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Number Skills

Practice exam-style IB Math AI questions for Number Skills, aligned with the syllabus and grouped by topic.

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Question 1
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A research station stores 3.84×1093.84\times10^9 bytes of satellite data. A further 2.75×1082.75\times10^8 bytes are added. The data are then transferred at a constant rate of 6.40×1066.40\times10^6 bytes per second.

A

the total amount of data. Give your answer to three significant figures in the form a×10ka\times10^k, where 1≤a<101\leq a<10 and k∈Zk\in\mathbb{Z}.

[2]
B

the time, in minutes, required to transfer all the data.

[3]
Question 2
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

Consider the expression

E=27×2−322E=\frac{2^7\times2^{-3}}{2^2}
A

Simplify EE, giving your answer as an integer.

[2]
B

10x=0.003710^x=0.0037.

[3]
Question 3
SL • Paper 1
Easy
Calculator Permitted
SL • Paper 1
Easy
Calculator Permitted

A laboratory sample contains 4.50×1084.50\times10^8 identical cells. The average volume of one cell is 2.80×10−152.80\times10^{-15} litres.

A

the total volume of the cells in litres. Give your answer in standard scientific notation.

[2]
B

Given that 11 litre is equal to 10610^6 microlitres, express the total volume in microlitres.

[2]
Question 4
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

At a concert, adult tickets cost 18 euros, student tickets cost 12 euros and child tickets cost 8 euros. A total of 180 tickets are sold for 2256 euros. The number of adult tickets sold is twice the number of child tickets sold.

Let xx, yy and zz be the numbers of adult, student and child tickets sold respectively.

A

three linear equations satisfied by xx, yy and zz.

[3]
B

Use technology to the number of tickets of each type sold.

[2]
C

the percentage of the tickets sold that were adult tickets.

[1]

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Question 5
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

During the first five hours of a test, the difference between the measured temperature and a reference temperature is modelled by

f(t)=t3−9t2+20t−12f(t)=t^3-9t^2+20t-12

where tt is the time in hours and 0≤t≤50\leq t\leq5.

A

Use technology to all the roots of the equation f(t)=0f(t)=0.

[3]
B

which roots are valid within the time interval of the model.

[1]
C

the length of time for which f(t)>0f(t)>0 during the interval 0≤t≤50\leq t\leq5.

[1]
Question 6
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The sound level LL, in decibels, is given by

L=10log⁡(II0)L=10\log\left(\frac{I}{I_0}\right)

where II is the sound intensity and I0=1.00×10−12 W m−2I_0=1.00\times10^{-12}\ \text{W m}^{-2}.

A

machine produces a sound level of 73.073.0 decibels. its sound intensity.

[3]
B

A second machine has four times the sound intensity of the first machine. the sound level produced by the second machine.

[2]
Question 7
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Consider the equation

ln⁡(x−1)+ln⁡(x+3)=ln⁡20\ln(x-1)+\ln(x+3)=\ln20
A

the domain restriction on xx.

[1]
B

the equation, giving your answer exactly.

[4]
Question 8
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Let x>0x>0 and

E=(16x6)342x12E=\frac{(16x^6)^{\frac34}}{2x^{\frac12}}
A

Simplify EE, giving your answer with an integer exponent.

[3]
B

E=324E=324.

[2]

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Question 9
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Positive real numbers xx and yy satisfy

log⁡x+log⁡y=2\log x+\log y=2

and

log⁡x−log⁡y=0.6\log x-\log y=0.6
A

the values of log⁡x\log x and log⁡y\log y.

[2]
B

xx and yy.

[2]
C

Use a logarithm law to xy=100xy=100.

[1]
Question 10
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Let C>0C>0. A dimensionless index RR is defined by

R=C23C−16R=\frac{C^{\frac23}}{C^{-\frac16}}
A

Express RR as a single power of CC.

[2]
B

Given that R=32R=32, for CC.

[2]
C

why the equation C56=32C^{\frac56}=32 has only one solution under the given restriction.

[1]
Question 11
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

The radius of a circular garden is measured as 8.48.4 m, correct to the nearest 0.10.1 m. A landscape designer calculates its area using A=π(8.4)2A=\pi(8.4)^2.

A

the lower and upper bounds for the actual radius rr.

[2]
B

the lower and upper bounds for the actual area of the garden.

[2]
C

the maximum possible percentage error in the area calculated by the designer.

[2]
Question 12
SL • Paper 1
Medium
Calculator Permitted
SL • Paper 1
Medium
Calculator Permitted

A rectangular solar panel has a measured length of 1.801.80 m and a measured width of 0.950.95 m. Both measurements are correct to the nearest 0.010.01 m. The area calculated from the measured dimensions is 1.71 m21.71\ \text{m}^2.

A

the interval containing the actual length and the interval containing the actual width.

[2]
B

the lower and upper bounds for the actual area.

[2]
C

the maximum possible percentage error in the stated area of 1.71 m21.71\ \text{m}^2.

[1]

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Question 13
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The pH of a solution with hydrogen ion concentration HH is given by

pH=−log⁡H\text{pH}=-\log H

Solution A has concentration HA=3.20×10−5H_A=3.20\times10^{-5} and solution B has concentration HB=8.00×10−7H_B=8.00\times10^{-7}.

A

Using the laws of logarithms, pHB−pHA\text{pH}_B-\text{pH}_A.

[3]
B

A third solution has concentration HC=HAHBH_C=\sqrt{H_AH_B}. its pH is the mean of pHA\text{pH}_A and pHB\text{pH}_B.

[2]
Question 14
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The mass MM, in grams, of an artificial model is related to its length LL, in centimetres, by

M=kL53M=kL^{\frac53}

where kk is a positive constant. A model of length 88 cm has mass 9696 g.

A

the value of kk.

[2]
B

A second model has mass 162162 g. its length.

[3]
C

the percentage increase in length from the first model to the second model.

[1]
Question 15
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

The magnitude MM of an earthquake is modelled by

M=log⁡(AA0)M=\log\left(\frac{A}{A_0}\right)

where AA is its measured amplitude and A0A_0 is a reference amplitude. Earthquake P has amplitude 2.50×104A02.50\times10^4A_0, and earthquake Q has amplitude 4.00×106A04.00\times10^6A_0.

A

the magnitudes of earthquakes P and Q.

[2]
B

Using a logarithm law, MQ−MPM_Q-M_P.

[2]
C

Earthquake R has a magnitude equal to the mean of the magnitudes of P and Q. the amplitude of earthquake R in terms of A0A_0.

[2]
Question 16
HL • Paper 1
Medium
Calculator Permitted
HL • Paper 1
Medium
Calculator Permitted

Consider the equation

2log⁡x−log⁡(x−2)=12\log x-\log(x-2)=1
A

the domain of the equation.

[1]
B

Express the left-hand side as the logarithm of a single expression.

[2]
C

the equation, giving both solutions exactly.

[3]

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Question 17
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

A space probe collects 3.75×10223.75\times10^{22} dust particles. The mean mass of one particle is 6.40×10−17 kg6.40\times10^{-17}\ \text{kg}. The probe also collects 8.60×105 kg8.60\times10^5\ \text{kg} of larger fragments.

A
I.

the total mass of the dust particles. Give your answer in standard scientific notation.

[2]
II.

the total mass collected by the probe. Give your answer in standard scientific notation.

[2]
B
I.

the percentage of the total collected mass that consists of dust particles.

[2]
II.

Containers can each hold 2.75×103 kg2.75\times10^3\ \text{kg}. A reserve capacity of 6%6\% is required. the minimum number of containers needed. If you did not obtain 3.26×1063.26\times10^6 in part (a)(ii), use this value.

[2]
Question 18
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

The response DD of a light sensor is modelled by
D=log⁡(II0)D=\log\left(\frac{I}{I_0}\right)
where II is the light intensity and I0=2.50×10−6 W m−2I_0=2.50\times10^{-6}\ \text{W m}^{-2}.

A
I.

(a)(i) DD when I=1.80×10−2 W m−2I=1.80\times10^{-2}\ \text{W m}^{-2}.

[2]
II.

(a)(ii) A second light source produces a response of D=2.90D=2.90. its intensity to 3 significant figures, retaining the unrounded calculator value for use in subsequent parts.

[2]
B
I.

(b)(i) the ratio of the first intensity to the second intensity.

[2]
II.

(b)(ii) the percentage decrease in intensity from the first light source to the second.

[2]
Question 19
SL • Paper 2
Medium
Calculator Permitted
SL • Paper 2
Medium
Calculator Permitted

An uncompressed digital map contains 2302^{30} bytes. A compression process multiplies its size by 2−82^{-8}, after which indexing data multiplies the resulting size by 232^3.

A
I.

Express the final size as a single power of 22.

[2]
II.

Write the final size in standard scientific notation.

[2]
B
I.

The map is transmitted at 2152^{15} bytes per second. the transmission time in minutes.

[2]
II.

Each additional compression round halves the final file size. the minimum number of additional rounds needed to reduce the size to at most 1.00×1061.00\times10^6 bytes.

[2]
Question 20
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A liquid medicine has a measured volume of 12.4 mL12.4\ \text{mL}, correct to the nearest 0.1 mL0.1\ \text{mL}. Its measured concentration is 2.35 mg mL−12.35\ \text{mg mL}^{-1}, correct to the nearest 0.01 mg mL−10.01\ \text{mg mL}^{-1}. A pharmacist reports that the dose contains 29.1 mg29.1\ \text{mg} of medicine.

A
I.

the interval containing the actual volume VV.

[2]
II.

the interval containing the actual concentration CC.

[2]
B
I.

the lower and upper bounds for the actual mass of medicine in the dose.

[2]
II.

the upper limit of the percentage error in the pharmacist's reported mass.

[2]

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Question 21
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A relief organization orders three types of supply crate: A, B and C. Let xx, yy and zz be the numbers of each type ordered. There are 120120 crates in total. The numbers of food units and medical units give the equations
2x+y+3z=2202x+y+3z=220
and
x+4y+2z=300x+4y+2z=300
Crates A, B and C cost 18, 24 and 31 euros respectively.

A
I.

the third equation satisfied by xx, yy and zz.

[1]
II.

Use technology to xx, yy and zz.

[3]
B
I.

the total cost of the order.

[2]
II.

All crate prices increase by 7.5%7.5\%. whether a budget of 3000 euros is sufficient, justifying your answer.

[2]
Question 22
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

During a 4.54.5-hour calibration test, the signed positioning error of a robotic arm is modelled by
p(t)=t4−10t3+35t2−50t+24p(t)=t^4-10t^3+35t^2-50t+24
where tt is measured in hours and 0≤t≤4.50\le t\le4.5. A positive value of p(t)p(t) means that the arm is above its target position.

A
I.

Use technology to all the roots of p(t)=0p(t)=0.

[2]
II.

the intervals during which the robotic arm is above its target position.

[2]
B
I.

the total length of time for which the arm is above its target position.

[2]
II.

the percentage of the calibration test for which the arm is not above its target position.

[2]
Question 23
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

A cylindrical storage vessel has measured diameter 24.6 cm24.6\ \text{cm}, correct to the nearest 0.1 cm0.1\ \text{cm}, and measured height 38 cm38\ \text{cm}, correct to the nearest centimetre. Its volume is calculated using
V=π(d2)2hV=\pi\left(\frac d2\right)^2h

A
AI.

the lower and upper bounds for dd and hh.

[2]
AII.

the volume using the measured dimensions. Give your answer in standard scientific notation.

[2]
B
BI.

the lower and upper bounds for the actual volume.

[2]
BII.

the maximum possible percentage error in the volume calculated from the measured dimensions, using the actual volume as the reference value. Use the unrounded endpoint values from part (b)(i).

[2]
Question 24
SL • Paper 2
Hard
Calculator Permitted
SL • Paper 2
Hard
Calculator Permitted

Three currents I1I_1, I2I_2 and I3I_3, measured in amperes, satisfy Kirchhoff's equations
I1−I2−I3=0I_1-I_2-I_3=0
2I1+4I2=182I_1+4I_2=18
2I1+5I3=252I_1+5I_3=25

A
I.

Use technology to the three currents.

[2]
II.

that the currents satisfy the junction equation.

[1]
B
I.

The modelled total power is P=2I12+4I22+5I32P=2I_1^2+4I_2^2+5I_3^2. PP.

[2]
II.

A power meter gives an actual reading of 108 W108\ \text{W}. the percentage error in the modelled power and whether the model overestimates or underestimates the actual power.

[3]

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Question 25
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The signal loss LL, in decibels, across a component is modelled by
L=20log⁡(VinVout)L=20\log\left(\frac{V_{\mathrm{in}}}{V_{\mathrm{out}}}\right)
where VinV_{\mathrm{in}} and VoutV_{\mathrm{out}} are positive voltages.

A
I.

the signal loss when Vin=5.00 VV_{\mathrm{in}}=5.00\ \text{V} and Vout=0.0200 VV_{\mathrm{out}}=0.0200\ \text{V}.

[2]
II.

Two components have voltage ratios 1212 and 88, respectively. their individual signal losses.

[2]
B
I.

Using a logarithm law, the total loss across the two components is equal to the loss corresponding to a single voltage ratio of 9696.

[2]
II.

An additional component is required so that the total loss is 60 dB60\ \text{dB}. the voltage ratio rr of this component.

[3]
Question 26
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The daily energy requirement EE, in kilojoules, of an experimental organism is modelled by
E=km34E=km^{\frac34}
where m>0m>0 is its mass in grams and kk is a positive constant. An organism of mass 16 g16\ \text{g} requires 240 kJ240\ \text{kJ}.

A
I.

the value of kk.

[2]
II.

the mass of an organism requiring 405 kJ405\ \text{kJ} per day.

[2]
B
I.

doubling the energy requirement multiplies the mass by 2432^{\frac43}.

[2]
II.

the percentage increase in mass when the energy requirement is doubled.

[2]
Question 27
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The apparent magnitude mm of a star is modelled by
m=−2.5log⁡(FF0)m=-2.5\log\left(\frac{F}{F_0}\right)
where FF is the observed flux and F0F_0 is a reference flux. Star A has flux 4.00×10−5F04.00\times10^{-5}F_0 and star B has flux 2.50×10−7F02.50\times10^{-7}F_0.

A
I.

the apparent magnitude of each star.

[2]
II.

mB−mAm_B-m_A.

[2]
B
I.

Use logarithm laws to
mB−mA=−2.5log⁡(FBFA)m_B-m_A=-2.5\log\left(\frac{F_B}{F_A}\right)

[2]
II.

Two stars differ in apparent magnitude by exactly 33. the ratio of the brighter star's flux to the dimmer star's flux.

[2]
Question 28
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

An observatory records 360360 images. Each image contains 2.40×1082.40\times10^8 pixels and each pixel requires 1414 bits. The recorded data are compressed to 17.5%17.5\% of their original size. The compressed data are transmitted through a link advertised at 6.25×1076.25\times10^7 bits per second, but the link operates at 82.0%82.0\% of this rate.

A
I.

the total number of bits before compression. Give your answer in standard scientific notation.

[2]
II.

the compressed data size in standard scientific notation. If you did not obtain an answer to part (a)(i), use 1.2096×10121.2096\times10^{12} bits.

[2]
B

the time required to transmit the compressed data. Give your answer in minutes.

[2]
C

Each storage module can hold 3.20×10103.20\times10^{10} bits. the minimum number of modules required, explaining why the answer must be rounded in the chosen direction.

[2]

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Question 29
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A filtration process begins with a particle concentration of 6.40×1056.40\times10^5 particles per millilitre. During each filtration cycle, the concentration is multiplied by 8−128^{-\frac12}. After nn cycles, the concentration is CnC_n.

A
I.

Express 8−128^{-\frac12} in the form 1a2\frac{1}{a\sqrt2}, where aa is an integer.

[2]
II.

an expression for CnC_n.

[2]
B

The water is considered safe when Cn<250C_n<250. the minimum number of complete filtration cycles required.

[2]
C

A technician claims that seven cycles are sufficient because the calculated boundary is closer to 77 than to 88. Evaluate this claim.

[2]
Question 30
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A rectangular water tank has measured length 1.841.84 m and measured width 0.960.96 m, each correct to the nearest 0.010.01 m. Its measured depth is 0.4250.425 m, correct to the nearest 0.0010.001 m. The volume calculated from the reported measurements is 0.75072 m30.75072\ \text{m}^3.

A
I.

the intervals containing the actual length ll and width ww.

[2]
II.

the interval containing the actual depth dd.

[2]
B

the lower and upper bounds for the actual volume of the tank.

[2]
C

the maximum possible percentage error in the stated volume 0.75072 m30.75072\ \text{m}^3.

[2]
Question 31
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A recycling plant produces a 100100 kg batch by mixing three materials, AA, BB and CC. Their metal contents are 80%80\%, 50%50\% and 20%20\%, respectively. Their processing costs are 77, 55 and 33 euros per kilogram, respectively. The batch contains 51.551.5 kg of metal and costs 510510 euros to process. Let xx, yy and zz be the masses, in kilograms, of AA, BB and CC.

A
I.

three linear equations satisfied by xx, yy and zz.

[2]
II.

Use technology to xx, yy and zz, given that y=1.5xy=1.5x.

[2]
B

The plant doubles the mass of each material to produce a second batch. Without resolving a system, the metal content and processing cost of the second batch. your answer.

[2]
C

A manager proposes a 100100 kg batch containing 6060 kg of AA, 1010 kg of BB and 3030 kg of CC, while claiming that it satisfies the original metal-content condition. whether the claim is correct.

[2]
Question 32
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The signed calibration error of a sensor is modelled by
f(t)=t3−7t2+14t−8f(t)=t^3-7t^2+14t-8
where tt is the time in hours and 0≤t≤50\le t\le5. A negative value indicates that the sensor reads below the reference value.

A
I.

Use technology to all roots of f(t)=0f(t)=0.

[2]
II.

the intervals during which the sensor reads below the reference value.

[2]
B

the total duration for which the sensor reads below the reference value.

[2]
C

A second sensor has error g(t)=f(t−0.5)g(t)=f(t-0.5). the three times at which its error is zero, without solving another cubic.

[2]

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Question 33
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The apparent brightness index mm of an astronomical object is defined by
m=−2.5log⁡(FF0)m=-2.5\log\left(\frac{F}{F_0}\right)
where FF is the observed flux and F0F_0 is a reference flux. Two unresolved stars have fluxes FA=3.20×10−8F0F_A=3.20\times10^{-8}F_0 and FB=7.50×10−9F0F_B=7.50\times10^{-9}F_0.

A
I.

the brightness index of star AA.

[2]
II.

the brightness index when the two stars are observed as one object.

[2]
B

Using logarithm laws,
mA+B−mA=−2.5log⁡(1+FBFA)m_{A+B}-m_A=-2.5\log\left(1+\frac{F_B}{F_A}\right)

[2]
C

a formula for the brightness index of nn identical unresolved stars, each having individual brightness index mm.

[2]
Question 34
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The strength index SS of a composite material is modelled by
S=kρ32d−12S=k\rho^{\frac32}d^{-\frac12}
where ρ>0\rho>0 is its relative density, d>0d>0 is its grain-size index and k>0k>0 is a constant. A sample with ρ=4\rho=4, d=9d=9 and S=80S=80 is used for calibration.

A
I.

Evaluate 4329−124^{\frac32}9^{-\frac12}.

[2]
II.

kk.

[2]
B

SS for a material with ρ=6.25\rho=6.25 and d=16d=16.

[2]
C

The relative density is multiplied by aa and the grain-size index is multiplied by bb, where a>0a>0 and b>0b>0. the factor by which SS changes.

[2]
Question 35
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A distribution centre sends parcels along three routes, XX, YY and ZZ. During one hour, a total of 120120 parcels are sent. A workload index gives the equations
3x+2y+z=2303x+2y+z=230
and
x+4y+2z=310x+4y+2z=310
where xx, yy and zz are the numbers of parcels sent along the routes.

A
I.

the third equation needed to form a system for xx, yy and zz.

[2]
II.

Use technology to the system.

[2]
B

In a second hour, every route total is increased by 20%20\%. the three new route totals without solving another system.

[2]
C

why multiplying the right-hand side of every equation in the original system by 1.201.20 produces the same new solution.

[2]
Question 36
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The wingbeat frequency ff of a mechanical flying model is represented by
f=km−13L−12f=km^{-\frac13}L^{-\frac12}
where m>0m>0 is a normalized mass, L>0L>0 is a normalized wing length and k>0k>0 is a constant. A model with m=8m=8, L=9L=9 and f=10f=10 is used for calibration.

A
I.

Evaluate 8−139−128^{-\frac13}9^{-\frac12}.

[2]
II.

kk.

[2]
B

the wingbeat frequency when m=27m=27 and L=16L=16.

[2]
C

multiplying the mass by 88 and dividing the wing length by 44 leaves the wingbeat frequency unchanged.

[2]

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Question 37
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A dimensionless security parameter xx satisfies
ln⁡(x−2)+2ln⁡(x+1)=ln⁡108\ln(x-2)+2\ln(x+1)=\ln108

A
I.

the domain restriction on xx.

[1]
II.

Use logarithm laws to express the equation as a polynomial equation.

[3]
B
I.

Use technology to the equation, rejecting any invalid solutions.

[2]
II.

The right-hand side is increased from ln⁡108\ln108 to ln⁡108+ln⁡8\ln108+\ln8. the effect on the value of (x−2)(x+1)2(x-2)(x+1)^2.

[2]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A cube has measured volume 125 cm3125\ \text{cm}^3, correct to the nearest cubic centimetre. Its side length is s=V13s=V^{\frac13} and its surface area is A=6V23A=6V^{\frac23}.

A
I.

the interval containing the actual volume.

[1]
II.

the lower and upper bounds for the side length.

[3]
B
I.

the lower and upper bounds for the surface area.

[2]
II.

The surface area calculated from the measured volume is 150 cm2150\ \text{cm}^2. its maximum possible percentage error.

[2]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The maximum load LL, in newtons, supported by a prototype column is modelled by
L=kd52L=kd^{\frac52}
where d>0d>0 is its diameter in centimetres. A column of diameter 4 cm4\ \text{cm} supports 960 N960\ \text{N}. The amount of material used is proportional to d32d^{\frac32}.

A
I.

kk.

[2]
II.

the diameter required to support 1500 N1500\ \text{N}.

[2]
B
I.

the ratio of material used by the two columns can be written as
(1500960)35\left(\frac{1500}{960}\right)^{\frac35}

[2]
II.

the percentage increase in material used.

[2]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A computational efficiency index is defined by
C=3log⁡n−log⁡tC=3\log n-\log t
where n>0n>0 is the number of operations, in millions, and t>0t>0 is the processing time in seconds.

A
I.

Express CC as the logarithm of a single expression.

[2]
II.

Given that C=8C=8 and t=125t=125, nn.

[2]
B
I.

The number of operations is doubled and the processing time is tripled. the change in the index is
log⁡(83)\log\left(\frac83\right)

[3]
II.

the increase in the index.

[2]

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Question 41
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The attenuation DD, in decibels, caused by a filter is defined by
D=10log⁡(PinPout)D=10\log\left(\frac{P_{\mathrm{in}}}{P_{\mathrm{out}}}\right)
A layer of material AA transmits 72%72\% of the power incident on it, while a layer of material BB transmits 45%45\%. Successive layers act independently.

A
I.

the attenuation caused by one layer of material AA.

[2]
II.

the total attenuation caused by three layers of AA and two layers of BB.

[2]
B

Use logarithm laws to why the attenuation of successive independent layers is the sum of their individual attenuations.

[2]
C

A filter contains one layer of BB and nn layers of AA. the minimum value of nn needed for a total attenuation of at least 1515 dB.

[2]
Question 42
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A spherical metal nanoparticle has measured diameter 8585 nm, correct to the nearest nanometre. The measured density of the metal is 19.3 g cm−319.3\ \text{g cm}^{-3}, correct to the nearest 0.1 g cm−30.1\ \text{g cm}^{-3}. Use 1 nm=10−71\ \text{nm}=10^{-7} cm and m=π6ρd3m=\frac{\pi}{6}\rho d^3.

A
I.

Express the measured diameter in centimetres using standard scientific notation.

[2]
II.

the mass obtained from the reported measurements. Give your answer in standard scientific notation.

[2]
B

lower and upper bounds for the actual mass of one nanoparticle.

[2]
C

A sample has exact total mass 2.50×10−62.50\times10^{-6} g. the greatest possible number of complete nanoparticles that the sample could contain.

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

An open box is made by cutting squares of side xx cm from each corner of an 1818 cm by 1212 cm rectangular sheet and folding up the sides. Its volume is
V=x(18−2x)(12−2x)V=x(18-2x)(12-2x)
where 0<x<60<x<6.

A
I.

a volume of 160 cm3160\ \text{cm}^3 leads to the equation
x3−15x2+54x−40=0x^3-15x^2+54x-40=0

[2]
II.

Use technology to this polynomial equation.

[2]
B

which solutions produce a box of volume 160 cm3160\ \text{cm}^3, giving a reason.

[2]
C

A designer claims that specifying the volume uniquely determines the cut size. Evaluate this claim using the result for 160 cm3160\ \text{cm}^3.

[2]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A rectangular cuboid has measured mass 2.362.36 kg, correct to the nearest 0.010.01 kg. Its measured dimensions are 25.425.4 cm, 8.68.6 cm and 4.24.2 cm, each correct to the nearest 0.10.1 cm. Density is calculated using ρ=m/V\rho=m/V.

A
I.

the density using the reported measurements, in g cm−3\text{g cm}^{-3}.

[2]
II.

the lower and upper bounds for the mass in grams.

[2]
B

lower and upper bounds for the actual density.

[2]
C

A database classifies a material as type PP when its density is at least 2.55 g cm−32.55\ \text{g cm}^{-3}. whether the measurements guarantee that the block is type PP.

[2]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A dimensionless computing index is defined by
K=log⁡(N2T)K=\log\left(\frac{N^2}{T}\right)
where N>0N>0 is a normalized operation count and T>0T>0 is a normalized processing time. System AA has N=2.40×107N=2.40\times10^7 and T=3.00×10−4T=3.00\times10^{-4}.

A
I.

Express KK as a sum and difference of logarithms.

[2]
II.

KK for system AA.

[2]
B

System BB has twice the operation count and one quarter of the processing time of system AA. KB−KAK_B-K_A without calculating KBK_B directly.

[2]
C

The operation count is multiplied by rr and the processing time is divided by rr. rr so that the index increases by exactly 22.

[2]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A calibration parameter x>0x>0 satisfies
log⁡(x+4)−2log⁡x=c\log(x+4)-2\log x=c
where cc is a constant determined by the calibration setting.

A
I.

Express the left-hand side as the logarithm of a single expression.

[2]
II.

For c=log⁡(34)c=\log\left(\frac34\right), the equation reduces to 3x2−4x−16=03x^2-4x-16=0.

[2]
B

the calibration equation exactly.

[2]
C

why the negative quadratic root cannot be accepted, even though it satisfies the quadratic equation.

[2]
Question 47
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Two positive dimensionless calibration constants aa and bb satisfy
ln⁡a+ln⁡b=ln⁡12\ln a+\ln b=\ln12
and
2ln⁡a−12ln⁡b=ln⁡(163)2\ln a-\frac12\ln b=\ln\left(\frac{16}{\sqrt3}\right)

A
I.

Use logarithm laws to express the two equations without logarithms.

[2]
II.

aa and bb.

[2]
B
I.

The product abab remains equal to 1212, but the quantity a2b−12a^2b^{-\frac12} is doubled. the new value of aa is 4×2254\times2^{\frac25}.

[3]
II.

the percentage increase in aa and the new value of bb.

[2]
Question 48
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A sensor reports a positive length as 12.012.0 mm, correct to the nearest 0.10.1 mm. A derived response is calculated using
R=s32R=s^{\frac32}
where ss is the actual length in millimetres.

A
I.

the interval containing ss.

[2]
II.

lower and upper bounds for RR.

[2]
B

The reported response is calculated as R0=12.03/2R_0=12.0^{3/2}. the maximum possible percentage error of the reported value R0R_0, taking the percentage error relative to R0R_0.

[2]
C

A positive quantity QQ is measured with absolute uncertainty hh, where 0≤h<Q0\le h<Q, and a derived quantity is Y=QrY=Q^r, where r>0r>0. expressions for the ratios of the upper and lower derived bounds to the reported value Y0=QrY_0=Q^r.

[2]

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