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## SS2: MATHEMATICS - 1ST TERM

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Lesson 6, Topic 4
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# Limits of Accuracy

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#### 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:

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 cm2

UB of area = UB × UB

= 14.65 × 11.85

= 173.6025 cm2

limit of accuracy for area

170.9625 cm2 ≤ area < 173.6025cm2

### 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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