Unit 1 Exam Practice
Numbers, Approximation and the GDC
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 marks a short question (like Section A) and Paper 2 marks a longer, structured question. In Applications and Interpretation, a graphic display calculator (GDC) is allowed on both papers. Try each question under exam conditions first, then open the markscheme.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
| (M1), (A1) | Implied mark: it is also awarded if this line is not written but the correct answer is seen |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise. Money is given to 2 decimal places. Never round in the middle of a calculation: use the full GDC value.
SL questions
SL 1 Paper 1
One mole of carbon contains \(6.022 \times 10^{23}\) atoms and has a mass of \(12.01\) g.
(a) Find the mass, in grams, of one carbon atom. Give your answer in the form \(a \times 10^{k}\), where \(1 \leq a < 10\) and \(k \in \mathbb{Z}\). [2]
A small diamond is made of pure carbon and has a mass of \(0.200\) g.
(b) Find the number of carbon atoms in the diamond. Give your answer in the form \(a \times 10^{k}\), where \(1 \leq a < 10\) and \(k \in \mathbb{Z}\). [2]
The diameter of a carbon atom is about \(1.5 \times 10^{-10}\) m. Imagine that all the atoms in the diamond are placed side by side in one straight line.
(c) Find the length of this line in kilometres. Give your answer in standard form. [2]
(d) The mean distance from the Earth to the Moon is \(3.84 \times 10^{5}\) km. Find how many times longer than this distance the line in (c) is. [1]
Total 7 marks
(a)
\(\dfrac{12.01}{6.022 \times 10^{23}}\) M1
\(= 1.99 \times 10^{-23}\) g \(\;(1.99435\ldots \times 10^{-23})\) A1
Do not accept calculator notation such as 1.99E-23.
[2 marks]
(b)
\(\dfrac{0.200}{1.99435\ldots \times 10^{-23}}\) \(\;\) or \(\;\dfrac{0.200}{12.01} \times 6.022 \times 10^{23}\) M1
\(= 1.00 \times 10^{22}\) atoms \(\;(1.00283\ldots \times 10^{22})\) A1
Accept \(1.01 \times 10^{22}\) from using the rounded mass \(1.99 \times 10^{-23}\) g.
[2 marks]
(c)
\(1.00283\ldots \times 10^{22} \times 1.5 \times 10^{-10} = 1.504\ldots \times 10^{12}\) m, then divide by \(1000\) M1
\(= 1.50 \times 10^{9}\) km A1
[2 marks]
(d)
\(\dfrac{1.504\ldots \times 10^{9}}{3.84 \times 10^{5}} = 3920\) times \(\;(3917.3\ldots)\) A1
Accept answers from \(3910\) to \(3930\) from rounded values in (b) or (c).
[1 mark]
See 1.1.
SL 2 Paper 1
A swimming pool is a cuboid. It is \(25\) m long, correct to the nearest metre, and \(12.5\) m wide, correct to 1 decimal place (d.p.). It is filled with water to a depth of \(1.8\) m, correct to 1 d.p. A pump fills the pool at a rate of \(0.45 \text{ m}^3\) per minute, correct to 2 d.p.
(a) Find the lower bound of the volume of water in the full pool. [2]
(b) Find the lower bound of the time needed to fill the pool, in minutes. [2]
The manager says that the pool will certainly be full within \(20\) hours of starting the pump.
(c) Determine whether the manager’s statement is correct. Justify your answer. [3]
Total 7 marks
(a)
\(24.5 \times 12.45 \times 1.75\) M1
\(= 534 \text{ m}^3\) \(\;(533.79375)\) A1
[2 marks]
(b)
\(\dfrac{\text{LB of volume}}{\text{UB of rate}} = \dfrac{533.79375}{0.455}\) M1
\(= 1170\) minutes \(\;(1173.17\ldots)\) A1
[2 marks]
(c)
\(\text{UB of time} = \dfrac{25.5 \times 12.55 \times 1.85}{0.445} = \dfrac{592.04625}{0.445}\) M1
\(= 1330\) minutes \(\;(1330.44\ldots)\), which is \(22.2\) hours A1
the time could be more than \(20\) hours (\(1200\) minutes), so the statement is not certainly correct R1
Award R1 only for a conclusion based on the upper bound of the time.
[3 marks]
See 1.2.
SL 3 Paper 1
The time for one complete swing of a pendulum, \(T\) seconds, is given by \[ T = 2\pi\sqrt{\frac{L}{9.81}} , \] where \(L\) metres is the length of the pendulum. A student measures the length as \(L = 0.64\) m, correct to 2 decimal places.
(a) Calculate the value of \(T\) given by the measured length. [2]
(b) Find the maximum percentage error in your answer to (a). [4]
(c) The student also measures \(T\) with a stopwatch that shows times correct to the nearest \(0.1\) s. Explain why it is better to time \(20\) swings and divide by \(20\) than to time a single swing. [1]
Total 7 marks
(a)
\(T = 2\pi\sqrt{\dfrac{0.64}{9.81}}\) M1
\(= 1.60\) s \(\;(1.60485\ldots)\) A1
[2 marks]
(b)
\(0.635 \leq L < 0.645\) A1
\(T\) at the bounds: \(1.59857\ldots\) and \(1.61111\ldots\) M1
\(\left| \dfrac{1.60485\ldots - 1.59857\ldots}{1.59857\ldots} \right| \times 100\%\) \(\;\) (and the same with \(1.61111\ldots\)) M1
\(0.393\%\) and \(0.388\%\), so the maximum percentage error is \(0.393\%\) A1
Do not award the final A1 if the rounded value \(1.60\) is used for \(v_A\) (this gives \(0.0893\%\) and \(0.690\%\)).
[4 marks]
(c)
the error of up to \(0.05\) s is shared by \(20\) swings, so the error in the time of one swing is at most \(0.0025\) s (the percentage error is \(20\) times smaller) R1
[1 mark]
SL 4 Paper 2
An electricity company charges each household a fixed monthly charge of \(\$F\), plus \(\$d\) for each kilowatt-hour (kWh) used in the day and \(\$n\) for each kWh used at night. The table shows the electricity used in one month by three households, and their bills.
| Household | Day (kWh) | Night (kWh) | Bill (\(\$\)) |
|---|---|---|---|
| A | \(210\) | \(150\) | \(114.30\) |
| B | \(340\) | \(95\) | \(145.45\) |
| C | \(125\) | \(260\) | \(108.00\) |
(a) Write down a system of three linear equations in \(F\), \(d\) and \(n\). [2]
(b) Find the values of \(F\), \(d\) and \(n\). [2]
(c) Interpret the value of \(F\) in this context. [1]
Household D used \(280\) kWh in the day that month, and its bill was \(\$151.90\).
(d) Find the amount of electricity that household D used at night. [2]
A second company charges \(\$0.29\) per kWh at all times, with no fixed charge. A household uses \(u\) kWh in a month: \(60\%\) in the day and \(40\%\) at night.
(e) Show that the household’s monthly bill from the first company is \(\$(18.60 + 0.268u)\). [2]
(f) Find the values of \(u\) for which the first company is cheaper. [3]
(g) State one assumption that you made in (f). [1]
Total 13 marks
(a)
\(F + 210d + 150n = 114.30\) M1
\(F + 340d + 95n = 145.45\) and \(F + 125d + 260n = 108.00\) A1
Award M1 for one correct equation, and A1 for all three equations correct.
[2 marks]
(b)
(using a GDC) \(F = 18.6\), \(\;d = 0.32\), \(\;n = 0.19\) A2
Award A1 for two correct values.
[2 marks]
(c)
every household pays \(\$18.60\) each month, even if it uses no electricity R1
[1 mark]
(d)
\(18.60 + 280(0.32) + 0.19x = 151.90\) (M1)
\(x = 230\) kWh A1
[2 marks]
(e)
\(18.60 + 0.32(0.6u) + 0.19(0.4u)\) M1
\(= 18.60 + 0.192u + 0.076u = 18.60 + 0.268u\) A1 AG
[2 marks]
(f)
\(18.60 + 0.268u < 0.29u\) M1
\(0.022u > 18.60\), so \(u = 845.45\ldots\) at the boundary A1
\(u > 845\) kWh A1
Accept an equation for M1. Award the final A1 only for the correct direction.
[3 marks]
(g)
for example: the prices stay the same; the household always uses exactly \(40\%\) at night R1
[1 mark]
See 1.4.
SL 5 Paper 2
A planter is an open box (no lid) with a square base of side \(x\) cm and a height of \(h\) cm. It is made from \(2400 \text{ cm}^2\) of wood: this is the total area of the base and the four sides. Ignore the thickness of the wood.
(a) Show that \(h = \dfrac{2400 - x^{2}}{4x}\). [2]
(b) Show that the volume of the planter, \(V \text{ cm}^3\), is given by \(V = 600x - \dfrac{x^{3}}{4}\). [2]
(c) Explain why \(0 < x < \sqrt{2400}\). [2]
The planter must hold \(8000 \text{ cm}^3\) of soil.
(d) Find the possible values of \(x\). [4]
(e) The planter is for herbs, which need a depth of at least \(15\) cm. Find the dimensions of the planter. Justify your answer. [2]
Total 12 marks
(a)
area \(= x^{2} + 4xh = 2400\) M1
\(4xh = 2400 - x^{2}\), so \(h = \dfrac{2400 - x^{2}}{4x}\) A1 AG
[2 marks]
(b)
\(V = x^{2}h = x^{2} \times \dfrac{2400 - x^{2}}{4x}\) M1
\(= \dfrac{x(2400 - x^{2})}{4} = 600x - \dfrac{x^{3}}{4}\) A1 AG
[2 marks]
(c)
\(x\) is a length, so \(x > 0\) R1
\(h > 0\) needs \(2400 - x^{2} > 0\), so \(x < \sqrt{2400}\) \(\;(48.98\ldots)\) R1
[2 marks]
(d)
\(600x - \dfrac{x^{3}}{4} = 8000\), or \(x^{3} - 2400x + 32\,000 = 0\) (M1)
(using a GDC) \(x = 14.6\) \(\;(14.6410\ldots)\) A1
\(x = 40\) A1
\(x = -54.6\) is rejected because \(x > 0\) R1
Award R1 only if the negative root is seen and rejected, or the domain is clearly used.
[4 marks]
(e)
\(x = 40\) gives \(h = 5\) cm, and \(x = 14.641\ldots\) gives \(h = 37.3\) cm \(\;(37.3205\ldots)\) A1
only \(h = 37.3 \geq 15\), so the planter is \(14.6\) cm by \(14.6\) cm by \(37.3\) cm R1
[2 marks]
See 1.5.
HL questions
Unit 1 has no AHL content, so the HL questions test the same syllabus content as SL, with richer contexts and longer chains of reasoning.
HL 1 Paper 1
During a storm, \(38\) mm of rain falls on an island. The area of the island is \(7.10 \times 10^{2} \text{ km}^2\).
(a) Find the volume of rain that falls on the island, in \(\text{m}^3\). Give your answer in the form \(a \times 10^{k}\), where \(1 \leq a < 10\) and \(k \in \mathbb{Z}\). [3]
The island has \(6.2 \times 10^{5}\) people. On average, each person uses \(150\) litres of water per day. \(1 \text{ m}^3 = 1000\) litres.
(b) Find the number of days for which the rain from this storm could supply the island’s water. [3]
(c) Explain why your answer to (b) is likely to be an overestimate. [1]
Total 7 marks
(a)
\(38 \text{ mm} = 0.038\) m \(\;(3.8 \times 10^{-2}\) m\()\) M1
\(7.10 \times 10^{2} \text{ km}^2 = 7.10 \times 10^{8} \text{ m}^2\) M1
\(V = 0.038 \times 7.10 \times 10^{8} = 2.70 \times 10^{7} \text{ m}^3\) \(\;(2.698 \times 10^{7})\) A1
Award the second M1 only for \(\times 10^{6}\) (not \(\times 10^{3}\)) for \(\text{km}^2\) to \(\text{m}^2\).
[3 marks]
(b)
rain \(= 2.698 \times 10^{7} \times 1000 = 2.698 \times 10^{10}\) litres M1
daily use \(= 6.2 \times 10^{5} \times 150 = 9.3 \times 10^{7}\) litres M1
\(\dfrac{2.698 \times 10^{10}}{9.3 \times 10^{7}} = 290\) days \(\;(290.10\ldots)\) A1
[3 marks]
(c)
not all the rain can be collected and stored (some evaporates or runs into the sea) R1
[1 mark]
See 1.1.
HL 2 Paper 1
The speed of a car increases from \(u = 12 \text{ m s}^{-1}\) to \(v = 27 \text{ m s}^{-1}\) in a time of \(t = 5.0\) s. The speeds are correct to the nearest \(\text{m s}^{-1}\), and the time is correct to 1 d.p. The average acceleration of the car is \(a = \dfrac{v - u}{t}\).
(a) Calculate \(a\), using the given values. [1]
(b) Find the upper and lower bounds of \(a\). [4]
(c) Hence give the value of \(a\) to an appropriate degree of accuracy. Justify your answer. [2]
Total 7 marks
(a)
\(a = \dfrac{27 - 12}{5.0} = 3 \text{ m s}^{-2}\) A1
[1 mark]
(b)
\(\text{UB} = \dfrac{27.5 - 11.5}{4.95}\) (largest \(v\), smallest \(u\), smallest \(t\)) M1
\(= 3.23 \text{ m s}^{-2}\) \(\;(3.2323\ldots)\) A1
\(\text{LB} = \dfrac{26.5 - 12.5}{5.05}\) (smallest \(v\), largest \(u\), largest \(t\)) M1
\(= 2.77 \text{ m s}^{-2}\) \(\;(2.7722\ldots)\) A1
[4 marks]
(c)
to 2 s.f. the bounds are \(3.2\) and \(2.8\) (different); to 1 s.f. both bounds round to \(3\) R1
\(a = 3 \text{ m s}^{-2}\) (1 s.f.) A1
Do not award A1 without R1.
[2 marks]
See 1.2.
HL 3 Paper 1
A speed camera measures the speed of cars on a road with a speed limit of \(60 \text{ km h}^{-1}\). The speed measured by the camera, \(v_A\), has a percentage error of at most \(3\%\) compared with the true speed \(v_E\).
(a) The camera measures a car’s speed as \(63 \text{ km h}^{-1}\). Determine whether the car was certainly travelling above the speed limit. [3]
(b) Find the greatest measured speed for which the true speed could be \(60 \text{ km h}^{-1}\) or less. [2]
The camera displays the measured speed rounded to the nearest \(\text{km h}^{-1}\).
(c) The display shows \(62 \text{ km h}^{-1}\). Determine whether the car was certainly travelling above the speed limit. Justify your answer. [3]
Total 8 marks
(a)
\(\dfrac{|63 - v_E|}{v_E} \leq 0.03\), and the smallest \(v_E\) satisfies \(63 = 1.03\,v_E\) M1
\(v_E \geq \dfrac{63}{1.03} = 61.2\) \(\;(61.165\ldots)\) A1
\(61.2 > 60\), so the car was certainly above the speed limit R1
[3 marks]
(b)
\(v_A = 1.03 \times 60\) M1
\(= 61.8 \text{ km h}^{-1}\) A1
[2 marks]
(c)
the measured speed is at least \(61.5\) (lower bound of \(62\) to the nearest whole number) M1
\(v_E \geq \dfrac{61.5}{1.03} = 59.7\) \(\;(59.708\ldots)\) A1
\(59.7 < 60\), so the car was not certainly above the speed limit R1
Accept the argument that \(61.5 < 61.8\) from (b), with a conclusion.
[3 marks]
HL 4 Paper 2
A drug is given by injection. The concentration of the drug in a patient’s blood, \(C\) mg per litre, \(t\) hours after the injection, is modelled by \[ C(t) = at^{3} + bt^{2} + ct, \qquad t \geq 0 . \] The table shows three measurements.
| \(t\) (hours) | \(1\) | \(2\) | \(4\) |
|---|---|---|---|
| \(C\) (mg per litre) | \(6.3\) | \(8.7\) | \(5.4\) |
(a) Explain why it is sensible that the model has no constant term. [1]
(b) Write down a system of three linear equations in \(a\), \(b\) and \(c\). [2]
(c) Find the values of \(a\), \(b\) and \(c\). [2]
(d) Find the positive values of \(t\) for which \(C(t) = 0\). [3]
(e) Hence explain why the model can only be used for \(0 \leq t \leq 5.35\). [2]
The drug is effective when its concentration is at least \(4\) mg per litre.
(f) Find the length of time for which the drug is effective. [4]
(g) A nurse measures the concentration \(3\) hours after the injection as \(7.0\) mg per litre. Find the percentage error of the model’s value, taking the nurse’s measurement as the exact value. [2]
Total 16 marks
(a)
\(C(0)\) must be \(0\): there is no drug in the blood at the moment of the injection R1
[1 mark]
(b)
\(a + b + c = 6.3\) M1
\(8a + 4b + 2c = 8.7\) and \(64a + 16b + 4c = 5.4\) A1
Award M1 for one correct equation, and A1 for all three equations correct.
[2 marks]
(c)
(using a GDC) \(a = 0.15\), \(\;b = -2.4\), \(\;c = 8.55\) A2
Award A1 for two correct values.
[2 marks]
(d)
\(0.15t^{3} - 2.4t^{2} + 8.55t = 0\) (using a GDC) (M1)
\(t = 5.35\) \(\;(5.35424\ldots)\) A1
\(t = 10.6\) \(\;(10.6457\ldots)\) A1
[3 marks]
(e)
for \(5.35 < t < 10.6\) the model gives \(C < 0\), but a concentration cannot be negative R1
for \(t > 10.6\) the model gives a concentration that increases again, with no new injection R1
Accept any two valid reasons based on the graph or the zeros of \(C\).
[2 marks]
(f)
\(0.15t^{3} - 2.4t^{2} + 8.55t = 4\) (for example, the intersection of \(y = C(t)\) and \(y = 4\) on a GDC) M1
\(t = 0.550\) \(\;(0.549759\ldots)\) A1
\(t = 4.38\) \(\;(4.38275\ldots)\) A1
\(4.38275\ldots - 0.549759\ldots = 3.83\) hours \(\;(3.83299\ldots)\) A1
The third root \(t = 11.1\) is outside the domain in (e), so it is not used.
[4 marks]
(g)
\(C(3) = 8.1\), so \(\left| \dfrac{8.1 - 7.0}{7.0} \right| \times 100\%\) M1
\(= 15.7\%\) \(\;(15.714\ldots)\) A1
[2 marks]
HL 5 Paper 2
Ceres is the largest object in the asteroid belt. In this question Ceres is modelled as a sphere. Its mass is \(9.4 \times 10^{20}\) kg and its radius is \(470\) km, both correct to 2 significant figures. The volume of a sphere is \(V = \dfrac{4}{3}\pi r^{3}\), and density \(= \dfrac{\text{mass}}{\text{volume}}\).
(a) Show that the volume of Ceres, using the given radius, is \(4.35 \times 10^{17} \text{ m}^3\), correct to 3 s.f. [2]
(b) Find the density of Ceres in \(\text{kg m}^{-3}\), using the given values. [2]
(c) Write down the lower and upper bounds of the mass, in kg, and of the radius, in m. [2]
(d) Find the upper and lower bounds of the density of Ceres. [4]
(e) Find the maximum percentage error in your answer to (b). [3]
In a simple model, Ceres is made of rock, with density \(3000 \text{ kg m}^{-3}\), and ice, with density \(920 \text{ kg m}^{-3}\). If \(p\) is the fraction of the volume of Ceres that is rock, then the density of Ceres is \(\rho = 920(1 - p) + 3000p\).
(f) A scientist claims that more than half of the volume of Ceres is rock. Determine whether the data are accurate enough to support this claim. [3]
Total 16 marks
(a)
\(r = 470 \text{ km} = 4.7 \times 10^{5}\) m, so \(V = \dfrac{4}{3}\pi\left(4.7 \times 10^{5}\right)^{3}\) M1
\(= 4.3489\ldots \times 10^{17} = 4.35 \times 10^{17} \text{ m}^3\) A1 AG
[2 marks]
(b)
\(\dfrac{9.4 \times 10^{20}}{4.3489\ldots \times 10^{17}}\) M1
\(= 2160 \text{ kg m}^{-3}\) \(\;(2161.45\ldots)\) A1
[2 marks]
(c)
\(9.35 \times 10^{20} \leq m < 9.45 \times 10^{20}\) A1
\(4.65 \times 10^{5} \leq r < 4.75 \times 10^{5}\) A1
[2 marks]
(d)
\(\text{UB} = \dfrac{9.45 \times 10^{20}}{\frac{4}{3}\pi\left(4.65 \times 10^{5}\right)^{3}}\) (largest mass, smallest radius) M1
\(= 2240 \text{ kg m}^{-3}\) \(\;(2243.80\ldots)\) A1
\(\text{LB} = \dfrac{9.35 \times 10^{20}}{\frac{4}{3}\pi\left(4.75 \times 10^{5}\right)^{3}}\) (smallest mass, largest radius) M1
\(= 2080 \text{ kg m}^{-3}\) \(\;(2082.77\ldots)\) A1
[4 marks]
(e)
\(\left| \dfrac{2161.45\ldots - 2082.77\ldots}{2082.77\ldots} \right| \times 100\%\) \(\;\) (or the same with \(2243.80\ldots\)) M1
\(3.78\%\) \(\;(3.7775\ldots)\) and \(3.67\%\) \(\;(3.6700\ldots)\) A1
maximum percentage error \(= 3.78\%\) A1
[3 marks]
(f)
\(p = \dfrac{\rho - 920}{2080}\), and the smallest \(p\) comes from the lower bound of \(\rho\) M1
\(p \geq \dfrac{2082.77\ldots - 920}{2080} = 0.559\) \(\;(0.55902\ldots)\) A1
even the lower bound of \(p\) is greater than \(0.5\), so the data support the claim (if the model is correct) R1
The upper bound of \(p\) is \(0.636\); it is not needed for the conclusion.
[3 marks]