Cambridge Mathematics 7 — Units 8 and 15 Comprehensive Assessment — Paper B

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CAMBRIDGE MATHEMATICS 7

UNITS 8 AND 15 — COMPREHENSIVE ASSESSMENT

Paper B — Academic Year 2026–2027

Time allowed: 150 minutes

Total: 100 marks

INSTRUCTIONS

Answer all 30 questions.

Section I contains 10 multiple-choice questions worth 20 marks.
Section II contains 10 written-response questions worth 40 marks.
Section III contains 5 explanation questions worth 15 marks.
Section IV contains 5 reasoning questions worth 25 marks.

Choose one answer for each multiple-choice question.

For all other questions, show your calculations and explain your reasoning when requested.

Include appropriate units in every measurement answer.

A ruler, a pencil and a calculator may be used. Use squared paper for nets and views where appropriate.

Diagrams are not necessarily drawn to scale. When considering symmetry, ignore vertex labels and dimension labels.

The order of rotational symmetry is the number of times a shape fits exactly onto itself during one complete turn, including its final position.

Illustrations for the written-response questions are provided in the illustrated solutions after the assessment.

SECTION I. MULTIPLE-CHOICE QUESTIONS — 20 MARKS

Choose the correct answer. Each question is worth 2 marks.

Question 1 (2 marks). The outline below is a rectangle that is not a square. Which statement describes its symmetry?

[Hình vẽ]

A. One line of symmetry and rotational symmetry of order 2.

B. Two lines of symmetry and rotational symmetry of order 2.

C. Two lines of symmetry and rotational symmetry of order 4.

D. Four lines of symmetry and rotational symmetry of order 4.

Question 2 (2 marks). The circle has centre OO. Points AA, BB, CC and DD lie on the circle, and ABAB passes through OO. Which labelled segment is a chord but not a diameter?

[Hình vẽ]

A. OAOA

B. ABAB

C. CDCD

D. OCOC

Question 3 (2 marks). The outline below is a regular hexagon. What is the smallest positive angle through which it can be rotated about its centre to fit exactly onto itself?

[Hình vẽ]

A. 30∘30^\circ

B. 45∘45^\circ

C. 60∘60^\circ

D. 120∘120^\circ

Question 4 (2 marks). Triangles ABCABC and PQRPQR are congruent, with corresponding vertices in the order shown by △ABC≅△PQR\triangle ABC \cong \triangle PQR. Given that AB=6AB=6 cm, BC=8BC=8 cm and AC=10AC=10 cm, what is the length of PRPR?

[Hình vẽ]

A. 66 cm

B. 88 cm

C. 1010 cm

D. 2424 cm

Question 5 (2 marks). The solid shown is a tetrahedron. How many faces, vertices and edges does it have?

[Hình vẽ]

A. 4 faces, 4 vertices and 6 edges.

B. 4 faces, 6 vertices and 4 edges.

C. 5 faces, 6 vertices and 9 edges.

D. 6 faces, 8 vertices and 12 edges.

Question 6 (2 marks). A cylinder stands upright on a horizontal table. What shape is its top view?

[Hình vẽ]

A. A triangle.

B. A circle.

C. A rectangle.

D. A regular hexagon.

Question 7 (2 marks). Square ABCDABCD has side length 0.60.6 m. What is its area in square centimetres?

[Hình vẽ]

A. 3636 cm²

B. 360360 cm²

C. 36003600 cm²

D. 36 00036\,000 cm²

Question 8 (2 marks). In triangle ABCABC, BC=12BC=12 cm and the perpendicular height AH=5AH=5 cm. What is the area of the triangle?

[Hình vẽ]

A. 1717 cm²

B. 3030 cm²

C. 6060 cm²

D. 120120 cm²

Question 9 (2 marks). A cuboid has length 1010 cm, width 44 cm and height 33 cm. What is its volume?

[Hình vẽ]

A. 1717 cm³

B. 4040 cm³

C. 120120 cm³

D. 164164 cm³

Question 10 (2 marks). A cube has side length 55 cm. What is its total surface area?

[Hình vẽ]

A. 2525 cm²

B. 100100 cm²

C. 125125 cm²

D. 150150 cm²

SECTION II. WRITTEN-RESPONSE AND PRACTICAL QUESTIONS — 40 MARKS

Show your working. Each question is worth 4 marks.

Question 1 (4 marks). Draw a square ABCDABCD with side length 44 cm.
a) Draw all its lines of symmetry. (2 marks)
b) State its order of rotational symmetry and its smallest positive angle of rotational symmetry. (2 marks)

Question 2 (4 marks). Draw a circle with centre OO and radius 33 cm. Draw a diameter ABAB and a chord CDCD that does not pass through OO.
a) State the length of ABAB. (1 mark)
b) Explain why CDCD is a chord but not a diameter. (1 mark)
c) Draw a tangent at AA. Mark the angle between the tangent and OAOA, and state its size. (2 marks)

Question 3 (4 marks). Sketch a regular octagon.
a) State how many equal sides and equal angles it has. (1 mark)
b) State its number of lines of symmetry. (1 mark)
c) State its order of rotational symmetry. (1 mark)
d) Explain what makes it a regular polygon, rather than simply an eight-sided polygon. (1 mark)

Question 4 (4 marks). Triangles ABCABC and DEFDEF are congruent, with corresponding vertices given by △ABC≅△DEF\triangle ABC \cong \triangle DEF. Triangle ABCABC is right-angled at BB, and AB=6AB=6 cm, BC=8BC=8 cm and AC=10AC=10 cm.
a) State the lengths of DEDE, EFEF and DFDF. (1 mark)
b) State the size of DEF^\widehat{DEF}. (1 mark)
c) Calculate the perimeter of triangle DEFDEF. (1 mark)
d) Calculate the area of triangle DEFDEF. (1 mark)

Question 5 (4 marks). A cuboid rests on a horizontal table. Its front face is 88 cm wide and 33 cm high. The cuboid extends 55 cm from front to back.
a) Draw and label its front view. (1 mark)
b) Draw and label its top view. (1 mark)
c) Draw and label its right-side view. (1 mark)
d) State the number of faces, vertices and edges of the cuboid. (1 mark)

Question 6 (4 marks). A rectangular card is 1212 cm long and 88 mm wide.
a) Calculate its area in square millimetres. (1 mark)
b) Convert this area to square centimetres. (1 mark)
c) Convert 2.752.75 m² to square centimetres. (1 mark)
d) Convert 650650 mm² to square centimetres. (1 mark)

Question 7 (4 marks). A rectangular field is 600600 m long and 250250 m wide.
a) Calculate its area in square metres. (1 mark)
b) Express its area in hectares. (1 mark)
c) The field is divided into two parts in the ratio 2:32:3. Calculate the area of the larger part in hectares and in square metres. (2 marks)

Question 8 (4 marks). A sign is made from a rectangle with a triangle attached along its entire top edge. The rectangle is 1212 cm wide and 55 cm high. The triangle has base 1212 cm and perpendicular height 44 cm. A rectangular opening measuring 22 cm by 11 cm is cut entirely inside the rectangular part.
a) Calculate the area of the rectangular part before the opening is cut. (1 mark)
b) Calculate the area of the triangular part. (1 mark)
c) Calculate the remaining area of the sign after the opening is cut. (2 marks)

Question 9 (4 marks). A closed cuboid-shaped box has length 1212 cm, width 88 cm and height 2525 mm.
a) Convert the height to centimetres. (1 mark)
b) Calculate the volume of the box in cubic centimetres. (1 mark)
c) Calculate its total surface area in square centimetres. (2 marks)

Question 10 (4 marks). A cube is made by folding a net of six congruent squares. Each square has side length 44 cm.
a) Draw one possible net of the cube. (2 marks)
b) Calculate the total surface area of the cube. (1 mark)
c) Calculate the volume of the cube. (1 mark)

SECTION III. EXPLANATION AND JUSTIFICATION — 15 MARKS

Give clear mathematical explanations. Each question is worth 3 marks.

Question 1 (3 marks). Two equilateral triangles have side lengths 44 cm and 66 cm respectively.

A student says, “Both triangles have three angles of 60∘60^\circ, so they must be congruent.”

Is the student correct? Explain what congruence requires and use the given measurements to support your answer.

Question 2 (3 marks). Rectangle AA measures 66 cm by 44 cm. Rectangle BB measures 88 cm by 22 cm.

A student says, “The rectangles have the same perimeter, so they must be congruent and have the same area.”

Calculate their perimeters and areas. Explain why the conclusion is incorrect.

Question 3 (3 marks). In triangle ABCABC, BC=14BC=14 cm, AC=10AC=10 cm and the perpendicular height from AA to BCBC is AH=6AH=6 cm.

A student writes:

Area=12×14×10=70.\text{Area}=\dfrac12\times14\times10=70.

Identify the mistake, explain which measurement should be used, and calculate the correct area with its unit.

Question 4 (3 marks). A square panel has side length 0.60.6 m, so its area is 0.360.36 m².

A student says, “To change square metres to square centimetres, I multiply by 100100. Therefore, the area is 3636 cm².”

Explain why the conversion factor is incorrect and find the correct area in square centimetres.

Question 5 (3 marks). Cube AA has side length 33 cm. Cube BB has side length 66 cm.

A student says, “The side length has doubled, so both the volume and the total surface area have doubled.”

Calculate the volume and total surface area of each cube. Explain the actual multiplication factor for each quantity.

SECTION IV. REASONING AND DEDUCTION — 25 MARKS

Show a complete chain of reasoning. Each question is worth 5 marks.

Question 1 (5 marks). Nine congruent square tiles form a 3×33\times3 square. Five tiles are dark and four are white.

The arrangement must have rotational symmetry of order 4, and tile colours must match after rotation.
a) Explain why the centre tile must be dark. (1 mark)
b) Draw the two possible colour arrangements. (2 marks)
c) State the number of lines of symmetry of each arrangement. (1 mark)
d) Explain why there are no other possible arrangements. (1 mark)

Question 2 (5 marks). A right prism has two congruent regular polygonal bases and rectangular side faces. It has 18 edges.
a) Determine the number of sides of each base. Explain how you use the edge count. (2 marks)
b) Name the prism and state its number of faces and vertices. (1 mark)
c) State the number of lines of symmetry of one base and its smallest positive angle of rotational symmetry. (2 marks)

Question 3 (5 marks). A rectangular site measures 800800 m by 600600 m. A rectangle measuring 300300 m by 200200 m is removed from its top-right corner, leaving an L-shaped site.

The land costs 2500 dollars per hectare. A fence around the entire boundary of the remaining site costs 8 dollars per metre. A buyer has a total budget of 130 000 dollars.

a) Calculate the area of the remaining site in hectares. (2 marks)
b) Calculate the total length of fence required. (1 mark)
c) Determine whether the budget covers both the land and the fence. State the amount left over or the shortfall. (2 marks)

Question 4 (5 marks). An open-top tank is a cuboid with length 22 m, width 1.51.5 m and height 11 m. Ignore the thickness of its walls.

Only the outside of its base and four walls will be painted. The tank needs two coats of paint.

One litre of paint covers 44 m² for one coat. Paint is sold only in 0.750.75-litre tins costing 18 dollars each. Assume there is no paint wastage.

a) Calculate the volume of the tank in cubic metres. (1 mark)
b) Calculate the outside area to be painted for one coat. (2 marks)
c) Calculate the minimum number of tins required for two coats and the total cost. (2 marks)

Question 5 (5 marks). A solid metal model consists of a cuboid measuring 1212 cm by 88 cm by 44 cm and a cube of side length 44 cm attached on top.

The cube stands at one corner of the cuboid’s top face. Its entire bottom face is in contact with the cuboid. There are no gaps or overlaps.

a) Calculate the volume of the complete model. (1 mark)
b) Calculate its total exposed surface area. Explain how the contact area is treated. (2 marks)
c) The model is melted down and recast into cubes of side length 22 cm. Assuming no metal is lost, calculate how many complete cubes can be made. (2 marks)

--------- END OF ASSESSMENT ---------

Student name: ................................................ Candidate number: ........................

ANSWER KEY AND ILLUSTRATED SOLUTIONS

The illustrations below accompany the written-response questions. Alternative correct drawings, methods and explanations are acceptable.

SECTION I. MULTIPLE-CHOICE ANSWERS

Question 1: B. A non-square rectangle has two lines of symmetry, through the midpoints of opposite sides, and rotational symmetry of order 2.

Question 2: C. Both endpoints of CDCD lie on the circle, but CDCD does not pass through the centre.

Question 3: C. The smallest positive rotation is 360∘÷6=60∘360^\circ\div6=60^\circ.

Question 4: C. Side PRPR corresponds to ACAC, so PR=10PR=10 cm.

Question 5: A. A tetrahedron has 4 triangular faces, 4 vertices and 6 edges.

Question 6: B. Viewed vertically from above, an upright cylinder has a circular outline.

Question 7: C. The side length is 6060 cm, so the area is 60×60=360060\times60=3600 cm².

Question 8: B. The area is 12×12×5=30\dfrac12\times12\times5=30 cm².

Question 9: C. The volume is 10×4×3=12010\times4\times3=120 cm³.

Question 10: D. The total surface area is 6×52=1506\times5^2=150 cm².

SECTION II. WRITTEN-RESPONSE AND PRACTICAL SOLUTIONS

Question 1

[Hình vẽ]

a) The four lines of symmetry are the two diagonals and the two lines joining the midpoints of opposite sides.

b) The order of rotational symmetry is 4.

The smallest positive angle of rotational symmetry is:

360∘÷4=90∘.360^\circ\div4=90^\circ.

Question 2

[Hình vẽ]

a) The diameter is twice the radius:

AB=2×3=6.AB=2\times3=6.

Therefore, AB=6AB=6 cm.

b) Segment CDCD joins two points on the circle, so it is a chord. It does not pass through OO, so it is not a diameter.

c) The tangent at AA is perpendicular to radius OAOA. The angle is 90∘90^\circ.

Question 3

[Hình vẽ]

a) It has 8 equal sides and 8 equal interior angles.

b) It has 8 lines of symmetry.

c) Its order of rotational symmetry is 8.

d) A regular polygon has all sides equal and all interior angles equal. Having eight sides alone does not guarantee either property.

Question 4

[Hình vẽ]

a) Corresponding sides are equal:

DE=6,EF=8,DF=10.DE=6,\qquad EF=8,\qquad DF=10.

All three lengths are in centimetres.

b) Angle DEF^\widehat{DEF} corresponds to angle ABC^\widehat{ABC}, so DEF^=90∘\widehat{DEF}=90^\circ.

c) The perimeter is:

6+8+10=24.6+8+10=24.

The perimeter is 2424 cm.

d) The two perpendicular sides are DEDE and EFEF:

Area=12×6×8=24.\text{Area}=\dfrac12\times6\times8=24.

The area is 2424 cm².

Question 5

[Hình vẽ]

a) The front view is a rectangle measuring 88 cm by 33 cm.

b) The top view is a rectangle measuring 88 cm by 55 cm.

c) The right-side view is a rectangle measuring 55 cm by 33 cm.

d) A cuboid has 6 faces, 8 vertices and 12 edges.

Question 6

[Hình vẽ]

The card has length AB=12AB=12 cm and width BC=8BC=8 mm.

a) Convert the length first: 1212 cm =120=120 mm.

Area=120×8=960.\text{Area}=120\times8=960.

The area is 960960 mm².

b) Since 11 cm² =100=100 mm²:

960÷100=9.6.960\div100=9.6.

The area is 9.69.6 cm².

c) Since 11 m² =10 000=10\,000 cm²:

2.75×10 000=27 500.2.75\times10\,000=27\,500.

Therefore, 2.752.75 m² =27 500=27\,500 cm².

d)

650÷100=6.5.650\div100=6.5.

Therefore, 650650 mm² =6.5=6.5 cm².

Question 7

[Hình vẽ]

Here AB=600AB=600 m and BC=250BC=250 m. The division shown gives areas in the ratio 2:32:3.

a)

Area=600×250=150 000.\text{Area}=600\times250=150\,000.

The field has area 150 000150\,000 m².

b)

150 000÷10 000=15.150\,000\div10\,000=15.

The area is 1515 ha.

c) There are 2+3=52+3=5 equal ratio parts. The larger portion contains three parts:

35×15=9.\dfrac35\times15=9.

The larger portion is 99 ha, or 90 00090\,000 m².

Question 8

[Hình vẽ]

The rectangle is 1212 cm by 55 cm, the triangular height is 44 cm, and the opening is 22 cm by 11 cm.

a)

Rectangle area=12×5=60.\text{Rectangle area}=12\times5=60.

The rectangular part has area 6060 cm².

b)

Triangle area=12×12×4=24.\text{Triangle area}=\dfrac12\times12\times4=24.

The triangular part has area 2424 cm².

c) The opening has area 2×1=22\times1=2 cm².

Remaining area=60+24−2=82.\text{Remaining area}=60+24-2=82.

The remaining area is 8282 cm².

Question 9

[Hình vẽ]

a) The height is 25÷10=2.525\div10=2.5 cm.

b)

V=12×8×2.5=240.V=12\times8\times2.5=240.

The volume is 240240 cm³.

c)

S=2(lw+lh+wh).S=2(lw+lh+wh).

S=2(12×8+12×2.5+8×2.5)=2(96+30+20)=292.S=2(12\times8+12\times2.5+8\times2.5) =2(96+30+20)=292.

The total surface area is 292292 cm².

Question 10

[Hình vẽ]

a) One valid net is shown. It contains six squares that fold to form the six faces without overlapping.

b)

S=6×42=96.S=6\times4^2=96.

The total surface area is 9696 cm².

c)

V=43=64.V=4^3=64.

The volume is 6464 cm³.

SECTION III. EXPLANATION AND JUSTIFICATION SOLUTIONS

Question 1

[Hình vẽ]

The student is incorrect.

Congruent shapes must have the same shape and the same size. Their corresponding sides and angles must be equal.

Although both equilateral triangles have angles of 60∘60^\circ, their corresponding sides measure 44 cm and 66 cm. Their sizes differ, so they are not congruent.

Question 2

[Hình vẽ]

The left rectangle measures 66 cm by 44 cm. The right rectangle measures 88 cm by 22 cm.

Their perimeters are:

PA=2(6+4)=20,PB=2(8+2)=20.P_A=2(6+4)=20,\qquad P_B=2(8+2)=20.

Both perimeters are 2020 cm.

Their areas are:

AA=6×4=24,AB=8×2=16.A_A=6\times4=24,\qquad A_B=8\times2=16.

Their areas are 2424 cm² and 1616 cm².

Equal perimeters do not imply congruence or equal areas. The rectangles have different side lengths and different areas.

Question 3

[Hình vẽ]

The student used the sloping side AC=10AC=10 cm as the height.

The height corresponding to base BCBC must be perpendicular to BCBC. The correct height is AH=6AH=6 cm.

Therefore:

Area=12×14×6=42.\text{Area}=\dfrac12\times14\times6=42.

The correct area is 4242 cm².

Question 4

[Hình vẽ]

The side length is 0.60.6 m, which is 6060 cm.

A square metre contains 100×100=10 000100\times100=10\,000 square centimetres. The linear conversion factor must be squared:

1 m2=10 000 cm2.1\text{ m}^2=10\,000\text{ cm}^2.

Therefore:

0.36×10 000=3600.0.36\times10\,000=3600.

The correct area is 36003600 cm².

This also agrees with 60×60=360060\times60=3600 cm².

Question 5

[Hình vẽ]

Cube AA:

VA=33=27,SA=6×32=54.V_A=3^3=27,\qquad S_A=6\times3^2=54.

Its volume is 2727 cm³ and its total surface area is 5454 cm².

Cube BB:

VB=63=216,SB=6×62=216.V_B=6^3=216,\qquad S_B=6\times6^2=216.

Its volume is 216216 cm³ and its total surface area is 216216 cm².

The volume multiplication factor is:

216÷27=8=23.216\div27=8=2^3.

The surface-area multiplication factor is:

216÷54=4=22.216\div54=4=2^2.

Doubling every length multiplies volume by 8 and surface area by 4, not by 2.

SECTION IV. REASONING AND DEDUCTION SOLUTIONS

Question 1

[Hình vẽ]

a) A quarter-turn leaves the centre tile in the same position.

The four corner positions cycle through one another, so they must all have the same colour. The four edge-middle positions also cycle through one another and must all have the same colour.

Dark tiles outside the centre therefore occur in groups of four. Since there are five dark tiles, the centre must be dark.

b) The other four dark tiles must occupy either all four corners or all four edge-middle positions. These give the two patterns shown.

c) Each arrangement has 4 lines of symmetry: one horizontal, one vertical and two diagonal.

d) Once the centre is dark, exactly four dark tiles remain. Rotational symmetry requires them to fill one complete group of four positions. There are only two such groups, so there are no other arrangements.

Question 2

[Hình vẽ]

a) Let each base have nn sides.

The two bases contribute 2n2n edges, and there are nn edges joining corresponding vertices. Therefore:

3n=18,n=6.3n=18,\qquad n=6.

Each base has 6 sides.

b) The solid is a hexagonal prism.

It has 6+2=86+2=8 faces and 2×6=122\times6=12 vertices.

c) A regular hexagonal base has 6 lines of symmetry.

Its smallest positive angle of rotational symmetry is:

360∘÷6=60∘.360^\circ\div6=60^\circ.

Question 3

[Hình vẽ]

The original rectangle has dimensions 800800 m by 600600 m. The removed corner DCGEDCGE measures 300300 m by 200200 m.

a) The remaining area is:

800×600−300×200=480 000−60 000=420 000.800\times600-300\times200 =480\,000-60\,000 =420\,000.

The area is 420 000420\,000 m².

In hectares:

420 000÷10 000=42.420\,000\div10\,000=42.

The remaining site has area 4242 ha.

b) The boundary lengths, in metres, are:

800, 400, 300, 200, 500, 600.800,\ 400,\ 300,\ 200,\ 500,\ 600.

Therefore:

P=800+400+300+200+500+600=2800.P=800+400+300+200+500+600=2800.

The fence must be 28002800 m long.

Removing the corner does not change the total perimeter: the removed outer lengths are replaced by equal inner lengths.

c) The land cost is:

42×2500=105 000.42\times2500=105\,000.

The fence cost is:

2800×8=22 400.2800\times8=22\,400.

The combined cost is:

105 000+22 400=127 400.105\,000+22\,400=127\,400.

The budget is sufficient, with:

130 000−127 400=2600.130\,000-127\,400=2600.

The buyer has 2600 dollars left over.

Question 4

[Hình vẽ]

a)

V=2×1.5×1=3.V=2\times1.5\times1=3.

The tank volume is 33 m³.

b) The outside base has area:

2×1.5=3.2\times1.5=3.

Its area is 33 m².

The four outside walls have total area:

2(2×1)+2(1.5×1)=4+3=7.2(2\times1)+2(1.5\times1)=4+3=7.

Their area is 77 m².

There is no top to paint, so:

Area for one coat=3+7=10.\text{Area for one coat}=3+7=10.

The area is 1010 m².

c) Two coats require coverage of:

2×10=20.2\times10=20.

This is 2020 m² of single-coat coverage.

The paint required is:

20÷4=5.20\div4=5.

The tank requires 55 litres.

The number of tins must be a whole number:

5÷0.75=623.5\div0.75=6\dfrac23.

Therefore, 7 tins are required.

The total cost is:

7×18=126.7\times18=126.

The paint costs 126 dollars.

Question 5

[Hình vẽ]

a) The cuboid volume is:

12×8×4=384.12\times8\times4=384.

The cube volume is:

43=64.4^3=64.

Therefore:

Vtotal=384+64=448.V_{\text{total}}=384+64=448.

The total volume is 448448 cm³.

b) The full surface area of the cuboid is:

Scuboid=2(12×8+12×4+8×4)=2(96+48+32)=352.S_{\text{cuboid}}=2(12\times8+12\times4+8\times4) =2(96+48+32)=352.

This is 352352 cm².

The full surface area of the cube is:

Scube=6×42=96.S_{\text{cube}}=6\times4^2=96.

This is 9696 cm².

The contact area is 4×4=164\times4=16 cm². It is counted once on the cuboid and once on the cube, but neither contact surface is exposed.

Therefore:

Sexposed=352+96−2×16=416.S_{\text{exposed}}=352+96-2\times16=416.

The total exposed surface area is 416416 cm².

c) Each new cube has volume:

23=8.2^3=8.

This is 88 cm³.

The number of complete cubes is:

448÷8=56.448\div8=56.

Therefore, 56 complete cubes can be made.

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