#### Topic Content:

- Calculation Involving Lower and Upper Bounds

When a quantity is measured, there is always an error, no matter how the measurement is carried out. Note that if a measurement is given to the nearest unit, then the maximum possible error is half of that unit.

### Calculation Involving Lower and Upper Bounds:

**Addition and Multiplication:**

**1.** When two approximate values are added:

Lower bound of the outcome = LB + LB

Upper bound of the outcome = UB + UB

**2.** When two approximate values are multiplied

Lower bound of the outcome = LB × LB

Upper bound of the outcome = UB × UB

**Subtraction and Division**

**3. **When two approximate values are subtracted

Lower bound of the outcome = LB – UB

Upper bound of the outcome = UB – LB

**4.** When two approximate values are divided

Lower bound of the outcome = LB ÷ UB or \( \frac{LB}{UB} \)

Upper bound of the outcome = UB ÷ LB or \( \frac{UB}{LB} \)

(Where LB = Lower bound; UB = Upper bound)

### Example 6.4.1:

Two rods have lengths of 146 cm and 187 cm, both measured to the nearest cm.**1.** Find the lower bound of the total length of both rods **2. **Find the upper bound of the difference between the lengths of both rods when they are placed side by side.

**Solution:**

For the shorter length 146 cm maximum error = ± 0.5

Lower bound = 146 – 0.5 = 145.5 cm

Upper bound = 146 + 0.5 = 146.5 cm

**1.** For the length 187 cm

Lower bound = 187 – 0.5 = 186.5 cm

Upper bound = 187 + 0.5 = 187.5 cm

LB of total length = LB of 146 + LB of 187

= 145.5 + 186.5

= 332 cm

**2. **Upper bound of the difference in lengths

= Upper bound of 187 cm – Lower bound of 146 cm

= 187.5 – 145.5

= 42 cm

### Example 6.4.2:

A rectangular room has sides given as 144.6 cm by 11.8 cm, each length measured to 3 s.f. find the limits of accuracy of the area of the rectangle.

**Solution:**

For the lengths: error = \( \frac{1}{20}\) = 0.05

LB = 14.6 – 0.05 = 14.55 cm

UB = 14.6 + 0.05 = 14.65 cm

limit of accuracy = 14.55 ≤ length < 14.65

For the width

LB = 11.8 – 0.05 = 11.75 cm

UB = 14.6 + 0.05 = 11.85 cm

limit of accuracy = 11.75 ≤ width < 11.85

LB of area = LB × LB

= 14.55 × 11.75

= 170.9625 cm^{2}

UB of area = UB × UB

= 14.65 × 11.85

= 173.6025 cm^{2}

limit of accuracy for area

170.9625 cm^{2} ≤ area < 173.6025cm^{2 }

### Example 6.4.3:

A car travels 8000 m, to the nearest “m” in 5 minutes 28.6 seconds, to the nearest 0.1 seconds, find

**1.** the minimum average speed in m/s

**2.** the maximum average speed in m/s

**Solution**

For speed of the car:

LB = 8000 – 0.5 = 7999.5 cm

UB = 8000 + 0.5 = 8000.5 cm

For the time taken to travel:

5 mins 28.6 seconds = 328.6 sec

LB = 328.6 – 0.05 = 328.55 s

UB = 328.6 + 0.05 = 328.65 s

**1. **Minimum speed = \( \frac {LB\; for\; distance} {UB\; for\; time}\)

= \( \frac {7999.5} {328.65}\)

= 24.340 m/s

**2.** Maximum speed = \( \frac {UB\; for\; distance } {LB\; for\; time}\)

= \( \frac {8000.5}{328.55} \)

= 24.351 m/s

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