Free worked solutions · 2026 Edition

IB Maths AI SL Chapter 1: Number: standard form, logarithms and error worked solutions

Worked solutions to the practice questions in Chapter 1 of Photon Guidebook: Mathematics Applications and Interpretation SL for the IB Diploma (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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  • 24 practice questions
  • 1 Going further
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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[2 marks]
Omar maintains a habitat database that records 5 830 000 leaf scans.
Express this count in standard form.
[2]
Show solutionHide solution
Move the decimal six places to obtain a coefficient between 1 and 10.
M1
5.83\times10^6.
A1

Final answer

5.83\times10^6 scans.

Question 2

[3 marks]
Leila's scanner transfers 7.56\times10^8 bytes in 2.7\times10^2 seconds at a constant rate.
Find its transfer rate in standard form.
[3]
Show solutionHide solution
Rate =(7.56\times10^8)/(2.7\times10^2).
M1
7.56/2.7=2.8 and 10^8/10^2=10^6.
M1
2.8\times10^6 bytes per second.
A1

Final answer

2.8\times10^6 bytes per second.

Question 3

[2 marks]
Ben's timer records 0.009964 seconds.
Round to three significant figures.
[2]
Show solutionHide solution
0.00996 s.
A1
The fourth significant digit is 4, so the third stays unchanged.
R1

Final answer

0.00996 s.

Question 4

[2 marks]
A dose prepared by Mei is 8.24 ml to two decimal places. Halfway values round upward.
Find the lower and upper bounds of the actual dose.
[2]
Show solutionHide solution
Half-unit =0.005 ml.
M1
8.235\le V<8.245 ml.
A1

Final answer

8.235\le V<8.245 mL.

Question 5

[3 marks]
A sheet used by Alex is reported as 28 cm long and 17 cm wide, each to the nearest centimetre. Halfway values round upward.
Find bounds for its perimeter.
[3]
Show solutionHide solution
27.5\le L<28.5, 16.5\le W<17.5.
A1
P=2(L+W).
M1
88\le P<92 cm.
A1

Final answer

88\le P<92 cm.

Question 6

[3 marks]
Sara's reference volume is exactly 375 ml. An approximate calculation gives 390 ml. Percentage error is 100 times absolute error divided by the exact value.
Find the percentage error.
[3]
Show solutionHide solution
Absolute error =|390-375|=15 ml.
A1
100(15/375).
M1
4%.
A1

Final answer

4%.

Question 7

[2 marks]
For a nonzero base, b^{-n}=1/b^n. Ravi's filter ratio is 5^{-3}.
Write the ratio as a fraction and a decimal.
[2]
Show solutionHide solution
1/125.
A1
0.008.
A1

Final answer

1/125=0.008.

Question 8

[3 marks]
Nora's device assigns the number L=\log_{10}275 to a positive measurement.
Use a GDC to find L to 3 significant figures and check that it lies between 2 and 3.
[3]
Show solutionHide solution
Use the base-ten logarithm key.
M1
L=2.44.
A1
10^2<275<10^3, so 2<L<3.
R1

Final answer

2.44; 100<275<1000.

Question 9

[5 marks]
Emil's calibration formula is D=8+5\log_{10}x for positive dimensionless x.
(a) Find D when x=1000.
[2]
(b) Find x when D=18.
[3]
Show solutionHide solution
(a) \log_{10}1000=3.
A1
D=8+5(3)=23.
A1
(b) 18=8+5\log_{10}x gives \log_{10}x=2.
M1
x=10^2.
M1
x=100.
A1

Final answer

  • (a) 23
  • (b) 100.

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 10

[5 marks]
Jules reads a town population of 48 000, rounded to the nearest thousand.
(a) State both population bounds.
[2]
(b) A river length L=1200 km is rounded to 2 significant figures. A student gives 1100\le L<1300. Explain the error.
[1]
(c) State the correct interval for L.
[2]
Show solutionHide solution
(a) Rounding unit 1000, half-unit 500: lower bound 47 500, upper bound 48 500. A1A1
[2]
(b) To 2 s.f. the rounding unit is 100, and the bounds are half a unit (not a whole unit) either side; the student used \pm100 instead of \pm50.
R1[1]
(c) 1150\le L<1250 A1A1
[2]

Final answer

  • (a) 47 500, 48 500
  • (b) half-unit is 50, not 100
  • (c) 1150\le L<1250 km.

Question 11

[6 marks]
Siti's fair has 26 stalls, each receiving 53 visitors per hour for 5 hours. Each visitor spends an average S$2.80 at a stall.
(a) Round every input to 1 significant figure and estimate total spending.
[3]
(b) Calculate the exact total to the nearest Singapore dollar.
[1]
(c) Find the estimate's percentage error.
[2]
Show solutionHide solution
(a) 26\to30, 53\to50, 5\to5, 2.80\to3
M1[1]
Estimate =30\times50\times5\times3
M1[1]
=\text{S\$}22\,500
A1[1]
(b) 26\times53\times5\times2.80=\text{S\$}19\,292
A1[1]
(c) \dfrac{|22\,500-19\,292|}{19\,292}\times100
M1[1]
=16.6\%
A1[1]

Final answer

  • (a) S$22 500
  • (b) S$19 292
  • (c) 16.6%.

Question 12

[7 marks]
Asha checks a stadium's actual capacity of 56 238 seats. Its website reports 56 000.
(a) Find the absolute error.
[2]
(b) Find the percentage error.
[2]
(c) How many significant figures did the website retain?
[1]
(d) Another website rounds to 1 significant figure. Find that figure's percentage error.
[2]
Show solutionHide solution
(a) \varepsilon=|56\,000-56\,238|
M1[1]
=238
A1[1]
(b) \dfrac{238}{56\,238}\times100
M1[1]
=0.4232\ldots=0.423\%
A1[1]
(c) 56 000 is 56 238 rounded to 2 significant figures.
A1[1]
(d) To 1 s.f. the figure is 60 000: \dfrac{|60\,000-56\,238|}{56\,238}\times100
M1[1]
=6.69\%
A1[1]

Final answer

  • (a) 238 seats
  • (b) 0.423%
  • (c) 2 s.f.
  • (d) 6.69%.

Question 13

[7 marks]
Ivan's cylindrical tin has radius 3.4 cm and height 9.8 cm, both rounded to 1 decimal place; V=\pi r^2h.
(a) State both measurement intervals.
[2]
(b) Find the lower and upper volume bounds.
[3]
(c) Is the label's claim “at least 340 mL” certainly true? Justify your decision. Use 1\text{ cm}^3=1 mL.
[2]
Show solutionHide solution
(a) 3.35\le r<3.45
A1[1]
9.75\le h<9.85
A1[1]
(b) V increases with both r and h, so use the smallest values for the lower bound and the largest for the upper bound.
M1[1]
Lower bound =\pi(3.35)^2(9.75)=343.75\ldots=344 cm^3
A1[1]
Upper bound =\pi(3.45)^2(9.85)=368.32\ldots=368 cm^3
A1[1]
(c) The least possible volume is 343.75 mL, which is greater than 340 mL.
M1[1]
So every possible volume exceeds 340 mL and the statement is certainly true.
R1[1]

Final answer

  • (a) 3.35\le r<3.45, 9.75\le h<9.85 cm
  • (b) 344, 368 cm^3
  • (c) certainly true: lower bound exceeds 340 mL.

Question 14

[8 marks]
Lena models registered app users by U=250e^{kt}, with t in hours since launch. After 6 hours there are 4100 users.
(a) Find k.
[3]
(b) Using unrounded k, predict users after 10 hours.
[2]
(c) When does the model reach one million users?
[2]
(d) Give a reason the model fails for large t.
[1]
Show solutionHide solution
(a) 250\mathrm{e}^{6k}=4100.
M1[1]
Solve with the GDC (graph y=250\mathrm{e}^{6x} and y=4100, or numeric solve) (M1)
[1]
to obtain k=0.466213\ldots=\boxed{0.466}.
A1[1]
(b) U=250\mathrm{e}^{10k} with k stored in the GDC
M1[1]
=\boxed{26\,500} registered users.
A1[1]
(c) Graph y=250\mathrm{e}^{kx} and y=1\,000\,000.
M1[1]
The intersection is at x=\boxed{17.8} hours.
A1[1]
(d) The number of users cannot grow without limit: it is bounded by the number of people who could ever use the app.
R1[1]

Final answer

  • (a) k=0.466
  • (b) 26 500 users
  • (c) 17.8 h
  • (d) finite potential users.

Question 15

[7 marks]
Nurse Fatima removes a vaccine vial from a cold box at t=0. In a room at 26\,{}^{\circ}C, its temperature after t minutes is T=26-24e^{-0.07t}, for t\ge0, in degrees Celsius.
(a) State its cold-box temperature.
[1]
(b) Find T(10).
[1]
(c) When does it reach 10\,{}^{\circ}C?
[2]
(d) It must be returned before exceeding 8\,{}^{\circ}C. When is that temperature reached?
[2]
(e) State its long-term model temperature.
[1]
Show solutionHide solution
(a) T(0)=26-24=\boxed{2\,{}^{\circ}\mathrm{C}}.
A1[1]
(b) T=26-24\mathrm{e}^{-0.7}=14.08\ldots=\boxed{14.1\,{}^{\circ}\mathrm{C}}.
A1[1]
(c) Graph y=26-24\mathrm{e}^{-0.07x} and y=10.
M1[1]
The intersection is at x=5.79\ldots=\boxed{5.79} minutes.
A1[1]
(d) Graph y=26-24\mathrm{e}^{-0.07x} and y=8.
M1[1]
The intersection is at x=4.109\ldots=\boxed{4.11} minutes.
A1[1]
(e) As t\to\infty, \mathrm{e}^{-0.07t}\to 0, so T\to\boxed{26\,{}^{\circ}\mathrm{C}} (room temperature).
A1[1]

Final answer

  • (a) 2\,{}^{\circ}C
  • (b) 14.1\,{}^{\circ}C
  • (c) 5.79 min
  • (d) 4.11 min
  • (e) 26\,{}^{\circ}C.

Question 16

[6 marks]
Technician Joel makes a buffer with \mathrm{pH}=4.76+\log_{10}([\mathrm B]/[\mathrm A]). Acid and salt concentrations [\mathrm A] and [\mathrm B] are in mol L^{-1}.
(a) Find pH for [\mathrm A]=0.20, [\mathrm B]=0.35.
[2]
(b) For pH 5.10 and [\mathrm A]=0.15, find [\mathrm B].
[2]
(c) Find [\mathrm B]/[\mathrm A] at pH 4.00 and 5.00. How many times larger is the second ratio?
[2]
Show solutionHide solution
(a) \mathrm{pH}=4.76+\log\!\left(\dfrac{0.35}{0.20}\right)
M1[1]
=5.003\ldots=\boxed{5.00}.
A1[1]
(b) 5.10=4.76+\log\!\left(\dfrac{[\mathrm{B}]}{0.15}\right), so \log\!\left(\dfrac{[\mathrm{B}]}{0.15}\right)=0.34 and \dfrac{[\mathrm{B}]}{0.15}=10^{0.34}
M1[1]
[\mathrm{B}]=0.15\times 10^{0.34}=0.3281\ldots=\boxed{0.328} moles per litre.
A1[1]
(c) pH 4.00: \log(\text{ratio})=-0.76, so the ratio is 10^{-0.76}=0.1738\ldots (0.174). pH 5.00: \log(\text{ratio})=0.24, so the ratio is 10^{0.24}=1.738\ldots (1.74).
A1[1]
The second ratio is \dfrac{1.738}{0.1738}=\boxed{10} times as large (a pH difference of 1 is a factor of 10).
A1[1]

Final answer

  • (a) 5.00
  • (b) 0.328 mol L^{-1}
  • (c) 0.174, 1.74; 10 times.

Question 17

[8 marks]
Esi tests sanitisers using \mathrm{LRV}=\log_{10}(N_0/N), with bacteria counts N_0 before and N after treatment.
(a) Find LRV for N_0=4.5\times10^6, N=1.2\times10^3.
[2]
(b) For LRV 4.6 and N_0=8.0\times10^7, find N to the nearest whole number.
[2]
(c) A and B have LRVs 2.6 and 4.1. With equal starting counts, how many times more bacteria remain after A?
[2]
(d) Approval needs LRV at least 3.8. For N_0=6.5\times10^6, find the greatest allowable whole-number N.
[2]
Show solutionHide solution
(a) \mathrm{LRV}=\log\!\left(\dfrac{4.5\times 10^{6}}{1.2\times 10^{3}}\right)=\log 3750
M1[1]
=3.574\ldots=\boxed{3.57}.
A1[1]
(b) 4.6=\log\!\left(\dfrac{8.0\times 10^{7}}{N}\right), so \dfrac{8.0\times 10^{7}}{N}=10^{4.6}
M1[1]
N=\dfrac{8.0\times 10^{7}}{10^{4.6}}=2009.5\ldots=\boxed{2010} bacteria.
A1[1]
(c) With the same starting number N_{0}: N_{A}=N_{0}\times 10^{-2.6} and N_{B}=N_{0}\times 10^{-4.1}.
M1[1]
Evaluating each on the GDC: 10^{-2.6}=2.512\times 10^{-3} and 10^{-4.1}=7.943\times 10^{-5}, so \dfrac{N_{A}}{N_{B}}=\dfrac{2.512\times 10^{-3}}{7.943\times 10^{-5}}=31.62\ldots=\boxed{31.6} times as many bacteria remain after using A.
A1[1]
(d) \log\!\left(\dfrac{6.5\times 10^{6}}{N}\right)\ge 3.8, so N\le\dfrac{6.5\times 10^{6}}{10^{3.8}}=1030.18\ldots
M1[1]
The greatest whole number is \boxed{1030} bacteria.
A1[1]

Final answer

  • (a) 3.57
  • (b) 2010 bacteria
  • (c) 31.6 times
  • (d) 1030 bacteria.

Question 18

[6 marks]
Mara's fishing boat has a fuel tank holding 3500 US gallons, correct to the nearest 100 gallons. One US gallon is 3.79 litres, correct to 3 significant figures.
(a) Write down the lower and upper bounds of the number of litres in one US gallon.
[2]
(b) Find the smallest possible capacity of the tank, in litres.
[2]
(c) A pump empties the tank at exactly 250 litres per minute. Find the time, in whole minutes, that is certain to be enough to empty a full tank.
[2]
Show solutionHide solution
(a) 3.785\le f<3.795 litres per US gallon (A1 A1). (b) Both lower bounds: 3450\times3.785=13\,058.25 litres (M1), 13 100 litres to 3 s.f. (A1). (c) Upper capacity bound 3550\times3.795=13\,472.25 litres (M1). This needs less than 13\,472.25/250=53.889 minutes, so 54 whole minutes is certain to be enough (A1).

Final answer

  • (a) 3.785\le f<3.795 L per US gallon
  • (b) 13 100 L
  • (c) 54 min.

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 19

[8 marks]
Rosa reads a city council's report of its annual population as 148 000, correct to the nearest thousand. The reported land area of the city is 62\text{ km}^{2}, correct to the nearest \text{km}^{2}. The population density is D=\dfrac{\text{population}}{\text{area}}, in people per \text{km}^{2}.
(a) Write down the lower and upper bounds of
(i) the population;
[1]
(ii) the land area.
[1]
(b) Find the lower bound of the population density, correct to 3 significant figures.
[3]
(c) Find the upper bound of the population density, correct to 3 significant figures.
[3]
Show solutionHide solution
(a) (i) 147\,500\le P<148\,500
[1]
(ii) 61.5\le A<62.5\text{ km}^{2}
[1]
(b) Least density uses least population and greatest area: D_{\min}=\dfrac{147\,500}{62.5}=2360\text{ people km}^{-2} (3 s.f.)
[3]
(c) Greatest density uses greatest population and least area: D_{\max}=\dfrac{148\,500}{61.5}=2410\text{ people km}^{-2} (3 s.f.)
[3]

Final answer

  • (a) 147\,500\le P<148\,500, 61.5\le A<62.5 km^2
  • (b) 2360
  • (c) 2410 people km^{-2}.

Question 20

[11 marks]
Designer Yusuf models a decorative pyramid-shaped ornament as a cuboid box for shipping. The box is measured as length 5.2 cm, width 3.8 cm and height 2.1 cm, each correct to one decimal place.
(a) Write down the lower and upper bounds of the height.
[2]
(b) Calculate the lower bound of the volume of the box.
[3]
(c) Calculate the upper bound of the volume of the box.
[3]
(d) The nominal (rounded) volume is V=5.2\times 3.8\times 2.1. Find the percentage error if the true volume is actually equal to the upper bound found in part (c). Give your answer correct to 3 significant figures.
[3]
Show solutionHide solution
(a) 2.05\le h<2.15 cm
[2]
(b) Lower bounds throughout: V_{\min}=5.15\times 3.75\times 2.05=39.6\text{ cm}^{3} (3 s.f.)
[3]
(c) Upper bounds throughout: V_{\max}=5.25\times 3.85\times 2.15=43.5\text{ cm}^{3} (3 s.f.)
[3]
(d) Nominal V=5.2\times 3.8\times 2.1=41.496\text{ cm}^{3}; true =43.4569\text{ cm}^{3}. \varepsilon=\dfrac{|41.496-43.4569|}{43.4569}\times 100\approx 4.51\%
[3]

Final answer

  • (a) 2.05\le h<2.15 cm
  • (b) 39.6 cm^3
  • (c) 43.5 cm^3
  • (d) 4.51%.

Question 21

[9 marks]
Mina models a colony of bacteria in a lab dish by P(t) = 400 \times (1.15)^{t}, where P is the number of bacteria and t is the time in hours.
(a) State the initial number of bacteria.
[1]
(b) Find the percentage growth rate per hour.
[1]
(c) Find the number of bacteria after 8 hours, giving your answer to the nearest whole number.
[2]
(d) Use your GDC to find, to the nearest hour, the time taken for the population to first exceed 2000.
[3]
(e) Give one reason why this model may not be realistic for large values of t.
[2]
Show solutionHide solution
(a) P(0) = 400 \times (1.15)^{0} = 400 bacteria.
[1]
(b) 1.15 = 1 + 0.15, so the growth rate is 15% per hour.
[1]
(c) P(8) = 400 \times (1.15)^{8} = 400 \times 3.0590 = 1223.6 \approx 1224 bacteria.
[2]
(d) Solve 400 \times (1.15)^{t} = 2000, i.e. (1.15)^{t} = 5. Using a GDC, t = 11.5 hours, so the population first exceeds 2000 after 12 hours.
[3]
(e) Exponential growth is unbounded, but real space, food and nutrients are limited, so the population cannot keep growing at 15% per hour indefinitely.
[2]

Final answer

  • (a) 400
  • (b) 15%
  • (c) 1224 bacteria
  • (d) 12 h
  • (e) limited nutrients prevent indefinite growth.

Question 22

[10 marks]
Astronomer Pavel measures the apparent brightness of stars using the magnitude scale. If two stars have apparent magnitudes m_1 and m_2, the ratio of their brightnesses b_1 and b_2 satisfies
m_1 - m_2 = -2.5\log_{10}\left(\dfrac{b_1}{b_2}\right).
A smaller magnitude means a brighter star.
(a) Star A has magnitude 2.1 and Star B has magnitude 5.6. Show that Star A is about 25 times brighter than Star B.
[3]
(b) A telescope can just detect stars as faint as magnitude 12.5. The faintest star visible to the unaided eye has magnitude 6.0. Find how many times fainter the telescope's limit is than the unaided-eye limit.
[3]
(c) A variable star changes brightness so that at its brightest it is 4.0 times as bright as at its faintest. Find the difference between its faintest and brightest magnitudes, correct to two decimal places.
[4]
Show solutionHide solution
(a) m_A - m_B = 2.1 - 5.6 = -3.5. So -3.5 = -2.5\log_{10}\left(\dfrac{b_A}{b_B}\right), giving \log_{10}\left(\dfrac{b_A}{b_B}\right) = 1.4.
Then \dfrac{b_A}{b_B} = 10^{1.4} = 25.1 \approx 25. Star A is about 25 times brighter.
[3]
(b) Magnitude difference = 12.5 - 6.0 = 6.5. Here 6.5 = -2.5\log_{10}\left(\dfrac{b_{\text{tel}}}{b_{\text{eye}}}\right), so \log_{10}\left(\dfrac{b_{\text{tel}}}{b_{\text{eye}}}\right) = -2.6.
Thus \dfrac{b_{\text{eye}}}{b_{\text{tel}}} = 10^{2.6} = 398. The telescope's limit is about 398 times fainter.
[3]
(c) With \dfrac{b_{\text{bright}}}{b_{\text{faint}}} = 4.0, the magnitude difference is
m_{\text{faint}} - m_{\text{bright}} = 2.5\log_{10}(4.0) = 2.5\times 0.6021 = 1.51.
The faintest magnitude is 1.51 greater than the brightest.
[4]

Final answer

  • (a) 25.1 times, about 25
  • (b) 398 times
  • (c) 1.51 magnitudes.

Question 23

[7 marks]
Seismologist Ada uses a regional model for the expected number N of earthquakes per year with magnitude at least M: \log_{10}N=5.2-0.9M, for 2\le M\le8.
(a) Find the expected number of earthquakes per year with magnitude at least 4.
[2]
(b) The equivalent model has the form N=p\times10^{-0.9M}. Find p by evaluating the model at M=0.
[2]
(c) Find the magnitude M for which one earthquake per year is expected.
[2]
(d) Explain why the model should not be used for earthquakes of magnitude at least 9.
[1]
Show solutionHide solution
(a) \log_{10}N=5.2-0.9(4)=1.6 (M1), so N=10^{1.6}=39.8 earthquakes per year (A1). (b) For this algebraic calculation, M=0 gives \log_{10}p=5.2 (M1); evaluating 10^{5.2} gives p=1.58\times10^5 (A1). This evaluates the coefficient, not an earthquake prediction outside the stated domain. (c) N=1 gives 0=5.2-0.9M (M1), so M=5.78 (A1). (d) 9 is outside the stated domain 2\le M\le8; the model is not supported there (R1).

Final answer

  • (a) 39.8 per year
  • (b) p=1.58\times10^5
  • (c) M=5.78
  • (d) outside the domain.

Question 24

[15 marks]
Nadia grows a bacterial culture in a hospital laboratory. The number of bacteria t hours after the start is modelled by N=2.4\times10^5\times1.18^t.
(a) Write down the initial number of bacteria and the percentage increase per hour.
[2]
(b) Find the number of bacteria after 12 hours. Express this count in standard form.
[2]
(c) Find the number of complete hours after the start when the number of bacteria first exceeds 10^7.
[3]
After 12 hours a technician counts 1.8\times10^6 bacteria, correct to 2 significant figures.
(d) Write down the bounds of the true number of bacteria.
[1]
(e) Taking 1.8\times10^6 as exact, find the percentage error in the model's value from part (b).
[2]
(f) Determine whether it is certain that the model's value is within 5% of the true number.
[2]
A disinfectant has a log reduction value of 4.5, meaning \log_{10}(N_0/N_1)=4.5, where N_0 and N_1 are the numbers of bacteria before and after treatment.
(g) The number of bacteria in part (b) is treated with this disinfectant. Find the number that remain, to the nearest whole number.
[2]
(h) Suggest one reason why the growth model may not be suitable after many hours.
[1]
Show solutionHide solution
(a) 240 000 bacteria (A1); 18% increase per hour (A1). (b) N(12)=2.4\times10^5\times1.18^{12} (M1), giving 1.75\times10^6 bacteria (A1). Store the full result. (c) A table across the threshold (M1) gives N(22)=9.15\times10^6 and N(23)=1.08\times10^7 (A1), hence 23 complete hours (A1). (d) 1.75\times10^6\le N<1.85\times10^6 (A1). (e) |N(12)-1.8\times10^6|/(1.8\times10^6)\times100 (M1), giving 2.83% (A1). (f) At the upper bound the limiting error is |N(12)-1.85\times10^6|/(1.85\times10^6)\times100=5.46\% (M1); values close enough to that bound have error above 5%, so it is not certain (R1). (g) N_1=N(12)/10^{4.5} (M1), giving about 55 bacteria to the nearest whole number (A1). (h) Nutrients or space become limited and growth slows (R1).

Final answer

  • (a) 240 000, 18%
  • (b) 1.75\times10^6
  • (c) 23 h
  • (d) 1.75\times10^6\le N<1.85\times10^6
  • (e) 2.83%
  • (f) not certain, limiting error 5.46%
  • (g) 55 bacteria
  • (h) nutrients run out.

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Investigate a rounded pH

G1 Investigation

[12 marks]
Farah tests rainwater with pH 4.3, correct to one decimal place. Hydrogen ion concentration is C=10^{-\mathrm{pH}} mol L^{-1}.
(a) Write down the bounds of the pH.
[1]
(b) Find the bounds of C.
[2]
(c) The concentration is quoted using pH 4.3. Find the largest possible percentage error in that quoted concentration.
[3]
(d) Repeat parts (b) and (c) for pH 9.3, correct to one decimal place. Comment on the result.
[2]
(e) Find the largest possible percentage errors when pH is given to two and to three decimal places. Hence state the accuracy needed for an error below 1%.
[4]
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(a) 4.25\le\mathrm{pH}<4.35 (A1). (b) The concentration decreases as pH increases. Thus 10^{-4.35}<C\le10^{-4.25} (M1 A1), with boundary values approximately 4.47\times10^{-5} and 5.62\times10^{-5} mol L^{-1}. (c) The quoted value is Q=10^{-4.3}. Test both endpoints (M1): at the lower concentration boundary 100(Q-10^{-4.35})/10^{-4.35}=12.2\% (A1); at the upper endpoint 100(10^{-4.25}-Q)/10^{-4.25}=10.9\%. The limiting largest error is 12.2% (A1); it is approached as pH approaches the excluded upper bound. (d) 10^{-9.35}<C\le10^{-9.25}, with boundary values about 4.47\times10^{-10} and 5.62\times10^{-10} mol L^{-1} (A1). Both values are scaled by the same factor compared with part (b), so the endpoint percentage errors remain 12.2% and 10.9% (R1). (e) Two decimal places give half-width 0.005, so testing the concentration boundaries gives limiting error 1.16% (M1 A1). Three decimal places give half-width 0.0005 and limiting error 0.115% (A1). Three decimal places are therefore needed for less than 1% (R1). All endpoint values are evaluated numerically; no logarithm law is required.

Final answer

  • (a) 4.25\le\mathrm{pH}<4.35
  • (b) 10^{-4.35}<C\le10^{-4.25} mol L^{-1}
  • (c) limiting error 12.2\%
  • (d) 10^{-9.35}<C\le10^{-9.25}, same percentage errors
  • (e) 1.16\%, 0.115\%; 3 d.p.

From the book

Photon Guidebook: Mathematics Applications and Interpretation SL for the IB Diploma

Chapter 1 teaches number: standard form, logarithms and error from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

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