Mathematics questions, chapter 8

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Mathematical Modeling

44 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working for Section B.

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Question 1

What is a mathematical model?

A table of answers to textbook questions
A mathematical relation that describes a real-world situation
A diagram that is drawn to scale
A formula that is always correct for every situation

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A model represents a real-world situation by a mathematical relation, such as I = Prt. A model is a simplification, so it is not always correct for every situation.

For the teacher: Remembering, low difficulty, standard 8.1.1

Textbook p. 228

Question 2

Which of the following is not a component of the mathematical modeling process?

Making assumptions and identifying the variables
Refining the mathematical model
Memorising the answer from a similar problem
Reporting the findings

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The six components are: identify and define the problem; make assumptions and identify variables; apply mathematics; verify and interpret; refine the model; report the findings.

For the teacher: Remembering, low difficulty, standard 8.1.1

Textbook p. 229

Question 3

Which question is an open-ended problem that needs mathematical modeling?

How long does it take to clean up an oil spill in the ocean?
Calculate 45% of 40 000.
Find the area of a rectangle of 8 cm by 5 cm.
Solve 3x + 5 = 20.

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The oil spill question does not give enough information; we must make assumptions and find data. The other three give all the information and have one correct answer.

For the teacher: Understanding, low difficulty, standard 8.1.1

Textbook p. 229

Question 4

Diagram 1 shows the daily profit of a drinks stall at different selling prices. Which function is a suitable model for the data?

Points rising to a maximum near a price of RM5 and then falling, lying on a parabola — Diagram 1

Linear function
Exponential function
Direct variation y = kx
Quadratic function

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The profit rises and then falls in the shape of a parabola, which is the pattern of a quadratic function. A linear graph is a straight line; an exponential graph keeps rising or falling at a sharp rate.

For the teacher: Understanding, low difficulty, standard 8.1.2

Textbook p. 236

Question 5

The number of bacteria in a dish is 3, 6, 12, 24 and 48 at the end of hours 0, 1, 2, 3 and 4. Which model fits best?

Linear: y = 3 + 3x
Exponential: y = 3(2)^x
Quadratic: y = 3x²
Direct variation: y = 3x

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The number doubles every hour, so it is multiplied by the same factor 2 each time: y = 3(2)^x. Linear growth would add the same amount each hour.

For the teacher: Understanding, low difficulty, standard 8.1.2

Textbook p. 236

Question 6

A car uses 45 litres of petrol for 405 km. Assuming the petrol used varies directly with the distance, how many litres are needed for 270 km?

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y = kx, so k = 45 ÷ 405 = 19. For x = 270, y = 19 × 270 = 30 litres.

For the teacher: Applying, low difficulty, standard 8.1.2

Textbook p. 236

Question 7

Mariam saves RM5 000 at 6% per annum, compounded once a year. How much will she have after 2 years?

RM5 618.00
RM5 600.00
RM5 636.00
RM6 000.00

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A = 5 000(1.06)² = 5 000 × 1.1236 = RM5 618.00. RM5 600 is the simple interest answer, which gives no interest on interest.

For the teacher: Applying, low difficulty, standard 8.1.2

Textbook p. 240

Question 8

RM10 000 is saved for 1 year at 6% per annum, compounded semi-annually. The amount at the end of the year is

RM10 600.00
RM10 300.00
RM10 060.00
RM10 609.00

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n = 2, rate per period = 3%, periods = 2: A = 10 000(1.03)² = RM10 609.00. RM10 600 is compounding once a year.

For the teacher: Applying, medium difficulty, standard 8.1.2

Textbook p. 240

Question 9

A shop 12 km farther away sells a pack of rice for RM16.50 instead of RM18.90. Rani buys 3 packs. Her car uses 1 litre per 15 km and petrol costs RM2.00 per litre. Using S = (P₁ − P₂)n and T = sm × H × 2, what is her net saving by driving to the farther shop?

RM7.20
RM6.40
RM4.00
RM3.20

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S = (18.90 − 16.50) × 3 = RM7.20. T = 1215 × 2.00 × 2 = RM3.20. Net saving = 7.20 − 3.20 = RM4.00. RM7.20 ignores the petrol; RM3.20 is the petrol only.

For the teacher: Analysing, high difficulty, standard 8.1.2

Textbook p. 232

Question 10

In the petrol model y = 19x, the 405 km route is through towns and the 198 km route is on a highway. Why might the model be unsuitable?

The petrol is used at different rates on the two routes, so the constant k is not the same
The distance 198 km is too short for any model
Petrol cannot be measured in litres
A linear model is never suitable for petrol

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Direct variation assumes one constant rate. Town driving uses more petrol per km than highway driving, so the assumption fails when the solution is verified in the real situation.

For the teacher: Analysing, medium difficulty, standard 8.1.2

Textbook p. 237

Question 11

An amount is invested at 8% per annum compounded quarterly for t years. Which model gives the amount, A(t)?

A(t) = P(1.08)^t
A(t) = P(1.02)^t
A(t) = P(1.08)^4t
A(t) = P(1.02)^4t

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The rate per period is 8% ÷ 4 = 2% and there are 4t periods, so A(t) = P(1 + 0.02)^4t.

For the teacher: Applying, medium difficulty, standard 8.1.2

Textbook p. 240

Question 12

The model y = 5(3)^x is written in the form y = Ca^x. What does C = 5 represent?

The growth factor per unit of x
The initial value, when x = 0
The value of y when x = 1
The gradient of the graph

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When x = 0, y = 5(3)⁰ = 5. C is the initial value; a = 3 is the factor by which y is multiplied for each increase of 1 in x.

For the teacher: Understanding, low difficulty, standard 8.1.2

Textbook p. 236

Section B: structured questions

32 marks

Answer all questions. Show your working clearly.

Question 13

Diagram 2 shows three graphs, A, B and C, that are drawn for three sets of data.

Graph A is a straight line, graph B is a parabola rising then falling, graph C is a curve rising slowly then very steeply — Diagram 2

[3 marks] (a)State the type of function that models each graph: A, B and C.

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A is linear [1]1
B is quadratic [1]1
C is exponential [1]1

Tick each point you got.

[1 mark] (b)The function for graph C is written as y = Ca^x. State what C represents.

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The initial value (value of y when x = 0) [1]1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 8.1.2

Textbook p. 236

Question 14

A canal has a width of 20 m. The depth, y m, is measured at distance x m from one bank: y = 0 at x = 0, y = 2.4 at x = 10 and y = 0 at x = 20. The data suggest a quadratic model y = ax² + bx + c.

[1 mark] (a)State the value of c. Give a reason.

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c = 0, because y = 0 when x = 0 [1]1

Tick each point you got.

[3 marks] (b)Find the values of a and b.

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2.4 = 100a + 10b and 0 = 400a + 20b [1]1
b = −20a, so 2.4 = 100a − 200a [1]1
a = −0.024, b = 0.48 [1]1

Tick each point you got.

[1 mark] (c)Use your model to find the depth of the canal 5 m from the bank.

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y = −0.024(25) + 0.48(5) = 1.8 m [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 8.1.2

Textbook p. 238

Question 15

Encik Hamid saves RM15 000 in an account with an interest rate of 5% per annum. The interest is compounded once a year.

[1 mark] (a)Write a model for his savings, A(t), after t years.

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A(t) = 15 000(1.05)^t [1]1

Tick each point you got.

[2 marks] (b)Calculate his savings after 3 years.

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15 000(1.05)³ [1]1
RM17 364.38 [1]1

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[2 marks] (c)How much more interest does he earn in 3 years than from simple interest at the same rate?

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Compound interest 17 364.38 − 15 000 = 2 364.38; simple interest 15 000 × 0.05 × 3 = 2 250 [1]1
Difference = RM114.38 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 8.1.2

Textbook p. 240

Question 16

Lina saves RM8 000 at 6% per annum, compounded quarterly, for 2 years.

[2 marks] (a)State the interest rate per period and the number of periods.

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Rate per period = 6% ÷ 4 = 1.5% [1]1
Number of periods = 4 × 2 = 8 [1]1

Tick each point you got.

[2 marks] (b)Calculate the total savings at the end of 2 years.

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8 000(1.015)⁸ [1]1
RM9 011.94 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 8.1.2

Textbook p. 240

Question 17

A market 9 km farther away sells chicken at RM11.80 per pack instead of RM13.50 at the nearby shop. Kamal buys 4 packs. His car travels 12 km per litre and petrol costs RM2.05 per litre. The model is S = (P₁ − P₂)n and T = sm × H × 2.

[1 mark] (a)Calculate S, the saving on the chicken.

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(13.50 − 11.80) × 4 = RM6.80 [1]1

Tick each point you got.

[2 marks] (b)Calculate T, the extra petrol cost.

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912 × 2.05 × 2 [1]1
RM3.08 (3.075) [1]1

Tick each point you got.

[1 mark] (c)Is it worth driving to the market? Give a reason.

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Yes, because S > T (6.80 > 3.08) [1]1

Tick each point you got.

[2 marks] (d)Find the greatest extra distance s for which the trip is still worth it.

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6.80 > s12 × 2.05 × 2 [1]1
s < 19.9 km [1]1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 8.1.2

Textbook p. 232

Question 18

A hot sample at 200°C is left to cool. The temperature is 200°C at t = 0 s, 140°C at 40 s, 100°C at 80 s and 75°C at 120 s. The room temperature is 25°C.

[2 marks] (a)Explain why a linear model is not suitable.

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The drops in temperature are 60, 40 and 25: not equal [1]1
A linear model drops by the same amount in equal times, and would go below room temperature [1]1

Tick each point you got.

[2 marks] (b)Which is a better model: quadratic or exponential? Explain.

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Exponential [1]1
The temperature decreases quickly at first, then more slowly towards a limit; a quadratic curve would turn and rise again [1]1

Tick each point you got.

For the teacher: Evaluating, medium difficulty, standard 8.1.2

Textbook p. 242

Question 19

A runner trains 32 km per week. He plans to increase his weekly distance by 10% each week. Let D(x) km be the weekly distance after x weeks of increases.

[2 marks] (a)Derive a mathematical model for D(x).

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Each week is multiplied by 1.10 [1]1
D(x) = 32(1.10)^x [1]1

Tick each point you got.

[1 mark] (b)Calculate his weekly distance after 4 weeks of increases. Give your answer to 2 decimal places.

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32(1.1)⁴ = 46.85 km [1]1

Tick each point you got.

[1 mark] (c)State one weakness of this model.

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Any one: a runner cannot keep increasing by 10% each week for ever; the model ignores rest, injury or illness [1]1

Tick each point you got.

For the teacher: Creating, medium difficulty, standard 8.1.2

Textbook p. 241

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