Teacher Resource • Answer Key
Benchmark Q13–Q15
Topics 4, 5, 6 (Problems 4A–6B)
Standard G.GF.1 Algebraic Mastery: Part + Part = Whole equation setup, solving linear equations, and back-substituting.
Topic 4: Solving for \(x\) (Benchmark Q13 Alignment) Matches Test Q13
Problem 4A Solution:
\(AB + BC = AC\)
\(3x + (5x + 4) = 10x + 2\)
\(8x + 4 = 10x + 2 \implies 2x = 2 \implies \mathbf{x = 1}\)
Answer: \(x = 1\) (Exact Benchmark Q13)
Problem 4B Solution:
\(PQ + QR = PR\)
\((4x - 1) + (2x + 7) = 8x - 12\)
\(6x + 6 = 8x - 12 \implies 2x = 18 \implies \mathbf{x = 9}\)
Answer: \(x = 9\)
Topic 5: Evaluating Segment Lengths (Benchmark Q14 Alignment) Matches Test Q14
Problem 5A Solution:
Substitute \(x = 1\) into \(AC = 10x + 2\):
\(AC = 10(1) + 2 = \mathbf{12}\)
Check: \(AB = 3(1)=3\), \(BC = 5(1)+4=9\); \(3+9=12\) ✓
Answer: \(AC = 12\) (Exact Benchmark Q14)
Problem 5B Solution:
Substitute \(x = 9\) into \(PQ = 4x - 1\):
\(PQ = 4(9) - 1 = 36 - 1 = \mathbf{35}\)
Verification: \(QR=25\), \(PR=60\); \(35+25=60\) ✓
Answer: \(PQ = 35\)
Topic 6: Re-Ordered Points & Midpoint Check (Benchmark Q15 Alignment) Matches Test Q15
Problem 6A Solution:
\(R\) is between \(S\) and \(T \implies SR + RT = ST\)
\((15x + 1) + (3x + 5) = 20x + 2\)
\(18x + 6 = 20x + 2 \implies 2x = 4 \implies \mathbf{x = 2}\)
\(SR = 15(2) + 1 = \mathbf{31}\)
Answer: \(x = 2, \; SR = 31\) (Exact Benchmark Q15)
Problem 6B Solution:
\((6x - 2) + (4x + 8) = 36 \implies 10x + 6 = 36 \implies \mathbf{x = 3}\)
\(KL = 6(3) - 2 = \mathbf{16}\); \(LM = 4(3) + 8 = \mathbf{20}\)
Since \(KL \neq LM\) (\(16 \neq 20\)), \(L\) does not bisect.
Answer: [X] No; \(KL \neq LM\)
The "R Between S and T" Order Trap:
Benchmark Q15 deliberately states that \(R\) is between \(S\) and \(T\). Students who casually write \(RS + ST = RT\) will fail to solve the equation. Encourage students to draw a quick 3-point sketch before writing their equation!
Teacher Guide & Key • Page 2 of 6 Standard G.GF.1: Topics 4–6 Solutions
Teacher Resource • Answer Key
Benchmark Q5, Q20, Q9
Topics 7, 8, 9 (Problems 7A–9B)
Standard G.GF.2 Inductive Logic: Recursive sequences, second differences, isolating hypothesis (\(p\)) and conclusion (\(q\)).
Topic 7: Fibonacci & Recursive Sequences (Benchmark Q5 Alignment) Matches Test Q5
Problem 7A Solution:
1, 1, 2, 3, 5, 8, 13, ___
Rule: \(1+1=2, 1+2=3, 2+3=5, 3+5=8, 5+8=13\)
Term 8: \(8 + 13 = \mathbf{21}\) | Term 9: \(13 + 21 = \mathbf{34}\)
Answer: 21 and 34 (Test Q5 Answer = 21)
Problem 7B Solution:
2, 3, 5, 8, 13, 21, 34, ___
Rule: Add preceding two terms: \(21 + 34 = \mathbf{55}\)
Answer: 8th term = 55
Topic 8: Growing Differences (Benchmark Q20 Alignment) Matches Test Q20
Problem 8A Solution:
2, 8, 15, 23, 32, ______
Differences: \(+6, +7, +8, +9 \implies \text{next is } \mathbf{+10}\)
Next number: \(32 + 10 = \mathbf{42}\)
Answer: 42 (Exact Benchmark Q20 Answer)
Problem 8B Solution:
5, 9, 14, 20, 27, ______
Differences: \(+4, +5, +6, +7 \implies \text{next is } \mathbf{+8}\)
Next number: \(27 + 8 = \mathbf{35}\)
Answer: 35 (Constant 2nd difference = +1)
Topic 9: Isolating Hypothesis & Conclusion (Benchmark Q9 Alignment) Matches Test Q9
Problem 9A Solution:
Hypothesis: A ray bisects an angle
Conclusion: it divides the angle into two congruent angles
Key Trap: The word "If" is NOT part of hypothesis; "Then" is NOT part of conclusion!
Problem 9B Solution:
Hypothesis (\(p\)): two lines are perpendicular
Conclusion (\(q\)): they intersect to form right angles
Format: In \(p \to q\), \(p\) is hypothesis, \(q\) is conclusion.
Multiple Choice Elimination for Test Q9:
On Test Q9, choice A is "A ray bisects an angle." (correct hypothesis), while choice B is "A ray divides the angle into two congruent angles." (the conclusion), and choices C and D are biconditional/contrapositive forms. Remind students: hypothesis = what is given/assumed.
Teacher Guide & Key • Page 3 of 6 Standard G.GF.2: Topics 7–9 Solutions
Teacher Resource • Answer Key
Benchmark Q6–Q8
Topics 10, 11, 12 (Problems 10A–12B)
Standard G.GF.2 Conditional Variations: Converse (\(Q \to P\)), Inverse (\(\sim P \to \sim Q\)), Contrapositive (\(\sim Q \to \sim P\)).
Topic 10: Writing the Converse (\(Q \to P\)) (Benchmark Q7 Alignment) Matches Test Q7
Problem 10A Solution:
"If it is raining, then today is Tuesday."
Truth Value: [X] Cannot be determined (it can rain on other days).
Exact match to Test Q7 Choice B.
Problem 10B Solution:
"If the sum of interior angles of a figure is \(180^\circ\), then it is a triangle."
Is converse true? [X] Yes (by polygon interior angle formula).
Topic 11: Writing the Inverse (\(\sim P \to \sim Q\)) (Benchmark Q8 Alignment) Matches Test Q8
Problem 11A Solution:
"If a ray does not bisect an angle, then it does not divide the angle into two congruent angles."
Rule: Negate hypothesis, negate conclusion. Do not change order.
Exact match to Test Q8 Choice B.
Problem 11B Solution:
"If \(x \neq 7\), then \(3x + 1 \neq 22\)."
Symbolic check: Original was \(P \to Q\); Inverse is \(\sim P \to \sim Q\).
Topic 12: Writing the Contrapositive (\(\sim Q \to \sim P\)) (Benchmark Q6 Alignment) Matches Test Q6
Problem 12A Solution:
"If it is not raining, then today is not Tuesday."
Rule: Switch clauses AND negate both (\(\sim Q \to \sim P\)).
Exact match to Test Q6 Choice D.
Problem 12B Solution:
"If two angles are not congruent, then they are not vertical."
Key Concept: Since the vertical angles theorem is true, its contrapositive is also always true!
The Benchmark Reference Box Trick:
Page 3 of the benchmark test prints a reference box: \(P \to Q\) Conditional, \(Q \to P\) Converse, \(-P \to -Q\) Inverse, \(-Q \to -P\) Contrapositive. Train students to match their answer choice symbol-by-symbol against this printed box!
Teacher Guide & Key • Page 4 of 6 Standard G.GF.2: Topics 10–12 Solutions
Teacher Resource • Answer Key
Benchmark Q1–Q4
Topics 13, 14, 15, 16 (Problems 13A–16B)
Standard G.GF.3 Deductive Reasoning: 2-Column tables: Def. Midpoint, Def. \(\perp\), Segment Subtraction, Halves Postulate.
Topic 13: Midpoint Deductions (Benchmark Q3 Alignment) Matches Test Q3
Problem 13A Solution:
Conclusion: \(\overline{AE} \cong \overline{EB}\) | Reason: Def. of Midpoint
Why not Segment Bisector? A point is given as the midpoint, not a line or segment doing the bisecting. Exact match to Test Q3 Choice C!
Problem 13B Solution:
Reason: Definition of Midpoint
\(3x - 5 = x + 11 \implies 2x = 16 \implies \mathbf{x = 8}\)
Topic 14: Perpendicular Deductions (Benchmark Q4 Alignment) Matches Test Q4
Problem 14A Solution:
Conclusion: \(\angle CEB\) is a right angle
Reason: Defn. Perpendicular Lines | Measure = \(90^\circ\)
Exact match to Test Q4 Choice A.
Problem 14B Solution:
[X] They intersect to form \(90^\circ\) right angles (Def. \(\perp\))
Lines do not have midpoints because lines extend infinitely.
Topic 15: Segment Subtraction (Benchmark Q2 Alignment) Matches Test Q2
Problem 15A Solution:
[X] Conclusion: \(\overline{YS} \cong \overline{TX}\) | Reason: Segment Subtraction Postulate
Wholes are congruent (\(TR \cong RS\)); congruent parts are subtracted (\(XR \cong YR\)). Exact match to Test Q2 Choice A!
Problem 15B Solution:
Reason: Segment Subtraction Postulate
Subtracting the common segment \(BC\) from equal segments \(AC\) and \(BD\) leaves congruent segments \(AB \cong CD\).
Topic 16: Halves Postulate (Benchmark Q1 Alignment) Matches Test Q1
Problem 16A Solution:
Conclusion: \(\angle DAC \cong \angle CAB\) | Reason: Halves Postulate
Since \(\angle DAB \cong \angle DCB\) and both are bisected, all resulting halves are congruent. Exact match to Test Q1 Choice D!
Problem 16B Solution:
"Halves of congruent quantities are congruent."
If \(m\angle A = m\angle B = 80^\circ\), then \(\frac{1}{2}m\angle A = \frac{1}{2}m\angle B = 40^\circ\).
The Benchmark Table Structure:
Benchmark Questions 1, 2, 3, and 4 all use a 2-column format: Conclusion and Reason. Warn students that test distractors pair a correct conclusion with an invalid reason (e.g., claiming "Def. Segment Bisector" instead of "Def. Midpoint").
Teacher Guide & Key • Page 5 of 6 Standard G.GF.3: Topics 13–16 Solutions
Teacher Resource • Answer Key
Benchmark Q16–Q19
Topics 17, 18, 19, 20 (Problems 17A–20B)
Standard G.GF.7 Coordinate Calculation: Midpoint formula \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) and Distance formula \(d = \sqrt{\Delta x^2 + \Delta y^2}\).
Topic 17: Midpoint Calculations (Benchmark Q16 & Q18 Alignment) Matches Test Q16 & Q18
Problem 17A Solution:
\(M = \left(\frac{5 + 2}{2}, \; \frac{10 + (-2)}{2}\right) = \left(\frac{7}{2}, \; \frac{8}{2}\right) = \mathbf{(3.5, 4)}\)
Answer: (3.5, 4) — Exact match to Test Q16 Choice A!
Problem 17B Solution:
\(E = \left(\frac{5 + 11}{2}, \; \frac{1 + 1}{2}\right) = \left(\frac{16}{2}, \; \frac{2}{2}\right) = \mathbf{(8, 1)}\)
Answer: (8, 1) — Exact match to Test Q18 Choice B!
Topic 18: Missing Endpoint Given Midpoint Algebraic Midpoint
Problem 18A Solution:
\(\frac{1 + x}{2} = 4 \implies 1 + x = 8 \implies \mathbf{x = 7}\)
\(\frac{-3 + y}{2} = 2 \implies -3 + y = 4 \implies \mathbf{y = 7}\)
Answer: Endpoint \(B = (7, 7)\)
Problem 18B Solution:
\(\frac{-6 + x}{2} = 0 \implies \mathbf{x = 6}\); \(\frac{2 + y}{2} = 5 \implies \mathbf{y = 8}\)
Answer: Endpoint \(D = (6, 8)\)
Topic 19: Distance Formula (Benchmark Q17 Alignment) Matches Test Q17
Problem 19A Solution:
\(d = \sqrt{(2 - 6)^2 + (-2 - 10)^2} = \sqrt{(-4)^2 + (-12)^2}\)
\(d = \sqrt{16 + 144} = \sqrt{160} \approx \mathbf{12.649 \dots \approx 12.6}\)
Answer: 12.6 — Exact match to Test Q17 Choice C!
Problem 19B Solution:
\(d = \sqrt{(9 - 1)^2 + (8 - 2)^2} = \sqrt{8^2 + 6^2}\)
\(d = \sqrt{64 + 36} = \mathbf{\sqrt{100} = 10}\)
Answer: Exact \(\sqrt{100}\), Value = 10
Topic 20: Distance with Negatives (Benchmark Q19 Alignment) Matches Test Q19
Problem 20A Solution:
\(d = \sqrt{(-6 - (-4))^2 + (8 - 5)^2} = \sqrt{(-2)^2 + 3^2}\)
\(d = \sqrt{4 + 9} = \sqrt{13} \approx \mathbf{3.605 \dots \approx 3.6}\)
Answer: 3.6 — Exact match to Test Q19 Choice A!
Problem 20B Solution:
\(d = \sqrt{(2 - (-3))^2 + (-7 - (-1))^2} = \sqrt{5^2 + (-6)^2}\)
\(d = \sqrt{25 + 36} = \sqrt{61} \approx \mathbf{7.810 \dots \approx 7.8}\)
Answer: 7.8
Distance Formula Squaring Traps:
When students compute \((-4)^2\) or \((-2)^2\), they must understand that the squared value is ALWAYS positive (+16, +4). A frequent incorrect distractor comes from subtracting under the radical (e.g., \(144 - 16 = 128\)).
Teacher Guide & Key • Page 6 of 6 Standard G.GF.7: Topics 17–20 Solutions
