Tuesday, September 28, 2021

# Algebra 1 PARCC question: shirts and ties

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## 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?

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.