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Logarithms of Numbers Less Than One2 Topics
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Theory of Logarithms6 Topics1 Quiz
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Approximation & Accuracy5 Topics1 Quiz
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Sequence2 Topics
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Series2 Topics1 Quiz
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Geometric Progression2 Topics1 Quiz
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Quadratic Equations II2 Topics
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Quadratic Equations III1 Topic
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Quadratic Equations IV3 Topics1 Quiz
Topic Content:
- Calculations 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.
Calculations 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 3.4.1:
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 3.4.2:
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 3.4.3:
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