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Algebra 1 Honors FRQ practice

6 problems · Worked solutions · 9-point rubrics · Mathematics

Code Scholars practice problems. These are not official exam prompts.

Problem 1 · Real Numbers and Algebraic Expressions · 9 points

Compare two discount offers

A jacket has price p>0\displaystyle p>0. Store A takes 20% off, then subtracts 5 dollars. Store B subtracts 5 dollars, then takes 20% off. Ignore tax.

  1. A. Write and simplify both price expressions.
  2. B. Find the difference and evaluate both at p = 60.
  3. C. State the price range where both final prices are nonnegative and explain why order matters.
Show worked solution and rubric
  1. A. A: 0.8p−5\displaystyle 0.8p-5. B: 0.8(p−5)=0.8p−4\displaystyle 0.8(p-5)=0.8p-4. Multiplication distributes over both terms in B.
  2. B. B costs 1\displaystyle 1 more for every allowed p\displaystyle p. At p=60\displaystyle p=60, A costs 43 dollars and B costs 44 dollars.
  3. C. A requires p≥6.25\displaystyle p\geq 6.25; B requires p≥5\displaystyle p\geq 5. Both are nonnegative for p≥6.25\displaystyle p\geq 6.25. Taking 20% off the five-dollar reduction saves only four dollars in B.

Scoring rubric

  • 3 points: A: 0.8p−5\displaystyle 0.8p-5
  • 3 points: B costs 1\displaystyle 1 more for every allowed p\displaystyle p
  • 3 points: A requires p≥6.25\displaystyle p\geq 6.25; B requires p≥5\displaystyle p\geq 5
Review this unit

Problem 2 · Linear Equations and Inequalities · 9 points

Plan an hourly rental

A bicycle rental costs 18 dollars plus 7 dollars per hour. The budget is at least 60 dollars and at most 95 dollars. Only whole hours may be rented.

  1. A. Form and solve a compound inequality.
  2. B. List all allowed whole-hour values.
  3. C. Explain what changes if the upper budget is strictly less than 95 dollars.
Show worked solution and rubric
  1. A. 60≤18+7h≤95\displaystyle 60\leq18+7h\leq95 gives 42≤7h≤77\displaystyle 42\leq7h\leq77, hence 6≤h≤11\displaystyle 6\leq h\leq11.
  2. B. The allowed values are 6,7,8,9,10,11\displaystyle 6,7,8,9,10,11 hours. Both endpoints satisfy the original budget.
  3. C. Then h<11\displaystyle h<11, so the whole-hour values become 6,7,8,9,10\displaystyle 6,7,8,9,10. The value 11 costs exactly 95 dollars.

Scoring rubric

  • 3 points: 60≤18+7h≤95\displaystyle 60\leq18+7h\leq95 gives 42≤7h≤77\displaystyle 42\leq7h\leq77, hence 6≤h≤11\displaystyle 6\leq h\leq11.
  • 3 points: The allowed values are 6,7,8,9,10,11\displaystyle 6,7,8,9,10,11 hours
  • 3 points: Then h<11\displaystyle h<11, so the whole-hour values become 6,7,8,9,10\displaystyle 6,7,8,9,10
Review this unit

Problem 3 · Linear Functions and Coordinate Geometry · 9 points

Build a line from measurements

A line passes through (−1,4)\displaystyle (-1,4) and (2,−2)\displaystyle (2,-2).

Problem diagram
-2-11234-6-4-2246xy
  1. A. Find the slope and equation.
  2. B. Find both intercepts and explain the sign of the slope.
  3. C. Write the equation of the perpendicular line through (2, -2).
Show worked solution and rubric
  1. A. m=−2−42−(−1)=−2\displaystyle m=\frac{-2-4}{2-(-1)}=-2. Using y=mx+b\displaystyle y=mx+b gives b=2\displaystyle b=2, so y=−2x+2\displaystyle y=-2x+2.
  2. B. The intercepts are (0,2)\displaystyle (0,2) and (1,0)\displaystyle (1,0). Each increase of one in x\displaystyle x lowers y\displaystyle y by two.
  3. C. The perpendicular slope is 12\displaystyle \frac12, since (−2)(12)=−1\displaystyle (-2)(\frac12)=-1. Thus y+2=12(x−2)\displaystyle y+2=\frac12(x-2), or y=12x−3\displaystyle y=\frac12x-3.
Solution diagram
-2-11234-6-4-2246xy

Scoring rubric

  • 3 points: m=−2−42−(−1)=−2\displaystyle m=\frac{-2-4}{2-(-1)}=-2
  • 3 points: The intercepts are (0,2)\displaystyle (0,2) and (1,0)\displaystyle (1,0)
  • 3 points: The perpendicular slope is 12\displaystyle \frac12, since (−2)(12)=−1\displaystyle (-2)(\frac12)=-1
Review this unit

Problem 4 · Systems of Equations and Inequalities · 9 points

Sell tickets for a school event

A club sells 40 tickets. Adult tickets cost 12 dollars and student tickets cost 8 dollars. Revenue is 400 dollars.

  1. A. Define variables and write two equations.
  2. B. Solve the system and verify revenue.
  3. C. If capacity becomes at most 40 and revenue must be at least 400 dollars, state the inequalities, including domain restrictions.
Show worked solution and rubric
  1. A. Let a\displaystyle a and s\displaystyle s be adult and student tickets. Then a+s=40\displaystyle a+s=40 and 12a+8s=400\displaystyle 12a+8s=400.
  2. B. Subtract 8(a+s)=320\displaystyle 8(a+s)=320 from the revenue equation: 4a=80\displaystyle 4a=80, so a=20,s=20\displaystyle a=20,s=20. Revenue is 240+160=400\displaystyle 240+160=400.
  3. C. a+s≤40\displaystyle a+s\leq40, 12a+8s≥400\displaystyle 12a+8s\geq400, with nonnegative integers a,s\displaystyle a,s. A feasible ticket count must satisfy all constraints.

Scoring rubric

  • 3 points: Let a\displaystyle a and s\displaystyle s be adult and student tickets
  • 3 points: Subtract 8(a+s)=320\displaystyle 8(a+s)=320 from the revenue equation: 4a=80\displaystyle 4a=80, so a=20,s=20\displaystyle a=20,s=20
  • 3 points: a+s≤40\displaystyle a+s\leq40, 12a+8s≥400\displaystyle 12a+8s\geq400, with nonnegative integers a,s\displaystyle a,s
Review this unit

Problem 5 · Polynomials and Factoring · 9 points

Find a rectangular garden

A garden has area x2+5x+6\displaystyle x^2+5x+6 square meters and one side x+2\displaystyle x+2 meters, where x>0\displaystyle x>0.

  1. A. Factor the area and find the other side.
  2. B. Write and simplify the perimeter.
  3. C. Find x when the area is 30 square meters; justify rejecting any root.
Show worked solution and rubric
  1. A. x2+5x+6=(x+2)(x+3)\displaystyle x^2+5x+6=(x+2)(x+3), so the other side is x+3\displaystyle x+3 meters.
  2. B. P=2(x+2)+2(x+3)=4x+10\displaystyle P=2(x+2)+2(x+3)=4x+10 meters.
  3. C. x2+5x−24=(x+8)(x−3)=0\displaystyle x^2+5x-24=(x+8)(x-3)=0. The roots are −8,3\displaystyle -8,3; only x=3\displaystyle x=3 fits x>0\displaystyle x>0. The sides are 5 and 6 meters.

Scoring rubric

  • 3 points: x2+5x+6=(x+2)(x+3)\displaystyle x^2+5x+6=(x+2)(x+3), so the other side is x+3\displaystyle x+3 meters.
  • 3 points: P=2(x+2)+2(x+3)=4x+10\displaystyle P=2(x+2)+2(x+3)=4x+10 meters.
  • 3 points: x2+5x−24=(x+8)(x−3)=0\displaystyle x^2+5x-24=(x+8)(x-3)=0
Review this unit

Problem 6 · Quadratic Equations and Models · 9 points

Analyze a thrown ball

A ball has height h(t)=−5t2+20t+1\displaystyle h(t)=-5t^2+20t+1 meters for time t≥0\displaystyle t\geq0 seconds, until it hits the ground.

Problem diagram
0.511.522.533.544.5-224681012141618202224xy
  1. A. Complete the square and find the maximum height and time.
  2. B. Find the exact time the ball hits the ground.
  3. C. State the physical domain and when the ball is rising.
Show worked solution and rubric
  1. A. h(t)=−5(t−2)2+21\displaystyle h(t)=-5(t-2)^2+21, so the maximum is 21 meters at 2 seconds.
  2. B. 5t2−20t−1=0\displaystyle 5t^2-20t-1=0 gives t=2±1055\displaystyle t=2\pm\frac{\sqrt{105}}5. Only 2+1055\displaystyle 2+\frac{\sqrt{105}}5 is nonnegative.
  3. C. The physical domain is 0≤t≤2+1055\displaystyle 0\leq t\leq2+\frac{\sqrt{105}}5. The ball rises for 0≤t<2\displaystyle 0\leq t<2 and falls after the vertex.
Solution diagram
0.511.522.533.544.5-224681012141618202224xy

Scoring rubric

  • 3 points: h(t)=−5(t−2)2+21\displaystyle h(t)=-5(t-2)^2+21, so the maximum is 21 meters at 2 seconds.
  • 3 points: 5t2−20t−1=0\displaystyle 5t^2-20t-1=0 gives t=2±1055\displaystyle t=2\pm\frac{\sqrt{105}}5
  • 3 points: The physical domain is 0≤t≤2+1055\displaystyle 0\leq t\leq2+\frac{\sqrt{105}}5
Review this unit

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