tie” start with “t.” When the letter appears in the equation, students are required to look a few times to make sure it’s representing the price for a shirt and tie, rather than the number of items purchased by Stephen or Ted. I only bring this up because keeping variables straight like that, in the context of the problem, is unrelated to the construct being assessed with the question, which is solving a system of linear equations (2 equations in 2 unknowns). Introducing a tricky naming convention is unwise and may mislead students, causing them to answer the question in the wrong frame of mind or reference.

#### About the Author

##### Paul Katula
Paul Katula is the executive editor of the Voxitatis Research Foundation, which publishes this blog. For more information, see the About page.

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