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Number - Rounding, Estimation, and Upper/Lower Bounds

Grade 8IGCSE

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

🔑Concepts

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Rounding to Decimal Places (dp): Keeping a specific number of digits after the decimal point.

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Rounding to Significant Figures (sf): Identifying the first non-zero digit as the first significant figure.

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Estimation: Approximating calculations by rounding every number to 1 significant figure first.

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Lower and Upper Bounds: The range of possible values a number could have been before it was rounded.

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Error Intervals: Representing the limits of accuracy using inequalities, usually in the form LB≤x<UBLB \le x < UB.

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Calculations with Bounds: Finding the maximum or minimum possible result of an arithmetic operation involving rounded numbers.

📐Formulae

Upper Bound (UB)=Value+12(unit of accuracy)\text{Upper Bound (UB)} = \text{Value} + \frac{1}{2}(\text{unit of accuracy})

Lower Bound (LB)=Value−12(unit of accuracy)\text{Lower Bound (LB)} = \text{Value} - \frac{1}{2}(\text{unit of accuracy})

Error Interval:LB≤x<UB\text{Error Interval}: LB \le x < UB

Max Addition=UB1+UB2\text{Max Addition} = UB_1 + UB_2

Min Subtraction=LB1−UB2\text{Min Subtraction} = LB_1 - UB_2

Max Division=UB1LB2\text{Max Division} = \frac{UB_1}{LB_2}

💡Examples

Problem 1:

Round 0.0045097 to 3 significant figures.

Solution:

0.00451

Explanation:

The first significant figure is the first non-zero digit (4). The second is 5, and the third is 0. The digit following 0 is 9, which is 5 or greater, so we round the 0 up to 1.

Problem 2:

Estimate the value of 48.7×10.34.92\frac{48.7 \times 10.3}{4.92}.

Solution:

100

Explanation:

Round each number to 1 significant figure: 48.7≈5048.7 \approx 50, 10.3≈1010.3 \approx 10, and 4.92≈54.92 \approx 5. The calculation becomes 50×105=5005=100\frac{50 \times 10}{5} = \frac{500}{5} = 100.

Problem 3:

A mass mm is measured as 120g, correct to the nearest 10g. Write down the error interval for mm.

Solution:

115≤m<125115 \le m < 125

Explanation:

The unit of accuracy is 10g. Half of this unit is 5g. Lower Bound = 120−5=115120 - 5 = 115. Upper Bound = 120+5=125120 + 5 = 125. The value can be exactly the lower bound but must be strictly less than the upper bound.

Problem 4:

A rectangle has a length of L=10L = 10 cm and width W=5W = 5 cm, both rounded to the nearest cm. Calculate the lower bound for the area.

Solution:

42.75 cm2cm^2

Explanation:

To find the minimum area, multiply the lower bounds of both dimensions. LLB=9.5L_{LB} = 9.5 cm, WLB=4.5W_{LB} = 4.5 cm. Min Area=9.5×4.5=42.75\text{Min Area} = 9.5 \times 4.5 = 42.75.