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Number and Algebra - Number systems

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Classification of numbers into sets: Natural numbers (N={0,1,2,3,...}\mathbb{N} = \{0, 1, 2, 3, ...\}), Integers (Z\mathbb{Z}), Rational numbers (Q\mathbb{Q}), and Real numbers (R\mathbb{R}).

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Approximation and Rounding: Rounding to a specific number of decimal places (dpdp) or significant figures (sfsf). Significant figures start counting from the first non-zero digit.

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Scientific Notation (Standard Form): Expressing numbers in the form a×10ka \times 10^k, where 1≤∣a∣<101 \le |a| < 10 and k∈Zk \in \mathbb{Z}.

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Absolute and Relative Error: Measures of how much an approximate value (vav_a) differs from the exact value (vev_e).

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Percentage Error: The relative error expressed as a percentage of the exact value.

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SI Units and Unit Conversions: Converting between different units of length, area, volume, and mass, including compound units like speed.

📐Formulae

Absolute Error=∣va−ve∣\text{Absolute Error} = |v_a - v_e|

Percentage Error=ϵ=∣va−veve∣×100%\text{Percentage Error} = \epsilon = \left| \frac{v_a - v_e}{v_e} \right| \times 100\%

a×10k, where 1≤a<10,k∈Za \times 10^k, \text{ where } 1 \le a < 10, k \in \mathbb{Z}

💡Examples

Problem 1:

A cylindrical water tank has an exact volume of 125.46125.46 m3m^3. A surveyor estimates the volume to be 120120 m3m^3. Calculate the percentage error of this estimate.

Solution:

Given ve=125.46v_e = 125.46 and va=120v_a = 120. ϵ=∣120−125.46125.46∣×100%\epsilon = \left| \frac{120 - 125.46}{125.46} \right| \times 100\% ϵ=∣−5.46125.46∣×100%\epsilon = \left| \frac{-5.46}{125.46} \right| \times 100\% ϵ≈4.35198...\epsilon \approx 4.35198... The percentage error is 4.35%4.35\% (to 33 sfsf).

Explanation:

Substitute the exact and approximate values into the percentage error formula, take the absolute value (making it positive), and multiply by 100100.

Problem 2:

Calculate the value of 4.5×1061.2×10−3\frac{4.5 \times 10^6}{1.2 \times 10^{-3}} and express the result in scientific notation.

Solution:

4.51.2×106−(−3)\frac{4.5}{1.2} \times 10^{6 - (-3)} 3.75×1093.75 \times 10^9

Explanation:

Divide the coefficients (4.5/1.24.5 / 1.2) and subtract the exponents of 1010 according to the laws of indices (6−(−3)=96 - (-3) = 9).

Problem 3:

Round the number 0.00408520.0040852 to 33 significant figures.

Solution:

The first significant figure is 44. The third significant figure is 88. The next digit is 55, so we round up. Result: 0.004090.00409.

Explanation:

Leading zeros are not significant. We count from the first non-zero digit (44). The third digit is 88, and since the following digit is 55, we increment 88 to 99.

Problem 4:

A square garden has a side length measured as 12.412.4 mm, correct to 11 decimal place. Calculate the upper bound for the area of the garden.

Solution:

If the side s=12.4s = 12.4 mm is correct to 11 dpdp, the error interval is ±0.05\pm 0.05. The upper bound for the side is 12.4512.45 mm. $$ \text{Upper Bound Area} = (12.45)^2 = 155.0025 \text{ m}^2

Explanation:

To find the upper bound of a measurement, add half of the degree of accuracy (0.1/2=0.050.1 / 2 = 0.05) to the measured value. Then calculate the area using this upper bound side length.