PARCC Advanced Mathematics Practice Questions 2 — Questions and Answers

Question 1: Simplify: (3x²y)(−4xy³)

−12x³y⁴ (Correct answer)
−12x²y⁴
12x³y³
−7x³y⁴

Correct answer: −12x³y⁴

Multiply coefficients: 3 × (−4) = −12. Add exponents: x² · x¹ = x³, y¹ · y³ = y⁴. Result: −12x³y⁴.

This problem tests multiplication of monomials with multiple variables and negative coefficients — a core PARCC algebra/advanced math concept. Step 1: Multiply the coefficients (the numerical parts): 3 × (−4) = −12. Step 2: Apply the product rule of exponents for each variable. For x: x² × x¹ = x^(2+1) = x³. For y: y¹ × y³ = y^(1+3) = y⁴. Step 3: Combine: −12x³y⁴. Common mistakes: - Forgetting to carry the negative sign from −4. - Multiplying exponents instead of adding them (a common error: x² × x = x², but the correct answer is x³). - Misreading implicit exponents (y has an exponent of 1 even when not written). Rule: When multiplying like bases, ADD the exponents. When raising a power to a power, MULTIPLY the exponents. Never confuse these two operations.

Question 2: What is the slope of a line perpendicular to the line 2x − 5y = 10?

2/5
−5/2 (Correct answer)
5/2
−2/5

Correct answer: −5/2

The given line has slope 2/5. The perpendicular slope is the negative reciprocal: −5/2.

This problem tests the relationship between slopes of perpendicular lines, an important PARCC geometry and algebra concept. Step 1: Rewrite 2x − 5y = 10 in slope-intercept form (y = mx + b). 2x − 5y = 10 −5y = −2x + 10 y = (2/5)x − 2 The slope of the given line is m = 2/5. Step 2: Two lines are perpendicular if and only if their slopes are negative reciprocals of each other: m₁ × m₂ = −1 (2/5) × m₂ = −1 m₂ = −5/2 The perpendicular slope is −5/2. Common mistakes: - Taking only the reciprocal but forgetting to negate: getting 5/2 instead of −5/2. - Forgetting to convert to y = mx + b form and misreading the coefficient. - Confusing perpendicular (negative reciprocal) with parallel (same slope). Memory tip: Parallel lines → same slope. Perpendicular lines → slopes multiply to −1 (negative reciprocals).

Question 3: Solve: |2x − 3| = 7

x = 5 only
x = −2 only
x = 5 or x = −2 (Correct answer)
x = 2 or x = 5

Correct answer: x = 5 or x = −2

Set up two equations: 2x − 3 = 7 (x = 5) and 2x − 3 = −7 (x = −2). Both solutions are valid.

Absolute value equations are a standard PARCC advanced mathematics topic. The key concept: |expression| = k has TWO solutions when k > 0. Step 1: Set up two cases. Case 1: 2x − 3 = 7 → 2x = 10 → x = 5 Case 2: 2x − 3 = −7 → 2x = −4 → x = −2 Step 2: Check both solutions. For x = 5: |2(5) − 3| = |10 − 3| = |7| = 7 ✓ For x = −2: |2(−2) − 3| = |−4 − 3| = |−7| = 7 ✓ Both solutions check out. Common mistakes: - Only solving one case (forgetting that the expression inside can be negative too). - Distributing the absolute value (you CANNOT split |2x − 3| into |2x| − |3|). - Making a sign error in Case 2: students sometimes write 2x − 3 = +7 for both cases. Special cases: if k = 0, there is exactly one solution. If k < 0, there are no solutions (absolute value is never negative).

Question 4: A car travels from City A to City B at 60 mph and returns at 40 mph. What is the car's average speed for the entire trip?

50 mph
48 mph (Correct answer)
52 mph
45 mph

Correct answer: 48 mph

Use the harmonic mean: 2(60)(40)/(60+40) = 4800/100 = 48 mph.

This is a classic PARCC rate problem that tests conceptual understanding, not just calculation. Many students instinctively average 60 and 40 to get 50, but that is wrong because the car spends more TIME at the slower speed. Correct approach — use the harmonic mean formula for equal distances: Average speed = 2(r₁)(r₂) / (r₁ + r₂) = 2(60)(40) / (60 + 40) = 4800 / 100 = 48 mph Why this works (proof using actual distance d): - Time going: d/60 - Time returning: d/40 - Total time: d/60 + d/40 = 2d/120 + 3d/120 = 5d/120 = d/24 - Total distance: 2d - Average speed = 2d ÷ (d/24) = 2d × 24/d = 48 mph ✓ Common mistake: arithmetic mean (50 mph) — this is wrong because the car spends MORE time at 40 mph than at 60 mph, pulling the average below 50. Lesson: average speed = total distance ÷ total time. Never average speeds directly when distances are equal but times are different.

Question 5: Which expression is equivalent to (x + 3)² − (x − 3)²?

12x (Correct answer)
2x² + 18
6x
x² − 9

Correct answer: 12x

(x+3)² = x²+6x+9 and (x−3)² = x²−6x+9. Subtracting: (x²+6x+9) − (x²−6x+9) = 12x.

This problem tests polynomial expansion and the difference of squares pattern — core PARCC algebra skills. Method 1 — Direct expansion: (x + 3)² = x² + 2(x)(3) + 3² = x² + 6x + 9 (x − 3)² = x² − 2(x)(3) + 3² = x² − 6x + 9 Subtracting: (x² + 6x + 9) − (x² − 6x + 9) = x² + 6x + 9 − x² + 6x − 9 = 12x Method 2 — Difference of squares shortcut: a² − b² = (a+b)(a−b) Here: a = (x+3), b = (x−3) (x+3)² − (x−3)² = [(x+3)+(x−3)] × [(x+3)−(x−3)] = [2x] × [6] = 12x Common mistakes: - Forgetting to distribute the negative sign when subtracting the second expansion (this would give x² + 6x + 9 − x² − 6x + 9 = 18, which is wrong). - Squaring each term separately: (x)² + (3)² instead of using the full binomial expansion formula. The difference-of-squares approach is faster and should be recognized as a pattern on timed tests.

Question 6: A store marks up a wholesale price by 40%, then offers a 15% discount on the marked price. What is the net percentage change from the original wholesale price?

19% increase (Correct answer)
25% increase
19% decrease
25% decrease

Correct answer: 19% increase

1.40 × 0.85 = 1.19. This means the final price is 119% of the original — a 19% increase overall.

This is a compound percentage problem, a common PARCC advanced mathematics application. The key insight is that percentages compound multiplicatively, not additively. A 40% increase followed by a 15% decrease is NOT a 25% net change. Step 1: Convert each percentage change to a multiplier. - 40% increase → multiply by 1.40 - 15% discount (decrease) → multiply by 0.85 (keeping 85% of the price) Step 2: Apply both changes by multiplying the multipliers. 1.40 × 0.85 = 1.19 Step 3: Interpret the result. 1.19 = 119% of the original price → a 19% net increase. Why not 25%? Because you don't add/subtract percentages directly. The 15% discount applies to the already-marked-up price, not the original price. 15% of a higher number reduces the original by more than 15% of the base — but combined with the 40% markup, the net effect is still +19%. Verification with $100 example: $100 → +40% → $140 → −15% → $140 × 0.85 = $119. Net change: $19 increase = 19%. This type of problem appears in PARCC extended response sections and requires showing work.

PARCC Advanced Mathematics Practice Questions 2

Simplify: (3x²y)(−4xy³)