PARCC Algebra Practice Test 3 — Questions and Answers

Question 1: Which of the following represents the graph of y = |x − 2| + 1?

A V-shaped graph with vertex at (2, 1) (Correct answer)
A V-shaped graph with vertex at (−2, 1)
A V-shaped graph with vertex at (2, −1)
A parabola with vertex at (2, 1)

Correct answer: A V-shaped graph with vertex at (2, 1)

y = |x − 2| + 1 has the vertex at (2, 1): the graph shifts 2 units right and 1 unit up from y = |x|.

Absolute value functions produce V-shaped graphs, and understanding transformations of parent functions is a key PARCC algebra standard. Parent function: y = |x| has a vertex at the origin (0, 0) and opens upward. Transformation rules for y = |x − h| + k: - h represents a horizontal shift. If h > 0, shift RIGHT by h. If h < 0, shift LEFT. - k represents a vertical shift. If k > 0, shift UP by k. If k < 0, shift DOWN. For y = |x − 2| + 1: - h = 2 → shift 2 units to the RIGHT - k = 1 → shift 1 unit UP - Vertex: (h, k) = (2, 1) Common mistakes: - Confusing the direction of horizontal shift. Because of the minus sign in |x − h|, x − 2 = 0 when x = 2, so the vertex is at x = 2 (right shift), NOT x = −2. - Confusing absolute value graphs (V-shape) with parabolas (U-shape). Absolute value functions have straight-line arms meeting at a vertex; parabolas are curved. To confirm: at x = 2, y = |2 − 2| + 1 = 0 + 1 = 1 → vertex (2, 1). At x = 0, y = |0 − 2| + 1 = 2 + 1 = 3. At x = 4, y = |4 − 2| + 1 = 2 + 1 = 3 — symmetric arms.

Question 2: For what value of k does the system 2x + ky = 8 and 4x + 6y = 16 have infinitely many solutions?

k = 3 (Correct answer)
k = 6
k = 2
k = 12

Correct answer: k = 3

Multiply the first equation by 2: 4x + 2ky = 16. For this to match 4x + 6y = 16, we need 2k = 6, so k = 3.

Systems of equations with special cases (infinitely many solutions or no solution) is an important PARCC algebra concept. For a system to have infinitely many solutions, the two equations must be dependent — one must be a scalar multiple of the other (they represent the same line). Given: Equation 1: 2x + ky = 8 Equation 2: 4x + 6y = 16 Notice that 4x + 6y = 16 can be written as 2(2x + 3y) = 2(8), which simplifies to 2x + 3y = 8. For Equation 1 to be equivalent to 2x + 3y = 8, we need: 2x + ky = 8 to equal 2x + 3y = 8 Therefore k = 3. Alternative approach using ratios: For a₁/a₂ = b₁/b₂ = c₁/c₂ (condition for infinitely many solutions): 2/4 = k/6 = 8/16 1/2 = k/6 k = 3 ✓ Contrast: for NO solution (parallel lines), the ratio a₁/a₂ = b₁/b₂ ≠ c₁/c₂. For a UNIQUE solution, a₁/a₂ ≠ b₁/b₂. Think of it graphically: infinitely many solutions = same line; no solution = parallel lines; one solution = lines that intersect at a single point.

Question 3: The sum of three consecutive odd integers is 57. What is the largest integer?

21 (Correct answer)
19
23
17

Correct answer: 21

n + (n+2) + (n+4) = 57 → 3n + 6 = 57 → n = 17. Largest = 17 + 4 = 21.

Setting up and solving equations from verbal descriptions is a critical PARCC algebra skill. This type of problem requires correctly defining variables. Key concept: Consecutive odd integers differ by 2 (e.g., 3, 5, 7 or 11, 13, 15). Let the three consecutive odd integers be: n, n + 2, n + 4 Set up the equation: n + (n + 2) + (n + 4) = 57 3n + 6 = 57 3n = 51 n = 17 The three integers are: 17, 19, 21 Verify: 17 + 19 + 21 = 57 ✓ The largest integer is 21. Common mistakes: - Using n, n+1, n+2 for consecutive ODD integers. This would give consecutive integers (not necessarily odd). Odd integers differ by 2. - Solving for n (17) and reporting it as the answer instead of finding the largest (21). - Arithmetic errors in the linear equation step. Extension: for consecutive even integers, the same approach applies (n, n+2, n+4). The difference between consecutive even and odd integers is only in what value n represents — odd integers require n to be odd.

Question 4: Which expression represents the product (2x + 5)(2x − 5)?

4x² − 25 (Correct answer)
4x² + 25
4x² − 20x − 25
4x² + 10x − 25

Correct answer: 4x² − 25

(2x + 5)(2x − 5) = (2x)² − (5)² = 4x² − 25.

The difference of squares pattern is one of the most important factoring and multiplication identities in algebra, frequently tested on PARCC. Pattern: (a + b)(a − b) = a² − b² This works because: (a + b)(a − b) = a² − ab + ab − b² = a² − b² The middle terms cancel. Applying to (2x + 5)(2x − 5): a = 2x, b = 5 (2x)² − (5)² = 4x² − 25 Verification by FOIL: (2x + 5)(2x − 5) = (2x)(2x) + (2x)(−5) + (5)(2x) + (5)(−5) = 4x² − 10x + 10x − 25 = 4x² − 25 ✓ The middle terms (+10x and −10x) cancel, which is why the difference of squares pattern produces a binomial (no x term). Common mistakes: - Option B (4x² + 25): forgetting the negative sign in (a − b) — the result is always a² MINUS b², never plus. - Options C and D: applying FOIL incorrectly, as if the second factor were (2x + 5) instead of (2x − 5), generating a middle term that should cancel.

Question 5: A train leaves Station A traveling at 80 mph. Two hours later, a second train leaves the same station traveling in the same direction at 100 mph. How many hours after the second train departs will it catch up to the first train?

8 hours (Correct answer)
10 hours
6 hours
4 hours

Correct answer: 8 hours

80(t + 2) = 100t → 80t + 160 = 100t → 160 = 20t → t = 8 hours after the second train departs.

Distance-rate-time problems are a classic PARCC algebra application. The formula is: Distance = Rate × Time. Let t = the number of hours after Train 2 departs when it catches Train 1. Then Train 1 has been traveling for (t + 2) hours (it left 2 hours earlier). Train 2 has been traveling for t hours. When Train 2 catches Train 1, they have traveled the same distance: Distance of Train 1 = Distance of Train 2 80(t + 2) = 100t Solve: 80t + 160 = 100t 160 = 20t t = 8 hours Verify: Train 1 distance: 80 × (8 + 2) = 80 × 10 = 800 miles Train 2 distance: 100 × 8 = 800 miles ✓ Common mistakes: - Setting up the time as (t) for Train 1 and (t + 2) for Train 2 instead of the reverse. The first train left EARLIER, so it has had MORE time. - Stopping after finding t = 8 and being confused about what the question asked. The question asks for hours after the second train departs, which is exactly t = 8. - Forgetting to verify by checking that both distances are equal.

Question 6: If f(x) = 3x − 1 and g(x) = x² + 2, find f(g(2)).

17 (Correct answer)
11
35
5

Correct answer: 17

g(2) = 2² + 2 = 6. Then f(6) = 3(6) − 1 = 17.

Composite functions — evaluating f(g(x)) — are a key PARCC algebra concept. The notation f(g(2)) means 'first apply g to 2, then apply f to the result.' Step 1: Evaluate the inner function g at x = 2. g(2) = (2)² + 2 = 4 + 2 = 6 Step 2: Use the result as input for the outer function f. f(g(2)) = f(6) = 3(6) − 1 = 18 − 1 = 17 So f(g(2)) = 17. Common mistakes: - Computing f(2) first and then g of that result: f(2) = 3(2) − 1 = 5, then g(5) = 25 + 2 = 27. This is g(f(2)), not f(g(2)) — order matters for composition. - Multiplying the two functions: f(x) × g(x) = (3x − 1)(x² + 2), which is not what f(g(x)) means. General rule: f(g(x)) ≠ g(f(x)) in most cases. Composition is not commutative. For the general composition: f(g(x)) = f(x² + 2) = 3(x² + 2) − 1 = 3x² + 6 − 1 = 3x² + 5. Verify: f(g(2)) = 3(4) + 5 = 12 + 5 = 17 ✓

PARCC Algebra Practice Test 3

Which of the following represents the graph of y = |x − 2| + 1?