PARCC Practice Test Question 20 (Day 165)

I know that I've been distracted lately. My most recent posts ended up being more about books that I'm reading than about the PARCC Practice Test. But this week marks the start of the more difficult questions on the PARCC. Every question from this point on has multiple parts, with some parts involving different sections of the U of Chicago text. So I want to make sure that every post this week is devoted mainly to PARCC.

By the way, most high schools don't have PARCC or SBAC testing this week or last week because of the AP testing. Early Start schools leaned towards having the Common Core tests before the AP, while Labor Day Start schools will most likely take the Common Core tests next week -- and for them, this discussion of the multi-part questions is timely.

Question 20 of the PARCC Practice Test is about a steel pipe. Students will have to determine, among other things, its lateral area.

A steel pipe in the shape of a right circular cylinder is used for drainage under a road. The length of the pipe is 12 feet, and its diameter is 36 inches. The pipe is open at both ends.

Part A

How many square feet is the outer surface of the pipe?

Give your answer to the nearest integer.

Part B

A wire screen in the shape of a square is attached at one end of the pipe to allow water to flow through but to keep animals from getting inside the pipe. The lengths of the diagonals of the screen are equal to the diameter of the pipe. The figure represents the placement of the screen at the end of the pipe.

(The figure shows the square screen inscribed in the circular pipe.)

What are the perimeter and area of the screen?

(A) The perimeter of the screen is approximately 72 inches, and the area of the screen is 324 square inches.
(B) The perimeter of the screen is approximately 72 inches, and the area of the screen is 648 square inches.
(C) The perimeter of the screen is approximately 102 inches, and the area of the screen is 648 square inches.
(D) The perimeter of the screen is approximately 125 inches, and the area of the screen is 1,018 square inches.

Part A of this section asks for the lateral area of the pipe. That the lateral area and not the full surface area of the pipe could not be made more explicit: "The pipe is open at both ends" and "How many square feet is the outer surface of the pipe?" (Emphasis mine.) Since the pipe is cylindrical, we use the formula for the lateral area of a right prism or cylinder. It is given in Section 10-1 of the U of Chicago text as L.A. = ph. For cylinders, we use the fact that the circumference of a circle is pi * d to obtain the perimeter. Since the diameter is 36 inches or 3 feet, the perimeter is 3pi feet. So the lateral area of the pipe is 3pi * 12 = 36pi square feet -- this works out to be 113.1, or to the nearest integer, 113 square feet.

Now let's move on to Part B, which asks for the perimeter and area of the screen. We have a square whose diagonal is given to be the diameter of the pipe, 36 inches. To find the length of a side, we note that the diagonal divides the square into two congruent 45-45-90 triangles. We learn about such isosceles right triangles in Section 14-1 of the U of Chicago text, which gives the hypotenuse to be sqrt(2) times the length of a leg. Since the hypotenuse, or diagonal, is 36 inches, the leg, or side of the square, is 36/sqrt(2) inches. Then the perimeter of the square must be four times this, or 144/sqrt(2) inches -- which is about 101.8 inches -- and the area is the square of the side, which is 1,296/2 = 648 square inches. The correct answer is (C).

There are several places where students are likely to make mistakes. Here are some very common mistakes students may make in Part A:

-- confusing lateral area with surface area
-- confusing feet with inches
-- confusing diameter with radius

And here is a common mistake for Part B:

-- confusing 45-45-90 triangles with 30-60-90 triangles

Indeed, we see that two of the choices have 72 instead of 102 inches as the perimeter. This error occurs if students find the leg to be half of the hypotenuse (as the shorter leg of a 30-60-90 triangle would be). If they make that mistake only when calculating the perimeter, then they obtain (B), but if they repeat that error when finding the area, then they obtain (A). Choice (D) is chosen by students who calculate the side of the square as the longer, rather than shorter, side of a 30-60-90 triangle.

All of the questions during this part of the test will be heavy on the Uses section of SPUR. This section requires students to use material learned in four different sections of the U of Chicago text -- Section 10-1 (lateral area of a cylinder), 14-1 (sides of a 45-45-90 triangle), 8-1 (perimeter of a square), and 8-3 (area of a square). When we see such tricky questions like this, students and teachers often wonder, is this a good question to include on the PARCC, or is it an unfair question.

My criterion for whether a word problem should be included on the PARCC is authenticity. We imagine that we are a construction worker or plumber working with this pipe, and we wonder whether the common mistakes that students make are issues that may come up for this worker. For example, it's easier to measure the diameter of a pipe than to measure its radius, so the diameter-radius issue can come up in real life. Similarly, one may measure longer distances, such as the length of a pipe, in feet and shorter distances, such as the width of a pipe, in inches.

If a word problem is authentic -- that is, that it includes issues that may confuse students but arise in the real world -- then it should be included on the test. Students who would have earned a high score without such a problem but a low score with the problem deserve the lower score, for they are the type of student who learns a lot of math but is stumped when needing to apply it to the real world. On the other hand, if a word problem is inauthentic -- that is, that it includes issues that confuse students and don't arise in the real world -- then it should not be included on this test. Students who would have earned a high score without such a problem but a low score with the problem deserve the higher score, as this question is designed only to keep the smart students down. Students and teachers would be correct to protest a test that includes this sort of question.

In order for students to be successful in this sort of question, attention to detail is key. In Part A, for example, some key words are cylinder, feet, and inches. These words should be enough to alert the student at least to consider the formulas for cylinders -- not, say, cones -- and seeing the question ask for "square inches" should be enough to eliminate the volume formula. The tricky part would be to distinguish lateral from surface area -- especially with the word "surface" mentioned above. The students have to know that the words "outer surface" refer to the lateral area, since the ends of the pipe are open.

Recall that in the worksheets, there will be a single question with each of its parts similar to the corresponding part of the original PARCC question.

PARCC Practice EOY Exam Question 20

U of Chicago Correspondence: Section 10-2, Surface Areas of Prisms and Cylinders

Key Theorem: Right Prism-Cylinder Lateral Area Formula

The lateral area L.A. of a right prism (or cylinder) is the product of its height h and the perimeter (circumference) of its base.
L.A. = ph

Common Core Standard:
CCSS.MATH.CONTENT.HSG.GMD.A.3
Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

Commentary: There are many word problems in the U of Chicago text, but of course none of them are exactly like this PARCC Question. In Section 10-1, Questions 16 and 21 are word problems involving the surface area of a cylinder, with Question 36 in the SPUR "Uses" section also on the surface area of a cylinder. Some of these questions require that one of the bases be included in the answer, rather than none (lateral area) or both (surface area). In Section 14-1, Question 13 is on the sides of an isosceles right triangle given the hypotenuse.