IB Mathematics: Applications and Interpretation Quiz: Number Sets And Approximation
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Number Sets And ApproximationQuestion 1 of 20

A digital scale displays the mass of a rock as 2.50 kg. Let the exact mass of the rock be mm kg. Which statement provides the most accurate classification for the number mm?

m=2.50m = 2.50, so mm must be a rational number.
mm must be a rational number, but not necessarily 2.50.
mm must be an integer because mass is a discrete quantity.
mm is a real number that is best approximated by 2.50.
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Number Sets And Approximation

Practice Number Sets And Approximation in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Number Sets And Approximation, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A digital scale displays the mass of a rock as 2.50 kg. Let the exact mass of the rock be mm kg. Which statement provides the most accurate classification for the number mm?

  1. m=2.50m = 2.50, so mm must be a rational number.
  2. mm must be a rational number, but not necessarily 2.50.
  3. mm must be an integer because mass is a discrete quantity.
  4. mm is a real number that is best approximated by 2.50. (correct answer)
Explanation: A physical quantity like mass is continuous and is represented by a real number (R\mathbb{R}). The exact mass mm could be rational or irrational. The digital scale provides a rounded approximation. Therefore, the most accurate statement is that mm is a real number. A says the measurement is exact. B incorrectly assumes the true value must be rational. C is incorrect as mass is not a discrete (integer) quantity.

Question 2

A circular garden has a radius of exactly 100 metres. Its circumference is calculated using two different approximations for π\pi: πA=3.14\pi_A = 3.14 and πB=227\pi_B = \frac{22}{7}. What is the absolute difference between the percentage errors of the two resulting circumference calculations?

  1. 0.000%
  2. 0.011% (correct answer)
  3. 0.091%
  4. 0.571%
Explanation: The percentage error in the circumference calculation will be the same as the percentage error in the approximation for π\pi, since the radius is an exact number. The exact value is π\pi. Percentage error for πA=3.14\pi_A = 3.14: εA=∣3.14−ππ∣×100%≈0.05069%\varepsilon_A = \left| \frac{3.14 - \pi}{\pi} \right| \times 100\% \approx 0.05069\%. Percentage error for πB=227\pi_B = \frac{22}{7}: εB=∣22/7−ππ∣×100%≈0.04025%\varepsilon_B = \left| \frac{22/7 - \pi}{\pi} \right| \times 100\% \approx 0.04025\%. The absolute difference between these two percentage errors is ∣εA−εB∣=∣0.05069%−0.04025%∣=0.01044%|\varepsilon_A - \varepsilon_B| = |0.05069\% - 0.04025\%| = 0.01044\%. The closest answer is 0.011%.

Question 3

The distance for a sprint is measured as 50.0±0.150.0 \pm 0.1 m and the time is measured as 4.00±0.054.00 \pm 0.05 s. What is the total range of possible average speeds (i.e., the difference between the maximum and minimum possible speeds)?

  1. 0.18 m/s
  2. 0.26 m/s
  3. 0.36 m/s (correct answer)
  4. 1.25 m/s
Explanation: Average speed s=d/ts = d/t. The minimum possible speed is smin=dmintmax=50.0−0.14.00+0.05=49.94.05≈12.321s_{min} = \frac{d_{min}}{t_{max}} = \frac{50.0 - 0.1}{4.00 + 0.05} = \frac{49.9}{4.05} \approx 12.321 m/s. The maximum possible speed is smax=dmaxtmin=50.0+0.14.00−0.05=50.13.95≈12.684s_{max} = \frac{d_{max}}{t_{min}} = \frac{50.0 + 0.1}{4.00 - 0.05} = \frac{50.1}{3.95} \approx 12.684 m/s. The range of possible speeds is the difference between the maximum and minimum: Range=smax−smin=12.684−12.321=0.363Range = s_{max} - s_{min} = 12.684 - 12.321 = 0.363 m/s. The closest answer is 0.36 m/s.

Question 4

The volume of a metal cube is calculated, and the result is found to have a percentage error of -4%. Assuming the error is small, what is the approximate percentage error in the measurement of the cube's side length?

  1. -12.0%
  2. -4.0%
  3. -2.7%
  4. -1.3% (correct answer)
Explanation: For a cube with side length LL, the volume is V=L3V = L^3. For small percentage errors, the relationship is εV≈3×εL\varepsilon_V \approx 3 \times \varepsilon_L, where εV\varepsilon_V is the percentage error in volume and εL\varepsilon_L is the percentage error in side length. We are given εV=−4%\varepsilon_V = -4\%. Therefore, εL≈εV3=−4%3≈−1.33%\varepsilon_L \approx \frac{\varepsilon_V}{3} = \frac{-4\%}{3} \approx -1.33\%. So, -1.3% is the closest answer.

Question 5

A scientific instrument measures the diameter of a microscopic particle to be 0.0006328 mm. For a report, this value must be simplified. What is the absolute difference between the value rounded to three significant figures and the value rounded to three decimal places?

  1. 0.0000002
  2. 0.0003670 (correct answer)
  3. 0.0003672
  4. 0.0010000
Explanation: The original value is 0.0006328. Rounding to three significant figures: The first non-zero digit is 6. The first three significant figures are 6, 3, and 2. The next digit is 8, so we round up. The value becomes 0.000633. Rounding to three decimal places: We look at the fourth decimal place. 0.000_6_... Since 6 is 5 or greater, we round up the third decimal place. The value becomes 0.001. The absolute difference is ∣0.001−0.000633∣=0.000367|0.001 - 0.000633| = 0.000367.

Question 6

The dimensions of a rectangular plot of land are measured as 12.7 m by 8.2 m. For a preliminary sketch, a landscape designer rounds both dimensions to one significant figure before calculating the area. What is the percentage error in the designer's area calculation?

  1. 0.13%
  2. 7.8%
  3. 23.2% (correct answer)
  4. 30.2%
Explanation: First, calculate the exact area: Aexact=12.7×8.2=104.14A_{exact} = 12.7 \times 8.2 = 104.14 m². Next, round the dimensions to one significant figure: 12.7 m rounds to 10 m, and 8.2 m rounds to 8 m. Then, calculate the approximate area using the rounded dimensions: Aapprox=10×8=80A_{approx} = 10 \times 8 = 80 m². Finally, calculate the percentage error using the formula ε=∣Aapprox−AexactAexact∣×100%\varepsilon = \left| \frac{A_{approx} - A_{exact}}{A_{exact}} \right| \times 100\%: ε=∣80−104.14104.14∣×100%≈23.18%\varepsilon = \left| \frac{80 - 104.14}{104.14} \right| \times 100\% \approx 23.18\%. This rounds to 23.2%.

Question 7

A savings account with a balance of $15,200 is compounded daily. The daily interest rate is 0.021%. The bank calculates the daily interest and rounds it to the nearest cent before adding it to the principal. After 10 days, what is the absolute difference between the bank's balance and a balance calculated without any intermediate rounding?

  1. $0.000
  2. $0.010
  3. $0.014 (correct answer)
  4. $0.054
Explanation: Unrounded calculation: The balance after 10 days is 15200×(1+0.00021)10≈15231.954015200 \times (1 + 0.00021)^{10} \approx 15231.9540. Rounded calculation (requires a GDC loop or spreadsheet): Day 1: Interest = round(15200 × 0.00021, 2) = $3.19. Balance = 15203.19. Day 2: Interest = round(15203.19 × 0.00021, 2) = $3.19. Balance = 15206.38. ... continuing this for 10 days... The final balance with rounding is $15231.94. Absolute difference = ∣15231.9540−15231.94∣≈0.014|15231.9540 - 15231.94| \approx 0.014. Distractor D comes from comparing the unrounded value to a simple interest calculation.

Question 8

A rectangular sports field is measured as 110 m long and 70 m wide, with both measurements correct to the nearest 10 metres. What is the maximum possible area of the field?

  1. 7175 m²
  2. 7700 m²
  3. 8250 m²
  4. 8625 m² (correct answer)
Explanation: To find the maximum possible area, we need to find the upper bounds for both the length and the width. A measurement correct to the nearest 10 metres has a tolerance of ±5\pm 5 m. The upper bound for the length is 110+5=115110 + 5 = 115 m. The upper bound for the width is 70+5=7570 + 5 = 75 m. The maximum possible area is the product of the upper bounds: Areamax=115×75=8625Area_{max} = 115 \times 75 = 8625 m².

Question 9

The radius of a sphere is measured as 5.2 mm, when its true radius is exactly 5.0 mm. What is the percentage error in the calculated volume based on the incorrect measurement?

  1. 4.00%
  2. 8.16%
  3. 11.1%
  4. 12.5% (correct answer)
Explanation: The formula for the volume of a sphere is V=43πr3V = \frac{4}{3}\pi r^3. Exact Volume: VE=43π(5.0)3=5003πV_E = \frac{4}{3}\pi (5.0)^3 = \frac{500}{3}\pi. Approximate Volume: VA=43π(5.2)3=43π(140.608)V_A = \frac{4}{3}\pi (5.2)^3 = \frac{4}{3}\pi (140.608). Percentage error ε=∣VA−VEVE∣×100%\varepsilon = \left| \frac{V_A - V_E}{V_E} \right| \times 100\%. The 43π\frac{4}{3}\pi terms cancel, so we can use ε=∣5.23−5.035.03∣×100%\varepsilon = \left| \frac{5.2^3 - 5.0^3}{5.0^3} \right| \times 100\%. ε=∣140.608−125125∣×100%=15.608125×100%=12.4864%≈12.5%\varepsilon = \left| \frac{140.608 - 125}{125} \right| \times 100\% = \frac{15.608}{125} \times 100\% = 12.4864\% \approx 12.5\%.

Question 10

A savings account with a balance of $15,200 is compounded daily. The daily interest rate is 0.021%. The bank calculates the daily interest and rounds it to the nearest cent before adding it to the principal. After 10 days, what is the absolute difference between the bank's balance and a balance calculated without any intermediate rounding?

  1. $0.000
  2. $0.010
  3. $0.014 (correct answer)
  4. $0.054
Explanation: Unrounded calculation: The balance after 10 days is 15200×(1+0.00021)10≈15231.954015200 \times (1 + 0.00021)^{10} \approx 15231.9540. Rounded calculation (requires a GDC loop or spreadsheet): Day 1: Interest = round(15200 × 0.00021, 2) = $3.19. Balance = 15203.19. Day 2: Interest = round(15203.19 × 0.00021, 2) = $3.19. Balance = 15206.38. ... continuing this for 10 days... The final balance with rounding is $15231.94. Absolute difference = ∣15231.9540−15231.94∣≈0.014|15231.9540 - 15231.94| \approx 0.014. Distractor D comes from comparing the unrounded value to a simple interest calculation.

Question 11

The volume of a metal cube is calculated, and the result is found to have a percentage error of -4%. Assuming the error is small, what is the approximate percentage error in the measurement of the cube's side length?

  1. -12.0%
  2. -4.0%
  3. -2.7%
  4. -1.3% (correct answer)
Explanation: For a cube with side length LL, the volume is V=L3V = L^3. For small percentage errors, the relationship is εV≈3×εL\varepsilon_V \approx 3 \times \varepsilon_L, where εV\varepsilon_V is the percentage error in volume and εL\varepsilon_L is the percentage error in side length. We are given εV=−4%\varepsilon_V = -4\%. Therefore, εL≈εV3=−4%3≈−1.33%\varepsilon_L \approx \frac{\varepsilon_V}{3} = \frac{-4\%}{3} \approx -1.33\%. So, -1.3% is the closest answer.

Question 12

The distance for a sprint is measured as 50.0±0.150.0 \pm 0.1 m and the time is measured as 4.00±0.054.00 \pm 0.05 s. What is the total range of possible average speeds (i.e., the difference between the maximum and minimum possible speeds)?

  1. 0.18 m/s
  2. 0.26 m/s
  3. 0.36 m/s (correct answer)
  4. 1.25 m/s
Explanation: Average speed s=d/ts = d/t. The minimum possible speed is smin=dmintmax=50.0−0.14.00+0.05=49.94.05≈12.321s_{min} = \frac{d_{min}}{t_{max}} = \frac{50.0 - 0.1}{4.00 + 0.05} = \frac{49.9}{4.05} \approx 12.321 m/s. The maximum possible speed is smax=dmaxtmin=50.0+0.14.00−0.05=50.13.95≈12.684s_{max} = \frac{d_{max}}{t_{min}} = \frac{50.0 + 0.1}{4.00 - 0.05} = \frac{50.1}{3.95} \approx 12.684 m/s. The range of possible speeds is the difference between the maximum and minimum: Range=smax−smin=12.684−12.321=0.363Range = s_{max} - s_{min} = 12.684 - 12.321 = 0.363 m/s. The closest answer is 0.36 m/s.

Question 13

The dimensions of a rectangular prism are measured as length = 15.2 cm, width = 8.4 cm, and height = 5.0 cm. According to scientific conventions for reporting results from measurements, to how many significant figures should the calculated volume be stated?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: When multiplying or dividing measurements, the result should be rounded to the same number of significant figures as the measurement with the fewest significant figures. The significant figures for each measurement are: Length = 15.2 cm (3 significant figures) Width = 8.4 cm (2 significant figures) Height = 5.0 cm (2 significant figures) The minimum number of significant figures is 2. Therefore, the calculated volume (15.2×8.4×5.0=638.415.2 \times 8.4 \times 5.0 = 638.4 cm³) should be rounded to 2 significant figures, which is 640 cm³. The question asks for the number of significant figures, which is 2.

Question 14

Two lengths are measured. Measurement A is 120.0±0.5120.0 \pm 0.5 cm. Measurement B is 5.0±0.55.0 \pm 0.5 cm. Which statement accurately compares the percentage (relative) errors of the two measurements?

  1. The percentage errors are equal because the absolute uncertainties are equal.
  2. The percentage error of A is larger than the percentage error of B.
  3. The percentage error of B is more than 20 times the percentage error of A. (correct answer)
  4. The absolute error of A is smaller than the absolute error of B.
Explanation: Percentage error is calculated as (uncertaintymeasurement)×100%\left( \frac{\text{uncertainty}}{\text{measurement}} \right) \times 100\%. For measurement A: εA=0.5120.0×100%≈0.417%\varepsilon_A = \frac{0.5}{120.0} \times 100\% \approx 0.417\%. For measurement B: εB=0.55.0×100%=10%\varepsilon_B = \frac{0.5}{5.0} \times 100\% = 10\%. Comparing the two, the percentage error of B is much larger than that of A. To see by how much, we can calculate the ratio: εBεA=100.417≈24\frac{\varepsilon_B}{\varepsilon_A} = \frac{10}{0.417} \approx 24. Thus, the percentage error of B is more than 20 times that of A.

Question 15

A circular garden has a radius of exactly 100 metres. Its circumference is calculated using two different approximations for π\pi: πA=3.14\pi_A = 3.14 and πB=227\pi_B = \frac{22}{7}. What is the absolute difference between the percentage errors of the two resulting circumference calculations?

  1. 0.000%
  2. 0.011% (correct answer)
  3. 0.091%
  4. 0.571%
Explanation: The percentage error in the circumference calculation will be the same as the percentage error in the approximation for π\pi, since the radius is an exact number. The exact value is π\pi. Percentage error for πA=3.14\pi_A = 3.14: εA=∣3.14−ππ∣×100%≈0.05069%\varepsilon_A = \left| \frac{3.14 - \pi}{\pi} \right| \times 100\% \approx 0.05069\%. Percentage error for πB=227\pi_B = \frac{22}{7}: εB=∣22/7−ππ∣×100%≈0.04025%\varepsilon_B = \left| \frac{22/7 - \pi}{\pi} \right| \times 100\% \approx 0.04025\%. The absolute difference between these two percentage errors is ∣εA−εB∣=∣0.05069%−0.04025%∣=0.01044%|\varepsilon_A - \varepsilon_B| = |0.05069\% - 0.04025\%| = 0.01044\%. The closest answer is 0.011%.

Question 16

From a wooden beam of length x=4.8x = 4.8 m, a piece of length y=1.9y = 1.9 m is cut off. Both measurements are correct to one decimal place. Find the minimum possible length of the remaining piece of wood.

  1. 2.80 m (correct answer)
  2. 2.85 m
  3. 2.90 m
  4. 3.00 m
Explanation: The length of the remaining piece is L=x−yL = x - y. To find the minimum possible length, LminL_{min}, we must subtract the maximum possible value of yy from the minimum possible value of xx. Since the measurements are correct to one decimal place, the uncertainty is ±0.05\pm 0.05 m. Lower bound of xx: xmin=4.8−0.05=4.75x_{min} = 4.8 - 0.05 = 4.75 m. Upper bound of yy: ymax=1.9+0.05=1.95y_{max} = 1.9 + 0.05 = 1.95 m. Minimum remaining length: Lmin=xmin−ymax=4.75−1.95=2.80L_{min} = x_{min} - y_{max} = 4.75 - 1.95 = 2.80 m.

Question 17

An ecologist is creating a computer model to simulate the number of bears, BB, in a national park. The model calculates the population at the end of each year. Which number set is the most appropriate for the domain of the variable BB?

  1. Natural numbers, N\mathbb{N} (correct answer)
  2. Integers, Z\mathbb{Z}
  3. Rational numbers, Q\mathbb{Q}
  4. Real numbers, R\mathbb{R}
Explanation: The number of bears must be a whole, non-negative number. The set of natural numbers, N={0,1,2,3,...}\mathbb{N} = \{0, 1, 2, 3, ...\}, is the most appropriate set for counting discrete entities like animals. Integers (Z\mathbb{Z}) include negative values, which are not possible. Rational (Q\mathbb{Q}) and real (R\mathbb{R}) numbers include fractions and irrational values, which are also not possible for a count of individual bears.

Question 18

A rectangular floor measures 2.85 m by 3.60 m. It needs to be covered with square tiles of side 30 cm. The tiles are sold in boxes of 12. Considering that tiles can be cut and used, how many boxes of tiles must be purchased?

  1. 9
  2. 10 (correct answer)
  3. 114
  4. 120
Explanation: First, calculate the area of the floor in m²: Areafloor=2.85×3.60=10.26Area_{floor} = 2.85 \times 3.60 = 10.26 m². Next, calculate the area of one tile in m². The side length is 30 cm = 0.30 m. Areatile=0.302=0.09Area_{tile} = 0.30^2 = 0.09 m². Calculate the total number of tiles needed by dividing the floor area by the tile area: Number of tiles=10.260.09=114Number \ of \ tiles = \frac{10.26}{0.09} = 114 tiles. Finally, calculate the number of boxes. Since each box contains 12 tiles: Number of boxes=11412=9.5Number \ of \ boxes = \frac{114}{12} = 9.5. Since it is not possible to buy half a box, the contractor must round up to the next whole number and purchase 10 boxes.

Question 19

The theoretical probability that a manufactured component is defective is exactly 1150\frac{1}{150}. An engineer records this probability in a report, rounded to three significant figures. What is the percentage error introduced by this rounding?

  1. 0.0033%
  2. 0.050% (correct answer)
  3. 0.67%
  4. 5.0%
Explanation: The exact value is vE=1150=0.006666...v_E = \frac{1}{150} = 0.006666.... The value rounded to three significant figures is vA=0.00667v_A = 0.00667. The percentage error is ε=∣vA−vEvE∣×100%\varepsilon = \left| \frac{v_A - v_E}{v_E} \right| \times 100\%. ε=∣0.00667−1/1501/150∣×100%=∣0.006671/150−1∣×100%\varepsilon = \left| \frac{0.00667 - 1/150}{1/150} \right| \times 100\% = \left| \frac{0.00667}{1/150} - 1 \right| \times 100\%. ε=∣(0.00667×150)−1∣×100%=∣1.0005−1∣×100%=0.0005×100%=0.05%\varepsilon = |(0.00667 \times 150) - 1| \times 100\% = |1.0005 - 1| \times 100\% = 0.0005 \times 100\% = 0.05\%.

Question 20

The dimensions of a rectangular prism are measured as length = 15.2 cm, width = 8.4 cm, and height = 5.0 cm. According to scientific conventions for reporting results from measurements, to how many significant figures should the calculated volume be stated?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. 4
Explanation: When multiplying or dividing measurements, the result should be rounded to the same number of significant figures as the measurement with the fewest significant figures. The significant figures for each measurement are: Length = 15.2 cm (3 significant figures) Width = 8.4 cm (2 significant figures) Height = 5.0 cm (2 significant figures) The minimum number of significant figures is 2. Therefore, the calculated volume (15.2×8.4×5.0=638.415.2 \times 8.4 \times 5.0 = 638.4 cm³) should be rounded to 2 significant figures, which is 640 cm³. The question asks for the number of significant figures, which is 2.