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Number - Lower and upper bounds

Grade 11IGCSE

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

🔑Concepts

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Lower Bound (LB): The smallest value that a number could have been before it was rounded.

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Upper Bound (UB): The smallest value that would round up to the next unit (the limit of the range).

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Degree of Accuracy: The unit to which a number is rounded (e.g., nearest 10, nearest 0.1, 2 significant figures).

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Error Interval: The range of possible values, expressed as LB≤x<UBLB \le x < UB.

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Operation Rules for Bounds: To maximize a result (Addition/Multiplication), use Upper Bounds. To maximize a subtraction (a−ba - b), use UBa−LBbUB_a - LB_b. To maximize a division (a/ba / b), use UBa/LBbUB_a / LB_b.

📐Formulae

Margin of Error=Degree of Accuracy2\text{Margin of Error} = \frac{\text{Degree of Accuracy}}{2}

Upper Bound (UB)=Measured Value+Margin of Error\text{Upper Bound (UB)} = \text{Measured Value} + \text{Margin of Error}

Lower Bound (LB)=Measured Value−Margin of Error\text{Lower Bound (LB)} = \text{Measured Value} - \text{Margin of Error}

Maximum (A−B)=UBA−LBB\text{Maximum } (A - B) = UB_A - LB_B

Minimum (A−B)=LBA−UBB\text{Minimum } (A - B) = LB_A - UB_B

Maximum (A/B)=UBA/LBB\text{Maximum } (A / B) = UB_A / LB_B

Minimum (A/B)=LBA/UBB\text{Minimum } (A / B) = LB_A / UB_B

💡Examples

Problem 1:

A mass mm is measured as 70 kg, correct to the nearest 10 kg. Find the error interval for mm.

Solution:

65≤m<7565 \le m < 75

Explanation:

The degree of accuracy is 10 kg. The margin of error is 10/2=510 / 2 = 5 kg. Lower Bound = 70−5=6570 - 5 = 65. Upper Bound = 70+5=7570 + 5 = 75. We use ≤\le for the lower bound and << for the upper bound.

Problem 2:

A field is a rectangle with length L=100L = 100 m and width W=50W = 50 m, both correct to the nearest 5 meters. Calculate the upper bound for the area of the field.

Solution:

5512.5 m25512.5 \text{ m}^2

Explanation:

Degree of accuracy is 5m, so margin is 2.52.5m. UBL=100+2.5=102.5UB_L = 100 + 2.5 = 102.5m. UBW=50+2.5=52.5UB_W = 50 + 2.5 = 52.5m. Area UB = UBL×UBW=102.5×52.5=5512.5UB_L \times UB_W = 102.5 \times 52.5 = 5512.5.

Problem 3:

Calculate the lower bound for the value of vv if v=s/tv = s / t, where s=160s = 160 m correct to the nearest 10 m and t=8.2t = 8.2 s correct to 1 decimal place.

Solution:

18.7918.79 (to 2 d.p.)

Explanation:

To find the minimum (LB) of a division, use LBnumerator/UBdenominatorLB_{numerator} / UB_{denominator}. LBs=160−5=155LB_s = 160 - 5 = 155. UBt=8.2+0.05=8.25UB_t = 8.2 + 0.05 = 8.25. Therefore, LBv=155/8.25≈18.7878...LB_v = 155 / 8.25 \approx 18.7878...