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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Session: November 2021
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2021 Global paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

25 marks

Graph algorithms with an unknown edge weight: bounds on a tour and route inspection

Number and algebra

(a)

1 mark

Visit every vertex once and return. For example:

A B C D E G H F A

(b)(i)

5 marks

Without A there are 7 vertices, so the tree has 6 edges. Prim's algorithm from B takes the cheapest edge joining the tree to a new vertex each time.
The weight of the tree is the sum.

BC 6, CD 5, DE 6, BH 7, GH 4, FG 5 6 + 5 + 6 + 7 + 4 + 5 = 33

(b)(ii)

2 marks

Add back the two cheapest edges at A: AB with weight 3 and AF with weight $x$ , which is smaller than every other edge at A when $x < 9$ .

lower bound = 33 + 3 + x = 36 + x

(c)(i)

1 mark

The shortest route from C to G is C–H–G.

p = 9 + 4 = 13

(c)(ii)

1 mark

The shortest route from D to F that doesn't use AF is D–E–F.

q = 6 + 11 = 17

(c)(iii)

1 mark

The route from D to F through A is D to A (14) and then AF, written as $14 + x$ .

r = 14

(d)(i)

3 marks

With $x = 3$ the nearest neighbour algorithm from A has a tie, since AB and AF are both 3. Choosing AB first gives the cycle A, B, C, D, E, G, H, F, A. Choosing AF first gives A, F, G, H, B, C, D, E, then back to A by the shortest route E to F to A.

ABCDEGHFA : 3 + 6 + 5 + 6 + 8 + 4 + 8 + 3 = 43 AFGHBCDE(F)A : 3 + 5 + 4 + 7 + 6 + 5 + 6 + 14 = 50

(d)(ii)

2 marks

The best upper bound is the shorter of the cycles.

43

(e)(i)

2 marks

For $x > 3$ the algorithm starts A, B, C, D, E, G, H, F. The return from F to A can use the edge AF with weight $x$ , or the route F–H–B–A with weight $8 + 7 + 3 = 18$ . AF will not definitely be avoided until it's longer than 18.

x \geq 19

(e)(ii)

2 marks

With the route via B used for the return, the cycle has length

3 + 6 + 5 + 6 + 8 + 4 + 8 + 18 = 58

(f)

5 marks

Corinne must walk every edge, starting and finishing at H. The odd-degree vertices are B, D, G and H. The edges to repeat come from the cheapest pairing.
The cheapest is BD with GH, adding 15 minutes. The total time is 120 minutes.

BH + DG : 7 + 14 = 21, BD + GH : 11 + 4 = 15, BG + DH : 11 + 10 = 21 (92 + x) + 15 = 120 \Rightarrow x = 13

Question 2

30 marks

Water draining from a cylinder: regression, differential equation and two containers

Statistics and probability
Calculus

(a)(i)

2 marks

Regression of $h$ on $t$ using the five readings.

h (t) = - 0.134 t + 3.1

(a)(ii)

1 mark

The parameter $a = - 0.134$ is the rate of change of the water height: the height falls by about 0.134 m each minute.

(a)(iii)

1 mark

Using the equation to predict when $h = 0$ goes beyond the data, which only runs to 20 minutes. The rate of fall may not stay constant as the container empties, so the prediction could be unreliable.

(b)(i)

1 mark

Quadratic regression on the same data.

h (t) = 0.002 t^{2} - 0.174 t + 3.2

(b)(ii)

2 marks

Solve $0.002 t^{2} - 0.174 t + 3.2 = 0$ and take the smaller root, since the quadratic turns back up later.

k \approx 26.4

(b)(iii)

1 mark

The function is a sensible model from the start until the water has run out.

0 \leq t \leq 26.4

(c)

3 marks

The container has radius 1, so $V = π (1)^{2} h = π h$ , and $\frac{d V}{d t} = π \frac{d h}{d t}$ . Equate this with the given rate.

π \frac{d h}{d t} = - π R^{2} 70560 h \Rightarrow \frac{d h}{d t} = - R^{2} 70560 h

(d)

5 marks

Separate the variables.
Write the constant as $C = 70560 c$ , then square both sides.

\int \frac{1}{h} d h = - R^{2} 70560 \int d t \Rightarrow 2 h = - R^{2} 70560 t + C h = \frac{70560}{2} (c - R^{2} t) \Rightarrow h = \frac{70560}{4} (c - R^{2} t)^{2} = 17640 (c - R^{2} t)^{2}

(e)

4 marks

Use $h (0) = 3.2$ to find $c$ .
The container is empty when $h = 0$ , so $c - R^{2} T = 0$ .

3.2 = 17640 c^{2} \Rightarrow c \approx 0.013469 T = \frac{c}{R ^{2}} = \frac{0.013469}{0.02 3 ^{2}} \approx 25.5 minutes

(f)

3 marks

For the water to run out in 15 minutes, $c^{'} = 15 R^{2} = 0.007935$ . The new height is $h (0) = 17640 c^{'2}$ .

h (0) = 17640 (15 \times 0.02 3^{2})^{2} \approx 1.11 m

(g)(i)

4 marks

Take the upper container with $h = 17640 (c - R^{2} t)^{2}$ and $c = 0.013469$ . The height of water in it is falling, so $70560 h = 35280 (c - R^{2} t)$ . Its volume flows out at rate $π R^{2} 70560 h$ into the lower container, which also has base area $π$ . So the height gain in the lower container from this inflow is $R^{2} \cdot 35280 (c - R^{2} t)$ . The lower container also drains at $R^{2} 70560 H$ .
Put in $R^{2} = 0.02 3^{2} = 0.000529$ and $c = 0.013469$ .

\frac{d H}{d t} = 35280 R^{2} (c - R^{2} t) - R^{2} 70560 H 35280 R^{2} c \approx 0.2514, 35280 R^{4} \approx 0.009873, R^{2} 70560 \approx 0.1405 \frac{d H}{d t} \approx 0.2514 - 0.009873 t - 0.1405 H

(g)(ii)

3 marks

Euler's method with step 0.5 minutes: start at $t = 0$ , $H = 1$ and use $H_{n + 1} = H_{n} + 0.5 \frac{d H}{d t}$ repeatedly. The height rises at first, reaches a peak, and then falls as the upper container empties.

H_{m a x} \approx 1.45 m

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra25 marks · 45%
Q1
Calculus22 marks · 40%
Q2
Statistics and probability8 marks · 15%
Q2

Mentor insights

Traps to watch for

01
Cases for the unknown weight · Q1
The edge AF with weight $x$ may or may not be used. State the range of $x$ for which each bound applies, and check the tie at $x = 18$ .
02
The lower bound adds two edges back · Q1
After deleting A, the minimum spanning tree plus the two smallest edges at A gives the bound. With $x < 9$ the two smallest are AB (3) and AF ( $x$ ).
03
Linear then quadratic then theory · Q2
The question moves from fitting data to a model from physics. Use the regression only inside the data range, and use the differential equation to extend it.
04
Don't forget the initial condition · Q2
In (d) the general solution has a constant. Use $h (0) = 3.2$ in (e) and the new time in (f).
05
Show-that parts give you the target · Q2
In (g)(i), build $\frac{d H}{d t}$ as the inflow from the upper container minus the outflow from the lower one, both as rates of change of height, since each container has base area $π$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Session: November 2021
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the November 2021 Global paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

25 marks

Graph algorithms with an unknown edge weight: bounds on a tour and route inspection

Number and algebra

(a)

1 mark

Visit every vertex once and return. For example:

A B C D E G H F A

(b)(i)

5 marks

Without A there are 7 vertices, so the tree has 6 edges. Prim's algorithm from B takes the cheapest edge joining the tree to a new vertex each time.
The weight of the tree is the sum.

BC 6, CD 5, DE 6, BH 7, GH 4, FG 5 6 + 5 + 6 + 7 + 4 + 5 = 33

(b)(ii)

2 marks

Add back the two cheapest edges at A: AB with weight 3 and AF with weight $x$ , which is smaller than every other edge at A when $x < 9$ .

lower bound = 33 + 3 + x = 36 + x

(c)(i)

1 mark

The shortest route from C to G is C–H–G.

p = 9 + 4 = 13

(c)(ii)

1 mark

The shortest route from D to F that doesn't use AF is D–E–F.

q = 6 + 11 = 17

(c)(iii)

1 mark

The route from D to F through A is D to A (14) and then AF, written as $14 + x$ .

r = 14

(d)(i)

3 marks

ABCDEGHFA : 3 + 6 + 5 + 6 + 8 + 4 + 8 + 3 = 43 AFGHBCDE(F)A : 3 + 5 + 4 + 7 + 6 + 5 + 6 + 14 = 50

(d)(ii)

2 marks

The best upper bound is the shorter of the cycles.

43

(e)(i)

2 marks

x \geq 19

(e)(ii)

2 marks

With the route via B used for the return, the cycle has length

3 + 6 + 5 + 6 + 8 + 4 + 8 + 18 = 58

(f)

5 marks

Corinne must walk every edge, starting and finishing at H. The odd-degree vertices are B, D, G and H. The edges to repeat come from the cheapest pairing.
The cheapest is BD with GH, adding 15 minutes. The total time is 120 minutes.

BH + DG : 7 + 14 = 21, BD + GH : 11 + 4 = 15, BG + DH : 11 + 10 = 21 (92 + x) + 15 = 120 \Rightarrow x = 13

Question 2

30 marks

Water draining from a cylinder: regression, differential equation and two containers

Statistics and probability
Calculus

(a)(i)

2 marks

Regression of $h$ on $t$ using the five readings.

h (t) = - 0.134 t + 3.1

(a)(ii)

1 mark

The parameter $a = - 0.134$ is the rate of change of the water height: the height falls by about 0.134 m each minute.

(a)(iii)

1 mark

Using the equation to predict when $h = 0$ goes beyond the data, which only runs to 20 minutes. The rate of fall may not stay constant as the container empties, so the prediction could be unreliable.

(b)(i)

1 mark

Quadratic regression on the same data.

h (t) = 0.002 t^{2} - 0.174 t + 3.2

(b)(ii)

2 marks

Solve $0.002 t^{2} - 0.174 t + 3.2 = 0$ and take the smaller root, since the quadratic turns back up later.

k \approx 26.4

(b)(iii)

1 mark

The function is a sensible model from the start until the water has run out.

0 \leq t \leq 26.4

(c)

3 marks

The container has radius 1, so $V = π (1)^{2} h = π h$ , and $\frac{d V}{d t} = π \frac{d h}{d t}$ . Equate this with the given rate.

π \frac{d h}{d t} = - π R^{2} 70560 h \Rightarrow \frac{d h}{d t} = - R^{2} 70560 h

(d)

5 marks

Separate the variables.
Write the constant as $C = 70560 c$ , then square both sides.

\int \frac{1}{h} d h = - R^{2} 70560 \int d t \Rightarrow 2 h = - R^{2} 70560 t + C h = \frac{70560}{2} (c - R^{2} t) \Rightarrow h = \frac{70560}{4} (c - R^{2} t)^{2} = 17640 (c - R^{2} t)^{2}

(e)

4 marks

Use $h (0) = 3.2$ to find $c$ .
The container is empty when $h = 0$ , so $c - R^{2} T = 0$ .

3.2 = 17640 c^{2} \Rightarrow c \approx 0.013469 T = \frac{c}{R ^{2}} = \frac{0.013469}{0.02 3 ^{2}} \approx 25.5 minutes

(f)

3 marks

For the water to run out in 15 minutes, $c^{'} = 15 R^{2} = 0.007935$ . The new height is $h (0) = 17640 c^{'2}$ .

h (0) = 17640 (15 \times 0.02 3^{2})^{2} \approx 1.11 m

(g)(i)

4 marks

Take the upper container with $h = 17640 (c - R^{2} t)^{2}$ and $c = 0.013469$ . The height of water in it is falling, so $70560 h = 35280 (c - R^{2} t)$ . Its volume flows out at rate $π R^{2} 70560 h$ into the lower container, which also has base area $π$ . So the height gain in the lower container from this inflow is $R^{2} \cdot 35280 (c - R^{2} t)$ . The lower container also drains at $R^{2} 70560 H$ .
Put in $R^{2} = 0.02 3^{2} = 0.000529$ and $c = 0.013469$ .

\frac{d H}{d t} = 35280 R^{2} (c - R^{2} t) - R^{2} 70560 H 35280 R^{2} c \approx 0.2514, 35280 R^{4} \approx 0.009873, R^{2} 70560 \approx 0.1405 \frac{d H}{d t} \approx 0.2514 - 0.009873 t - 0.1405 H

(g)(ii)

3 marks

H_{m a x} \approx 1.45 m

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Number and algebra carries the most marks on this paper.

Number and algebra25 marks · 45%
Q1
Calculus22 marks · 40%
Q2
Statistics and probability8 marks · 15%
Q2

Mentor insights

Traps to watch for

01
Cases for the unknown weight · Q1
The edge AF with weight $x$ may or may not be used. State the range of $x$ for which each bound applies, and check the tie at $x = 18$ .
02
The lower bound adds two edges back · Q1
After deleting A, the minimum spanning tree plus the two smallest edges at A gives the bound. With $x < 9$ the two smallest are AB (3) and AF ( $x$ ).
03
Linear then quadratic then theory · Q2
The question moves from fitting data to a model from physics. Use the regression only inside the data range, and use the differential equation to extend it.
04
Don't forget the initial condition · Q2
In (d) the general solution has a constant. Use $h (0) = 3.2$ in (e) and the new time in (f).
05
Show-that parts give you the target · Q2
In (g)(i), build $\frac{d H}{d t}$ as the inflow from the upper container minus the outflow from the lower one, both as rates of change of height, since each container has base area $π$ .

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers