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Number - Upper and lower bounds-extended

Grade 10IB

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

🔑Concepts

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Upper and lower bounds represent the range of possible values that a number could have been before it was rounded to a specific degree of accuracy.

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The degree of accuracy is the unit to which the number was rounded (e.g., nearest 1010, nearest 0.10.1, or 22 significant figures).

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To find the bounds, take the degree of accuracy, divide it by 22, and then add it to the rounded value for the Upper Bound (UBUB) and subtract it for the Lower Bound (LBLB).

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The error interval is written as LB≤x<UBLB \le x < UB. Note that the Upper Bound is the value that the number is strictly less than.

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For addition: UB=UB1+UB2UB = UB_1 + UB_2 and LB=LB1+LB2LB = LB_1 + LB_2.

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For subtraction: UB=UB1−LB2UB = UB_1 - LB_2 and LB=LB1−UB2LB = LB_1 - UB_2.

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For multiplication: UB=UB1×UB2UB = UB_1 \times UB_2 and LB=LB1×LB2LB = LB_1 \times LB_2 (assuming positive values).

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For division: UB=UB1LB2UB = \frac{UB_1}{LB_2} and LB=LB1UB2LB = \frac{LB_1}{UB_2}.

📐Formulae

LB=Value−12(Degree of Accuracy)LB = \text{Value} - \frac{1}{2}(\text{Degree of Accuracy})

UB=Value+12(Degree of Accuracy)UB = \text{Value} + \frac{1}{2}(\text{Degree of Accuracy})

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

Maximum Value of ab=UBaLBb\text{Maximum Value of } \frac{a}{b} = \frac{UB_a}{LB_b}

Minimum Value of ab=LBaUBb\text{Minimum Value of } \frac{a}{b} = \frac{LB_a}{UB_b}

Maximum Value of (a−b)=UBa−LBb\text{Maximum Value of } (a - b) = UB_a - LB_b

Minimum Value of (a−b)=LBa−UBb\text{Minimum Value of } (a - b) = LB_a - UB_b

💡Examples

Problem 1:

A rectangular field has a length L=50 mL = 50\text{ m} and a width W=30 mW = 30\text{ m}, both measured to the nearest 5 m5\text{ m}. Calculate the upper bound for the area of the field.

Solution:

  1. Find the bounds for LL and WW: Degree of accuracy = 5 m5\text{ m}. Variation = 52=2.5 m\frac{5}{2} = 2.5\text{ m}. UBL=50+2.5=52.5 mUB_L = 50 + 2.5 = 52.5\text{ m}. UBW=30+2.5=32.5 mUB_W = 30 + 2.5 = 32.5\text{ m}.
  2. Calculate the upper bound for Area (A=L×WA = L \times W): UBA=UBL×UBWUB_A = UB_L \times UB_W UBA=52.5×32.5=1706.25 m2UB_A = 52.5 \times 32.5 = 1706.25\text{ m}^2.

Explanation:

To maximize a product, we multiply the upper bounds of both variables.

Problem 2:

Given x=12.4x = 12.4 correct to 11 decimal place and y=8y = 8 correct to the nearest integer. Find the lower bound of z=x−yz = x - y.

Solution:

  1. Find bounds for xx: Accuracy = 0.10.1. LBx=12.4−0.05=12.35LB_x = 12.4 - 0.05 = 12.35, UBx=12.4+0.05=12.45UB_x = 12.4 + 0.05 = 12.45.
  2. Find bounds for yy: Accuracy = 11. LBy=8−0.5=7.5LB_y = 8 - 0.5 = 7.5, UBy=8+0.5=8.5UB_y = 8 + 0.5 = 8.5.
  3. Find the lower bound for z=x−yz = x - y: LBz=LBx−UByLB_z = LB_x - UB_y LBz=12.35−8.5=3.85LB_z = 12.35 - 8.5 = 3.85.

Explanation:

To find the minimum result of a subtraction, subtract the largest possible value of the second number from the smallest possible value of the first number.

Problem 3:

Calculate the upper bound for vv if v=stv = \frac{s}{t}, where s=200 ms = 200\text{ m} correct to 22 significant figures and t=10.0 st = 10.0\text{ s} correct to 11 decimal place.

Solution:

  1. Find bounds for ss: s=200s = 200 to 22 sig figs means accuracy is to the nearest 1010. LBs=195,UBs=205LB_s = 195, UB_s = 205.
  2. Find bounds for tt: t=10.0t = 10.0 to 11 d.p. means accuracy is to the nearest 0.10.1. LBt=9.95,UBt=10.05LB_t = 9.95, UB_t = 10.05.
  3. Find UBvUB_v: UBv=UBsLBt=2059.95≈20.603 m/sUB_v = \frac{UB_s}{LB_t} = \frac{205}{9.95} \approx 20.603\text{ m/s}.

Explanation:

To maximize a fraction, use the largest possible numerator (Upper Bound) and the smallest possible denominator (Lower Bound).