Number Sets & Approximation - IB Mathematics Applications and Interpretation

1. 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?
   Answer: $0.014 Unrounded calculation: The balance after 10 days is $15200 \times (1 + 0.00021)^{10} \approx 15231.9540$. 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 = $|15231.9540 - 15231.94| \approx 0.014$. Distractor D comes from comparing the unrounded value to a simple interest calculation.
2. Two lengths are measured. Measurement A is $120.0 \pm 0.5$ cm. Measurement B is $5.0 \pm 0.5$ cm. Which statement accurately compares the percentage (relative) errors of the two measurements?
   Answer: The percentage error of B is more than 20 times the percentage error of A. Percentage error is calculated as $\left( \frac{\text{uncertainty}}{\text{measurement}} \right) \times 100\%$. For measurement A: $\varepsilon_A = \frac{0.5}{120.0} \times 100\% \approx 0.417\%$. For measurement B: $\varepsilon_B = \frac{0.5}{5.0} \times 100\% = 10\%$. 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: $\frac{\varepsilon_B}{\varepsilon_A} = \frac{10}{0.417} \approx 24$. Thus, the percentage error of B is more than 20 times that of A.
3. From a wooden beam of length $x = 4.8$ m, a piece of length $y = 1.9$ m is cut off. Both measurements are correct to one decimal place. Find the minimum possible length of the remaining piece of wood.
   Answer: 2.80 m The length of the remaining piece is $L = x - y$. To find the minimum possible length, $L_{min}$, we must subtract the maximum possible value of $y$ from the minimum possible value of $x$. Since the measurements are correct to one decimal place, the uncertainty is $\pm 0.05$ m. Lower bound of $x$: $x_{min} = 4.8 - 0.05 = 4.75$ m. Upper bound of $y$: $y_{max} = 1.9 + 0.05 = 1.95$ m. Minimum remaining length: $L_{min} = x_{min} - y_{max} = 4.75 - 1.95 = 2.80$ m.
4. The distance for a sprint is measured as $50.0 \pm 0.1$ m and the time is measured as $4.00 \pm 0.05$ s. What is the total range of possible average speeds (i.e., the difference between the maximum and minimum possible speeds)?
   Answer: 0.36 m/s Average speed $s = d/t$. The minimum possible speed is $s_{min} = \frac{d_{min}}{t_{max}} = \frac{50.0 - 0.1}{4.00 + 0.05} = \frac{49.9}{4.05} \approx 12.321$ m/s. The maximum possible speed is $s_{max} = \frac{d_{max}}{t_{min}} = \frac{50.0 + 0.1}{4.00 - 0.05} = \frac{50.1}{3.95} \approx 12.684$ m/s. The range of possible speeds is the difference between the maximum and minimum: $Range = s_{max} - s_{min} = 12.684 - 12.321 = 0.363$ m/s. The closest answer is 0.36 m/s.
5. 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?
   Answer: 2 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 ($15.2 \times 8.4 \times 5.0 = 638.4$ cm³) should be rounded to 2 significant figures, which is 640 cm³. The question asks for the number of significant figures, which is 2.
6. 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?
   Answer: 0.0003670 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 $|0.001 - 0.000633| = 0.000367$.
7. A car travels a distance $d = 250$ m (correct to the nearest 10 m) in a time $t = 9.8$ s (correct to one decimal place). Calculate the maximum possible average speed of the car, in m/s.
   Answer: 26.2 m/s Average speed is given by $s = d/t$. To find the maximum possible speed, we must divide the maximum possible distance by the minimum possible time. Distance measurement is correct to the nearest 10 m, so uncertainty is $\pm 5$ m. The upper bound for distance is $d_{max} = 250 + 5 = 255$ m. Time measurement is correct to one decimal place, so uncertainty is $\pm 0.05$ s. The lower bound for time is $t_{min} = 9.8 - 0.05 = 9.75$ s. Maximum speed: $s_{max} = \frac{d_{max}}{t_{min}} = \frac{255}{9.75} \approx 26.1538...$ m/s. To 3 significant figures, this is 26.2 m/s.
8. 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?
   Answer: 10 First, calculate the area of the floor in m²: $Area_{floor} = 2.85 \times 3.60 = 10.26$ m². Next, calculate the area of one tile in m². The side length is 30 cm = 0.30 m. $Area_{tile} = 0.30^2 = 0.09$ m². Calculate the total number of tiles needed by dividing the floor area by the tile area: $Number \ of \ tiles = \frac{10.26}{0.09} = 114$ tiles. Finally, calculate the number of boxes. Since each box contains 12 tiles: $Number \ of \ boxes = \frac{114}{12} = 9.5$. Since it is not possible to buy half a box, the contractor must round up to the next whole number and purchase 10 boxes.
9. 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?
   Answer: 8625 m² 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 $\pm 5$ m. The upper bound for the length is $110 + 5 = 115$ m. The upper bound for the width is $70 + 5 = 75$ m. The maximum possible area is the product of the upper bounds: $Area_{max} = 115 \times 75 = 8625$ m².
10. A chemist mixes three substances with masses $m_1 = 120.5$ g, $m_2 = 34.57$ g, and $m_3 = 8.6$ g. When reporting the total mass, what is the most appropriate degree of accuracy to use?
    Answer: To the nearest tenth of a gram The rule for addition of measurements is that the result should be stated to the same number of decimal places as the measurement with the fewest decimal places. $m_1 = 120.5$ g (1 decimal place) $m_2 = 34.57$ g (2 decimal places) $m_3 = 8.6$ g (1 decimal place) The fewest number of decimal places is one. Therefore, the total mass should be rounded to one decimal place, which is the nearest tenth of a gram. The calculated total mass is $120.5 + 34.57 + 8.6 = 163.67$ g. Rounded to one decimal place, this is 163.7 g.
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?
    Answer: -1.3% For a cube with side length $L$, the volume is $V = L^3$. For small percentage errors, the relationship is $\varepsilon_V \approx 3 \times \varepsilon_L$, where $\varepsilon_V$ is the percentage error in volume and $\varepsilon_L$ is the percentage error in side length. We are given $\varepsilon_V = -4\%$. Therefore, $\varepsilon_L \approx \frac{\varepsilon_V}{3} = \frac{-4\%}{3} \approx -1.33\%$. So, -1.3% is the closest answer.
12. The kinetic energy of an object is given by the formula $E = \frac{1}{2}mv^2$. The mass $m$ is measured as 4.5 kg and the velocity $v$ is measured as 12 m/s. Both measurements are correct to 2 significant figures. Find the maximum possible kinetic energy of the object, in Joules.
    Answer: 355 J To find the maximum possible kinetic energy, we must use the upper bounds for both mass and velocity. For mass $m = 4.5$ kg (correct to 2 s.f.), the interval is $[4.45, 4.55)$. So, $m_{max} = 4.55$ kg. For velocity $v = 12$ m/s (correct to 2 s.f.), the interval is $[11.5, 12.5)$. So, $v_{max} = 12.5$ m/s. Now, calculate the maximum energy: $E_{max} = \frac{1}{2} m_{max} (v_{max})^2 = \frac{1}{2} \times 4.55 \times (12.5)^2 = 0.5 \times 4.55 \times 156.25 = 355.46875$ J. The closest answer is 355 J.
13. An ecologist is creating a computer model to simulate the number of bears, $B$, 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 $B$?
    Answer: Natural numbers, $\mathbb{N}$ The number of bears must be a whole, non-negative number. The set of natural numbers, $\mathbb{N} = \{0, 1, 2, 3, ...\}$, is the most appropriate set for counting discrete entities like animals. Integers ($\mathbb{Z}$) include negative values, which are not possible. Rational ($\mathbb{Q}$) and real ($\mathbb{R}$) numbers include fractions and irrational values, which are also not possible for a count of individual bears.
14. The side length of a square-shaped solar panel is measured with a percentage error of +3%. What is the resulting percentage error in the calculated area of the panel?
    Answer: 6.09% Let the true side length be $L$. The measured length is $L_A = L(1 + 0.03) = 1.03L$. The true area is $A_E = L^2$. The measured area is $A_A = (1.03L)^2 = (1.03)^2 L^2 = 1.0609L^2$. The percentage error in the area is calculated as $\left| \frac{A_A - A_E}{A_E} \right| \times 100\%$. $\varepsilon = \left| \frac{1.0609L^2 - L^2}{L^2} \right| \times 100\% = |1.0609 - 1| \times 100\% = 0.0609 \times 100\% = 6.09\%$.
15. A circular garden has a radius of exactly 100 metres. Its circumference is calculated using two different approximations for $\pi$: $\pi_A = 3.14$ and $\pi_B = \frac{22}{7}$. What is the absolute difference between the percentage errors of the two resulting circumference calculations?
    Answer: 0.011% The percentage error in the circumference calculation will be the same as the percentage error in the approximation for $\pi$, since the radius is an exact number. The exact value is $\pi$. Percentage error for $\pi_A = 3.14$: $\varepsilon_A = \left| \frac{3.14 - \pi}{\pi} \right| \times 100\% \approx 0.05069\%$. Percentage error for $\pi_B = \frac{22}{7}$: $\varepsilon_B = \left| \frac{22/7 - \pi}{\pi} \right| \times 100\% \approx 0.04025\%$. The absolute difference between these two percentage errors is $|\varepsilon_A - \varepsilon_B| = |0.05069\% - 0.04025\%| = 0.01044\%$. The closest answer is 0.011%.
16. 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?
    Answer: 23.2% First, calculate the exact area: $A_{exact} = 12.7 \times 8.2 = 104.14$ 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: $A_{approx} = 10 \times 8 = 80$ m². Finally, calculate the percentage error using the formula $\varepsilon = \left| \frac{A_{approx} - A_{exact}}{A_{exact}} \right| \times 100\%$: $\varepsilon = \left| \frac{80 - 104.14}{104.14} \right| \times 100\% \approx 23.18\%$. This rounds to 23.2%.
17. A room's length is measured as 5.4 m and its width as 3.8 m. Both measurements are correct to one decimal place. What is the minimum possible perimeter of the room?
    Answer: 18.2 m To find the minimum possible perimeter, we need the lower bounds for both length and width. A measurement correct to one decimal place (0.1) has a tolerance of $\pm 0.05$. The lower bound for the length is $5.4 - 0.05 = 5.35$ m. The lower bound for the width is $3.8 - 0.05 = 3.75$ m. The perimeter is calculated as $P = 2(l + w)$. The minimum perimeter is $P_{min} = 2(5.35 + 3.75) = 2(9.1) = 18.2$ m.
18. The theoretical probability that a manufactured component is defective is exactly $\frac{1}{150}$. An engineer records this probability in a report, rounded to three significant figures. What is the percentage error introduced by this rounding?
    Answer: 0.050% The exact value is $v_E = \frac{1}{150} = 0.006666...$. The value rounded to three significant figures is $v_A = 0.00667$. The percentage error is $\varepsilon = \left| \frac{v_A - v_E}{v_E} \right| \times 100\%$. $\varepsilon = \left| \frac{0.00667 - 1/150}{1/150} \right| \times 100\% = \left| \frac{0.00667}{1/150} - 1 \right| \times 100\%$. $\varepsilon = |(0.00667 \times 150) - 1| \times 100\% = |1.0005 - 1| \times 100\% = 0.0005 \times 100\% = 0.05\%$.
19. A digital scale displays the mass of a rock as 2.50 kg. Let the exact mass of the rock be $m$ kg. Which statement provides the most accurate classification for the number $m$?
    Answer: $m$ is a real number that is best approximated by 2.50. A physical quantity like mass is continuous and is represented by a real number ($\mathbb{R}$). The exact mass $m$ could be rational or irrational. The digital scale provides a rounded approximation. Therefore, the most accurate statement is that $m$ 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.
20. 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?
    Answer: 12.5% The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$. Exact Volume: $V_E = \frac{4}{3}\pi (5.0)^3 = \frac{500}{3}\pi$. Approximate Volume: $V_A = \frac{4}{3}\pi (5.2)^3 = \frac{4}{3}\pi (140.608)$. Percentage error $\varepsilon = \left| \frac{V_A - V_E}{V_E} \right| \times 100\%$. The $\frac{4}{3}\pi$ terms cancel, so we can use $\varepsilon = \left| \frac{5.2^3 - 5.0^3}{5.0^3} \right| \times 100\%$. $\varepsilon = \left| \frac{140.608 - 125}{125} \right| \times 100\% = \frac{15.608}{125} \times 100\% = 12.4864\% \approx 12.5\%$.

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