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IB Maths Applications & Interpretation · SL and HL · Topic 3

Geometry and trigonometry

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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →

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+ Add to test✓ In your test1IBMI-032-00-01Right-angled trigonometry, bearings and the sine and cosine rules Paper SLEasy5 marks

A community centre is installing a straight wheelchair ramp. The ramp rises a vertical height of 0.45 m over a horizontal distance of 5.2 m.

(a)

Find the angle that the ramp makes with the horizontal.

(2)

(b)

Find the length of the sloping surface of the ramp.

(2)

(c)

Local building guidance states that a ramp of this kind should make an angle of no more than 4.8° with the horizontal. State, with a reason, whether this ramp meets the guidance.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses tan θ = 0.45 / 5.2 M1	✓ 1
θ = 4.95° A1	✓ 1
Part (b)
Uses Pythagoras: √(5.2² + 0.45²) M1	✓ 1
5.22 m A1	✓ 1
Part (c)
No, because 4.95° is greater than 4.8° (follow through from part (a)) R1	✓ 1

Exam tip: Make sure your calculator is in degree mode before using inverse trigonometric functions; an answer of 0.0863 here would be the angle in radians.

+ Add to test✓ In your test2IBMI-032-00-02Right-angled trigonometry, bearings and the sine and cosine rules Paper SLEasy6 marks

Amara stands on level ground 35 m from the foot of a vertical clock tower. Her eyes are 1.6 m above the ground. She measures the angle of elevation of the top of the tower as 38°.

(a)

Find the vertical distance from Amara's eye level to the top of the tower.

(2)

(b)

Hence find the height of the clock tower.

(1)

(c)

Amara walks directly towards the tower until the angle of elevation of the top is 55°. Find how far she walked.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses tan 38° = h / 35 M1	✓ 1
h = 27.3 m A1	✓ 1
Part (b)
27.345 + 1.6 = 28.9 m A1	✓ 1
Part (c)
New horizontal distance = 27.345 / tan 55° M1	✓ 1
= 19.1 m A1	✓ 1
Distance walked = 35 − 19.1 = 15.9 m A1	✓ 1

Common error: In part (c) the vertical distance in the triangle is still measured from eye level, so use 27.3 m, not the full tower height of 28.9 m.

+ Add to test✓ In your test3IBMI-032-00-03Right-angled trigonometry, bearings and the sine and cosine rules Paper SLEasy5 marks

A delivery drone takes off from a depot P and flies in a straight line for 4.2 km on a bearing of 125° to a drop-off point D.

(a)

Find how far east of P the point D is.

(2)

(b)

Find how far south of P the point D is.

(2)

(c)

Write down the bearing on which the drone must fly to return directly from D to P.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
4.2 sin 125° (or 4.2 cos 35°) M1	✓ 1
3.44 km A1	✓ 1
Part (b)
4.2 cos 55° (or −4.2 cos 125°) M1	✓ 1
2.41 km A1	✓ 1
Part (c)
305° A1	✓ 1

Exam tip: A back bearing differs from the original bearing by exactly 180°; add 180° if the original is less than 180°, subtract if it is more.

+ Add to test✓ In your test4IBMI-032-00-04Right-angled trigonometry, bearings and the sine and cosine rules Paper SLEasy6 marks

A triangular garden plot ABC has AB = 14 m, AC = 9.5 m and angle BAC = 72°.

(a)

Find the area of the plot.

(2)

(b)

Find the length BC.

(2)

(c)

The whole plot is to be covered with turf costing $6.40 per square metre. Turf can only be bought in whole square metres. Find the cost of the turf.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses ½ × 14 × 9.5 × sin 72° M1	✓ 1
63.2 m² A1	✓ 1
Part (b)
BC² = 14² + 9.5² − 2(14)(9.5)cos 72° M1	✓ 1
BC = 14.3 m A1	✓ 1
Part (c)
Rounds area up to 64 m² R1	✓ 1
64 × 6.40 = $409.60 A1	✓ 1

Common error: Rounding 63.2 m² to the nearest whole number gives 63 m², which would not cover the plot. In context the area must be rounded up.

+ Add to test✓ In your test5IBMI-032-06-01Volume and surface area in context Paper SLEasy5 marks

A farm stores rainwater in a closed cylindrical tank. The tank has internal radius 1.2 m and internal height 2.5 m.

(a)

Calculate the volume of the tank, in m³.

(2)

(b)

Write down the capacity of the tank in litres.

(1)

(c)

The empty tank is filled by a pump delivering 15 litres per minute. Find the time taken to fill the tank, giving your answer in minutes.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for substituting into $V=\pi r^2 h$: $\pi\times1.2^2\times2.5$	✓ 1
A1 for $V=11.3$ m³ (11.3097…)	✓ 1
Part (b)
A1 for 11 300 litres, using 1 m³ = 1000 litres; accept 11 310 or 11 309.7; follow through from their answer to part (a)	✓ 1
Part (c)
M1 for dividing their capacity by the rate, e.g. $11309.7\div15$	✓ 1
A1 for 754 minutes (753.98…), which is about 12.6 hours; accept 753 minutes as follow-through from 11 300 litres	✓ 1

Exam tip: Keep the unrounded volume (11.3097…) in your calculator and use it in later parts. A rounded value such as 11 300 litres gives 753 minutes rather than 754, and rounding early can cost accuracy marks when no follow-through is allowed.

+ Add to test✓ In your test6IBMI-032-06-02Volume and surface area in context Paper SLEasy4 marks

A traffic cone is modelled as a right circular cone with an open base. The base has diameter 36 cm and the slant height of the cone is 70 cm.

(a)

Find the vertical height of the cone.

(2)

(b)

The curved surface of the cone is painted with reflective paint. Find the area to be painted.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for Pythagoras with radius 18: $h=\sqrt{70^2-18^2}$	✓ 1
A1 for $h=67.6$ cm (67.646…)	✓ 1
Part (b)
M1 for using $\pi r l$: $\pi\times18\times70$	✓ 1
A1 for 3960 cm² (3958.4…)	✓ 1

Common error: Using 36 as the radius, or putting the vertical height into $\pi r l$ in place of the slant height, are both frequent slips. The formula needs the radius and the slant height.

+ Add to test✓ In your test7IBMI-032-06-03Volume and surface area in context Paper SLEasy5 marks

The roof of a planetarium is a hemispherical dome of radius 9 m. The outside curved surface of the dome is to be painted. One litre of paint covers 12 m². Paint is sold only in 5-litre tins costing $38.50 each.

(a)

Calculate the area of the curved surface of the dome.

(2)

(b)

Find the number of tins of paint that must be bought.

(2)

(c)

Find the total cost of the paint.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for half the surface area of a sphere: $\frac{1}{2}\times4\pi\times9^2$	✓ 1
A1 for 509 m² (508.938…)	✓ 1
Part (b)
M1 for $508.938\div12=42.4$ litres, then dividing by 5 to get 8.48 tins	✓ 1
A1 for 9 tins, rounded up in context	✓ 1
Part (c)
A1 for $9\times38.50=\$346.50$	✓ 1

Common error: Rounding 8.48 to 8 tins leaves part of the dome unpainted. In a context like this, always round up to a whole number of tins.

+ Add to test✓ In your test8IBMI-032-06-04Volume and surface area in context Paper SLEasy6 marks

A glass aquarium is a cuboid with no lid. Its internal dimensions are length 80 cm, width 35 cm and height 45 cm. Ignore the thickness of the glass.

(a)

Calculate the total area of glass used.

(2)

(b)

The aquarium is filled with water to 5 cm below the top. Find the volume of water in litres.

(2)

(c)

Find the length of the longest thin straight rod that can fit inside the empty aquarium.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for adding the base and four sides only: $80\times35+2(80\times45)+2(35\times45)$	✓ 1
A1 for 13 150 cm² (accept 13 200 cm²)	✓ 1
Part (b)
M1 for $80\times35\times40=112\,000$ cm³	✓ 1
A1 for 112 litres	✓ 1
Part (c)
M1 for the space diagonal: $\sqrt{80^2+35^2+45^2}$	✓ 1
A1 for 98.2 cm (98.234…)	✓ 1

Exam tip: The longest rod lies along the space diagonal, from one bottom corner to the opposite top corner. All three dimensions go into one square root.

+ Add to test✓ In your test9IBMI-032-05-01Right-angled trigonometry, bearings and the sine and cosine rules Paper SLEasy4 marks

A wheelchair ramp outside a library rises a vertical height of 0.45 m over a horizontal distance of 5.4 m.

(a)

Find the angle that the ramp makes with the horizontal.

(2)

(b)

Find the length of the sloping surface of the ramp.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan θ = 0.45 / 5.4 M1 ; θ = 4.76° (4.7636…) A1	✓ 2
Part (b)
√(5.4² + 0.45²) or equivalent trig method M1 ; 5.42 m (5.4187…) A1	✓ 2

Exam tip: Check that your calculator is in degree mode before using any trigonometric function. A ramp angle near 0.08 is a sign it is in radians.

+ Add to test✓ In your test10IBMI-032-05-02Right-angled trigonometry, bearings and the sine and cosine rules Paper SLEasy6 marks

A surveyor stands on level ground 85 m from the base of a vertical radio tower. Her eye level is 1.6 m above the ground. The angle of elevation from her eye to the top of the tower is 32°.

(a)

Find the height of the tower.

(3)

(b)

The surveyor walks directly towards the tower until the angle of elevation to the top is 50°. Find the distance she walks.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan 32° = h / 85 M1 ; h = 53.11 m above eye level A1 ; height = 53.11 + 1.6 = 54.7 m A1	✓ 3
Part (b)
d = 53.11 / tan 50° M1 ; d = 44.57 m A1 ; distance walked = 85 − 44.57 = 40.4 m A1	✓ 3

Common error: The 1.6 m eye height affects only the total tower height. In part (b) the right-angled triangle still uses 53.11 m, measured from eye level, not 54.7 m.

+ Add to test✓ In your test11IBMI-032-05-04Right-angled trigonometry, bearings and the sine and cosine rules Paper SLEasy4 marks

A triangular vegetable plot has two straight sides of length 14 m and 9 m, and the angle between these sides is 72°.

(a)

Find the length of the third side of the plot.

(2)

(b)

Find the area of the plot.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
c² = 14² + 9² − 2(14)(9)cos 72° M1 ; c = 14.1 m (14.111…) A1	✓ 2
Part (b)
Area = ½ × 14 × 9 × sin 72° M1 ; 59.9 m² A1	✓ 2

Common error: The area formula ½ab sin C needs the angle between the two given sides. Here that is 72°, so the third side is not needed for the area.

+ Add to test✓ In your test12IBMI-032-01-02Volume and surface area in context Paper SLEasy6 marks

A party hat is a right circular cone with no base. The base radius is 9 cm and the vertical height is 20 cm.

(a)

Find the slant height of the hat.

(2)

(b)

Find the volume of air inside the hat.

(2)

(c)

Find the area of card needed to make the hat, ignoring any overlap.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$l=\sqrt{9^2+20^2}$ M1 ; l = 21.9 cm (21.932…) A1	✓ 2
Part (b)
$V=\frac{1}{3}\pi(9^2)(20)$ M1 ; V = 1700 cm³ (1696.5…) A1	✓ 2
Part (c)
Curved surface area $=\pi r l=\pi(9)(21.932…)$ M1 ; 620 cm² (620.10…) A1	✓ 2

Exam tip: The curved surface area formula uses the slant height l, not the vertical height. Keep the unrounded value of l in your calculator for the next part.

+ Add to test✓ In your test13IBMI-032-01-03Volume and surface area in context Paper SLEasy6 marks

A gym orders spherical exercise balls, each with diameter 65 cm. Treat each ball as a perfect sphere and ignore the thickness of the rubber.

(a)

Find the volume of air in one ball, giving your answer in m³.

(2)

(b)

Find the surface area of one ball in m².

(2)

(c)

The rubber used costs 8.40 USD per m². Find the cost of the rubber needed to make 50 balls.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$V=\frac{4}{3}\pi(0.325)^3$ M1 ; 0.144 m³ (0.14379…) A1	✓ 2
Part (b)
$A=4\pi(0.325)^2$ M1 ; 1.33 m² (1.3273…) A1	✓ 2
Part (c)
1.3273… × 8.40 × 50 M1 ; 557.48 USD A1 (accept 557 or 557.5 USD; accept 558.60 USD from using 1.33)	✓ 2

Common error: Converting 65 cm to 0.65 m and then forgetting to halve it is a frequent slip. The radius is 0.325 m.

+ Add to test✓ In your test14IBMI-032-01-04Volume and surface area in context Paper SLEasy6 marks

A camping tent is a triangular prism. Each end is an isosceles triangle with base 2.4 m and vertical height 1.6 m. The tent is 3.0 m long. The tent has no fabric floor.

(a)

Find the volume of the tent.

(2)

(b)

Find the length of one sloping edge of a triangular end.

(2)

(c)

Find the total area of fabric used to make the two sloping sides and the two ends.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Area of end = ½ × 2.4 × 1.6 = 1.92 m² M1 ; V = 5.76 m³ A1	✓ 2
Part (b)
$\sqrt{1.2^2+1.6^2}$ M1 ; 2.0 m A1	✓ 2
Part (c)
2(2.0 × 3.0) + 2(1.92) M1 ; 15.8 m² (15.84) A1	✓ 2

Exam tip: Read the context carefully: the tent has no floor, so the rectangular base must not be included in the fabric area.

+ Add to test✓ In your test15IBMI-032-07-r2-01Voronoi diagrams Paper SLEasy6 marks

Two fire stations are located at A(2, 3) and B(8, 7) on a coordinate grid. Distances are in kilometres. The Voronoi diagram for the two stations has one edge, the perpendicular bisector of AB.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AB, giving your answer in the form $y = mx + c$.

(3)

(c)

A house is at H(4, 6). Find the distance from H to the nearest fire station.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(5, 5) A1	✓ 1
Part (b)
Gradient of AB $= \frac{7-3}{8-2} = \frac{2}{3}$ M1	✓ 1
Perpendicular gradient $= -\frac{3}{2}$ A1	✓ 1
$y - 5 = -1.5(x - 5)$ so $y = -1.5x + 12.5$ A1	✓ 1
Part (c)
Compares distances, e.g. HA $= \sqrt{13}$ and HB $= \sqrt{17}$, or uses the bisector to show H is in the cell of A M1	✓ 1
Nearest station is A, distance 3.61 km A1	✓ 1

Exam tip: A point lies in a site's Voronoi cell exactly when it is closer to that site than to any other, so comparing distances is a quick check on which side of an edge a point lies.

+ Add to test✓ In your test16IBMI-032-07-r2-02Voronoi diagrams Paper SLEasy6 marks

Three mobile phone masts are at P(0, 0), Q(6, 0) and R(0, 4), with coordinates in kilometres. A Voronoi diagram is drawn with the masts as sites.

(a)

Write down the equation of the perpendicular bisector of PQ and the equation of the perpendicular bisector of PR.

(2)

(b)

Hence write down the coordinates of the Voronoi vertex V.

(1)

(c)

Explain why the perpendicular bisector of QR must also pass through V.

(1)

(d)

Find the distance from V to each mast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 3$ A1	✓ 1
$y = 2$ A1	✓ 1
Part (b)
V(3, 2) A1	✓ 1
Part (c)
V is equidistant from P and Q and from P and R, so it is equidistant from Q and R and lies on the bisector of QR (or: the midpoint of QR is (3, 2)) R1	✓ 1
Part (d)
$\sqrt{3^2 + 2^2}$ M1	✓ 1
3.61 km A1	✓ 1

Common error: Writing $y = 3$ instead of $x = 3$ for the bisector of a horizontal segment. The bisector of a horizontal segment is a vertical line.

+ Add to test✓ In your test17IBMI-032-02-r2-01Voronoi diagrams Paper SLEasy6 marks

Two mobile phone masts, A and B, are located at A(2, 3) and B(8, 7) on a coordinate grid. Distances are in kilometres. A Voronoi diagram is used to model which mast each house connects to, assuming each house connects to its nearest mast.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AB. Give your answer in the form $y = mx + c$.

(3)

(c)

A house is located at H(4, 6). Determine which mast the house connects to, and find the distance from the house to this mast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(5, 5) A1	✓ 1
Part (b)
Gradient of AB = $\frac{7-3}{8-2} = \frac{2}{3}$ A1	✓ 1
Perpendicular gradient = $-\frac{3}{2}$ M1	✓ 1
$y - 5 = -\frac{3}{2}(x - 5)$, so $y = -1.5x + 12.5$ A1	✓ 1
Part (c)
HA = $\sqrt{2^2+3^2} = \sqrt{13}$ and HB = $\sqrt{4^2+1^2} = \sqrt{17}$, compared M1	✓ 1
Mast A is nearest, distance 3.61 km A1	✓ 1

Exam tip: The perpendicular bisector always passes through the midpoint, and its gradient is the negative reciprocal of the gradient joining the two sites.

+ Add to test✓ In your test18IBMI-032-02-r2-02Voronoi diagrams Paper SLEasy5 marks

Three fire stations are located at P(0, 0), Q(10, 0) and R(0, 6) on a map grid, where units are kilometres. A Voronoi diagram is drawn with the fire stations as sites.

(a)

Write down the equation of the perpendicular bisector of (i) PQ; (ii) PR.

(2)

(b)

Hence write down the coordinates of the vertex of the Voronoi diagram.

(1)

(c)

A fire breaks out at this vertex. Find the distance from the fire to the nearest fire station.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(i) $x = 5$ A1	✓ 1
(ii) $y = 3$ A1	✓ 1
Part (b)
(5, 3) A1	✓ 1
Part (c)
Distance = $\sqrt{5^2 + 3^2}$ M1	✓ 1
= 5.83 km (equidistant from all three stations) A1	✓ 1

Exam tip: A Voronoi vertex is equidistant from the three sites whose cells meet there, so you only need one distance calculation.

+ Add to test✓ In your test19IBMI-032-07-r1-02Voronoi diagrams Paper SLEasy8 marks

Three fire stations are located at F₁(0, 0), F₂(10, 0) and F₃(4, 8), where distances are in kilometres. A Voronoi diagram is constructed with the three fire stations as sites.

(a)

Write down the equation of the perpendicular bisector of F₁F₂.

(1)

(b)

Show that the perpendicular bisector of F₁F₃ has equation y = −0.5x + 5.

(3)

(c)

Hence find the coordinates of the vertex V of the Voronoi diagram.

(2)

(d)

Find the distance from V to F₁, and interpret the significance of V in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
x = 5 A1	✓ 1
Part (b)
Midpoint of F₁F₃ = (2, 4) A1	✓ 1
Gradient of F₁F₃ = 8/4 = 2, so perpendicular gradient = −0.5 M1	✓ 1
y − 4 = −0.5(x − 2), leading to y = −0.5x + 5 A1	✓ 1
Part (c)
Substituting x = 5: y = −0.5(5) + 5 M1	✓ 1
V(5, 2.5) A1	✓ 1
Part (d)
VF₁ = √(5² + 2.5²) = 5.59 km A1	✓ 1
V is equidistant (5.59 km) from all three fire stations, so a fire at V is equally close to every station. Accept: V is the point in the region farthest from its nearest station. R1	✓ 1

Common error: A Voronoi vertex is equidistant from three sites, not just two. Check your vertex by confirming that its distance to the third site matches. A mismatch shows an algebra slip.

+ Add to test✓ In your test20IBMI-032-02-r1-04Voronoi diagrams Paper SLEasy7 marks

Three weather stations are located on a coordinate grid where 1 unit represents 1 km. The rainfall recorded at each station during one week is shown.
Station A (1, 5): 42 mm
Station B (7, 1): 35 mm
Station C (9, 8): 51 mm
A Voronoi diagram is drawn with the three stations as sites. The rainfall at a farm X(5, 4) is to be estimated using nearest-neighbour interpolation.

(a)

Calculate the distance from X to each of the three stations.

(3)

(b)

Hence state which Voronoi cell contains X and estimate the rainfall at X.

(2)

(c)

A rain gauge at the farm actually recorded 39 mm. Taking this as the exact value, calculate the percentage error in the estimate.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Use of distance formula, e.g. XA = $\sqrt{4^2 + 1^2}$ M1	✓ 1
XA = $\sqrt{17}$ = 4.12 km, XB = $\sqrt{13}$ = 3.61 km A1	✓ 1
XC = $\sqrt{32}$ = 5.66 km A1	✓ 1
Part (b)
X is closest to B, so X lies in the cell of B R1	✓ 1
Estimated rainfall = 35 mm A1	✓ 1
Part (c)
$\frac{\|35 - 39\|}{39} \times 100$ M1	✓ 1
= 10.3% A1	✓ 1

Common error: Nearest-neighbour interpolation uses only the value at the site whose cell contains the point; do not average the values from several stations.

+ Add to test✓ In your test21IBMI-032-02-r1-01Voronoi diagrams Paper SLEasy7 marks

A town council models two public libraries as points on a coordinate grid, where 1 unit represents 1 km. Library A is at (2, 3) and library B is at (8, 7). The council draws a Voronoi diagram using A and B as sites.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the edge of the Voronoi diagram that separates the cells of A and B. Give your answer in the form $y = mx + c$.

(3)

(c)

A student lives at P(6, 4). Determine which library is nearer to P, and find the distance from P to that library.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(5, 5) A1	✓ 1
Part (b)
Gradient of AB = $\frac{7-3}{8-2} = \frac{2}{3}$ M1	✓ 1
Gradient of perpendicular bisector = $-\frac{3}{2}$ M1	✓ 1
$y - 5 = -1.5(x - 5)$ ⇒ $y = -1.5x + 12.5$ A1	✓ 1
Alternative: set $(x-2)^2+(y-3)^2=(x-8)^2+(y-7)^2$ (M1), simplify to $12x+8y=100$ (M1), giving $y=-1.5x+12.5$ (A1).	—
Part (c)
Valid method to decide which library is nearer M1 . Either find PA = $\sqrt{4^2+1^2} = \sqrt{17}$ ≈ 4.12 km and PB = $\sqrt{2^2+3^2} = \sqrt{13}$ and compare them. Or substitute x = 6 into the edge equation to get y = 3.5 < 4, so P lies on the same side of the edge as B.	✓ 1
P is nearer library B (P lies in B's cell) A1	✓ 1
Distance PB = $\sqrt{13}$ ≈ 3.61 km A1	✓ 1

Exam tip: A Voronoi edge between two sites is always the perpendicular bisector of the segment joining them: use the midpoint and the negative reciprocal of the gradient.

+ Add to test✓ In your test22IBMI-032-07-r1-01Voronoi diagrams Paper SLEasy6 marks

Two mobile phone masts, A and B, serve a rural area. On a map with coordinates in kilometres, A is at (2, 3) and B is at (8, 7). A Voronoi diagram is drawn with A and B as its sites.

(a)

Find the coordinates of the midpoint of AB.

(1)

(b)

Find the equation of the edge of the Voronoi diagram between the cells of A and B. Give your answer in the form y = mx + c.

(3)

(c)

A hiker stands at point P(6, 3). Determine which mast is closer to the hiker and find the distance from P to that mast.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint = ((2 + 8)/2, (3 + 7)/2) = (5, 5) A1	✓ 1
Part (b)
Gradient of AB = (7 − 3)/(8 − 2) = 2/3 M1	✓ 1
Gradient of perpendicular bisector = −3/2 M1	✓ 1
y − 5 = −1.5(x − 5), so y = −1.5x + 12.5 A1	✓ 1
Part (c)
Valid method to decide which mast is closer: either PA = √((6 − 2)² + (3 − 3)²) = 4 and PB = √((6 − 8)² + (3 − 7)²) = √20 ≈ 4.47, or at x = 6 the edge has y = 3.5, so P (with y = 3) lies on A's side of the edge M1	✓ 1
Mast A is closer, at a distance of 4 km A1	✓ 1

Exam tip: A Voronoi edge between two sites is always part of their perpendicular bisector. It passes through the midpoint and has gradient equal to the negative reciprocal of the gradient joining the sites.

+ Add to test✓ In your test23IBMI-032-00-05Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium8 marks

A ferry leaves harbour H and sails 18 km on a bearing of 040° to an island A. It then sails 25 km on a bearing of 115° to a second island B.

(a)

Show that angle HAB = 105°.

(2)

(b)

Find the distance HB.

(3)

(c)

Find the bearing of B from H.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Bearing of H from A is 220° (or uses alternate angles with the north line at A) M1	✓ 1
220° − 115° = 105° A1	✓ 1
Part (b)
HB² = 18² + 25² − 2(18)(25)cos 105° M1	✓ 1
HB² = 1182 (1181.9…) A1	✓ 1
HB = 34.4 km A1	✓ 1
Part (c)
sin AHB / 25 = sin 105° / 34.38 M1	✓ 1
Angle AHB = 44.6° A1	✓ 1
Bearing = 40° + 44.6° = 084.6° A1	✓ 1

Exam tip: Keep the unrounded value of HB in your calculator for part (c); using 34.4 still gives 084.6°, but early rounding can shift answers in longer chains.

+ Add to test✓ In your test24IBMI-032-00-06Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium7 marks

A coastguard stands at the top of a vertical cliff 62 m above sea level. Two buoys, X and Y, lie due east of the foot of the cliff, in a straight line with it, with X nearer the cliff. The angles of depression from the coastguard to X and to Y are 28° and 17° respectively.

(a)

Find the horizontal distance from the foot of the cliff to X.

(2)

(b)

Find the distance between the two buoys.

(3)

(c)

A kayak travels in a straight line from X to Y in 40 seconds. Find its average speed in km h⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan 28° = 62 / d, or equivalent M1	✓ 1
d = 117 m (116.6…) A1	✓ 1
Part (b)
Distance to Y = 62 / tan 17° (= 202.8… m) M1	✓ 1
Subtracts the distance to X from the distance to Y: 202.8 − 116.6 M1	✓ 1
XY = 86.2 m. Accept 86 m from rounded values (203 − 117) A1	✓ 1
Part (c)
Speed = their XY / 40 (= 2.155 m s⁻¹) and converts by × 3.6, or equivalent M1	✓ 1
7.76 km h⁻¹. Accept 7.74 from XY = 86 m; follow through from their XY A1	✓ 1

Exam tip: Keep full calculator values for the distances to X and Y, and round only at the end. Early rounding changes the final speed in the third significant figure. The angle of depression equals the angle of elevation from the buoy, so use it directly with tan.

+ Add to test✓ In your test25IBMI-032-00-07Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium6 marks

A rectangular storeroom is a cuboid measuring 6.0 m long, 4.5 m wide and 2.8 m high. A straight cable is fixed from one top corner of the room to the bottom corner diagonally opposite it.

(a)

Find the length of the diagonal of the floor.

(2)

(b)

Find the length of the cable.

(2)

(c)

Find the angle between the cable and the floor.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
√(6.0² + 4.5²) M1	✓ 1
7.50 m A1	✓ 1
Part (b)
√(7.5² + 2.8²) M1	✓ 1
8.01 m A1	✓ 1
Part (c)
tan θ = 2.8 / 7.5 (or equivalent with sin or cos) M1	✓ 1
θ = 20.5° A1	✓ 1

Exam tip: For the angle between a line and a plane, identify the right-angled triangle formed by the line, its projection onto the plane, and the vertical drop.

+ Add to test✓ In your test26IBMI-032-00-09Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium7 marks

A triangular sail for a small dinghy has sides of length 4.8 m, 3.6 m and 5.5 m.

(a)

Find the size of the largest angle of the sail.

(3)

(b)

Find the area of the sail.

(2)

(c)

Sailcloth costs $34.75 per square metre. The sailmaker buys 15% more cloth than the area of the sail to allow for waste. Find the cost of the cloth.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Identifies the angle opposite the 5.5 m side R1	✓ 1
cos θ = (4.8² + 3.6² − 5.5²) / (2 × 4.8 × 3.6) M1	✓ 1
θ = 80.4° A1	✓ 1
Part (b)
½ × 4.8 × 3.6 × sin 80.4° M1	✓ 1
8.52 m² A1	✓ 1
Part (c)
8.52 × 1.15 × 34.75 M1	✓ 1
$340.46 (accept $340.48 from 8.52) A1	✓ 1

Common error: The largest angle is always opposite the longest side; choosing the wrong side in the cosine rule gives one of the smaller angles.

+ Add to test✓ In your test27IBMI-032-00-10Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium7 marks

Two hikers leave a campsite C at the same time. Hiker A walks 3.5 km on a bearing of 030°. Hiker B walks 5.2 km on a bearing of 120°.

(a)

Explain why angle ACB is a right angle.

(1)

(b)

Find the distance between the two hikers.

(2)

(c)

Find angle CAB.

(2)

(d)

Find the bearing of hiker B from hiker A.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
120° − 30° = 90° R1	✓ 1
Part (b)
√(3.5² + 5.2²) M1	✓ 1
6.27 km A1	✓ 1
Part (c)
tan CAB = 5.2 / 3.5 M1	✓ 1
CAB = 56.1° A1	✓ 1
Part (d)
Bearing of C from A is 210° M1	✓ 1
210° − 56.1° = 154° A1	✓ 1

Exam tip: Sketch the north line at A and mark the back bearing to C first; then decide whether to add or subtract the triangle angle.

+ Add to test✓ In your test28IBMI-032-06-05Volume and surface area in context Paper SLMedium8 marks

A ridge tent is modelled as a triangular prism of length 3 m. Its cross-section is an isosceles triangle. The two sloping sides are each 2.1 m long, and the angle between them at the ridge is 70°. The tent is made entirely of fabric, including the two triangular ends and a rectangular groundsheet.

(a)

Find the area of the triangular cross-section.

(2)

(b)

Hence find the volume of the tent.

(1)

(c)

Find the width of the base of the tent.

(2)

(d)

Find the total area of fabric needed.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for using $\frac{1}{2}ab\sin C$: $\frac{1}{2}\times2.1^2\times\sin70°$	✓ 1
A1 for 2.07 m² (2.07202…)	✓ 1
Part (b)
A1 for $2.07202\times3=6.22$ m³	✓ 1
Part (c)
M1 for the cosine rule, $\sqrt{2.1^2+2.1^2-2(2.1)(2.1)\cos70°}$, or for $2\times2.1\sin35°$	✓ 1
A1 for 2.41 m (2.40902…)	✓ 1
Part (d)
M1 for finding the three rectangular face areas: two sloping rectangles $2\times2.1\times3=12.6$ and the groundsheet $2.40902\times3=7.227$	✓ 1
M1 for adding the two triangular ends, $2\times2.07202=4.144$, to the rectangular areas, so that all five faces are summed	✓ 1
A1 for 24.0 m² (23.971…)	✓ 1

Common error: The usual way to lose marks here is to leave out the groundsheet or one of the triangular ends. List all five faces of the prism before adding them.

+ Add to test✓ In your test29IBMI-032-06-06Volume and surface area in context Paper SLMedium7 marks

A grain silo is modelled as a cylinder of radius 3.5 m and height 12 m, with a hemispherical roof of the same radius on top. The silo stands on a concrete base.

(a)

Find the total volume of the silo.

(3)

(b)

The silo is filled to 80% of its volume with grain of density 750 kg m⁻³. Find the mass of grain in the silo.

(2)

(c)

The outer wall and roof are to be given a weatherproof coating. The base is not coated. Find the area to be coated.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for the cylinder volume, $\pi\times3.5^2\times12=461.8$	✓ 1
M1 for the hemisphere volume, $\frac{2}{3}\pi\times3.5^3=89.80$	✓ 1
A1 for 552 m³ (551.61…)	✓ 1
Part (b)
M1 for $0.8\times551.61\times750$	✓ 1
A1 for 331 000 kg (331 tonnes)	✓ 1
Part (c)
M1 for adding the curved cylinder area and half a sphere: $2\pi(3.5)(12)+2\pi(3.5)^2$	✓ 1
A1 for 341 m² (340.86…)	✓ 1

Exam tip: Only the exposed surfaces are coated. The circle where the roof meets the cylinder is not a surface, and the base is excluded, so do not add any $\pi r^2$ terms.

+ Add to test✓ In your test30IBMI-032-06-07Volume and surface area in context Paper SLMedium10 marks

A glass entrance to a museum is a right pyramid with a square base of side 30 m. Its apex is 20 m vertically above the centre of the base. The four triangular faces are glass; the base is the floor.

(a)

Find the height of one triangular face, measured from the apex to the midpoint of a base edge.

(2)

(b)

Find the total area of glass.

(2)

(c)

Find the volume of air enclosed by the pyramid.

(2)

(d)

Find the angle between a triangular face and the floor.

(2)

(e)

Find the angle between a sloping edge of the pyramid and the floor.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\sqrt{20^2+15^2}$	✓ 1
A1 for 25 m	✓ 1
Part (b)
M1 for $4\times\frac{1}{2}\times30\times25$	✓ 1
A1 for 1500 m²	✓ 1
Part (c)
M1 for $\frac{1}{3}\times30^2\times20$	✓ 1
A1 for 6000 m³	✓ 1
Part (d)
M1 for $\tan\theta=\frac{20}{15}$	✓ 1
A1 for 53.1°	✓ 1
Part (e)
M1 for using half the diagonal of the base, $15\sqrt2=21.21$, in $\tan\theta=\frac{20}{21.21}$	✓ 1
A1 for 43.3°	✓ 1

Common error: The angle a face makes with the floor uses half the side (15 m). The angle an edge makes with the floor uses half the diagonal (21.2 m). Swapping these two lengths is a frequent mistake.

+ Add to test✓ In your test31IBMI-032-06-08Volume and surface area in context Paper SLMedium5 marks

Three tennis balls, each a sphere of diameter 6.7 cm, are sold in a closed cylindrical tube. The balls fit exactly: they touch the curved wall, and the stack touches both ends of the tube.

(a)

Find the volume of the tube.

(2)

(b)

Find the percentage of the tube's volume that is empty space.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for radius 3.35 and height 3 × 6.7 = 20.1 substituted in $\pi r^2h$	✓ 1
A1 for 709 cm³ (708.656…)	✓ 1
Part (b)
M1 for the volume of the balls: $3\times\frac{4}{3}\pi(3.35)^3=472.44$ cm³	✓ 1
M1 for $\frac{708.66-472.44}{708.66}\times100$ (or equivalent, including use of 709)	✓ 1
A1 for 33.3% (exactly one third); accept 33.4% from using 709	✓ 1

Exam tip: The height of the tube is three diameters, 20.1 cm. Keep unrounded values in your calculator until the final step so your percentage does not drift.

+ Add to test✓ In your test32IBMI-032-06-10Volume and surface area in context Paper SLMedium6 marks

A water trough for livestock is a prism of length 4 m. Its cross-section is an isosceles trapezium, with an open top of width 0.9 m, a base of width 0.5 m and a vertical depth of 0.4 m.

(a)

Find the area of the cross-section.

(2)

(b)

Find the capacity of the trough in litres.

(2)

(c)

Find the angle that each sloping side makes with the horizontal.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for $\frac{1}{2}(0.9+0.5)\times0.4$	✓ 1
A1 for 0.28 m²	✓ 1
Part (b)
M1 for $0.28\times4=1.12$ m³	✓ 1
A1 for 1120 litres	✓ 1
Part (c)
M1 for a horizontal offset of 0.2 m on each side, with $\tan\theta=\frac{0.4}{0.2}$	✓ 1
A1 for 63.4°	✓ 1

Exam tip: Each sloping side covers half of the 0.4 m difference in widths, so the horizontal offset is 0.2 m, not 0.4 m.

+ Add to test✓ In your test33IBMI-032-05-05Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium11 marks

A ferry leaves port P and sails 18 km on a bearing of 065° to a buoy A. It then turns and sails 25 km on a bearing of 150° to an island B.

(a)

Show that angle PAB = 95°.

(2)

(b)

Find the distance PB.

(3)

(c)

Find the bearing of B from P.

(4)

(d)

The ferry returns directly from B to P at an average speed of 14 km h⁻¹. Find the time taken, in minutes, to the nearest minute.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Bearing of P from A is 065° + 180° = 245° (or the angle between north at A and AP is 65° by alternate angles) M1 ; ∠PAB = 245° − 150° = 95° A1	✓ 2
Part (b)
Use of the cosine rule M1 ; PB² = 18² + 25² − 2(18)(25)cos 95° A1 ; PB = 32.1 km (32.054…) A1	✓ 3
Part (c)
sin(∠APB) / 25 = sin 95° / 32.054 M1 ; ∠APB = 51.0° A1 ; bearing = 065° + 51.0° M1 ; 116° A1	✓ 4
Part (d)
time = 32.054 / 14 × 60 M1 ; 137 minutes (2 h 17 min) A1	✓ 2

Exam tip: Draw a north line at every turning point. Bearings are measured clockwise from north, so the interior angles of the triangle come from the differences between the north lines and the paths.

+ Add to test✓ In your test34IBMI-032-05-06Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium6 marks

From the top of a vertical cliff 72 m above sea level, a coastguard sees two kayaks. Both kayaks are directly out to sea, in a straight line with the foot of the cliff. The angles of depression of the kayaks are 41° and 28°.

(a)

Find the horizontal distance from the foot of the cliff to the nearer kayak.

(2)

(b)

Find the distance between the two kayaks.

(3)

(c)

State one assumption you have made in your calculations.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
tan 41° = 72 / d M1 ; d = 82.8 m A1	✓ 2
Part (b)
Distance to the further kayak = 72 / tan 28° M1 ; = 135.4 m A1 ; separation = 135.41 − 82.83 = 52.6 m A1	✓ 3
Part (c)
For example: the cliff is vertical / the sea is level / the kayaks and the foot of the cliff lie on one straight line / the kayaks are treated as points R1	✓ 1

Common error: The angle of depression is measured down from the horizontal at the observer. By alternate angles it equals the angle of elevation at the kayak, not the angle at the top of the cliff inside the triangle.

+ Add to test✓ In your test35IBMI-032-05-09Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium10 marks

Coastguard station B is 12 km due east of coastguard station A. A fishing boat S is out at sea to the north of the line AB. The bearing of S from A is 052° and the bearing of S from B is 318°.

(a)

Find angle ASB.

(3)

(b)

Find the distances AS and BS.

(3)

(c)

The boat sails due south until it reaches the line AB. Find the distance it sails.

(2)

(d)

Find the distance of the boat from B when it reaches the line AB.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
∠SAB = 90° − 52° = 38° A1 ; ∠SBA = 318° − 270° = 48° A1 ; ∠ASB = 180° − 38° − 48° = 94° A1	✓ 3
Part (b)
Sine rule, e.g. AS / sin 48° = 12 / sin 94° (or equivalent for BS) M1 ; AS = 8.94 km A1 ; BS = 7.41 km A1	✓ 3
Part (c)
distance = AS sin 38° (or BS sin 48°) M1 ; 5.50 km A1	✓ 2
Part (d)
Valid method, e.g. 12 − AS cos 38° = 12 − 7.04, or BS cos 48° M1 ; 4.96 km A1	✓ 2

Common error: A bearing of 318° is not the angle inside the triangle. Find the angle between the bearing and the direction from B to A, which has bearing 270°.

+ Add to test✓ In your test36IBMI-032-05-10Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium6 marks

The inside of a shipping container is a cuboid 12.0 m long, 2.35 m wide and 2.39 m high. A company wants to transport long steel rods inside it.

(a)

Find the length of the diagonal of the floor of the container.

(2)

(b)

Find the length of the longest straight rod that fits inside the container.

(2)

(c)

Find the angle between this rod and the floor of the container.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
√(12.0² + 2.35²) M1 ; 12.2 m (12.228…) A1	✓ 2
Part (b)
√(12.228² + 2.39²) or √(12.0² + 2.35² + 2.39²) M1 ; 12.5 m (12.459…) A1	✓ 2
Part (c)
tan θ = 2.39 / 12.228 (or equivalent) M1 ; θ = 11.1° A1	✓ 2

Exam tip: In 3D problems, find a right-angled triangle that contains the line you need. The angle between a line and a plane uses the line's projection onto the plane, which here is the floor diagonal.

+ Add to test✓ In your test37IBMI-032-05-11Right-angled trigonometry, bearings and the sine and cosine rules Paper SLMedium7 marks

A vertical wind-turbine tower TF has height 60 m, where F is its foot on horizontal ground. Two walkers stand at points A and B on the ground. The angle of elevation of T from A is 35° and from B is 22°. The angle AFB is 70°.

(a)

Find the distance AF.

(2)

(b)

Find the distance BF.

(2)

(c)

Find the distance between the two walkers.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
AF = 60 / tan 35° M1 ; 85.7 m A1	✓ 2
Part (b)
BF = 60 / tan 22° M1 ; 149 m (148.50…) A1	✓ 2
Part (c)
Use of the cosine rule in the horizontal triangle AFB M1 ; AB² = 85.69² + 148.50² − 2(85.69)(148.50)cos 70° A1 ; AB = 144 m A1	✓ 3

Common error: The 70° angle lies in the horizontal plane at F. Use it with the horizontal distances AF and BF, not with the sloping lines of sight TA and TB.

+ Add to test✓ In your test38IBMI-032-01-05Volume and surface area in context Paper SLMedium6 marks

A grain silo consists of a cylinder of radius 3.5 m and height 12 m, topped by a hemisphere of the same radius.

(a)

Find the total volume of the silo.

(3)

(b)

The silo is filled to 85% of its capacity with wheat of density 760 kg m⁻³. Find the mass of the wheat in tonnes.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Cylinder: π(3.5)²(12) = 461.81… m³ M1 ; Hemisphere: $\frac{2}{3}\pi(3.5)^3$ = 89.80… m³ M1 ; Total = 552 m³ (551.61…) A1	✓ 3
Part (b)
Volume of wheat: 0.85 × 551.61… = 468.87… m³ M1 ; Mass: 468.87… × 760 = 356 341 kg (approx.) M1 ; 356 tonnes A1 (accept 357 tonnes by follow-through from 552 m³)	✓ 3

Common error: Using the full sphere formula for the dome doubles its contribution. A hemisphere is half of $\frac{4}{3}\pi r^3$. Keep full calculator values between parts to avoid rounding drift.

+ Add to test✓ In your test39IBMI-032-01-07Volume and surface area in context Paper SLMedium7 marks

A drinks company makes closed cylindrical cans that hold 330 cm³. The current design has radius 3.2 cm. Assume the metal used for every can has the same uniform thickness, so the amount of metal is proportional to the total surface area.

(a)

Find the height of the current can.

(2)

(b)

Find the total surface area of the current can.

(2)

(c)

A new design with radius 3.8 cm also holds 330 cm³. Determine whether the new design uses less metal, justifying your answer.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
π(3.2)²h = 330 M1 ; h = 10.3 cm (10.258…) A1	✓ 2
Part (b)
2π(3.2)² + 2π(3.2)(10.258…) M1 ; 271 cm² (270.59…) A1	✓ 2
Part (c)
New height = 330/(π × 3.8²) = 7.274… cm M1 ; surface area = 2π(3.8)² + 2π(3.8)(7.274…) = 264 cm² (264.41…) A1 ; 264 < 271, so the new design uses less metal R1	✓ 3

Exam tip: A closed can has two circular ends. In a comparison question, state a clear conclusion that refers to both values.

+ Add to test✓ In your test40IBMI-032-01-08Volume and surface area in context Paper SLMedium6 marks

A water trough for cattle is a prism 2.5 m long. Its cross-section is an isosceles trapezium with a horizontal base of width 50 cm, an open top of width 80 cm and a depth of 30 cm.

(a)

Find the capacity of the trough in litres.

(3)

(b)

The water in the trough is 20 cm deep. Find the volume of water as a percentage of the capacity of the trough.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Area = ½(0.8 + 0.5)(0.3) = 0.195 m² M1 ; × 2.5 = 0.4875 m³ M1 ; 488 litres (487.5) A1	✓ 3
Part (b)
Width of water surface = 50 + 20 = 70 cm M1 ; volume = ½(0.5 + 0.7)(0.2)(2.5) = 0.3 m³ M1 ; 61.5% A1	✓ 3

Common error: Assuming the trough is two-thirds full because the water is two-thirds of the depth ignores the narrowing cross-section. The width at the water surface is less than 80 cm.

+ Add to test✓ In your test41IBMI-032-01-10Volume and surface area in context Paper SLMedium11 marks

A shipping crate is a cuboid with internal dimensions 2.4 m long, 1.5 m wide and 1.2 m high.

(a)

Find the length of the longest straight rod that fits inside the crate.

(2)

(b)

Find the angle between this rod and the base of the crate.

(2)

(c)

The crate is packed with upright cylindrical drums, each of radius 0.28 m and height 1.15 m, arranged in rows parallel to the sides. Find the maximum number of drums that fit in the crate.

(2)

(d)

Find the volume of empty space in the crate when it contains this number of drums.

(3)

(e)

Find the percentage of the crate's volume occupied by the drums.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\sqrt{2.4^2+1.5^2+1.2^2}$ M1 ; 3.07 m (3.0741…) A1	✓ 2
Part (b)
Base diagonal = $\sqrt{2.4^2+1.5^2}$ = 2.830… and tan θ = 1.2/2.830… M1 ; θ = 23.0° A1	✓ 2
Part (c)
2.4 ÷ 0.56 = 4.28… so 4 along the length and 1.5 ÷ 0.56 = 2.67… so 2 across, one layer since 1.15 < 1.2 M1 ; 8 drums A1	✓ 2
Part (d)
One drum: π(0.28)²(1.15) = 0.2832… m³ M1 ; crate volume 4.32 m³ and 4.32 − 8(0.2832…) M1 ; 2.05 m³ (2.054…) A1	✓ 3
Part (e)
2.2660… ÷ 4.32 × 100 M1 ; 52.5% A1	✓ 2

Common error: Dividing the crate volume by the drum volume gives 15.25, but drums cannot be cut up. Count how many whole diameters fit along each edge.

+ Add to test✓ In your test42IBMI-032-07-r1-03Voronoi diagrams Paper SLMedium6 marks

Four villages are located at P(0, 0), Q(8, 0), R(8, 6) and S(0, 10), with coordinates in kilometres. The Voronoi diagram for the four villages has exactly two vertices, V₁(4, 3) and V₂(2.5, 5). A new landfill site must be built inside the quadrilateral PQRS, and the council will only consider the two Voronoi vertices as possible locations.

(a)

Verify that V₁ is equidistant from P, Q and R.

(2)

(b)

Find the distance from V₂ to the nearest village.

(2)

(c)

The landfill should be as far as possible from the nearest village. State which vertex should be chosen, justifying your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses distance formula, e.g. V₁P = √(4² + 3²) M1	✓ 1
V₁P = V₁Q = V₁R = 5 km A1	✓ 1
Part (b)
V₂P = √(2.5² + 5²) (= V₂R = V₂S), V₂Q = √(5.5² + 5²) = 7.43 M1	✓ 1
Nearest distance = 5.59 km A1	✓ 1
Part (c)
V₂ A1	✓ 1
Its nearest village is 5.59 km away, which is greater than 5 km for V₁ R1	✓ 1

Exam tip: For a 'furthest from all sites' problem, compare the distance from each candidate point to its NEAREST site, then choose the largest of these minimum distances.

+ Add to test✓ In your test43IBMI-032-07-r1-05Voronoi diagrams Paper SLMedium9 marks

A rectangular park occupies the region 0 ≤ x ≤ 10, 0 ≤ y ≤ 6, where one unit represents 100 m. Two ice-cream kiosks are located at K₁(2, 1) and K₂(6, 5). Visitors always walk to the nearest kiosk.

(a)

Find the equation of the Voronoi edge separating the cells of K₁ and K₂.

(3)

(b)

Find the coordinates of the points where this edge meets the boundary of the park.

(2)

(c)

Find the area of the park that is closer to K₁, giving your answer in m².

(2)

(d)

Assuming visitors are spread evenly across the park, find the percentage of visitors expected to use K₂, and comment on your answer.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of K₁K₂ = (4, 3) A1	✓ 1
Gradient K₁K₂ = 1, so perpendicular gradient = −1 M1	✓ 1
y = −x + 7 A1	✓ 1
Part (b)
At y = 0: (7, 0) A1	✓ 1
At y = 6: (1, 6) A1	✓ 1
Part (c)
Cell of K₁ is the trapezium (0,0), (7,0), (1,6), (0,6): area = ½(7 + 1)(6) = 24 units² M1	✓ 1
= 24 × 10 000 = 240 000 m² (2.40 × 10⁵ m²) A1	✓ 1
Part (d)
(60 − 24)/60 × 100 = 60% A1	✓ 1
K₂ should serve more visitors than K₁ (60% vs 40%), so it may need more stock/staff R1	✓ 1

Common error: Remember to convert area units: one grid unit is 100 m, so one square unit is 100 × 100 = 10 000 m², not 100 m².

+ Add to test✓ In your test44IBMI-032-07-r2-03Voronoi diagrams Paper SLMedium8 marks

A village has three wells at A(1, 2), B(5, 2) and C(3, 6), with coordinates in kilometres. Each household collects water from its nearest well. The perpendicular bisector of AB has equation $x = 3$.

(a)

Show that the perpendicular bisector of AC has equation $y = -0.5x + 5$.

(3)

(b)

Find the coordinates of the Voronoi vertex V where the three cells meet.

(2)

(c)

Explain why V is the same distance from all three wells.

(1)

(d)

A fourth well is to be dug inside triangle ABC as far as possible from the existing wells. It is placed at V. Find its distance from the nearest existing well.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of AC is (2, 4) A1	✓ 1
Gradient of AC $= 2$, so perpendicular gradient $= -0.5$ M1	✓ 1
$y - 4 = -0.5(x - 2)$ leading to $y = -0.5x + 5$ A1	✓ 1
Part (b)
Substitutes $x = 3$: $y = -1.5 + 5$ M1	✓ 1
V(3, 3.5) A1	✓ 1
Part (c)
V lies on the bisectors of AB and AC, so VA = VB and VA = VC R1	✓ 1
Part (d)
$\sqrt{(3-1)^2 + (3.5-2)^2}$ M1	✓ 1
2.5 km A1	✓ 1

Exam tip: A 'show that' answer must show every step, from the midpoint through the perpendicular gradient to the rearranged equation. Simply checking that the given equation works earns fewer marks.

+ Add to test✓ In your test45IBMI-032-07-r2-04Voronoi diagrams Paper SLMedium9 marks

Four weather stations record daily rainfall. Their coordinates (in km) and the rainfall recorded on one day are:
S₁(2, 1): 12 mm; S₂(7, 3): 18 mm; S₃(4, 8): 9 mm; S₄(9, 9): 15 mm.
Rainfall at other locations is estimated by nearest-neighbour interpolation, using a Voronoi diagram of the stations.

(a)

Find the equation of the perpendicular bisector of S₂S₃.

(4)

(b)

The town T is at (6, 6). Determine which station's cell contains T and hence estimate the rainfall at T.

(2)

(c)

Comment on the reliability of this estimate.

(1)

(d)

A farm F at $(k, 5)$ is the same distance from S₂ and S₃. Find $k$.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint (5.5, 5.5) A1	✓ 1
Gradient S₂S₃ $= -\frac{5}{3}$ M1	✓ 1
Perpendicular gradient $\frac{3}{5}$ A1	✓ 1
$y = 0.6x + 2.2$ A1	✓ 1
Part (b)
Distances from T: S₃ $\sqrt{8} = 2.83$ is the smallest (S₂ 3.16, S₄ 4.24, S₁ 6.40), so T is in the cell of S₃ R1	✓ 1
Estimated rainfall 9 mm A1	✓ 1
Part (c)
Valid comment, e.g. T is almost as close to S₂ (18 mm) as to S₃, so the estimate may be unreliable because rainfall varies continuously R1	✓ 1
Part (d)
$5 = 0.6k + 2.2$ M1	✓ 1
$k = 4.67$ A1	✓ 1

Common error: Using the gradient of S₂S₃ itself in the bisector equation. Take the negative reciprocal to get the perpendicular gradient.

+ Add to test✓ In your test46IBMI-032-07-r2-06Voronoi diagrams Paper SLMedium10 marks

A town is modelled by the square $0 \le x \le 8$, $0 \le y \le 8$, with distances in km. Its population is spread evenly at 150 people per km². There are two pharmacies, at P(1, 2) and Q(5, 4). Each resident uses the nearer pharmacy.

(a)

Find the equation of the perpendicular bisector of PQ.

(4)

(b)

Find the coordinates of the points where this bisector meets the edges $y = 8$ and $y = 0$ of the town.

(2)

(c)

Find the area of the Voronoi cell of P within the town.

(2)

(d)

Hence find the number of residents expected to use pharmacy Q.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint (3, 3) A1	✓ 1
Gradient PQ $= 0.5$ M1	✓ 1
Perpendicular gradient $-2$ A1	✓ 1
$y = -2x + 9$ A1	✓ 1
Part (b)
(0.5, 8) A1	✓ 1
(4.5, 0) A1	✓ 1
Part (c)
Trapezium with parallel sides 0.5 and 4.5 and height 8: $\frac{1}{2}(0.5 + 4.5)(8)$ M1	✓ 1
20 km² A1	✓ 1
Part (d)
Area of Q's cell $= 64 - 20 = 44$ km² M1	✓ 1
$44 \times 150 = 6600$ residents A1	✓ 1

Common error: Assuming that the cells of two sites split a region in half. This is only true when the bisector passes through the centre of a symmetric region.

+ Add to test✓ In your test47IBMI-032-02-r1-02Voronoi diagrams Paper SLMedium7 marks

Two electric-vehicle charging stations, A and B, are modelled as sites of a Voronoi diagram on a grid where 1 unit represents 1 km. Station A is at (2, 1). The location of B has been lost from the records, but the edge separating the two cells is known to lie on the line $x + 2y = 9$.

(a)

Find the gradient of the edge, and hence write down the gradient of the line segment AB.

(2)

(b)

Write down the equation of the line through A and B.

(1)

(c)

Find the coordinates of M, the point where line AB meets the edge.

(2)

(d)

Hence find the coordinates of B.

(1)

(e)

Find the distance between the two charging stations.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$y = -\frac{1}{2}x + 4.5$, so gradient of edge = $-\frac{1}{2}$ A1	✓ 1
The edge is the perpendicular bisector of AB, so gradient of AB = 2 R1	✓ 1
Part (b)
$y = 2x - 3$ A1	✓ 1
Part (c)
Substituting: $x + 2(2x - 3) = 9$ M1	✓ 1
$x = 3$, $y = 3$, so M(3, 3) A1	✓ 1
Part (d)
M is the midpoint of AB, so B = (2×3 − 2, 2×3 − 1) = (4, 5) A1	✓ 1
Part (e)
AB = $\sqrt{2^2 + 4^2} = \sqrt{20}$ = 4.47 km A1	✓ 1

Common error: M is the midpoint of AB, not the position of B. B is found by going the same displacement again from M as from A to M.

+ Add to test✓ In your test48IBMI-032-02-r1-03Voronoi diagrams Paper SLMedium10 marks

A square area of farmland is modelled by the region $0 \le x \le 8$, $0 \le y \le 8$, where 1 unit represents 1 km. Two water wells are located at A(2, 2) and B(6, 4). Each point of the farmland is supplied by the nearer well.

(a)

Find the equation of the perpendicular bisector of AB.

(3)

(b)

Find the coordinates of the points where this bisector meets the boundary lines $y = 0$ and $y = 8$.

(2)

(c)

A storage shed is at F(5, 1). Show that F lies on the boundary between the two cells and interpret this result in context.

(2)

(d)

Find the area of farmland supplied by well A.

(2)

(e)

Find the percentage of the farmland supplied by well B.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint (4, 3) and gradient of AB = $\frac{2}{4} = \frac{1}{2}$ M1	✓ 1
Perpendicular gradient = −2 M1	✓ 1
$y = -2x + 11$ A1	✓ 1
Part (b)
$y = 0$: $x = 5.5$, giving (5.5, 0) A1	✓ 1
$y = 8$: $x = 1.5$, giving (1.5, 8) A1	✓ 1
Part (c)
At $x = 5$, $y = -2(5) + 11 = 1$, so F lies on the bisector (or FA = FB = $\sqrt{10}$) M1	✓ 1
F is equidistant from both wells, so either well is equally near to the shed R1	✓ 1
Part (d)
Region is the trapezium with vertices (0, 0), (5.5, 0), (1.5, 8), (0, 8): area = $\frac{5.5 + 1.5}{2} \times 8$ M1	✓ 1
= 28 km² A1	✓ 1
Part (e)
$\frac{64 - 28}{64} \times 100$ = 56.3% A1	✓ 1

Exam tip: When a Voronoi cell is clipped by a rectangular boundary, list the vertices of the cell first; the shape is often a trapezium or can be split into simple shapes.

+ Add to test✓ In your test49IBMI-032-02-r2-04Voronoi diagrams Paper SLMedium8 marks

A district is modelled by the square region $0 \le x \le 10$, $0 \le y \le 10$, with units in kilometres. Four libraries are located at A(2, 2), B(8, 2), C(2, 8) and D(8, 8). The Voronoi diagram for these four sites consists of the lines $x = 5$ and $y = 5$. A fifth library, E, is built at (5, 5) and the Voronoi diagram is updated.

(a)

Show that the perpendicular bisector of AE has equation $x + y = 7$.

(2)

(b)

Write down the equations of the perpendicular bisectors of BE, CE and DE.

(2)

(c)

Find the coordinates of the vertices of the Voronoi cell of library E.

(2)

(d)

Find the area of the region served by library E.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint (3.5, 3.5) and gradient of AE = 1, so perpendicular gradient = −1 M1	✓ 1
$y - 3.5 = -(x - 3.5)$ giving $x + y = 7$ A1	✓ 1
Part (b)
BE: $y = x - 3$; CE: $y = x + 3$ A1	✓ 1
DE: $x + y = 13$ A1	✓ 1
Part (c)
Solving pairs of bisector equations simultaneously M1	✓ 1
(5, 2), (8, 5), (5, 8), (2, 5) A1	✓ 1
Part (d)
The cell is a square (rhombus) with diagonals of length 6 km, area = $\frac{6 \times 6}{2}$ M1	✓ 1
= 18 km² A1	✓ 1

Exam tip: When a new site is added, only the bisectors between the new site and its neighbours are needed; symmetry can save a lot of working.

+ Add to test✓ In your test50IBMI-032-02-r2-05Voronoi diagrams Paper SLMedium9 marks

Four weather stations record the rainfall in one month. Their positions (in km) and readings are:
P(0, 0): 42 mm; Q(6, 2): 55 mm; R(1, 7): 38 mm; S(7, 8): 61 mm.
Rainfall at other locations is estimated using nearest-neighbour interpolation, based on the Voronoi diagram of the four stations.

(a)

Find the equation of the perpendicular bisector of QR.

(3)

(b)

A farm is located at T(4, 5). Show that T lies on the boundary between the cells of Q and R.

(2)

(c)

A village is located at U(3, 3). Estimate the rainfall at U.

(3)

(d)

Comment on the reliability of using nearest-neighbour interpolation to estimate the rainfall at T.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of QR = (3.5, 4.5) A1	✓ 1
Gradient of QR = $\frac{5}{-5} = -1$, so perpendicular gradient = 1 M1	✓ 1
$y = x + 1$ A1	✓ 1
Part (b)
TQ = TR = $\sqrt{13}$ = 3.61 km (or 5 = 4 + 1 so T is on the bisector) M1	✓ 1
TP = $\sqrt{41}$ = 6.40 km and TS = $\sqrt{18}$ = 4.24 km are both greater, so T is on the edge between Q and R R1	✓ 1
Part (c)
Distances from U: P 4.24, Q 3.16, R 4.47, S 6.40 km M1	✓ 1
Q is the nearest station A1	✓ 1
Estimated rainfall = 55 mm A1	✓ 1
Part (d)
T is equally close to Q and R, whose readings differ by 17 mm, so the estimate is ambiguous/unreliable (e.g. an average of 46.5 mm may be more sensible) R1	✓ 1

Exam tip: To decide which cell a point is in, compare its distances to all relevant sites; the smallest distance identifies the cell.

+ Add to test✓ In your test51IBMI-032-07-r1-04Voronoi diagrams Paper SLMedium8 marks

Three weather stations record monthly rainfall. Coordinates are in kilometres.
Station W₁ is at (2, 2) and recorded 34 mm.
Station W₂ is at (10, 4) and recorded 41 mm.
Station W₃ is at (6, 10) and recorded 28 mm.
Rainfall at other locations is estimated using nearest-neighbour interpolation, based on the Voronoi diagram of the three stations.

(a)

A farm is located at F(7, 5). Estimate the monthly rainfall at F, showing clear working.

(3)

(b)

Find the equation of the perpendicular bisector of W₂W₃.

(3)

(c)

A second farm G lies on the edge between the cells of W₂ and W₃ and has x-coordinate 11. Find the y-coordinate of G. Suggest, with a reason, a suitable rainfall estimate for G.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Calculates the distances from F to all three stations: FW₁ = √34 ≈ 5.83 km, FW₂ = √10 ≈ 3.16 km, FW₃ = √26 ≈ 5.10 km M1	✓ 1
F lies in the cell of W₂, the nearest station A1	✓ 1
Estimated rainfall = 41 mm A1	✓ 1
Part (b)
Midpoint = (8, 7) A1	✓ 1
Gradient of W₂W₃ = 6/(−4) = −1.5, so the perpendicular gradient = 2/3 M1	✓ 1
y − 7 = (2/3)(x − 8), i.e. y = (2/3)x + 5/3 (or 2x − 3y + 5 = 0) A1	✓ 1
Part (c)
y = (2/3)(11) + 5/3 = 9 A1	✓ 1
G is equidistant (√26 km) from W₂ and W₃, so nearest-neighbour interpolation does not give a single value. Award the mark for any sensible estimate with this reason, for example the mean (41 + 28)/2 = 34.5 mm, or either 41 mm or 28 mm with a stated tie-break rule R1	✓ 1

Exam tip: Nearest-neighbour interpolation assigns the value of the site whose cell contains the point. Justify the cell by comparing distances, not by judging position by eye. A point on a cell edge is equidistant from two sites, so state how you resolve the tie.

+ Add to test✓ In your test52IBMI-032-02-r1-05Voronoi diagrams Paper SLMedium7 marks

Three villages are located at A(0, 0), B(8, 0) and C(2, 6), where 1 unit represents 1 km. A recycling plant is to be built inside triangle ABC at the point that is as far as possible from the nearest village. This point is the vertex of the Voronoi diagram of the three villages.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Find the equation of the perpendicular bisector of AC.

(2)

(c)

Find the coordinates of the Voronoi vertex V.

(1)

(d)

Find the distance from V to each village.

(2)

(e)

Interpret your answer to the previous part in context.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 4$ A1	✓ 1
Part (b)
Midpoint (1, 3) and gradient of AC = 3, so perpendicular gradient $-\frac{1}{3}$ M1	✓ 1
$y = -\frac{1}{3}x + \frac{10}{3}$ (or equivalent, e.g. $x + 3y = 10$) A1	✓ 1
Part (c)
At $x = 4$: $y = -\frac{4}{3} + \frac{10}{3} = 2$, so V(4, 2) A1	✓ 1
Part (d)
VA = $\sqrt{4^2 + 2^2}$, VB = $\sqrt{4^2 + 2^2}$, VC = $\sqrt{2^2 + 4^2}$ M1	✓ 1
Each distance = $\sqrt{20}$ ≈ 4.47 km A1	✓ 1
Part (e)
The plant is 4.47 km from all three villages. Moving it anywhere else in the triangle brings it closer to at least one village, so 4.47 km is the greatest possible distance to the nearest village R1	✓ 1

Exam tip: The Voronoi vertex is equidistant from all three sites, so checking that VA = VB = VC is a quick way to verify your coordinates.

+ Add to test✓ In your test53IBMI-032-02-r2-03Voronoi diagrams Paper SLMedium8 marks

A supermarket chain has three stores in a city, at A(1, 1), B(9, 1) and C(5, 7), where units are kilometres. A rival company wants to open a new store inside triangle ABC. It chooses the point whose distance to the nearest existing store is as large as possible. For this triangle, that point is the vertex of the Voronoi diagram for A, B and C.

(a)

Show that the perpendicular bisector of AC has equation $2x + 3y = 18$.

(3)

(b)

Write down the equation of the perpendicular bisector of AB.

(1)

(c)

Find the coordinates of the vertex of the Voronoi diagram for the three stores.

(2)

(d)

The rival opens its store at this vertex. Find the distance from the new store to the nearest existing store.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of AC = (3, 4) A1	✓ 1
Gradient of AC = $\frac{6}{4} = \frac{3}{2}$, so the perpendicular gradient = $-\frac{2}{3}$ M1	✓ 1
$y - 4 = -\frac{2}{3}(x-3)$ leading to $3y - 12 = -2x + 6$, i.e. $2x + 3y = 18$ AG A1	✓ 1
Part (b)
$x = 5$ A1	✓ 1
Part (c)
Substituting $x = 5$ into $2x + 3y = 18$: $10 + 3y = 18$ M1	✓ 1
$\left(5, \frac{8}{3}\right)$ or (5, 2.67) A1	✓ 1
Part (d)
For example, the distance to A: $\sqrt{(5-1)^2 + \left(\frac{8}{3}-1\right)^2}$ M1	✓ 1
$= \frac{13}{3}$ km, or 4.33 km (3 s.f.) A1	✓ 1
The vertex is equidistant from A, B and C, so a correct distance to any one store earns full marks.	—

Common error: A common error is substituting the rounded value y = 2.67 into the distance calculation, which can give a slightly wrong final answer. Use the exact value 8/3 or the full calculator value.

+ Add to test✓ In your test54IBMI-032-00-11Right-angled trigonometry, bearings and the sine and cosine rules Paper SLHard13 marks

A rescue helicopter is based at B. It receives a call from a walker at P, which is 42 km from B on a bearing of 310°. A second call comes from a climber at Q, which is 35 km from B on a bearing of 025°.

(a)

Show that angle PBQ = 75°.

(1)

(b)

Find the distance PQ.

(3)

(c)

Find angle BPQ.

(3)

(d)

Find the bearing of Q from P.

(3)

(e)

The helicopter flies B → P → Q → B at a constant 210 km h⁻¹ and hovers for 12 minutes at each of P and Q. It can stay airborne for at most 60 minutes before refuelling. Determine whether the helicopter can complete the mission without refuelling.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
(360° − 310°) + 25° = 75° A1	✓ 1
Part (b)
PQ² = 42² + 35² − 2(42)(35)cos 75° M1	✓ 1
PQ² = 2228 (2228.07…) A1	✓ 1
PQ = 47.2 km A1	✓ 1
Part (c)
sin BPQ / 35 = sin 75° / 47.2 (or the cosine rule) M1	✓ 1
sin BPQ = 0.716 A1	✓ 1
BPQ = 45.7° (acute, since BQ < BP) A1	✓ 1
Part (d)
Bearing of B from P = 310° − 180° = 130° M1	✓ 1
Q lies anticlockwise of the line PB as seen from P, so subtract: 130° − 45.7° (or an equivalent valid method, e.g. coordinates) M1	✓ 1
= 084.3° A1	✓ 1
Part (e)
Total distance = 42 + 47.2 + 35 = 124.2 km; flying time = 124.2 ÷ 210 h (= 35.5 minutes) M1	✓ 1
Total time = 35.5 + 24 = 59.5 minutes A1	✓ 1
59.5 minutes < 60 minutes, so yes, it can complete the mission without refuelling R1	✓ 1

Exam tip: When a comparison is close, keep at least 3 significant figures in your total before concluding. State the comparison explicitly, e.g. 59.5 < 60.

+ Add to test✓ In your test55IBMI-032-00-12Right-angled trigonometry, bearings and the sine and cosine rules Paper SLHard12 marks

A vertical telecommunications mast FT stands on horizontal ground, with foot F and top T. Two engineers stand at points A and B on the same horizontal ground, where B is 180 m due east of A. The mast is on a bearing of 032° from A, and angle FBA = 71°. From A, the angle of elevation of T is 26°.

(a)

Show that angle AFB = 51°.

(2)

(b)

Find the distance AF.

(2)

(c)

Find the height of the mast.

(2)

(d)

A drone flies in a straight line from A to T. Find the distance it flies.

(2)

(e)

Find the angle of elevation of T from B.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
AB points due east, so it is perpendicular to north. Hence angle FAB = 90° − 32° = 58° A1	✓ 1
Angle AFB = 180° − 58° − 71° = 51° A1	✓ 1
Part (b)
$\frac{AF}{\sin 71°} = \frac{180}{\sin 51°}$ M1	✓ 1
AF = 219 m (219.0) A1	✓ 1
Part (c)
FT = 219.0 tan 26° M1	✓ 1
= 107 m (106.8) A1	✓ 1
Part (d)
AT = 219.0 / cos 26° (or √(219.0² + 106.8²)) M1	✓ 1
= 244 m A1	✓ 1
Part (e)
$\frac{BF}{\sin 58°} = \frac{180}{\sin 51°}$ (or the cosine rule in triangle ABF) M1	✓ 1
BF = 196 m (196.4) A1	✓ 1
tan θ = 106.8 / 196.4 M1	✓ 1
θ = 28.5° A1	✓ 1

Common error: Triangle ABF lies flat on the ground, and each angle of elevation belongs to a separate vertical triangle. Do not mix horizontal and vertical angles in the same sine rule. Carry unrounded values forward to avoid rounding errors in (e).

+ Add to test✓ In your test56IBMI-032-06-09Volume and surface area in context Paper SLHard8 marks

A metal bucket is the frustum of a right circular cone. It is open at the top and has a flat circular base. The top has radius 15 cm, the base has radius 10 cm, and the vertical depth is 24 cm. The bucket can be modelled as a large cone with a smaller, similar cone removed from its tip.

(a)

Show that the height of the large cone is 72 cm.

(2)

(b)

Find the capacity of the bucket in litres.

(3)

(c)

Find the total area of metal used to make the bucket.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for similar triangles: $\frac{H}{15}=\frac{H-24}{10}$	✓ 1
A1 for $10H=15H-360$, giving $H=72$	✓ 1
Part (b)
M1 for the large cone minus the small cone: $\frac{1}{3}\pi(15^2)(72)-\frac{1}{3}\pi(10^2)(48)$	✓ 1
A1 for 11 900 cm³ (11 938)	✓ 1
A1 for 11.9 litres	✓ 1
Part (c)
M1 for the slant height of the frustum, $\sqrt{24^2+5^2}=24.52$, or for the slant heights of both cones, 73.55 and 49.03	✓ 1
M1 for the curved area $\pi(15+10)(24.52)=1925$ plus the base $\pi\times10^2$	✓ 1
A1 for 2240 cm² (2239.6)	✓ 1

Common error: Taking the curved area of a frustum as $\pi R\times24$ is wrong for two reasons: it uses the vertical depth instead of a slant height, and it ignores the removed cone. Find the curved area as a difference of two cones, or use the frustum's slant height.

+ Add to test✓ In your test57IBMI-032-06-12Volume and surface area in context Paper SLHard10 marks

A drinks company is designing a closed cylindrical can that must hold 330 cm³. The can has radius r cm and height h cm.

(a)

Write down an expression for h in terms of r.

(1)

(b)

Show that the total surface area of the can is $A=2\pi r^2+\frac{660}{r}$.

(2)

(c)

Use your graphic display calculator to find the value of r that minimises A, and the minimum surface area.

(3)

(d)

In the factory, the curved surface is cut from a rectangular sheet with no waste. Each circular end is cut from a square sheet of side 2r, and the rest of each square is scrapped. For the can with minimum surface area, find the total area of metal needed, including the scrap.

(2)

(e)

Find the percentage of this metal that is scrapped.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 for $h=\frac{330}{\pi r^2}$	✓ 1
Part (b)
M1 for $A=2\pi r^2+2\pi rh$	✓ 1
A1 for substituting $h$ to give $2\pi r\times\frac{330}{\pi r^2}=\frac{660}{r}$, and hence the given result	✓ 1
Part (c)
M1 for sketching the graph of A on a GDC, or for any valid method of finding its minimum point	✓ 1
A1 for $r=3.74$ cm (3.7449…)	✓ 1
A1 for $A=264$ cm² (264.36…)	✓ 1
Part (d)
M1 for $2(2r)^2+\frac{660}{r}$, i.e. $8r^2+\frac{660}{r}$, with $r=3.7449…$	✓ 1
A1 for 288 cm² (288.43…)	✓ 1
Part (e)
M1 for a scrap area of $2r^2(4-\pi)=24.08$, divided by 288.43 and multiplied by 100	✓ 1
A1 for 8.35%	✓ 1

Exam tip: Store the unrounded value of r from your GDC and reuse it in the scrap calculations. Using r = 3.74 instead gives 8.33%, which changes the third significant figure of the final percentage.

+ Add to test✓ In your test58IBMI-032-05-07Right-angled trigonometry, bearings and the sine and cosine rules Paper SLHard10 marks

A park ranger is placing a marker post C on a straight footpath that starts at gate A. The footpath makes an angle of 38° with the straight fence AB, where AB = 340 m. The post must be exactly 250 m from B.

(a)

Find the two possible values of angle ACB.

(3)

(b)

Hence find the two possible distances AC.

(4)

(c)

The post is placed at the position closer to A. Find the area of triangle ABC.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
sin C / 340 = sin 38° / 250 M1 ; ∠ACB = 56.9° A1 ; or ∠ACB = 180° − 56.9° = 123.1° A1	✓ 3
Part (b)
∠ABC = 180° − 38° − 56.86° = 85.1° or 180° − 38° − 123.14° = 18.9° M1 ; AC / sin B = 250 / sin 38° (or a correct cosine-rule quadratic in AC) M1 ; AC = 405 m A1 ; AC = 131 m A1	✓ 4
Part (c)
Selects AC = 131.24 m, the shorter distance R1 ; Area = ½ × 340 × 131.24 × sin 38° M1 ; 13 700 m² (1.37 × 10⁴ m²) A1	✓ 3

Exam tip: When the sine rule is used to find an angle and the given side opposite the known angle is shorter than the other given side, check whether the obtuse angle 180° − θ also gives a valid triangle.

+ Add to test✓ In your test59IBMI-032-05-12Right-angled trigonometry, bearings and the sine and cosine rules Paper SLHard16 marks

An orienteering loop starts at S. Runners go 1.2 km on a bearing of 040° to checkpoint P, then 0.9 km on a bearing of 160° to checkpoint Q, and then run straight back to S.

(a)

Show that angle SPQ = 60°.

(2)

(b)

Find the distance SQ.

(3)

(c)

Find the bearing on which runners travel from Q back to S.

(4)

(d)

Find the area of ground enclosed by the loop.

(2)

(e)

A water station W is to be placed on the path SQ so that it is as close as possible to P. Find the distance PW.

(3)

(f)

One runner completes the whole loop in 22 minutes. Find her average speed in km h⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Bearing of S from P = 040° + 180° = 220° M1 ; ∠SPQ = 220° − 160° = 60° A1	✓ 2
Part (b)
Use of the cosine rule M1 ; SQ² = 1.2² + 0.9² − 2(1.2)(0.9)cos 60° = 1.17 A1 ; SQ = 1.08 km (1.0817…) A1	✓ 3
Part (c)
sin(∠PSQ) / 0.9 = sin 60° / 1.0817 M1 ; ∠PSQ = 46.1° A1 ; bearing of Q from S = 040° + 46.1° = 086.1°, so the bearing of S from Q = 086.1° + 180° M1 ; 266° A1	✓ 4
Part (d)
½ × 1.2 × 0.9 × sin 60° M1 ; 0.468 km² A1	✓ 2
Part (e)
PW is perpendicular to SQ R1 ; PW = 1.2 sin 46.10° (or 2 × area / SQ) M1 ; 0.865 km A1	✓ 3
Part (f)
(1.2 + 0.9 + 1.0817) / (22/60) M1 ; 8.68 km h⁻¹ A1	✓ 2

Exam tip: The shortest distance from a point to a straight path is along the perpendicular. Spotting this lets you use the right-angled triangle SPW instead of setting up a harder method.

+ Add to test✓ In your test60IBMI-032-01-11Volume and surface area in context Paper SLHard14 marks

A rectangular swimming pool is 25 m long and 10 m wide. The depth of the pool is 1.2 m at the shallow end and 2.8 m at the deep end, and the floor slopes uniformly along the length of the pool. The side walls are vertical trapeziums and the end walls are vertical rectangles.

(a)

Find the volume of the pool.

(2)

(b)

Find the length of the sloping floor, from the shallow end to the deep end.

(2)

(c)

Find the angle that the floor makes with the horizontal.

(2)

(d)

The floor and the four walls are to be tiled. Find the total area to be tiled.

(3)

(e)

Tiles cost 34.50 dollars per m², and 10% extra must be bought to allow for breakages. Find the cost of the tiles, to the nearest dollar.

(2)

(f)

The empty pool is filled until the water surface is 10 cm below the top of the walls. Water flows in at 0.6 m³ per minute. Find the time taken, in hours.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Cross-section area = ½(1.2 + 2.8)(25) = 50 m², or an equivalent method ; volume = 500 m³	✓ 2
Part (b)
$\sqrt{25^2+1.6^2}$ ; 25.1 m (25.051…)	✓ 2
Part (c)
tan θ = 1.6/25, or an equivalent trigonometric ratio using the answer to (b) ; θ = 3.66°	✓ 2
Part (d)
Floor = 25.051… × 10 = 250.5 m² ; side walls 2 × 50 = 100 m² and end walls 12 + 28 = 40 m² ; total = 391 m² (390.51…)	✓ 3
Part (e)
390.51… × 1.1 × 34.50 ; 14 820 dollars (accept 14 838 dollars from using 391 m²)	✓ 2
Part (f)
Unfilled layer = 25 × 10 × 0.1 = 25 m³, so water volume = 475 m³ ; 475 ÷ 0.6 = 791.7 minutes ; 13.2 hours	✓ 3

Exam tip: The water surface is horizontal, so the empty top layer is a 25 m × 10 m × 0.1 m cuboid, even though the floor slopes. Spotting this saves a lot of work.

+ Add to test✓ In your test61IBMI-032-01-12Volume and surface area in context Paper SLHard12 marks

A conical funnel in a laboratory is held with its vertex pointing downwards and its outlet closed. The funnel has rim radius 40 cm and vertical height 60 cm. When the water is h cm deep, the volume of water is V cm³.

(a)

Find the capacity of the funnel in litres.

(2)

(b)

Show that $V=\frac{4\pi h^3}{27}$.

(2)

(c)

Find the depth of water when the funnel is half full.

(3)

(d)

Find the area of the inner surface of the funnel that is in contact with the water when it is half full.

(3)

(e)

The outlet is opened when the funnel is full, and water drains at a constant 250 cm³ per second. Find the time taken for the funnel to become half full.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\frac{1}{3}\pi(40^2)(60)$ = 100 531 cm³ M1 ; 101 litres A1	✓ 2
Part (b)
Similar triangles give radius of water surface r = $\frac{40}{60}h=\frac{2h}{3}$ M1 ; $V=\frac{1}{3}\pi\left(\frac{2h}{3}\right)^2h=\frac{4\pi h^3}{27}$ A1	✓ 2
Part (c)
$\frac{4\pi h^3}{27}=16000\pi$ M1 ; h³ = 108 000 M1 ; h = 47.6 cm (47.622…) A1	✓ 3
Part (d)
r = $\frac{2}{3}$(47.622…) = 31.75 cm M1 ; slant height $\sqrt{47.622^2+31.748^2}$ = 57.23 cm M1 ; πrl = 5710 cm² (5708.5…) A1	✓ 3
Part (e)
50 265.5 ÷ 250 M1 ; 201 seconds A1	✓ 2

Common error: Half the volume does not mean half the depth. Because V is proportional to h³, a half-full cone is filled to about 79% of its height.

+ Add to test✓ In your test62IBMI-032-02-r1-06Voronoi diagrams Paper SLHard14 marks

A town is modelled by the rectangle $0 \le x \le 10$, $0 \le y \le 8$, where 1 unit represents 1 km. There are three fire stations, at A(2, 2), B(8, 2) and C(5, 6). Each point in the town is served by its nearest fire station, and a Voronoi diagram is constructed with the stations as sites.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Show that the perpendicular bisector of AC has equation $y = -0.75x + 6.625$.

(3)

(c)

Find the coordinates of the Voronoi vertex V.

(2)

(d)

Find the area of the town served by station A.

(3)

(e)

Find the percentage of the town served by station C.

(2)

(f)

A council officer claims that V is the point in the town that is furthest from its nearest fire station. By considering the corner of the town at (0, 8), show that this claim is incorrect.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 5$ A1	✓ 1
Part (b)
Gradient AC = $\frac{4}{3}$, so perpendicular gradient = $-\frac{3}{4}$ M1	✓ 1
Midpoint of AC = (3.5, 4), substituted: $y - 4 = -0.75(x - 3.5)$ M1	✓ 1
$y = -0.75x + 2.625 + 4 = -0.75x + 6.625$ A1	✓ 1
Part (c)
Substituting $x = 5$: $y = -3.75 + 6.625$ M1	✓ 1
V(5, 2.875) A1	✓ 1
Part (d)
Bisector of AC meets $x = 0$ at (0, 6.625) A1	✓ 1
Cell is the trapezium (0, 0), (5, 0), (5, 2.875), (0, 6.625): area = $\frac{6.625 + 2.875}{2} \times 5$ M1	✓ 1
= 23.75 km² (23.8 km²) A1	✓ 1
Part (e)
By symmetry B also serves 23.75 km², so C serves 80 − 2 × 23.75 = 32.5 km² M1	✓ 1
$\frac{32.5}{80} \times 100$ = 40.6% A1	✓ 1
Part (f)
Distance from (0, 8) to A = $\sqrt{40}$ = 6.32 km and to C = $\sqrt{29}$ M1	✓ 1
Nearest station to (0, 8) is C, at 5.39 km A1	✓ 1
Distance from V to its nearest station is 3.125 km; since 5.39 > 3.125 the corner is further from help than V, so the claim is incorrect R1	✓ 1

Exam tip: In a bounded region, the point furthest from every site can lie on the boundary or at a corner, not only at a Voronoi vertex. Check corners and edge intersections as well.

+ Add to test✓ In your test63IBMI-032-07-r2-05Voronoi diagrams Paper SLHard12 marks

A rectangular county is modelled by the region $0 \le x \le 10$, $0 \le y \le 8$, with distances in km. Three towns lie at A(2, 2), B(8, 2) and C(5, 7). A waste-recycling plant is to be built in the county as far as possible from the nearest town.

(a)

Write down the equation of the perpendicular bisector of AB.

(1)

(b)

Show that the perpendicular bisector of AC has equation $y = -0.6x + 6.6$.

(3)

(c)

Find the coordinates of the Voronoi vertex V, the point equidistant from all three towns.

(2)

(d)

Find the distance from V to the nearest town.

(2)

(e)

A planner claims that V is the best location for the plant. By considering the corners of the county, show that this claim is incorrect. State a better location and its distance from the nearest town.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$x = 5$ A1	✓ 1
Part (b)
Midpoint of AC is (3.5, 4.5) A1	✓ 1
Gradient AC $= \frac{5}{3}$, so perpendicular gradient $= -0.6$ M1	✓ 1
$y - 4.5 = -0.6(x - 3.5)$ leading to $y = -0.6x + 6.6$ A1	✓ 1
Part (c)
Substitutes $x = 5$ into $y = -0.6x + 6.6$ M1	✓ 1
V(5, 3.6) A1	✓ 1
Part (d)
$\sqrt{3^2 + 1.6^2}$ or $\sqrt{11.56}$, or $7 - 3.6$ (distance to C) M1	✓ 1
3.4 km A1	✓ 1
Part (e)
Considers a top corner, e.g. (0, 8), and identifies C as its nearest town (distance to A is 6.32 km, distance to B is 10.0 km) M1	✓ 1
Distance from (0, 8) to C $= \sqrt{26} = 5.10$ km A1	✓ 1
5.10 > 3.4, so V is not the furthest point from the towns and the claim is incorrect R1	✓ 1
Plant at (0, 8) or (10, 8), 5.10 km from the nearest town A1	✓ 1

Exam tip: For a largest-empty-circle problem in a bounded region, test the Voronoi vertices, the corners of the region and the points where Voronoi edges meet the boundary. The optimum is often on the boundary.

+ Add to test✓ In your test64IBMI-032-02-r2-06Voronoi diagrams Paper SLHard15 marks

A county is modelled by the rectangle $0 \le x \le 12$, $0 \le y \le 8$, with units in kilometres. Three towns are located at A(2, 2), B(10, 2) and C(5, 7). A waste incinerator is to be built anywhere in the county, including on its boundary. It must be at the point that is as far as possible from the nearest town.

(a)

Show that the perpendicular bisector of AC has equation $3x + 5y = 33$.

(3)

(b)

Find the equations of the perpendicular bisectors of AB and BC.

(2)

(c)

Find the coordinates of the vertex V of the Voronoi diagram and the distance from V to each town.

(3)

(d)

Find the coordinates of the three points where the edges of the Voronoi diagram meet the boundary of the county.

(3)

(e)

Determine the optimal location for the incinerator and state its distance from the nearest town. Justify your answer.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Midpoint of AC = (3.5, 4.5)	✓ 1
Gradient AC = $\frac{5}{3}$, so perpendicular gradient = $-\frac{3}{5}$	✓ 1
$y - 4.5 = -0.6(x - 3.5)$ leading to $5y = -3x + 33$, i.e. $3x + 5y = 33$	✓ 1
Part (b)
AB: $x = 6$	✓ 1
BC: midpoint (7.5, 4.5), perpendicular gradient 1, giving $y = x - 3$	✓ 1
Part (c)
Solves $x = 6$ with $3x + 5y = 33$ (or with $y = x - 3$)	✓ 1
V(6, 3)	✓ 1
Distance = $\sqrt{17}$ ≈ 4.12 km	✓ 1
Part (d)
(6, 0)	✓ 1
(11, 8)	✓ 1
(0, 6.6)	✓ 1
Part (e)
Within each Voronoi cell, the distance to that cell's town is greatest at a vertex of the cell. The candidates are therefore V, the points in (d) and the corners of the rectangle	✓ 1
Distances to the nearest town: V 4.12; (6, 0) 4.47; (11, 8) 6.08; (0, 6.6) 5.02; (0, 0) 2.83; (12, 0) 2.83; (0, 8) 5.10; (12, 8) 6.32	✓ 1
Optimal location is the corner (12, 8)	✓ 1
Distance to the nearest town (B) = $\sqrt{40}$ ≈ 6.32 km	✓ 1

Common error: The Voronoi vertex is not automatically the answer to a largest empty circle problem. The points where edges meet the boundary, and the corners of the region, must also be checked.

+ Add to test✓ In your test65IBMI-032-03-r2-01Vectors and kinematics Paper HLEasy6 marks

A fishing boat leaves a mooring and moves with constant velocity. Relative to a harbour at the origin O, its position vector $t$ hours after 08:00 is
$\mathbf{r} = \begin{pmatrix}2\\-1\end{pmatrix} + t\begin{pmatrix}3\\4\end{pmatrix}$
where distances are in kilometres, the first component is east of O and the second is north of O.

(a)

Find the speed of the boat.

(2)

(b)

Find the position vector of the boat at 10:30.

(2)

(c)

Find the distance of the boat from the harbour at 10:30.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Speed = $\sqrt{3^2+4^2}$ M1	✓ 1
= 5 km h⁻¹ A1	✓ 1
Part (b)
Uses $t = 2.5$ M1	✓ 1
$\mathbf{r} = \begin{pmatrix}9.5\\9\end{pmatrix}$ A1	✓ 1
Part (c)
$\sqrt{9.5^2+9^2}$ M1	✓ 1
= 13.1 km (13.086…) A1	✓ 1

Common error: Using t = 10.5 or t = 2.3 for 10:30. The parameter counts hours after 08:00, so 10:30 is 2.5 hours later.

+ Add to test✓ In your test66IBMI-032-03-r2-02Vectors and kinematics Paper HLEasy6 marks

A survey drone takes off from the roof of a building. Relative to a point O at ground level directly below the take-off point, the drone's position $t$ seconds after take-off is
$\mathbf{r} = \begin{pmatrix}0\\0\\20\end{pmatrix} + t\begin{pmatrix}4\\-3\\1.5\end{pmatrix}$
where distances are in metres and the third component is the height above the ground.

(a)

Find the speed of the drone.

(2)

(b)

Find the time at which the drone is 50 m above the ground.

(2)

(c)

Find the horizontal distance of the drone from O at this time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\sqrt{4^2+(-3)^2+1.5^2} = \sqrt{27.25}$ M1	✓ 1
= 5.22 m s⁻¹ A1	✓ 1
Part (b)
$20 + 1.5t = 50$ M1	✓ 1
$t = 20$ s A1	✓ 1
Part (c)
Horizontal components at $t=20$: 80 and −60, so $\sqrt{80^2+60^2}$ M1	✓ 1
= 100 m A1	✓ 1

Exam tip: Horizontal distance uses only the first two components. Include the vertical component only when the full straight-line distance is asked for.

+ Add to test✓ In your test67IBMI-032-04-r2-01Graph theory and networks Paper HLEasy6 marks

A road inspector is responsible for the roads joining four villages, A, B, C and D. There are direct roads between A and B, A and C, B and C, B and D, and C and D. There are no other roads. The road network is modelled by a graph G whose vertices are the villages and whose edges are the roads.

(a)

Write down the adjacency matrix M for G, with the vertices in the order A, B, C, D.

(2)

(b)

Find the number of different walks of length 4 that start at A and finish at D.

(2)

(c)

The inspector wants to drive along every road exactly once. Determine whether this is possible. If it is, state the villages where the route must start and finish.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$M=\begin{pmatrix}0&1&1&0\\1&0&1&1\\1&1&0&1\\0&1&1&0\end{pmatrix}$	—
A1 for a symmetric 4×4 matrix with zeros on the leading diagonal	✓ 1
A1 for all entries correct	✓ 1
Part (b)
M1 for using the (A, D) entry of $M^4$	✓ 1
A1 for 10 walks	✓ 1
Part (c)
R1 for degrees A = 2, B = 3, C = 3, D = 2, so exactly two vertices (B and C) have odd degree	✓ 1
A1 for: it is possible (an Eulerian trail exists), but the route must start at B and finish at C, or start at C and finish at B	✓ 1

Exam tip: Use your calculator's matrix mode to find $M^4$. Take the entry in row A and column D; do not add up a whole row.

+ Add to test✓ In your test68IBMI-032-04-r2-02Graph theory and networks Paper HLEasy6 marks

A college is laying fibre-optic cable to connect six buildings, A, B, C, D, E and F. The cable can only be laid along the following routes. Each length is given in metres.
AB 120, AC 90, AD 150, BC 70, BE 110, CD 100, CE 140, DE 80, DF 130, EF 60.
Every building must be connected to the network, and the total length of cable must be as small as possible.

(a)

Use Kruskal's algorithm to find a minimum spanning tree for the network. List the edges in the order in which they are added.

(3)

(b)

Write down the total length of cable required.

(1)

(c)

The cable costs \$18.50 per metre to lay. Find the total cost of laying the cable.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for considering edges in increasing order of length	✓ 1
A1 for EF, BC, DE added first, in that order	✓ 1
A1 for AC then CD, with the algorithm stopping once 5 edges are included	✓ 1
Part (b)
A1 for 400 m	✓ 1
Part (c)
M1 for 400 × 18.50	✓ 1
A1 for \$7400.00	✓ 1

Common error: A spanning tree on six vertices has exactly five edges. Once five edges have been added without forming a cycle, stop: adding a sixth edge always creates a cycle.

+ Add to test✓ In your test69IBMI-032-08-r1-01Vectors and kinematics Paper HLEasy5 marks

A rescue boat leaves a harbour buoy. Its position vector, in kilometres relative to a lighthouse at the origin O, is given by
$\mathbf{r} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} + t\begin{pmatrix} 3 \\ 4 \end{pmatrix}$
where t is the time in hours after leaving the buoy. The unit vectors point east and north respectively.

(a)

Find the speed of the boat.

(2)

(b)

Find the position vector of the boat when t = 2.5.

(1)

(c)

Find the distance of the boat from the lighthouse when t = 2.5.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: speed = $\sqrt{3^2+4^2}$	✓ 1
A1: 5 km h⁻¹	✓ 1
Part (b)
A1: $\begin{pmatrix} 9.5 \\ 9 \end{pmatrix}$	✓ 1
Part (c)
M1: $\sqrt{9.5^2+9^2}$	✓ 1
A1: 13.1 km (13.086…)	✓ 1

Exam tip: The speed comes from the direction (velocity) vector only; the distance from O comes from the position vector. Do not mix the two.

+ Add to test✓ In your test70IBMI-032-08-r1-02Vectors and kinematics Paper HLEasy6 marks

A survey drone takes off from a point with position vector $\begin{pmatrix} 10 \\ 5 \\ 0 \end{pmatrix}$ m, measured from a ground station. It flies with constant velocity $\begin{pmatrix} 4 \\ -2 \\ 1.5 \end{pmatrix}$ m s⁻¹, where the third component is vertical.

(a)

Calculate the speed of the drone.

(2)

(b)

Find the time taken for the drone to reach a height of 30 m.

(2)

(c)

Find the distance the drone has travelled at this time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\sqrt{4^2+(-2)^2+1.5^2} = \sqrt{22.25}$	✓ 1
A1: 4.72 m s⁻¹	✓ 1
Part (b)
M1: 1.5t = 30	✓ 1
A1: t = 20 s	✓ 1
Part (c)
M1: 20 × 4.717… or $\left\|20\begin{pmatrix} 4 \\ -2 \\ 1.5 \end{pmatrix}\right\|$	✓ 1
A1: 94.3 m	✓ 1

Common error: Finding the distance from the ground station instead of the distance travelled from the take-off point. The drone did not start at the origin.

+ Add to test✓ In your test71IBMI-032-08-r2-01Vectors and kinematics Paper HLEasy5 marks

A fishing boat leaves a harbour. Relative to a lighthouse at the origin O, the position vector of the boat t hours after leaving the harbour is
$\boldsymbol{r} = \begin{pmatrix}3\\-2\end{pmatrix} + t\begin{pmatrix}4\\3\end{pmatrix}$
where distances are in kilometres, $\boldsymbol{i}$ points east and $\boldsymbol{j}$ points north.

(a)

Find the speed of the boat.

(2)

(b)

Find the position vector of the boat 2.5 hours after it leaves the harbour.

(1)

(c)

Hence find the distance of the boat from the lighthouse at this time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Speed = $\sqrt{4^2+3^2}$ M1	✓ 1
= 5 km h⁻¹ A1	✓ 1
Part (b)
$\begin{pmatrix}3+10\\-2+7.5\end{pmatrix} = \begin{pmatrix}13\\5.5\end{pmatrix}$ A1	✓ 1
Part (c)
$\sqrt{13^2+5.5^2}$ M1	✓ 1
= 14.1 km (14.1157…) A1	✓ 1

Exam tip: Speed is the magnitude of the direction (velocity) vector only. The starting position vector plays no part in finding the speed.

+ Add to test✓ In your test72IBMI-032-08-r2-06Vectors and kinematics Paper HLEasy7 marks

A plane is pointed due north and flies at 300 km h⁻¹ relative to the air. A wind blows with velocity $\begin{pmatrix}40\\-30\end{pmatrix}$ km h⁻¹, where $\boldsymbol{i}$ points east and $\boldsymbol{j}$ points north.

(a)

Write down the velocity of the plane relative to the ground.

(1)

(b)

Find the ground speed of the plane.

(2)

(c)

Find the bearing of the plane's actual path.

(2)

(d)

The plane keeps this ground velocity for 500 km. Find the time taken, in hours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\begin{pmatrix}0\\300\end{pmatrix} + \begin{pmatrix}40\\-30\end{pmatrix} = \begin{pmatrix}40\\270\end{pmatrix}$ km h⁻¹ A1	✓ 1
Part (b)
$\sqrt{40^2+270^2} = \sqrt{74500}$ M1	✓ 1
= 273 km h⁻¹ A1	✓ 1
Part (c)
$\tan\theta = \frac{40}{270}$ M1	✓ 1
Bearing 008.4° A1	✓ 1
Part (d)
Time = 500 ÷ 272.95 M1	✓ 1
= 1.83 h A1	✓ 1

Common error: Do not subtract the wind or simply add the speeds 300 + 50. Velocities combine as vectors: add the components, then find the magnitude.

+ Add to test✓ In your test73IBMI-032-03-r1-01Vectors and kinematics Paper HLEasy5 marks

A delivery drone moves in a straight line after take-off. Relative to an origin at the base of its charging station, its position vector t seconds after take-off is
r = (3, −2, 1) + t(4, 5, 2),
where the components are in metres, x is east, y is north and z is vertically upwards. Ground level is z = 0.

(a)

Find the speed of the drone.

(2)

(b)

Find the position vector of the drone when t = 6.

(1)

(c)

The drone must reach a height of 15 m above the ground before it may cross a road. Find the time taken for the drone to reach this height.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: finds the magnitude of the velocity vector, $\sqrt{4^2+5^2+2^2}=\sqrt{45}$	✓ 1
A1: 6.71 m s⁻¹	✓ 1
Part (b)
A1: (27, 28, 13) m	✓ 1
Part (c)
M1: sets the vertical component equal to 15, 1 + 2t = 15	✓ 1
A1: t = 7 s	✓ 1

Exam tip: Speed is the magnitude of the direction (velocity) vector, not of the position vector. Height comes from the z-component only, including its starting value.

+ Add to test✓ In your test74IBMI-032-04-r1-01Graph theory and networks Paper HLEasy6 marks

Four villages, A, B, C and D, are linked by roads. There are direct roads between A and B, A and C, B and C, B and D, and C and D. There are no other direct roads. The road network is modelled by a graph G whose vertices are the villages and whose edges are the roads.

(a)

Write down the adjacency matrix M for G, with the vertices in the order A, B, C, D.

(2)

(b)

Find the entry in row B, column B of M², and interpret this value in context.

(2)

(c)

A delivery van travels along exactly 4 roads, starting at A and finishing at D. Roads and villages may be revisited. Find the number of different possible journeys.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\begin{pmatrix}0&1&1&0\\1&0&1&1\\1&1&0&1\\0&1&1&0\end{pmatrix}$	—
A1 symmetric matrix with zero leading diagonal	✓ 1
A1 all entries correct	✓ 1
Part (b)
Entry = 3	✓ 1
R1 there are 3 different journeys along exactly two roads that start and finish at B (equal to the number of roads meeting at B)	✓ 1
Part (c)
M1 uses the (A, D) entry of M⁴	✓ 1
A1 10 journeys	✓ 1

Exam tip: The number of walks of length n between two vertices is read straight from Mⁿ. Enter the matrix into your calculator and raise it to the power. Do not try to list the walks by hand.

+ Add to test✓ In your test75IBMI-032-04-r1-02Graph theory and networks Paper HLEasy7 marks

A university plans to connect six buildings, P, Q, R, S, T and U, with fibre-optic cable. Cable can only be laid along the routes listed below. The lengths are in metres.
PQ 120, PR 90, QR 70, QS 150, RS 110, RT 130, ST 80, SU 100, TU 95, QU 200
Every building must be connected to the network, and the total length of cable must be as small as possible.

(a)

Use Kruskal's algorithm to find a minimum spanning tree. List the edges in the order in which they are considered, and state clearly any edge that is rejected.

(4)

(b)

Write down the total length of cable required.

(1)

(c)

The cable costs $18.50 per metre to install. Find the total installation cost.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 edges considered in increasing order of length	✓ 1
A1 QR (70), ST (80), PR (90), TU (95) accepted	✓ 1
R1 SU (100) rejected because it would form a cycle S–T–U	✓ 1
A1 RS (110) accepted, and the tree is complete with 5 edges	✓ 1
Part (b)
A1 445 m	✓ 1
Part (c)
M1 445 × 18.50	✓ 1
A1 $8232.50	✓ 1

Common error: Stop as soon as you have accepted n − 1 edges for n vertices. When you reject an edge, name the cycle it would create, because a stated reason is what earns the R mark.

+ Add to test✓ In your test76IBMI-032-03-r1-02Vectors and kinematics Paper HLEasy6 marks

A ferry leaves a harbour, taken as the origin, and travels at a constant speed of 12 km h⁻¹ on a bearing of 060°. Position vectors are given as (east component, north component) in kilometres, and t is the time in hours after the ferry leaves the harbour. A lighthouse is located at (20, 4).

(a)

Find the velocity vector of the ferry, giving its components to 3 significant figures.

(2)

(b)

Find the position vector of the ferry after 1.5 hours.

(2)

(c)

Find the distance between the ferry and the lighthouse at this time.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: uses east = 12 sin 60°, north = 12 cos 60°	✓ 1
A1: (10.4, 6.00) km h⁻¹ (accept 6)	✓ 1
Part (b)
M1: multiplies the velocity vector by 1.5	✓ 1
A1: (15.6, 9.00) km (accept 9)	✓ 1
Part (c)
M1: finds the magnitude of the displacement from ferry to lighthouse, e.g. $\sqrt{(20-15.59)^2+(4-9)^2}$	✓ 1
A1: 6.67 km (unrounded values); accept 6.66 km from the rounded position (15.6, 9)	✓ 1

Common error: Bearings are measured clockwise from north, so the east component uses sine and the north component uses cosine. Swapping them gives (6, 10.4), a bearing of 030°. Keep unrounded values in your calculator for later parts.

+ Add to test✓ In your test77IBMI-032-06-11Volume and surface area in context Paper HLMedium8 marks

An architect models a triangular glass roof panel with vertices A(2, 0, 4), B(10, 1, 4) and C(5, 6, 7), where distances are in metres.

(a)

Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

(2)

(b)

Find $\overrightarrow{AB}\times\overrightarrow{AC}$.

(2)

(c)

Hence find the area of the panel.

(2)

(d)

Glass for the panel costs USD 85 per m². Find the cost of the glass, to the nearest cent.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 for $\overrightarrow{AB}=(8,1,0)$	✓ 1
A1 for $\overrightarrow{AC}=(3,6,3)$	✓ 1
Part (b)
M1 for a correct method for the vector product, e.g. at least two correct components	✓ 1
A1 for $(3,-24,45)$	✓ 1
Part (c)
M1 for $\frac{1}{2}\sqrt{3^2+(-24)^2+45^2}=\frac{1}{2}\sqrt{2610}$	✓ 1
A1 for 25.5 m² (25.5440…)	✓ 1
Part (d)
M1 for multiplying the unrounded area by 85, e.g. $\frac{1}{2}\sqrt{2610}\times85$ or $25.54408\ldots\times85$	✓ 1
A1 for USD 2171.25	✓ 1

Common error: Rounding the area to 25.544 before multiplying gives USD 2171.24, not USD 2171.25. Keep the full calculator value of the area when the answer is needed to the nearest cent. Also remember that the magnitude of the vector product gives the area of a parallelogram, so halve it for a triangle.

+ Add to test✓ In your test78IBMI-032-03-r2-03Vectors and kinematics Paper HLMedium7 marks

Two cargo ships, A and B, move with constant velocities. Relative to a fixed buoy, their position vectors $t$ hours after 13:00 are
$\mathbf{r}_A = t\begin{pmatrix}12\\5\end{pmatrix}$ and $\mathbf{r}_B = \begin{pmatrix}30\\-10\end{pmatrix} + t\begin{pmatrix}4\\9\end{pmatrix}$
where distances are in kilometres.

(a)

Show that the displacement vector from A to B at time $t$ is $\begin{pmatrix}30-8t\\-10+4t\end{pmatrix}$.

(2)

(b)

Find the time of day at which the ships are closest together.

(3)

(c)

Find the minimum distance between the ships.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses $\mathbf{r}_B - \mathbf{r}_A$ M1	✓ 1
$\begin{pmatrix}30+4t-12t\\-10+9t-5t\end{pmatrix}$ leading to the given vector A1	✓ 1
Part (b)
$d^2 = (30-8t)^2 + (-10+4t)^2 = 80t^2 - 560t + 1000$ M1	✓ 1
Minimises, e.g. $\frac{d(d^2)}{dt} = 160t - 560 = 0$ or uses GDC M1	✓ 1
$t = 3.5$, so the time is 16:30 A1	✓ 1
Part (c)
Substitutes $t = 3.5$: $\begin{pmatrix}2\\4\end{pmatrix}$, $d = \sqrt{2^2+4^2}$ M1	✓ 1
= 4.47 km ($\sqrt{20}$) A1	✓ 1

Exam tip: Minimising d² gives the same t as minimising d and avoids differentiating a square root. Give the answer as a time of day when the question asks for one.

+ Add to test✓ In your test79IBMI-032-03-r2-05Vectors and kinematics Paper HLMedium7 marks

An air traffic controller monitors two light aircraft. Relative to a radar station, their positions $t$ minutes after 14:00 are
$\mathbf{r}_A = \begin{pmatrix}0\\0\\6\end{pmatrix} + t\begin{pmatrix}3\\4\\0\end{pmatrix}$ and $\mathbf{r}_B = \begin{pmatrix}24\\10\\4.8\end{pmatrix} + t\begin{pmatrix}-2\\1\\0.2\end{pmatrix}$
where distances are in kilometres and the third component is altitude.

(a)

Show that the flight paths of the two aircraft intersect, and find the coordinates of the point of intersection.

(3)

(b)

Determine whether the two aircraft collide. Justify your answer.

(2)

(c)

Find the distance between the aircraft at 14:04.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Uses different parameters, e.g. $3\lambda = 24 - 2\mu$, $4\lambda = 10 + \mu$, $6 = 4.8 + 0.2\mu$ M1	✓ 1
$\mu = 6$, $\lambda = 4$, and all three equations are satisfied A1	✓ 1
Point (12, 16, 6) A1	✓ 1
Part (b)
A is at the point when t = 4, but B is there when t = 6 R1	✓ 1
They are not at the point at the same time, so they do not collide A1	✓ 1
Part (c)
$\mathbf{r}_B(4) = \begin{pmatrix}16\\14\\5.6\end{pmatrix}$, so $\mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix}4\\-2\\-0.4\end{pmatrix}$ M1	✓ 1
$\sqrt{20.16}$ = 4.49 km A1	✓ 1

Common error: Using the same parameter t for both lines when testing whether the paths intersect. Paths can cross at different times, so use two parameters. Only a collision needs the same t.

+ Add to test✓ In your test80IBMI-032-04-r2-03Graph theory and networks Paper HLMedium13 marks

A gritting lorry treats the roads that connect five junctions, P, Q, R, S and T. The roads and their lengths in km are:
PQ 4, PR 6, PS 5, QR 3, QT 7, RS 2, RT 4, ST 8.
There are no other roads. The lorry must travel along every road at least once.

(a)

State, with a reason, whether the network has an Eulerian circuit.

(2)

(b)

Find the length of the shortest route from P to T.

(2)

(c)

The lorry is kept at junction R and must start and finish there. Use the Chinese postman algorithm to find the length of the shortest possible route. State which roads must be travelled twice.

(6)

(d)

On a different day, the lorry may start and finish at any two junctions. Find the length of the shortest route that covers every road, and state where it must start and finish.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 for degrees P = 3, Q = 3, R = 4, S = 3, T = 3	✓ 1
R1 for: there is no Eulerian circuit, because some vertices (four of them) have odd degree	✓ 1
Part (b)
M1 for comparing possible routes, e.g. P–R–T, P–Q–T, P–S–R–T	✓ 1
A1 for 10 km via P–R–T	✓ 1
Part (c)
A1 for identifying the odd vertices P, Q, S, T	✓ 1
M1 for considering the three possible pairings	✓ 1
A1 for all three totals correct: PQ + ST = 4 + 6 = 10, PS + QT = 5 + 7 = 12, PT + QS = 10 + 5 = 15	✓ 1
A1 for total length of all roads = 39 km	✓ 1
A1 for minimum route length 39 + 10 = 49 km	✓ 1
A1 for roads repeated: PQ, and RS and RT (the shortest S–T path is S–R–T)	✓ 1
Part (d)
R1 for: start and finish at two of the odd vertices, and repeat the shortest path between the other two	✓ 1
M1 for choosing the smallest single shortest path between two odd vertices, PQ = 4	✓ 1
A1 for 39 + 4 = 43 km, starting at S and finishing at T (or the reverse)	✓ 1

Exam tip: When you pair odd vertices, use the shortest path between them, not just the direct road. Here the shortest path from S to T goes through R and is shorter than road ST.

+ Add to test✓ In your test81IBMI-032-04-r2-05Graph theory and networks Paper HLMedium10 marks

A small website has four pages, W, X, Y and Z. The one-way links between the pages are:
W links to X and Y; X links to Z; Y links to W, X and Z; Z links to W.
A visitor moves from page to page by clicking a link. On each page, the visitor picks one of that page's links at random, with each link equally likely.

(a)

Write down the transition matrix T. Each column represents the current page and each row the next page, both in the order W, X, Y, Z.

(2)

(b)

A visitor starts on page W. Find the probability that the visitor is on page Z after exactly three clicks.

(2)

(c)

Find the steady-state vector for T, giving each probability to 3 significant figures.

(4)

(d)

The website owner ranks the pages by their long-term popularity. Write down the ranking and interpret the value for page W in context.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$T=\begin{pmatrix}0&0&\frac13&1\\\frac12&0&\frac13&0\\\frac12&0&0&0\\0&1&\frac13&0\end{pmatrix}$	—
M1 for a matrix whose columns each sum to 1	✓ 1
A1 for all entries correct	✓ 1
Part (b)
M1 for calculating $T^3$ multiplied by the column vector (1, 0, 0, 0)	✓ 1
A1 for $\frac16$ ≈ 0.167	✓ 1
Part (c)
M1 for setting up Ts = s with the components summing to 1	✓ 1
M1 for solving the system, or for using a high power of T	✓ 1
A1 for W = 0.333 and Y = 0.167	✓ 1
A1 for X = 0.222 and Z = 0.278	✓ 1
Part (d)
A1 for the order W, Z, X, Y	✓ 1
R1 for: in the long run, about one third of a visitor's clicks land on page W	✓ 1

Exam tip: To check a steady-state vector, multiply it by T on your calculator. You should get the same vector back, and its entries should add up to 1.

+ Add to test✓ In your test82IBMI-032-08-r1-03Vectors and kinematics Paper HLMedium7 marks

Two ships are tracked by a port radar. Their positions, in kilometres relative to the radar, t hours after 09:00 are
Ship A: $\mathbf{r}_A = t\begin{pmatrix} 12 \\ 5 \end{pmatrix}$
Ship B: $\mathbf{r}_B = \begin{pmatrix} 30 \\ -10 \end{pmatrix} + t\begin{pmatrix} -6 \\ 9 \end{pmatrix}$
A collision-risk alert is triggered if the ships come within 3 km of each other.

(a)

Show that the position of ship B relative to ship A is $\begin{pmatrix} 30-18t \\ -10+4t \end{pmatrix}$.

(2)

(b)

Find the time at which the ships are closest together. Give your answer to the nearest minute.

(2)

(c)

Find the minimum distance between the ships and determine whether the alert is triggered.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\mathbf{r}_B - \mathbf{r}_A$ attempted	✓ 1
A1: $\begin{pmatrix} 30-6t-12t \\ -10+9t-5t \end{pmatrix}$ leading to given answer	✓ 1
Part (b)
M1: minimising $d^2=(30-18t)^2+(-10+4t)^2$ (by GDC or differentiation: 340t = 580)	✓ 1
A1: t = 1.706 h, i.e. 10:42	✓ 1
Part (c)
M1: substituting t = 1.706 into $\sqrt{(30-18t)^2+(-10+4t)^2}$	✓ 1
A1: 3.25 km	✓ 1
R1: 3.25 > 3 so the alert is not triggered	✓ 1

Exam tip: Keep the unrounded value of t in your calculator when finding the minimum distance; rounding t to 1.7 first changes the third significant figure.

+ Add to test✓ In your test83IBMI-032-08-r1-04Vectors and kinematics Paper HLMedium6 marks

A light aircraft flies with an airspeed of 220 km h⁻¹ on a heading of 060°. A steady wind of 40 km h⁻¹ blows from the north. Take the unit vectors $\mathbf{i}$ and $\mathbf{j}$ to point east and north respectively.

(a)

Write down the velocity of the aircraft relative to the ground as a column vector.

(2)

(b)

Find the ground speed of the aircraft.

(2)

(c)

Find the bearing of the aircraft's actual path over the ground.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: air velocity $\begin{pmatrix} 220\sin 60° \\ 220\cos 60° \end{pmatrix}$ added to wind $\begin{pmatrix} 0 \\ -40 \end{pmatrix}$	✓ 1
A1: $\begin{pmatrix} 191 \\ 70 \end{pmatrix}$ (190.53, 70)	✓ 1
Part (b)
M1: $\sqrt{190.53^2+70^2}$	✓ 1
A1: 203 km h⁻¹	✓ 1
Part (c)
M1: $\tan^{-1}\left(\frac{190.53}{70}\right)$ measured clockwise from north	✓ 1
A1: 069.8°	✓ 1

Common error: A wind 'from the north' blows towards the south, so its vector is (0, −40), not (0, 40).

+ Add to test✓ In your test84IBMI-032-08-r1-05Vectors and kinematics Paper HLMedium11 marks

Two delivery drones fly in straight lines at constant velocity above a warehouse yard. Positions are in metres and t is the time in seconds after both drones are launched.
Drone P: $\mathbf{r}_P = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$
Drone Q: $\mathbf{r}_Q = \begin{pmatrix} -7 \\ 9 \\ 3 \end{pmatrix} + t\begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix}$

(a)

Show that the flight paths of the two drones intersect and find the coordinates of the point of intersection.

(4)

(b)

Explain why the drones do not collide.

(2)

(c)

Find the acute angle between the two flight paths.

(3)

(d)

Find the distance between the drones when t = 2.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: using different parameters and equating, e.g. 1 + 2λ = −7 + 4μ, 2 − λ = 9 − 3μ	✓ 1
A1: λ = 2	✓ 1
A1: μ = 3	✓ 1
A1: third components both equal 6, so paths meet at (5, 0, 6)	✓ 1
Part (b)
R1: P reaches (5, 0, 6) at t = 2 but Q reaches it at t = 3	✓ 1
A1: they are never at the same point at the same time, so no collision	✓ 1
Part (c)
M1: scalar product 2(4) + (−1)(−3) + 3 = 14	✓ 2
M1: $\cos\theta = \frac{14}{\sqrt{14}\sqrt{26}}$	✓ 1
A1: 42.8°	✓ 1
Part (d)
M1: Q at (1, 3, 5), P at (5, 0, 6), difference $\begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix}$	✓ 1
A1: $\sqrt{26}$ = 5.10 m	✓ 1

Common error: Using the same parameter t for both lines when testing whether paths cross. That tests for a collision, not an intersection; use two different parameters first.

+ Add to test✓ In your test85IBMI-032-08-r2-03Vectors and kinematics Paper HLMedium7 marks

Two ships, A and B, are observed from a port at the origin. At time t hours after 09:00 their position vectors, in kilometres, are
$\boldsymbol{r}_A = t\begin{pmatrix}10\\5\end{pmatrix}$ and $\boldsymbol{r}_B = \begin{pmatrix}20\\-5\end{pmatrix} + t\begin{pmatrix}2\\9\end{pmatrix}$.

(a)

Find the position vector of B relative to A, $\boldsymbol{r}_B - \boldsymbol{r}_A$, in terms of t.

(2)

(b)

Show that the square of the distance between the ships is $d^2 = 80t^2 - 360t + 425$.

(2)

(c)

Find the minimum distance between the ships and the time of day at which it occurs.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Subtracts the position vectors M1	✓ 1
$\begin{pmatrix}20-8t\\-5+4t\end{pmatrix}$ A1	✓ 1
Part (b)
$d^2 = (20-8t)^2 + (4t-5)^2$ M1	✓ 1
$= 400 - 320t + 64t^2 + 16t^2 - 40t + 25 = 80t^2 - 360t + 425$ A1	✓ 1
Part (c)
Minimises $d^2$, e.g. $t = \frac{360}{2\times80}$ or using a GDC M1	✓ 1
t = 2.25 h, so the time is 11:15 A1	✓ 1
$d = \sqrt{80(2.25)^2 - 360(2.25) + 425} = \sqrt{20}$ = 4.47 km A1	✓ 1

Exam tip: Minimising $d^2$ gives the same time as minimising $d$, and the algebra is far easier. Remember to take the square root at the end and to convert t into a clock time.

+ Add to test✓ In your test86IBMI-032-08-r2-04Vectors and kinematics Paper HLMedium12 marks

Two delivery drones, P and Q, fly in straight lines at constant velocity above a warehouse. Their position vectors, in metres, at time t seconds are
$\boldsymbol{r}_P = \begin{pmatrix}0\\0\\50\end{pmatrix} + t\begin{pmatrix}4\\3\\1\end{pmatrix}$ and $\boldsymbol{r}_Q = \begin{pmatrix}52\\18\\63\end{pmatrix} + t\begin{pmatrix}-2\\2\\-0.5\end{pmatrix}$.

(a)

Find the speed of drone P.

(2)

(b)

Using different parameters s and u for the two paths, show that the flight paths intersect, and find the coordinates of the point of intersection.

(5)

(c)

State, with a reason, whether the drones collide.

(2)

(d)

Find the distance between the drones when P passes through the point of intersection.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\sqrt{4^2+3^2+1^2} = \sqrt{26}$ M1	✓ 1
= 5.10 m s⁻¹ A1	✓ 1
Part (b)
Sets up equations, e.g. 4s = 52 − 2u and 3s = 18 + 2u M1	✓ 1
s = 10 A1	✓ 1
u = 6 A1	✓ 1
Checks third component: 50 + 10 = 60 and 63 − 0.5(6) = 60, so the paths intersect R1	✓ 1
Point (40, 30, 60) A1	✓ 1
Part (c)
P reaches the point at t = 10 s but Q reaches it at t = 6 s R1	✓ 1
The drones are not there at the same time, so they do not collide A1	✓ 1
Part (d)
Substitutes t = 10 into $\boldsymbol{r}_Q$ M1	✓ 1
Q at (32, 38, 58); P at (40, 30, 60) A1	✓ 1
Distance = $\sqrt{8^2+8^2+2^2}$ = 11.5 m (11.489…) A1	✓ 1

Common error: Intersecting paths do not mean a collision. A collision needs the same position at the same value of t, so the two parameters must be equal.

+ Add to test✓ In your test87IBMI-032-03-r1-03Vectors and kinematics Paper HLMedium7 marks

Two fishing boats, A and B, move with constant velocities. At time t hours after 08:00, their position vectors relative to a harbour are
r_A = (2, 1) + t(3, 4) and r_B = (14, −3) + t(−1, 6),
where distances are in kilometres and the components are east and north respectively.

(a)

Find the distance between the boats at 08:00.

(1)

(b)

Show that the position of B relative to A at time t is (12 − 4t, −4 + 2t).

(2)

(c)

Find the time of day at which the boats are closest together, and the minimum distance between them.

(4)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1: $\sqrt{12^2+(-4)^2}=\sqrt{160}$ = 12.6 km	✓ 1
Part (b)
M1: subtracts, r_B − r_A = (14 − 2, −3 − 1) + t(−1 − 3, 6 − 4)	✓ 1
A1: simplifies correctly to (12 − 4t, −4 + 2t)	✓ 1
Part (c)
M1: valid method to minimise distance, e.g. minimises D² = (12 − 4t)² + (−4 + 2t)² by differentiating or using a GDC, or sets (12 − 4t, −4 + 2t)·(−4, 2) = 0	✓ 1
A1: t = 2.8	✓ 1
A1: time of day 10:48	✓ 1
A1: minimum distance 1.79 km	✓ 1

Exam tip: Convert the decimal time into hours and minutes: 0.8 h is 48 minutes, not 80 minutes. A final answer given in context, as a time of day, earns the interpretation mark.

+ Add to test✓ In your test88IBMI-032-03-r1-04Vectors and kinematics Paper HLMedium14 marks

Air traffic controllers model the motion of two aircraft, A and B, flying with constant velocities. Relative to a radar station, their positions t minutes after 14:00 are
r_A = (−20, 5, 8) + t(4, 1, 0)
r_B = (14, −19, 3) + t(−2, 6, 1)
where distances are in kilometres and the third component is altitude.

(a)

Find the speed of aircraft A in km h⁻¹.

(2)

(b)

Show that the flight paths of the two aircraft intersect, and find the coordinates of the point of intersection.

(5)

(c)

Explain why the aircraft do not collide.

(2)

(d)

Find the distance between the two aircraft at 14:05.

(2)

(e)

Find the angle between the directions of travel of the two aircraft.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\sqrt{4^2+1^2}=\sqrt{17}$ = 4.12 km per minute	✓ 1
A1: multiplies by 60 to get 247 km h⁻¹	✓ 1
Part (b)
M1: equates the two paths using different parameters, e.g. −20 + 4s = 14 − 2u, 5 + s = −19 + 6u, 8 = 3 + u	✓ 1
A1: u = 5	✓ 1
A1: s = 6	✓ 1
R1: checks the remaining equation: −20 + 24 = 4 and 14 − 10 = 4, so the equations are consistent and the paths intersect	✓ 1
A1: point of intersection (4, 11, 8)	✓ 1
Part (c)
R1: A reaches (4, 11, 8) at t = 6 but B reaches it at t = 5	✓ 1
A1: the aircraft are at the intersection point at different times, so they are never in the same place at the same time	✓ 1
Part (d)
M1: finds positions at t = 5, A at (0, 10, 8) and B at (4, 11, 8), and the magnitude of their difference	✓ 1
A1: $\sqrt{17}$ = 4.12 km	✓ 1
Part (e)
M1: uses the scalar product of the direction vectors, (4)(−2) + (6) + (0) = −2	✓ 3
M1: substitutes magnitudes into cos θ = $\frac{-2}{\sqrt{17}\sqrt{41}}$	✓ 1
A1: θ = 94.3°	✓ 1

Common error: Using the same parameter t for both lines when testing for intersection only checks for a collision. Use different parameters to find where the paths cross, then compare the times.

+ Add to test✓ In your test89IBMI-032-03-r1-06Vectors and kinematics Paper HLMedium7 marks

A hot-air balloon drifts with constant velocity. Relative to its launch site, with x east, y north and z vertically upwards and all distances in metres, it is observed at position (200, 150, 300) at t = 0 and at position (260, 70, 340) at t = 20 seconds. A monument stands on level ground at (329, −22, 0).

(a)

Find the velocity vector of the balloon.

(2)

(b)

Find the angle at which the balloon is rising, measured from the horizontal.

(2)

(c)

Show that the balloon passes directly above the monument, and find its height at that moment.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: displacement (60, −80, 40) divided by 20	✓ 1
A1: (3, −4, 2) m s⁻¹	✓ 1
Part (b)
M1: horizontal speed $\sqrt{3^2+4^2}$ = 5 and tan θ = $\frac{2}{5}$	✓ 1
A1: θ = 21.8°	✓ 1
Part (c)
M1: solves 200 + 3t = 329 and 150 − 4t = −22	✓ 1
R1: both give t = 43, so the balloon is directly above the monument at that time	✓ 1
A1: height = 300 + 2(43) = 386 m	✓ 1

Common error: Finding the time from the x-component alone is not enough. Both horizontal components must give the same t for the balloon to be directly above the point.

+ Add to test✓ In your test90IBMI-032-04-r1-03Graph theory and networks Paper HLMedium13 marks

A road-sweeping vehicle must travel along every street in a small town at least once. The street network is modelled by a weighted graph with vertices A, B, C, D and E. The edges and their lengths, in km, are:
AB 0.8, AC 1.2, AE 1.5, BC 0.5, BD 1.0, CD 0.7, CE 0.9, DE 0.6

(a)

Explain why the vehicle cannot sweep every street exactly once and return to its starting point.

(2)

(b)

Find the total length of all the streets.

(1)

(c)

The vehicle starts and finishes at its depot at A. By considering all possible pairings of the odd vertices, find which streets must be swept twice and the minimum total distance travelled.

(5)

(d)

Write down one possible route of this minimum length, starting and finishing at A.

(2)

(e)

The council decides that the vehicle may start and finish at different vertices. Find the vertices where it should start and finish, and the new minimum distance.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 vertices A, B, D and E have odd degree (each has degree 3)	✓ 1
R1 a closed route that uses every edge exactly once (an Eulerian circuit) requires every vertex to have even degree	✓ 1
Part (b)
A1 7.2 km	✓ 1
Part (c)
M1 considers all three pairings of A, B, D, E using shortest paths	✓ 1
A1 AB + DE = 0.8 + 0.6 = 1.4	✓ 1
A1 AD + BE = 1.8 + 1.4 = 3.2 (AD via B, BE via C)	✓ 1
A1 AE + BD = 1.5 + 1.0 = 2.5	✓ 1
A1 repeat AB and DE, so the minimum distance is 7.2 + 1.4 = 8.6 km	✓ 1
Part (d)
For example A–B–A–C–B–D–E–D–C–E–A	—
A1 every edge is used, with AB and DE each used twice	✓ 1
A1 the route is a valid closed trail starting and ending at A	✓ 1
Part (e)
M1 recognises that only one pair of odd vertices needs a repeated path, and chooses the pair whose repeat is shortest	✓ 1
A1 start at A and finish at B (or the reverse), repeating DE	✓ 1
A1 7.2 + 0.6 = 7.8 km	✓ 1

Exam tip: With four odd vertices there are exactly three pairings, and you must show all three to earn the method mark. Use the shortest path between each pair of vertices, which is not always the direct edge.

+ Add to test✓ In your test91IBMI-032-04-r1-04Graph theory and networks Paper HLMedium10 marks

A heritage park has four attractions, W, X, Y and Z, joined by footpaths. There are paths between W and X, W and Y, X and Y, and Y and Z only. A visitor at any attraction chooses one of the paths leaving it at random, with each path equally likely, and walks to the next attraction. This is modelled as a random walk with transition matrix T, where the columns represent the current attraction and the rows represent the next attraction, both in the order W, X, Y, Z.

(a)

Write down the matrix T.

(2)

(b)

A visitor starts at Z. Find the probability that the visitor is at W after exactly three moves.

(2)

(c)

Find the steady-state probability vector for this random walk.

(4)

(d)

The park manager wants to open one refreshment kiosk at the attraction that will be visited most often in the long run. State, with a reason, where it should be placed.

(1)

(e)

On a busy day there are 400 visitors, each moving independently in this way. After a long time, estimate the number of visitors at Y.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\mathbf{T}=\begin{pmatrix}0&\frac12&\frac13&0\\\frac12&0&\frac13&0\\\frac12&\frac12&0&1\\0&0&\frac13&0\end{pmatrix}$	—
A1 each column sums to 1 with the correct non-zero positions	✓ 1
A1 all entries correct	✓ 1
Part (b)
M1 calculates T³ applied to (0, 0, 0, 1)^T, or traces the paths Z→Y→X→W	✓ 1
A1 $\frac{1}{6}$ = 0.167	✓ 1
Part (c)
M1 sets up Tv = v, or finds a high power of T	✓ 1
M1 uses the condition that the components sum to 1	✓ 1
A1 W = 0.25, X = 0.25	✓ 1
A1 Y = 0.375, Z = 0.125	✓ 1
Part (d)
R1 at Y, because it has the largest long-run probability (0.375) of being occupied, as it is joined to the most paths	✓ 1
Part (e)
A1 400 × 0.375 = 150 visitors	✓ 1

Common error: Check which way round the matrix is set up. When the columns are 'from', each column must sum to 1. Entering the matrix the other way round gives wrong powers of T.

+ Add to test✓ In your test92IBMI-032-04-r1-05Graph theory and networks Paper HLMedium6 marks

A town park has 8 junctions, J₁ to J₈, joined by 12 walkways. The network is modelled by a connected simple graph. The degrees of J₁ to J₇ are 3, 3, 2, 4, 2, 3 and 4 respectively. The degree of J₈ is k.

(a)

Find the value of k.

(2)

(b)

A jogger wants to run along every walkway exactly once, without needing to return to the start. Explain why this is not possible.

(2)

(c)

To reduce maintenance, the council will close some walkways while making sure every junction can still be reached from every other junction. Find the greatest number of walkways that can be closed.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 uses the handshaking lemma: the sum of degrees = 2 × 12 = 24	✓ 1
A1 21 + k = 24, so k = 3	✓ 1
Part (b)
A1 J₁, J₂, J₆ and J₈ have odd degree, so there are four odd vertices	✓ 1
R1 an Eulerian trail exists only if there are 0 or 2 odd vertices	✓ 1
Part (c)
M1 the remaining network must be a spanning tree with 8 − 1 = 7 edges	✓ 1
A1 12 − 7 = 5 walkways	✓ 1

Exam tip: The handshaking lemma (sum of degrees = 2 × number of edges) and the fact that a tree on n vertices has n − 1 edges are quick results that each earn a method mark when you state them.

+ Add to test✓ In your test93IBMI-032-04-r2-04Graph theory and networks Paper HLMedium7 marks

A courier must visit five depots, A, B, C, D and E. The courier starts and finishes at A and visits every other depot exactly once. Every pair of depots is joined by a direct road. The road lengths in km are:
AB 14, AC 20, AD 17, AE 11, BC 13, BD 22, BE 18, CD 16, CE 19, DE 15.

(a)

Use the nearest neighbour algorithm, starting at A, to find an upper bound for the length of the courier's route.

(2)

(b)

Delete vertex A. Use the deleted vertex algorithm to find a lower bound for the length of the route.

(4)

(c)

Hence comment on the route found using the nearest neighbour algorithm.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 for applying nearest neighbour to give the cycle A–E–D–C–B–A	✓ 1
A1 for upper bound 11 + 15 + 16 + 13 + 14 = 69 km	✓ 1
Part (b)
M1 for finding a minimum spanning tree of B, C, D, E	✓ 1
A1 for edges BC, DE, CD with total weight 13 + 15 + 16 = 44	✓ 1
M1 for adding the two shortest edges from A, AE = 11 and AB = 14	✓ 1
A1 for lower bound 44 + 25 = 69 km	✓ 1
Part (c)
R1 for: the lower bound equals the upper bound, so 69 km is the minimum length and route A–E–D–C–B–A is optimal	✓ 1

Common error: In the deleted vertex algorithm, add the two shortest edges that join the deleted vertex to the tree. Do not add just one edge, and do not use any edge between the remaining vertices.

+ Add to test✓ In your test94IBMI-032-08-r2-02Vectors and kinematics Paper HLMedium6 marks

A light aircraft is climbing after take-off. Its velocity is constant and equal to
$\boldsymbol{v} = \begin{pmatrix}200\\150\\10\end{pmatrix}$ km h⁻¹
where $\boldsymbol{i}$ points east, $\boldsymbol{j}$ points north and $\boldsymbol{k}$ points vertically upwards.

(a)

Find the speed of the aircraft.

(2)

(b)

Find the angle at which the aircraft is climbing, measured from the horizontal.

(2)

(c)

Find the bearing on which the aircraft is travelling.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\|\boldsymbol{v}\| = \sqrt{200^2+150^2+10^2} = \sqrt{62600}$ M1	✓ 1
= 250 km h⁻¹ (3 s.f.) (accept 250.2) A1	✓ 1
Part (b)
Horizontal speed = $\sqrt{200^2+150^2}$ = 250, so $\tan\theta = \frac{10}{250}$ (or equivalent, e.g. $\sin\theta = \frac{10}{\sqrt{62600}}$) M1	✓ 1
θ = 2.29° A1	✓ 1
Part (c)
$\tan\alpha = \frac{200}{150}$, where α is the angle measured clockwise from north M1	✓ 1
Bearing = 053.1° A1	✓ 1

Common error: The angle of climb compares the vertical component with the horizontal speed $\sqrt{200^2+150^2}$. Using the full 3D speed in a tangent ratio gives the wrong angle. The full speed belongs in a sine ratio instead.

+ Add to test✓ In your test95IBMI-032-03-r2-04Vectors and kinematics Paper HLMedium13 marks

A ferry leaves port P at 09:00. Take P as the origin, with the first component of each vector measured east and the second measured north, in kilometres. From 09:00 until 10:30 the ferry has constant velocity $\begin{pmatrix}6\\8\end{pmatrix}$ km h⁻¹. At 10:30 it changes course, and from then on it has constant velocity $\begin{pmatrix}2\\10\end{pmatrix}$ km h⁻¹. Harbour H has position vector $\begin{pmatrix}13\\32\end{pmatrix}$.

(a)

Find the speed of the ferry before 10:30 and the bearing on which it travels, giving the bearing to the nearest 0.1°.

(4)

(b)

Write down the position vector of the ferry at 10:30.

(1)

(c)

Show that the ferry passes through H, and find the time at which it does so.

(3)

(d)

Find the total distance the ferry travels from P to H.

(3)

(e)

Find the angle through which the ferry turns at 10:30.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
$\sqrt{6^2+8^2}$ M1	✓ 1
= 10 km h⁻¹ A1	✓ 1
$\tan\theta = \frac{6}{8}$, with θ measured clockwise from north M1	✓ 1
Bearing 036.9° A1	✓ 1
Part (b)
$1.5\begin{pmatrix}6\\8\end{pmatrix} = \begin{pmatrix}9\\12\end{pmatrix}$ A1	✓ 1
Part (c)
$\begin{pmatrix}9\\12\end{pmatrix} + s\begin{pmatrix}2\\10\end{pmatrix} = \begin{pmatrix}13\\32\end{pmatrix}$, where s is hours after 10:30 M1	✓ 1
$s = 2$ from both components, so the ferry passes through H R1	✓ 1
Time 12:30 A1	✓ 1
Part (d)
First stage: 10 × 1.5 = 15 km A1	✓ 1
Second stage: speed × time, $2\sqrt{2^2+10^2} = 2\sqrt{104}$ (= 20.396…) km M1	✓ 1
Total = 35.4 km A1	✓ 1
Part (e)
$\cos\theta = \frac{6(2)+8(10)}{10\sqrt{104}} = \frac{92}{10\sqrt{104}}$ M1	✓ 1
θ = 25.6° A1	✓ 1

Common error: A common slip is to find |PH| as the straight-line distance from P to H. The ferry changes direction at 10:30, so add the lengths of the two stages.

+ Add to test✓ In your test96IBMI-032-08-r1-06Vectors and kinematics Paper HLHard12 marks

A coastguard station is at the origin O. At 14:00 a yacht in difficulty is drifting with position vector, in kilometres,
$\mathbf{r}_Y = \begin{pmatrix} 4 \\ 10 \end{pmatrix} + t\begin{pmatrix} 6 \\ -2 \end{pmatrix}$
where t is the time in hours after 14:00 and the unit vectors point east and north. At 14:00 a lifeboat leaves O, travelling in a straight line at a constant speed of 15 km h⁻¹, so as to intercept the yacht.

(a)

Find the speed of the yacht.

(2)

(b)

Find the bearing on which the yacht is drifting.

(2)

(c)

Show that the time of interception, t hours, satisfies $185t^2 - 8t - 116 = 0$.

(3)

(d)

Find the time at which the lifeboat reaches the yacht, to the nearest minute.

(2)

(e)

Find the bearing on which the lifeboat must travel.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\sqrt{6^2+(-2)^2}=\sqrt{40}$	✓ 1
A1: 6.32 km h⁻¹	✓ 1
Part (b)
M1: angle below east $\tan^{-1}\left(\frac{2}{6}\right)=18.4°$	✓ 1
A1: bearing 108° (108.4°)	✓ 1
Part (c)
M1: lifeboat distance 15t equals distance of yacht from O: $(4+6t)^2+(10-2t)^2=(15t)^2$	✓ 1
A1: expanding: 16 + 48t + 36t² + 100 − 40t + 4t² = 225t²	✓ 1
A1: rearranging to 185t² − 8t − 116 = 0	✓ 1
Part (d)
A1: t = 0.814 h (positive root, 0.81377…)	✓ 1
A1: 14:49	✓ 1
Part (e)
M1: interception point $\begin{pmatrix} 4+6(0.8138) \\ 10-2(0.8138) \end{pmatrix}$	✓ 1
A1: (8.88, 8.37)	✓ 1
A1: bearing $\tan^{-1}\left(\frac{8.88}{8.37}\right)$ = 046.7°	✓ 1

Exam tip: Reject the negative root of the quadratic with a reason in context: the interception must happen after the lifeboat sets off.

+ Add to test✓ In your test97IBMI-032-08-r2-05Vectors and kinematics Paper HLHard16 marks

A coastguard base is at the origin O. The vector $\boldsymbol{i}$ points east, $\boldsymbol{j}$ points north and distances are in kilometres. At time t hours after 12:00, a suspect boat S has position vector
$\boldsymbol{r}_S = \begin{pmatrix}10\\4\end{pmatrix} + t\begin{pmatrix}6\\8\end{pmatrix}$.
At 12:30 a patrol boat leaves O. It travels in a straight line at a constant speed of 20 km h⁻¹ and intercepts S at time t = T.

(a)

Find the speed of S and the bearing on which it travels.

(3)

(b)

Find the distance of S from O at 12:30.

(2)

(c)

Show that T satisfies $300T^2 - 584T - 16 = 0$.

(4)

(d)

Find the time of day at which the interception takes place.

(4)

(e)

Find the bearing on which the patrol boat must travel.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Speed = $\sqrt{6^2+8^2}$ = 10 km h⁻¹ A1	✓ 1
$\tan\alpha = \frac{6}{8}$ M1	✓ 1
Bearing 036.9° A1	✓ 1
Part (b)
Position at t = 0.5 is (13, 8) M1	✓ 1
Distance = $\sqrt{233}$ = 15.3 km A1	✓ 1
Part (c)
Patrol boat distance travelled = 20(T − 0.5) M1	✓ 1
$\|\boldsymbol{r}_S\|^2 = (10+6T)^2 + (4+8T)^2 = 100T^2 + 184T + 116$ M1	✓ 1
Equates the two distances (squared): $100T^2 + 184T + 116 = 400(T-0.5)^2$ M1	✓ 1
$400T^2 - 400T + 100 - 100T^2 - 184T - 116 = 0 \Rightarrow 300T^2 - 584T - 16 = 0$ A1	✓ 1
Part (d)
Solves the quadratic (GDC or formula) M1	✓ 1
T = 1.97 (1.97369…) A1	✓ 1
Rejects the negative root (T = −0.027), since the patrol boat leaves at t = 0.5 R1	✓ 1
Time ≈ 13:58 A1	✓ 1
Part (e)
Substitutes T into $\boldsymbol{r}_S$ M1	✓ 1
Interception point (21.8, 19.8) A1	✓ 1
Bearing = arctan(21.84/19.79) = 047.8° A1	✓ 1

Exam tip: The patrol boat starts half an hour late, so the distance it covers is 20(T − 0.5), not 20T. Always check that a root makes sense in context before using it.

+ Add to test✓ In your test98IBMI-032-04-r1-06Graph theory and networks Paper HLHard14 marks

An engineer must inspect five wind turbines, A, B, C, D and E. She starts and finishes at A and visits each other turbine exactly once. The shortest road distances between the turbines, in km, are:
AB 12, AC 18, AD 15, AE 10
BC 9, BD 14, BE 11
CD 8, CE 16
DE 13

(a)

Use the nearest neighbour algorithm, starting at A, to find an upper bound for the length of the engineer's route. State the route.

(3)

(b)

By deleting vertex A, use the deleted vertex algorithm to find a lower bound for the length of the route.

(4)

(c)

Find the lower bound obtained by deleting vertex C instead. Hence state the best interval you can give for the length L km of the optimal route.

(4)

(d)

A colleague suggests the route A–B–C–D–E–A. Find its length, and comment on what this tells you about L, given that all distances are whole numbers of kilometres.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1 at each step chooses the nearest unvisited turbine	✓ 1
A1 A–E–B–C–D–A	✓ 1
A1 10 + 11 + 9 + 8 + 15 = 53 km	✓ 1
Part (b)
M1 finds a minimum spanning tree of B, C, D, E	✓ 1
A1 CD + BC + BE = 8 + 9 + 11 = 28	✓ 1
M1 adds the two shortest edges from A: AE and AB	✓ 1
A1 28 + 10 + 12 = 50 km	✓ 1
Part (c)
A1 minimum spanning tree of A, B, D, E: AE + BE + DE = 10 + 11 + 13 = 34	✓ 1
A1 two shortest edges from C: CD + BC = 8 + 9 = 17	✓ 1
A1 lower bound 34 + 17 = 51 km	✓ 1
R1 the larger lower bound is the better one, so 51 ≤ L ≤ 53	✓ 1
Part (d)
A1 12 + 9 + 8 + 13 + 10 = 52 km	✓ 1
R1 this is a valid Hamiltonian cycle shorter than 53, so it gives an improved upper bound: 51 ≤ L ≤ 52	✓ 1
R1 so L is 51 or 52, and this route is at most 1 km longer than the optimal route	✓ 1

Common error: For a lower bound, choose the largest value from the deleted vertex algorithm. For an upper bound, choose the smallest valid tour length. Mixing these up gives an interval that is either impossible or too wide.

+ Add to test✓ In your test99IBMI-032-03-r2-06Vectors and kinematics Paper HLHard14 marks

A golf ball is struck from a point O on level ground. Take horizontal and vertical components in metres, and ignore air resistance. The velocity of the ball $t$ seconds after it is struck is
$\mathbf{v} = \begin{pmatrix}15\\20-9.8t\end{pmatrix}$ m s⁻¹.
A tree of height 17 m stands 40 m horizontally from O.

(a)

Find the initial speed of the ball and the angle its initial velocity makes with the horizontal.

(3)

(b)

Find an expression for the displacement vector $\mathbf{r}$ of the ball from O at time $t$.

(3)

(c)

Find the maximum height reached by the ball.

(3)

(d)

Determine whether the ball passes over the tree, stating the clearance or shortfall.

(3)

(e)

Find the speed of the ball as it passes the tree.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
Speed $\sqrt{15^2+20^2}$ = 25 m s⁻¹ A1	✓ 1
$\tan\theta = \frac{20}{15}$ M1	✓ 1
θ = 53.1° A1	✓ 1
Part (b)
Integrates each component M1	✓ 1
$\mathbf{r} = \begin{pmatrix}15t\\20t-4.9t^2\end{pmatrix} + \mathbf{c}$ A1	✓ 1
$\mathbf{r}=\mathbf{0}$ when $t=0$, so $\mathbf{c}=\mathbf{0}$ A1	✓ 1
Part (c)
Vertical velocity is zero: $20 - 9.8t = 0$ M1	✓ 1
$t = 2.04$ s (2.0408…) A1	✓ 1
Height $= 20(2.0408) - 4.9(2.0408)^2$ = 20.4 m A1	✓ 1
Part (d)
$15t = 40$ gives $t = \frac{8}{3}$ (2.67 s) M1	✓ 1
Height $= 20\left(\frac{8}{3}\right) - 4.9\left(\frac{8}{3}\right)^2$ = 18.5 m (18.489…) A1	✓ 1
18.5 > 17, so the ball clears the tree by 1.49 m (accept 1.5 m) R1	✓ 1
Part (e)
$\mathbf{v} = \begin{pmatrix}15\\-6.13\end{pmatrix}$, so speed $= \sqrt{15^2+6.133^2}$ M1	✓ 1
= 16.2 m s⁻¹ A1	✓ 1

Exam tip: To find when the ball reaches the tree, use the horizontal component, because it is linear in t. Then substitute that time into the vertical component to find the height.

+ Add to test✓ In your test100IBMI-032-03-r1-05Vectors and kinematics Paper HLHard13 marks

A coastguard station is at the origin. Position vectors are written as (east component, north component) in kilometres. At time t hours after an alert is raised, a drifting cargo ship has position
r_S = (10, 4) + t(−2, 5).
At t = 0 a rescue boat leaves the station and travels in a straight line at a constant speed of 15 km h⁻¹ so that it meets the ship.
Give non-exact answers to three significant figures and bearings to one decimal place.

(a)

Find the speed of the cargo ship.

(2)

(b)

Find the distance and the bearing of the ship from the station when the alert is raised.

(3)

(c)

The rescue boat meets the ship at time T hours. Show that 196T² = 116.

(3)

(d)

Find the time taken for the rescue boat to reach the ship, giving your answer in minutes.

(2)

(e)

Find the bearing on which the rescue boat must travel.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
M1: $\sqrt{(-2)^2+5^2}=\sqrt{29}$	✓ 1
A1: 5.39 km h⁻¹	✓ 1
Part (b)
A1: distance $\sqrt{10^2+4^2}=\sqrt{116}$ = 10.8 km	✓ 1
M1: uses tan θ = $\frac{10}{4}$ with the angle measured clockwise from north	✓ 1
A1: bearing 068.2°	✓ 1
Part (c)
M1: ship position at time T is (10 − 2T, 4 + 5T)	✓ 1
M1: sets the distance of this point from the origin equal to the distance travelled by the boat: (10 − 2T)² + (4 + 5T)² = (15T)²	✓ 1
A1: expands to 116 + 29T² = 225T², with the terms in T cancelling, giving 196T² = 116	✓ 1
Part (d)
M1: solves to get T = $\frac{\sqrt{116}}{14}$ = 0.769 h, rejecting the negative root	✓ 1
A1: 46.2 minutes, using the unrounded T (accept 46.1 from 0.769 × 60)	✓ 1
Part (e)
M1: substitutes T into the ship's position to find the meeting point	✓ 1
A1: meeting point (8.46, 7.85)	✓ 1
A1: bearing = arctan(east ÷ north) = 047.2° using unrounded values (accept 047.1° from the rounded coordinates)	✓ 1

Exam tip: The boat should be steered towards where the ship will be, not where it is now. Keep full calculator values until the final answer. To check your result, divide the meeting point by T to get the boat's velocity: its magnitude should be 15.

+ Add to test✓ In your test101IBMI-032-04-r2-06Graph theory and networks Paper HLHard14 marks

A park ranger must inspect five shelters, A, B, C, D and E. The ranger starts and finishes at A and visits every other shelter. Not every pair of shelters is joined by a direct path. The only paths and their lengths in km are:
AB 8, AC 5, BC 4, BD 6, CD 9, CE 7, DE 3.
The ranger may pass through a shelter more than once.

(a)

A table of least distances is needed. Find the least distance between each of these pairs: A and D, A and E, B and E.

(3)

(b)

Use the nearest neighbour algorithm on the table of least distances, starting at A, to find an upper bound for the length of the ranger's route. Write down the route the ranger actually follows along the paths.

(4)

(c)

Use the deleted vertex algorithm on the table of least distances to find a lower bound, first by deleting A and then by deleting D. Hence state the best interval for the length L km of the shortest route that you can obtain from your bounds.

(4)

(d)

Find a Hamiltonian cycle in the original path network that is shorter than the route found using the nearest neighbour algorithm. Hence improve the interval for L.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)
A1 for AD = 14 km (via B or via C)	✓ 1
A1 for AE = 12 km (A–C–E)	✓ 1
A1 for BE = 9 km (B–D–E)	✓ 1
Part (b)
M1 for applying nearest neighbour to the table of least distances	✓ 1
A1 for the cycle A–C–B–D–E–A	✓ 1
A1 for upper bound 5 + 4 + 6 + 3 + 12 = 30 km	✓ 1
A1 for the actual route A–C–B–D–E–C–A, because E to A goes through C	✓ 1
Part (c)
M1 for the method: find a minimum spanning tree of the remaining vertices, then add the two shortest edges from the deleted vertex	✓ 1
A1 for deleting A: MST DE, BC, BD = 13, plus 5 + 8, giving 26	✓ 1
A1 for deleting D: MST BC, AC, CE = 16, plus 3 + 6, giving 25	✓ 1
A1 for 26 ≤ L ≤ 30, using the larger lower bound	✓ 1
Part (d)
A1 for A–B–D–E–C–A (or its reverse)	✓ 1
A1 for length 8 + 6 + 3 + 7 + 5 = 29 km	✓ 1
A1 for the improved interval 26 ≤ L ≤ 29	✓ 1

Exam tip: Always give the best upper bound (the smallest) and the best lower bound (the largest). When an edge in the table of least distances is not a direct path, write out the actual path the ranger follows.

Marking point	Mark	Notes
Part (a)
M1 for substituting into \(V=\pi r^2 h\): \(\pi\times1.2^2\times2.5\)	✓ 1
A1 for \(V=11.3\) m³ (11.3097…)	✓ 1
Part (b)
A1 for 11 300 litres, using 1 m³ = 1000 litres; accept 11 310 or 11 309.7; follow through from their answer to part (a)	✓ 1
Part (c)
M1 for dividing their capacity by the rate, e.g. \(11309.7\div15\)	✓ 1
A1 for 754 minutes (753.98…), which is about 12.6 hours; accept 753 minutes as follow-through from 11 300 litres	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for Pythagoras with radius 18: \(h=\sqrt{70^2-18^2}\)	✓ 1
A1 for \(h=67.6\) cm (67.646…)	✓ 1
Part (b)
M1 for using \(\pi r l\): \(\pi\times18\times70\)	✓ 1
A1 for 3960 cm² (3958.4…)	✓ 1

Marking point	Mark	Notes
Part (a)
\(l=\sqrt{9^2+20^2}\) M1 ; l = 21.9 cm (21.932…) A1	✓ 2
Part (b)
\(V=\frac{1}{3}\pi(9^2)(20)\) M1 ; V = 1700 cm³ (1696.5…) A1	✓ 2
Part (c)
Curved surface area \(=\pi r l=\pi(9)(21.932…)\) M1 ; 620 cm² (620.10…) A1	✓ 2

Marking point	Mark	Notes
Part (a)
\(V=\frac{4}{3}\pi(0.325)^3\) M1 ; 0.144 m³ (0.14379…) A1	✓ 2
Part (b)
\(A=4\pi(0.325)^2\) M1 ; 1.33 m² (1.3273…) A1	✓ 2
Part (c)
1.3273… × 8.40 × 50 M1 ; 557.48 USD A1 (accept 557 or 557.5 USD; accept 558.60 USD from using 1.33)	✓ 2

Marking point	Mark	Notes
Part (a)
\(x = 3\) A1	✓ 1
\(y = 2\) A1	✓ 1
Part (b)
V(3, 2) A1	✓ 1
Part (c)
V is equidistant from P and Q and from P and R, so it is equidistant from Q and R and lies on the bisector of QR (or: the midpoint of QR is (3, 2)) R1	✓ 1
Part (d)
\(\sqrt{3^2 + 2^2}\) M1	✓ 1
3.61 km A1	✓ 1

Marking point	Mark	Notes
Part (a)
(i) \(x = 5\) A1	✓ 1
(ii) \(y = 3\) A1	✓ 1
Part (b)
(5, 3) A1	✓ 1
Part (c)
Distance = \(\sqrt{5^2 + 3^2}\) M1	✓ 1
= 5.83 km (equidistant from all three stations) A1	✓ 1

Marking point	Mark	Notes
Part (a)
Use of distance formula, e.g. XA = \(\sqrt{4^2 + 1^2}\) M1	✓ 1
XA = \(\sqrt{17}\) = 4.12 km, XB = \(\sqrt{13}\) = 3.61 km A1	✓ 1
XC = \(\sqrt{32}\) = 5.66 km A1	✓ 1
Part (b)
X is closest to B, so X lies in the cell of B R1	✓ 1
Estimated rainfall = 35 mm A1	✓ 1
Part (c)
\(\frac{\|35 - 39\|}{39} \times 100\) M1	✓ 1
= 10.3% A1	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for using \(\frac{1}{2}ab\sin C\): \(\frac{1}{2}\times2.1^2\times\sin70°\)	✓ 1
A1 for 2.07 m² (2.07202…)	✓ 1
Part (b)
A1 for \(2.07202\times3=6.22\) m³	✓ 1
Part (c)
M1 for the cosine rule, \(\sqrt{2.1^2+2.1^2-2(2.1)(2.1)\cos70°}\), or for \(2\times2.1\sin35°\)	✓ 1
A1 for 2.41 m (2.40902…)	✓ 1
Part (d)
M1 for finding the three rectangular face areas: two sloping rectangles \(2\times2.1\times3=12.6\) and the groundsheet \(2.40902\times3=7.227\)	✓ 1
M1 for adding the two triangular ends, \(2\times2.07202=4.144\), to the rectangular areas, so that all five faces are summed	✓ 1
A1 for 24.0 m² (23.971…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for the cylinder volume, \(\pi\times3.5^2\times12=461.8\)	✓ 1
M1 for the hemisphere volume, \(\frac{2}{3}\pi\times3.5^3=89.80\)	✓ 1
A1 for 552 m³ (551.61…)	✓ 1
Part (b)
M1 for \(0.8\times551.61\times750\)	✓ 1
A1 for 331 000 kg (331 tonnes)	✓ 1
Part (c)
M1 for adding the curved cylinder area and half a sphere: \(2\pi(3.5)(12)+2\pi(3.5)^2\)	✓ 1
A1 for 341 m² (340.86…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\sqrt{20^2+15^2}\)	✓ 1
A1 for 25 m	✓ 1
Part (b)
M1 for \(4\times\frac{1}{2}\times30\times25\)	✓ 1
A1 for 1500 m²	✓ 1
Part (c)
M1 for \(\frac{1}{3}\times30^2\times20\)	✓ 1
A1 for 6000 m³	✓ 1
Part (d)
M1 for \(\tan\theta=\frac{20}{15}\)	✓ 1
A1 for 53.1°	✓ 1
Part (e)
M1 for using half the diagonal of the base, \(15\sqrt2=21.21\), in \(\tan\theta=\frac{20}{21.21}\)	✓ 1
A1 for 43.3°	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for \(\frac{1}{2}(0.9+0.5)\times0.4\)	✓ 1
A1 for 0.28 m²	✓ 1
Part (b)
M1 for \(0.28\times4=1.12\) m³	✓ 1
A1 for 1120 litres	✓ 1
Part (c)
M1 for a horizontal offset of 0.2 m on each side, with \(\tan\theta=\frac{0.4}{0.2}\)	✓ 1
A1 for 63.4°	✓ 1

Marking point	Mark	Notes
Part (a)
\(\sqrt{2.4^2+1.5^2+1.2^2}\) M1 ; 3.07 m (3.0741…) A1	✓ 2
Part (b)
Base diagonal = \(\sqrt{2.4^2+1.5^2}\) = 2.830… and tan θ = 1.2/2.830… M1 ; θ = 23.0° A1	✓ 2
Part (c)
2.4 ÷ 0.56 = 4.28… so 4 along the length and 1.5 ÷ 0.56 = 2.67… so 2 across, one layer since 1.15 < 1.2 M1 ; 8 drums A1	✓ 2
Part (d)
One drum: π(0.28)²(1.15) = 0.2832… m³ M1 ; crate volume 4.32 m³ and 4.32 − 8(0.2832…) M1 ; 2.05 m³ (2.054…) A1	✓ 3
Part (e)
2.2660… ÷ 4.32 × 100 M1 ; 52.5% A1	✓ 2

Marking point	Mark	Notes
Part (a)
\(y = -\frac{1}{2}x + 4.5\), so gradient of edge = \(-\frac{1}{2}\) A1	✓ 1
The edge is the perpendicular bisector of AB, so gradient of AB = 2 R1	✓ 1
Part (b)
\(y = 2x - 3\) A1	✓ 1
Part (c)
Substituting: \(x + 2(2x - 3) = 9\) M1	✓ 1
\(x = 3\), \(y = 3\), so M(3, 3) A1	✓ 1
Part (d)
M is the midpoint of AB, so B = (2×3 − 2, 2×3 − 1) = (4, 5) A1	✓ 1
Part (e)
AB = \(\sqrt{2^2 + 4^2} = \sqrt{20}\) = 4.47 km A1	✓ 1

Marking point	Mark	Notes
Part (a)
Midpoint (4, 3) and gradient of AB = \(\frac{2}{4} = \frac{1}{2}\) M1	✓ 1
Perpendicular gradient = −2 M1	✓ 1
\(y = -2x + 11\) A1	✓ 1
Part (b)
\(y = 0\): \(x = 5.5\), giving (5.5, 0) A1	✓ 1
\(y = 8\): \(x = 1.5\), giving (1.5, 8) A1	✓ 1
Part (c)
At \(x = 5\), \(y = -2(5) + 11 = 1\), so F lies on the bisector (or FA = FB = \(\sqrt{10}\)) M1	✓ 1
F is equidistant from both wells, so either well is equally near to the shed R1	✓ 1
Part (d)
Region is the trapezium with vertices (0, 0), (5.5, 0), (1.5, 8), (0, 8): area = \(\frac{5.5 + 1.5}{2} \times 8\) M1	✓ 1
= 28 km² A1	✓ 1
Part (e)
\(\frac{64 - 28}{64} \times 100\) = 56.3% A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(x = 4\) A1	✓ 1
Part (b)
Midpoint (1, 3) and gradient of AC = 3, so perpendicular gradient \(-\frac{1}{3}\) M1	✓ 1
\(y = -\frac{1}{3}x + \frac{10}{3}\) (or equivalent, e.g. \(x + 3y = 10\)) A1	✓ 1
Part (c)
At \(x = 4\): \(y = -\frac{4}{3} + \frac{10}{3} = 2\), so V(4, 2) A1	✓ 1
Part (d)
VA = \(\sqrt{4^2 + 2^2}\), VB = \(\sqrt{4^2 + 2^2}\), VC = \(\sqrt{2^2 + 4^2}\) M1	✓ 1
Each distance = \(\sqrt{20}\) ≈ 4.47 km A1	✓ 1
Part (e)
The plant is 4.47 km from all three villages. Moving it anywhere else in the triangle brings it closer to at least one village, so 4.47 km is the greatest possible distance to the nearest village R1	✓ 1

Marking point	Mark	Notes
Part (a)
M1 for similar triangles: \(\frac{H}{15}=\frac{H-24}{10}\)	✓ 1
A1 for \(10H=15H-360\), giving \(H=72\)	✓ 1
Part (b)
M1 for the large cone minus the small cone: \(\frac{1}{3}\pi(15^2)(72)-\frac{1}{3}\pi(10^2)(48)\)	✓ 1
A1 for 11 900 cm³ (11 938)	✓ 1
A1 for 11.9 litres	✓ 1
Part (c)
M1 for the slant height of the frustum, \(\sqrt{24^2+5^2}=24.52\), or for the slant heights of both cones, 73.55 and 49.03	✓ 1
M1 for the curved area \(\pi(15+10)(24.52)=1925\) plus the base \(\pi\times10^2\)	✓ 1
A1 for 2240 cm² (2239.6)	✓ 1

Marking point	Mark	Notes
Part (a)
A1 for \(h=\frac{330}{\pi r^2}\)	✓ 1
Part (b)
M1 for \(A=2\pi r^2+2\pi rh\)	✓ 1
A1 for substituting \(h\) to give \(2\pi r\times\frac{330}{\pi r^2}=\frac{660}{r}\), and hence the given result	✓ 1
Part (c)
M1 for sketching the graph of A on a GDC, or for any valid method of finding its minimum point	✓ 1
A1 for \(r=3.74\) cm (3.7449…)	✓ 1
A1 for \(A=264\) cm² (264.36…)	✓ 1
Part (d)
M1 for \(2(2r)^2+\frac{660}{r}\), i.e. \(8r^2+\frac{660}{r}\), with \(r=3.7449…\)	✓ 1
A1 for 288 cm² (288.43…)	✓ 1
Part (e)
M1 for a scrap area of \(2r^2(4-\pi)=24.08\), divided by 288.43 and multiplied by 100	✓ 1
A1 for 8.35%	✓ 1

Marking point	Mark	Notes
Part (a)
\(\frac{1}{3}\pi(40^2)(60)\) = 100 531 cm³ M1 ; 101 litres A1	✓ 2
Part (b)
Similar triangles give radius of water surface r = \(\frac{40}{60}h=\frac{2h}{3}\) M1 ; \(V=\frac{1}{3}\pi\left(\frac{2h}{3}\right)^2h=\frac{4\pi h^3}{27}\) A1	✓ 2
Part (c)
\(\frac{4\pi h^3}{27}=16000\pi\) M1 ; h³ = 108 000 M1 ; h = 47.6 cm (47.622…) A1	✓ 3
Part (d)
r = \(\frac{2}{3}\)(47.622…) = 31.75 cm M1 ; slant height \(\sqrt{47.622^2+31.748^2}\) = 57.23 cm M1 ; πrl = 5710 cm² (5708.5…) A1	✓ 3
Part (e)
50 265.5 ÷ 250 M1 ; 201 seconds A1	✓ 2

Marking point	Mark	Notes
Part (a)
\(x = 5\) A1	✓ 1
Part (b)
Gradient AC = \(\frac{4}{3}\), so perpendicular gradient = \(-\frac{3}{4}\) M1	✓ 1
Midpoint of AC = (3.5, 4), substituted: \(y - 4 = -0.75(x - 3.5)\) M1	✓ 1
\(y = -0.75x + 2.625 + 4 = -0.75x + 6.625\) A1	✓ 1
Part (c)
Substituting \(x = 5\): \(y = -3.75 + 6.625\) M1	✓ 1
V(5, 2.875) A1	✓ 1
Part (d)
Bisector of AC meets \(x = 0\) at (0, 6.625) A1	✓ 1
Cell is the trapezium (0, 0), (5, 0), (5, 2.875), (0, 6.625): area = \(\frac{6.625 + 2.875}{2} \times 5\) M1	✓ 1
= 23.75 km² (23.8 km²) A1	✓ 1
Part (e)
By symmetry B also serves 23.75 km², so C serves 80 − 2 × 23.75 = 32.5 km² M1	✓ 1
\(\frac{32.5}{80} \times 100\) = 40.6% A1	✓ 1
Part (f)
Distance from (0, 8) to A = \(\sqrt{40}\) = 6.32 km and to C = \(\sqrt{29}\) M1	✓ 1
Nearest station to (0, 8) is C, at 5.39 km A1	✓ 1
Distance from V to its nearest station is 3.125 km; since 5.39 > 3.125 the corner is further from help than V, so the claim is incorrect R1	✓ 1

Marking point	Mark	Notes
Part (a)
Speed = \(\sqrt{3^2+4^2}\) M1	✓ 1
= 5 km h⁻¹ A1	✓ 1
Part (b)
Uses \(t = 2.5\) M1	✓ 1
\(\mathbf{r} = \begin{pmatrix}9.5\\9\end{pmatrix}\) A1	✓ 1
Part (c)
\(\sqrt{9.5^2+9^2}\) M1	✓ 1
= 13.1 km (13.086…) A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(\sqrt{4^2+(-3)^2+1.5^2} = \sqrt{27.25}\) M1	✓ 1
= 5.22 m s⁻¹ A1	✓ 1
Part (b)
\(20 + 1.5t = 50\) M1	✓ 1
\(t = 20\) s A1	✓ 1
Part (c)
Horizontal components at \(t=20\): 80 and −60, so \(\sqrt{80^2+60^2}\) M1	✓ 1
= 100 m A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(M=\begin{pmatrix}0&1&1&0\\1&0&1&1\\1&1&0&1\\0&1&1&0\end{pmatrix}\)	—
A1 for a symmetric 4×4 matrix with zeros on the leading diagonal	✓ 1
A1 for all entries correct	✓ 1
Part (b)
M1 for using the (A, D) entry of \(M^4\)	✓ 1
A1 for 10 walks	✓ 1
Part (c)
R1 for degrees A = 2, B = 3, C = 3, D = 2, so exactly two vertices (B and C) have odd degree	✓ 1
A1 for: it is possible (an Eulerian trail exists), but the route must start at B and finish at C, or start at C and finish at B	✓ 1

Marking point	Mark	Notes
Part (a)
M1: speed = \(\sqrt{3^2+4^2}\)	✓ 1
A1: 5 km h⁻¹	✓ 1
Part (b)
A1: \(\begin{pmatrix} 9.5 \\ 9 \end{pmatrix}\)	✓ 1
Part (c)
M1: \(\sqrt{9.5^2+9^2}\)	✓ 1
A1: 13.1 km (13.086…)	✓ 1

Marking point	Mark	Notes
Part (a)
M1: \(\sqrt{4^2+(-2)^2+1.5^2} = \sqrt{22.25}\)	✓ 1
A1: 4.72 m s⁻¹	✓ 1
Part (b)
M1: 1.5t = 30	✓ 1
A1: t = 20 s	✓ 1
Part (c)
M1: 20 × 4.717… or \(\left\|20\begin{pmatrix} 4 \\ -2 \\ 1.5 \end{pmatrix}\right\|\)	✓ 1
A1: 94.3 m	✓ 1

Marking point	Mark	Notes
Part (a)
Speed = \(\sqrt{4^2+3^2}\) M1	✓ 1
= 5 km h⁻¹ A1	✓ 1
Part (b)
\(\begin{pmatrix}3+10\\-2+7.5\end{pmatrix} = \begin{pmatrix}13\\5.5\end{pmatrix}\) A1	✓ 1
Part (c)
\(\sqrt{13^2+5.5^2}\) M1	✓ 1
= 14.1 km (14.1157…) A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(\begin{pmatrix}0\\300\end{pmatrix} + \begin{pmatrix}40\\-30\end{pmatrix} = \begin{pmatrix}40\\270\end{pmatrix}\) km h⁻¹ A1	✓ 1
Part (b)
\(\sqrt{40^2+270^2} = \sqrt{74500}\) M1	✓ 1
= 273 km h⁻¹ A1	✓ 1
Part (c)
\(\tan\theta = \frac{40}{270}\) M1	✓ 1
Bearing 008.4° A1	✓ 1
Part (d)
Time = 500 ÷ 272.95 M1	✓ 1
= 1.83 h A1	✓ 1

Marking point	Mark	Notes
Part (a)
\(\begin{pmatrix}0&1&1&0\\1&0&1&1\\1&1&0&1\\0&1&1&0\end{pmatrix}\)	—
A1 symmetric matrix with zero leading diagonal	✓ 1
A1 all entries correct	✓ 1
Part (b)
Entry = 3	✓ 1
R1 there are 3 different journeys along exactly two roads that start and finish at B (equal to the number of roads meeting at B)	✓ 1
Part (c)
M1 uses the (A, D) entry of M⁴	✓ 1
A1 10 journeys	✓ 1

Marking point	Mark	Notes
Part (a)
A1 for \(\overrightarrow{AB}=(8,1,0)\)	✓ 1
A1 for \(\overrightarrow{AC}=(3,6,3)\)	✓ 1
Part (b)
M1 for a correct method for the vector product, e.g. at least two correct components	✓ 1
A1 for \((3,-24,45)\)	✓ 1
Part (c)
M1 for \(\frac{1}{2}\sqrt{3^2+(-24)^2+45^2}=\frac{1}{2}\sqrt{2610}\)	✓ 1
A1 for 25.5 m² (25.5440…)	✓ 1
Part (d)
M1 for multiplying the unrounded area by 85, e.g. \(\frac{1}{2}\sqrt{2610}\times85\) or \(25.54408\ldots\times85\)	✓ 1
A1 for USD 2171.25	✓ 1

Marking point	Mark	Notes
Part (a)
Uses \(\mathbf{r}_B - \mathbf{r}_A\) M1	✓ 1
\(\begin{pmatrix}30+4t-12t\\-10+9t-5t\end{pmatrix}\) leading to the given vector A1	✓ 1
Part (b)
\(d^2 = (30-8t)^2 + (-10+4t)^2 = 80t^2 - 560t + 1000\) M1	✓ 1
Minimises, e.g. \(\frac{d(d^2)}{dt} = 160t - 560 = 0\) or uses GDC M1	✓ 1
\(t = 3.5\), so the time is 16:30 A1	✓ 1
Part (c)
Substitutes \(t = 3.5\): \(\begin{pmatrix}2\\4\end{pmatrix}\), \(d = \sqrt{2^2+4^2}\) M1	✓ 1
= 4.47 km (\(\sqrt{20}\)) A1	✓ 1

Marking point	Mark	Notes
Part (a)
Uses different parameters, e.g. \(3\lambda = 24 - 2\mu\), \(4\lambda = 10 + \mu\), \(6 = 4.8 + 0.2\mu\) M1	✓ 1
\(\mu = 6\), \(\lambda = 4\), and all three equations are satisfied A1	✓ 1
Point (12, 16, 6) A1	✓ 1
Part (b)
A is at the point when t = 4, but B is there when t = 6 R1	✓ 1
They are not at the point at the same time, so they do not collide A1	✓ 1
Part (c)
\(\mathbf{r}_B(4) = \begin{pmatrix}16\\14\\5.6\end{pmatrix}\), so \(\mathbf{r}_B - \mathbf{r}_A = \begin{pmatrix}4\\-2\\-0.4\end{pmatrix}\) M1	✓ 1
\(\sqrt{20.16}\) = 4.49 km A1	✓ 1