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Chapter 9 Practice Paper: Perimeter, Area, Volume and Surface Area — mark scheme

QuizQuipPractice Series
QQ/S1MATH/CH9

Secondary 1 · Lower Secondary Mathematics (G2 / G3)

Mark scheme

Perimeter, Area, Volume and Surface Area

Chapter 9 Practice Paper · Maximum mark: 30

Notes for markers

  • Award one mark for each point marked [1], up to the marks shown for the part.
  • Accept answers worded differently that show the same understanding.
  • Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

QuestionAnswer
1CArea = base × height, so the height is 36 ÷ 9 = 4 cm. The slanted sides are not involved.1
2A42 = ½ × 14 × height, so the height is 42 × 2 ÷ 14 = 6 cm.1
3DPerimeter = 2 × (7 + 3) = 20 cm. The value 21 cm² would be the area.1
4B72 = ½ × (10 + 14) × h = 12h, so h = 72 ÷ 12 = 6 cm.1
5CVolume = 3 × 3 × 3 = 27 cm³ — and volumes always carry cubic units.1
6AVolume of a prism = cross-section area × length = 18 × 12 = 216 cm³.1
7DSurface area = 2 × (5 × 4 + 5 × 3 + 4 × 3) = 2 × (20 + 15 + 12) = 94 cm².1
8B1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1 000 000 cm³, and 2.5 m³ = 2 500 000 cm³.1
9AVolume = pi × r² × h = (22/7) × 49 × 12 = 1 848 cm³.1
10CArea = 8 × 3 = 24 m², and the cost is 24 × $2.50 = $60.1

Section B: Structured questions[20 marks]

QuestionAnswer and marking pointsMarks
1(a)

12 × 8 = 96 m² [1]

1
1(b)

5 × 3 = 15 m² [1]

1
1(c)

96 - 15 = 81 m² [1]

1
1(d)

Cost = 81 × 12 = $972

  • Multiplies 81 by 12 [1]
  • $972 [1]
2
Total for Question 15
2(a)

Length = (x + 4) cm, and the area equation is x(x + 4) = 45.

  • Length x + 4 [1]
  • Equation x(x + 4) = 45 [1]
2
2(b)

Try x = 5: 5 × 9 = 45 ✓. So x = 5. (x = 6 gives 6 × 10 = 60, too big; x = 4 gives 4 × 8 = 32, too small.)

  • Tests at least two values or reasons about size [1]
  • x = 5 [1]

Any systematic search is accepted; the check of neighbouring values strengthens the argument.

2
2(c)

2 × (5 + 9) = 28 cm [1]

1
Total for Question 25
3(a)

6 × 4 × 2.5 = 60 cm³ [1]

1
3(b)

Surface area = 2 × (6 × 4 + 6 × 2.5 + 4 × 2.5) = 2 × (24 + 15 + 10) = 2 × 49 = 98 cm².

  • All three face products seen (24, 15, 10) [1]
  • 98 cm² [1]
2
3(c)

Box volume = 12 × 12 × 10 = 1 440 cm³, and 1 440 ÷ 60 = 24 blocks.

  • Box volume 1 440 cm³ [1]
  • 24 blocks [1]

Award full marks for a justified arrangement argument; the volume ratio must divide exactly, which it does here.

2
Total for Question 35
4(a)

Volume = pi × r² × h = (22/7) × 7 × 7 × 15 = 22 × 7 × 15 = 2 310 cm³.

  • Uses pi × r² × h [1]
  • 2 310 cm³ [1]
2
4(b)

Curved surface area = 2 × pi × r × h = 2 × (22/7) × 7 × 15 = 660 cm².

  • Uses 2 × pi × r × h [1]
  • 660 cm² [1]
2
4(c)

Volume counts how many unit cubes fill the space, so the unit is cubed; surface area counts how many unit squares cover the faces, so the unit is squared.

  • Cubes filling space vs squares covering surfaces [1]
1
Total for Question 45

— End of mark scheme —