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Number and Algebra - Numbers – rounding – scientific form

Grade 11IB_AA

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

🔑Concepts

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Rounding to nn decimal places (dp) involves looking at the digit in the (n+1)th(n+1)^{th} position. If that digit is 55 or greater, the nthn^{th} digit is rounded up by 11.

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Significant figures (sf) are counted starting from the first non-zero digit from the left. Leading zeros (e.g., 0.00...0.00...) are not significant, but trapped zeros (e.g., 105105) and trailing zeros in a decimal (e.g., 1.201.20) are significant.

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A number is in scientific form (standard form) when written as a×10ka \times 10^k, where 1≤∣a∣<101 \le |a| < 10 and kk is an integer (k∈Zk \in \mathbb{Z}).

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In the IB DP Analysis and Approaches course, final answers should generally be given exactly or to 33 significant figures unless specified otherwise.

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The absolute error is the magnitude of the difference between an exact value (vEv_E) and an approximate value (vAv_A).

📐Formulae

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

Percentage Error=∣vA−vEvE∣×100%\text{Percentage Error} = \left| \frac{v_A - v_E}{v_E} \right| \times 100\%

Absolute Error=∣vA−vE∣\text{Absolute Error} = |v_A - v_E|

💡Examples

Problem 1:

Express the number 0.000670820.00067082 in scientific form rounded to 33 significant figures.

Solution:

6.71×10−46.71 \times 10^{-4}

Explanation:

The first non-zero digit is 66. The first three significant figures are 6,7,06, 7, 0. We look at the fourth significant digit, which is 88. Since 8≥58 \ge 5, we round the third digit (00) up to 11. To place the decimal after the 66, we move it 44 places to the right, resulting in 10−410^{-4}.

Problem 2:

A student measures the length of a desk to be 120120 cm, but the exact length is 121.5121.5 cm. Calculate the absolute error and the percentage error, rounding the percentage to 33 significant figures.

Solution:

Absolute Error: 1.5 cm1.5 \text{ cm} Percentage Error: 1.23%1.23\%

Explanation:

First, find the absolute difference using vertical subtraction: 121.5−120.01.5\begin{array}{r} 121.5 \\ -120.0 \\ \hline 1.5 \end{array} The percentage error is calculated as ϵ=∣120−121.5121.5∣×100%=1.5121.5×100%≈1.2345...%\epsilon = \left| \frac{120 - 121.5}{121.5} \right| \times 100\% = \frac{1.5}{121.5} \times 100\% \approx 1.2345...\%. Rounding to 33 sf gives 1.23%1.23\%.

Problem 3:

Calculate (2.4×105)×(5.0×10−2)(2.4 \times 10^5) \times (5.0 \times 10^{-2}) and write the answer in scientific form.

Solution:

1.2×1041.2 \times 10^4

Explanation:

Multiply the coefficients: 2.4×5.0=122.4 \times 5.0 = 12. Multiply the powers of 1010: 105×10−2=105−2=10310^5 \times 10^{-2} = 10^{5-2} = 10^3. This gives 12×10312 \times 10^3. To convert to scientific form (1≤a<101 \le a < 10), we rewrite 1212 as 1.2×1011.2 \times 10^1. Thus, (1.2×101)×103=1.2×104(1.2 \times 10^1) \times 10^3 = 1.2 \times 10^4.

Numbers – rounding – scientific form Grade 11 Notes & Examples