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Number - Estimation and Limits of Accuracy

Grade 9IGCSE

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

🔑Concepts

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Rounding to Decimal Places (d.p.): Rounding a number based on the digits after the decimal point.

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Significant Figures (s.f.): Rounding to a specific number of meaningful digits, starting from the first non-zero digit.

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Estimation: Approximating a calculation by rounding every number to 1 significant figure before performing operations.

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

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Error Intervals: Representing the range of possible values using inequalities (e.g., LB≤x<UBLB \le x < UB).

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Accuracy in Calculations: Finding the upper and lower bounds of a result when performing operations (addition, subtraction, multiplication, division) on 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})

Maximum Sum=UB1+UB2\text{Maximum Sum} = UB_1 + UB_2

Minimum Difference=LB1−UB2\text{Minimum Difference} = LB_1 - UB_2

Maximum Product=UB1×UB2\text{Maximum Product} = UB_1 \times UB_2

Maximum Quotient=UBnumeratorLBdenominator\text{Maximum Quotient} = \frac{UB_{\text{numerator}}}{LB_{\text{denominator}}}

Minimum Quotient=LBnumeratorUBdenominator\text{Minimum Quotient} = \frac{LB_{\text{numerator}}}{UB_{\text{denominator}}}

💡Examples

Problem 1:

Estimate the value of 4.82×19.70.24\frac{4.82 \times 19.7}{0.24}.

Solution:

400

Explanation:

To estimate, round each number to 1 significant figure: 4.82≈54.82 \approx 5, 19.7≈2019.7 \approx 20, and 0.24≈0.20.24 \approx 0.2. The calculation becomes 5×200.2=1000.2=500\frac{5 \times 20}{0.2} = \frac{100}{0.2} = 500.

Problem 2:

A length LL is given as 12.412.4 cm, correct to 1 decimal place. Write down the error interval for LL.

Solution:

12.35≤L<12.4512.35 \le L < 12.45

Explanation:

The unit of accuracy is 0.1 cm. Half of this is 0.050.05. Lower Bound = 12.4−0.05=12.3512.4 - 0.05 = 12.35. Upper Bound = 12.4+0.05=12.4512.4 + 0.05 = 12.45. The interval includes the LB but excludes the UB.

Problem 3:

A field has a length x=50x = 50 m and width y=30y = 30 m, both rounded to the nearest 10 m. Calculate the upper bound for the area of the field.

Solution:

1925 m²

Explanation:

To find the upper bound of a product, multiply the upper bounds of the dimensions. For xx: 50±5→UB=5550 \pm 5 \rightarrow UB = 55. For yy: 30±5→UB=3530 \pm 5 \rightarrow UB = 35. Max Area = 55×35=192555 \times 35 = 1925.

Problem 4:

Calculate the lower bound for vv if v=stv = \frac{s}{t}, where s=100s = 100 m (to the nearest meter) and t=20t = 20 s (to the nearest second).

Solution:

4.85 m/s

Explanation:

To find the minimum quotient, divide the lower bound of the numerator by the upper bound of the denominator. LBs=99.5LB_s = 99.5, UBt=20.5UB_t = 20.5. Lower Bound of v=99.520.5≈4.8536...v = \frac{99.5}{20.5} \approx 4.8536... Rounding to 3 s.f. gives 4.854.85.