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Number - Upper and Lower Bounds

Grade 10IGCSE

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

🔑Concepts

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A measurement is only accurate to a certain degree (e.g., nearest cm, 1 decimal place, 2 significant figures).

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The Lower Bound (LB) is the smallest value that would round to the given measurement.

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The Upper Bound (UB) is the smallest value that would round to the next unit up (it acts as the limit).

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The 'Error Interval' represents the range of possible values, written as: LB≤x<UBLB \le x < UB.

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To find the bounds, determine the unit of accuracy and divide it by 2. Add/subtract this from the value.

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For calculations, to maximize a result (UB): Use UB for addition/multiplication, but for subtraction/division, use the UB of the first number and the LB of the second.

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For calculations, to minimize a result (LB): Use LB for addition/multiplication, but for subtraction/division, use the LB of the first number and the UB of the second.

📐Formulae

Upper Bound (UB)=Value+Degree of Accuracy2\text{Upper Bound (UB)} = \text{Value} + \frac{\text{Degree of Accuracy}}{2}

Lower Bound (LB)=Value−Degree of Accuracy2\text{Lower Bound (LB)} = \text{Value} - \frac{\text{Degree of Accuracy}}{2}

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

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

Maximum Difference=UB1−LB2\text{Maximum Difference} = UB_1 - LB_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}}}

💡Examples

Problem 1:

The length of a table is given as 140 cm140\text{ cm} correct to the nearest 10 cm10\text{ cm}. Find the error interval for the length LL.

Solution:

135≤L<145135 \le L < 145

Explanation:

The degree of accuracy is 10 cm10\text{ cm}. Half of this is 5 cm5\text{ cm}. Lower Bound: 140−5=135140 - 5 = 135. Upper Bound: 140+5=145140 + 5 = 145.

Problem 2:

A car travels a distance of d=200 kmd = 200\text{ km} (to the nearest 10 km10\text{ km}) in a time t=4.0 hourst = 4.0\text{ hours} (to the nearest 0.1 hour0.1\text{ hour}). Calculate the upper bound for the average speed.

Solution:

SpeedUB=2053.95≈51.8987 km/hSpeed_{UB} = \frac{205}{3.95} \approx 51.8987\text{ km/h}

Explanation:

To find the maximum speed, we need the largest distance divided by the shortest time (UBd/LBtUB_d / LB_t). Distance dd: 200±5→UB=205200 \pm 5 \rightarrow UB = 205. Time tt: 4.0±0.05→LB=3.954.0 \pm 0.05 \rightarrow LB = 3.95.

Problem 3:

A rectangle has a length of 8.4 cm8.4\text{ cm} and a width of 5.2 cm5.2\text{ cm}, both rounded to 11 decimal place. Calculate the lower bound of the area.

Solution:

AreaLB=8.35×5.15=43.0025 cm2Area_{LB} = 8.35 \times 5.15 = 43.0025\text{ cm}^2

Explanation:

To find the minimum area, multiply the lower bounds of both dimensions. Accuracy is 0.10.1, so we add/subtract 0.050.05. Length LB: 8.4−0.05=8.358.4 - 0.05 = 8.35. Width LB: 5.2−0.05=5.155.2 - 0.05 = 5.15.