Approximation and Bounds.
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Lesson Objectives
- Round numbers to a given number of decimal places
- Round numbers to a given number of significant figures
- Understand the concept of upper and lower bounds
- Find upper and lower bounds for numbers rounded to a given accuracy
- Calculate upper and lower bounds for sums, differences, products, and quotients
- Apply approximation and bounds to solve real-world problems
Introduction to Approximation and Bounds
In real life, measurements and calculations are rarely exact. We often need to approximate numbers to make them easier to work with. However, when a number has been rounded, the true value lies within a range. The smallest possible value is called the lower bound, and the largest possible value is called the upper bound. Understanding bounds helps us determine the possible range of answers in calculations involving rounded numbers.
Key Concepts
• Decimal Places (d.p.): Number of digits after the decimal point.
• Significant Figures (s.f.): Number of meaningful digits in a number.
• Lower Bound: The smallest possible value that rounds to the given number.
• Upper Bound: The largest possible value that rounds to the given number.
• Error Interval: The range between the lower and upper bounds.
• Decimal Places (d.p.): Number of digits after the decimal point.
• Significant Figures (s.f.): Number of meaningful digits in a number.
• Lower Bound: The smallest possible value that rounds to the given number.
• Upper Bound: The largest possible value that rounds to the given number.
• Error Interval: The range between the lower and upper bounds.
Key Definitions:
• Rounding: Replacing a number with an approximate value that has fewer digits.
• Decimal Place: A position to the right of the decimal point.
• Significant Figure: A digit that contributes to the precision of a number (non-zero digits are always significant; zeros between non-zero digits are significant; trailing zeros after a decimal point are significant).
• Upper Bound (UB): The maximum possible actual value.
• Lower Bound (LB): The minimum possible actual value.
• Error Interval: $LB \leq x < UB$ (for rounded numbers).
• Rounding: Replacing a number with an approximate value that has fewer digits.
• Decimal Place: A position to the right of the decimal point.
• Significant Figure: A digit that contributes to the precision of a number (non-zero digits are always significant; zeros between non-zero digits are significant; trailing zeros after a decimal point are significant).
• Upper Bound (UB): The maximum possible actual value.
• Lower Bound (LB): The minimum possible actual value.
• Error Interval: $LB \leq x < UB$ (for rounded numbers).
Rounding to Decimal Places
Rounding to a given number of decimal places means keeping that many digits after the decimal point. The digit in the next place determines whether to round up or stay the same.
Step-by-Step Method for Rounding to Decimal Places:
1. Identify the digit in the required decimal place.
2. Look at the digit immediately to the right (the next decimal place).
3. If that digit is 5 or greater, round up (increase the digit by 1).
4. If that digit is 4 or less, round down (keep the digit the same).
5. Remove all digits to the right of the rounding place.
1. Identify the digit in the required decimal place.
2. Look at the digit immediately to the right (the next decimal place).
3. If that digit is 5 or greater, round up (increase the digit by 1).
4. If that digit is 4 or less, round down (keep the digit the same).
5. Remove all digits to the right of the rounding place.
Example 1: Rounding to 1 Decimal Place
Problem: Round 3.47 to 1 decimal place.
Formula: Look at the second decimal digit to decide.
Solution:
Step 1: The first decimal digit is 4.
Step 2: Look at the second decimal digit: 7.
Step 3: 7 ≥ 5, so round up: 4 becomes 5.
Answer: 3.5
Problem: Round 3.47 to 1 decimal place.
Formula: Look at the second decimal digit to decide.
Solution:
Step 1: The first decimal digit is 4.
Step 2: Look at the second decimal digit: 7.
Step 3: 7 ≥ 5, so round up: 4 becomes 5.
Answer: 3.5
Example 2: Rounding to 2 Decimal Places
Problem: Round 4.563 to 2 decimal places.
Solution:
Step 1: The second decimal digit is 6.
Step 2: Look at the third decimal digit: 3.
Step 3: 3 < 5, so round down (keep 6).
Answer: 4.56
Problem: Round 4.563 to 2 decimal places.
Solution:
Step 1: The second decimal digit is 6.
Step 2: Look at the third decimal digit: 3.
Step 3: 3 < 5, so round down (keep 6).
Answer: 4.56
Example 3: Rounding with 9s
Problem: Round 2.998 to 2 decimal places.
Solution:
Step 1: The second decimal digit is 9.
Step 2: Look at the third decimal digit: 8 ≥ 5, so round up.
Step 3: 9 rounds up to 10, so carry over: 2.99 becomes 3.00.
Answer: 3.00
Problem: Round 2.998 to 2 decimal places.
Solution:
Step 1: The second decimal digit is 9.
Step 2: Look at the third decimal digit: 8 ≥ 5, so round up.
Step 3: 9 rounds up to 10, so carry over: 2.99 becomes 3.00.
Answer: 3.00
Practice for Decimal Places
- Round 5.678 to 1 decimal place.
- Round 7.342 to 2 decimal places.
- Round 0.9876 to 3 decimal places.
- Round 12.999 to 2 decimal places.
- Round 0.0056 to 3 decimal places.
Significant Figures
Significant figures indicate the precision of a measurement. Rules for identifying significant figures: non-zero digits are always significant; zeros between non-zero digits are significant; trailing zeros after a decimal point are significant; leading zeros are not significant.
Rules for Significant Figures:
1. Non-zero digits: Always significant (1-9).
2. Zeros between non-zero digits: Always significant (e.g., 105 has 3 s.f.).
3. Leading zeros: Never significant (e.g., 0.005 has 1 s.f.).
4. Trailing zeros after a decimal point: Always significant (e.g., 2.500 has 4 s.f.).
5. Trailing zeros in whole numbers without a decimal point: Ambiguous; use standard form to clarify.
1. Non-zero digits: Always significant (1-9).
2. Zeros between non-zero digits: Always significant (e.g., 105 has 3 s.f.).
3. Leading zeros: Never significant (e.g., 0.005 has 1 s.f.).
4. Trailing zeros after a decimal point: Always significant (e.g., 2.500 has 4 s.f.).
5. Trailing zeros in whole numbers without a decimal point: Ambiguous; use standard form to clarify.
Example 1: Rounding to 3 Significant Figures
Problem: Round 45,678 to 3 significant figures.
Solution:
Step 1: The first three digits are 4,5,6.
Step 2: Look at the fourth digit: 7 ≥ 5, so round up.
Step 3: 456 rounds up to 457.
Step 4: Replace remaining digits with zeros: 45,700.
Answer: 45,700
Problem: Round 45,678 to 3 significant figures.
Solution:
Step 1: The first three digits are 4,5,6.
Step 2: Look at the fourth digit: 7 ≥ 5, so round up.
Step 3: 456 rounds up to 457.
Step 4: Replace remaining digits with zeros: 45,700.
Answer: 45,700
Example 2: Rounding to 2 Significant Figures (Small Number)
Problem: Round 0.004567 to 2 significant figures.
Solution:
Step 1: Leading zeros are not significant. Start from first non-zero: 4.
Step 2: The first two significant digits are 4 and 5.
Step 3: Look at the third significant digit: 6 ≥ 5, so round up.
Step 4: 45 rounds up to 46.
Answer: 0.0046
Problem: Round 0.004567 to 2 significant figures.
Solution:
Step 1: Leading zeros are not significant. Start from first non-zero: 4.
Step 2: The first two significant digits are 4 and 5.
Step 3: Look at the third significant digit: 6 ≥ 5, so round up.
Step 4: 45 rounds up to 46.
Answer: 0.0046
Example 3: Rounding with Trailing Zeros
Problem: Round 2.049 to 2 significant figures.
Solution:
Step 1: The first two significant digits are 2 and 0 (zero between non-zero digits counts).
Step 2: Look at the third significant digit: 4 < 5, so round down.
Answer: 2.0
Problem: Round 2.049 to 2 significant figures.
Solution:
Step 1: The first two significant digits are 2 and 0 (zero between non-zero digits counts).
Step 2: Look at the third significant digit: 4 < 5, so round down.
Answer: 2.0
Example 4: Zero as a Significant Figure
Problem: Round 10.05 to 3 significant figures.
Solution:
Step 1: The digits 1,0,0 are significant (zero between non-zero digits).
Step 2: Look at the fourth digit: 5 ≥ 5, so round up.
Step 3: 100 rounds up to 101.
Answer: 10.1
Problem: Round 10.05 to 3 significant figures.
Solution:
Step 1: The digits 1,0,0 are significant (zero between non-zero digits).
Step 2: Look at the fourth digit: 5 ≥ 5, so round up.
Step 3: 100 rounds up to 101.
Answer: 10.1
Practice for Significant Figures
- Round 6,789 to 2 significant figures.
- Round 0.003456 to 2 significant figures.
- Round 12.34 to 3 significant figures.
- Round 1.099 to 2 significant figures.
- Round 500 to 1 significant figure.
Upper and Lower Bounds
When a number has been rounded, the true value lies within an interval. For a number rounded to the nearest unit (or a given degree of accuracy), the lower bound is halfway down, and the upper bound is halfway up (not inclusive).
Finding Bounds
If a number $x$ is rounded to $n$ decimal places or to the nearest $k$ units:
Lower Bound = Rounded value - $\frac{1}{2} \times$ (degree of accuracy)
Upper Bound = Rounded value + $\frac{1}{2} \times$ (degree of accuracy)
Error interval: $LB \leq x < UB$
If a number $x$ is rounded to $n$ decimal places or to the nearest $k$ units:
Lower Bound = Rounded value - $\frac{1}{2} \times$ (degree of accuracy)
Upper Bound = Rounded value + $\frac{1}{2} \times$ (degree of accuracy)
Error interval: $LB \leq x < UB$
Step-by-Step Method for Finding Bounds:
1. Identify the degree of accuracy (e.g., nearest 10, nearest 0.1, 2 decimal places).
2. The absolute error is half of the degree of accuracy.
3. Lower bound = given value - absolute error.
4. Upper bound = given value + absolute error.
5. Write the error interval as $LB \leq x < UB$.
1. Identify the degree of accuracy (e.g., nearest 10, nearest 0.1, 2 decimal places).
2. The absolute error is half of the degree of accuracy.
3. Lower bound = given value - absolute error.
4. Upper bound = given value + absolute error.
5. Write the error interval as $LB \leq x < UB$.
Example 1: Bounds for a Number Rounded to the Nearest 10
Problem: A number is 70 when rounded to the nearest 10. Find the bounds.
Formula: Degree of accuracy = 10, absolute error = 5
Solution:
Lower bound = 70 - 5 = 65
Upper bound = 70 + 5 = 75
Error interval: $65 \leq x < 75$
Answer: $65 \leq x < 75$
Problem: A number is 70 when rounded to the nearest 10. Find the bounds.
Formula: Degree of accuracy = 10, absolute error = 5
Solution:
Lower bound = 70 - 5 = 65
Upper bound = 70 + 5 = 75
Error interval: $65 \leq x < 75$
Answer: $65 \leq x < 75$
Example 2: Bounds for a Number Rounded to 1 Decimal Place
Problem: A measurement is 3.6 cm to 1 decimal place. Find the bounds.
Formula: Degree of accuracy = 0.1, absolute error = 0.05
Solution:
Lower bound = 3.6 - 0.05 = 3.55
Upper bound = 3.6 + 0.05 = 3.65
Error interval: $3.55 \leq x < 3.65$
Answer: $3.55 \leq x < 3.65$
Problem: A measurement is 3.6 cm to 1 decimal place. Find the bounds.
Formula: Degree of accuracy = 0.1, absolute error = 0.05
Solution:
Lower bound = 3.6 - 0.05 = 3.55
Upper bound = 3.6 + 0.05 = 3.65
Error interval: $3.55 \leq x < 3.65$
Answer: $3.55 \leq x < 3.65$
Example 3: Bounds for a Number Rounded to 2 Significant Figures
Problem: A number is 4,500 to 2 significant figures. Find the bounds.
Solution:
4,500 to 2 s.f. means the number is between 4,450 and 4,550.
Lower bound = 4,450
Upper bound = 4,550
Error interval: $4,450 \leq x < 4,550$
Answer: $4,450 \leq x < 4,550$
Problem: A number is 4,500 to 2 significant figures. Find the bounds.
Solution:
4,500 to 2 s.f. means the number is between 4,450 and 4,550.
Lower bound = 4,450
Upper bound = 4,550
Error interval: $4,450 \leq x < 4,550$
Answer: $4,450 \leq x < 4,550$
Example 4: Bounds for a Number Rounded to 3 Significant Figures
Problem: A number is 0.0450 to 3 significant figures. Find the bounds.
Solution:
0.0450 to 3 s.f. means the last significant digit is in the ten-thousandths place.
Degree of accuracy = 0.0001, absolute error = 0.00005
Lower bound = 0.0450 - 0.00005 = 0.04495
Upper bound = 0.0450 + 0.00005 = 0.04505
Error interval: $0.04495 \leq x < 0.04505$
Answer: $0.04495 \leq x < 0.04505$
Problem: A number is 0.0450 to 3 significant figures. Find the bounds.
Solution:
0.0450 to 3 s.f. means the last significant digit is in the ten-thousandths place.
Degree of accuracy = 0.0001, absolute error = 0.00005
Lower bound = 0.0450 - 0.00005 = 0.04495
Upper bound = 0.0450 + 0.00005 = 0.04505
Error interval: $0.04495 \leq x < 0.04505$
Answer: $0.04495 \leq x < 0.04505$
Watch Out!
For a number rounded to the nearest whole number, the bounds are $\pm 0.5$. For a number rounded to 1 decimal place, the bounds are $\pm 0.05$. The upper bound is always < (strictly less than), not ≤.
For a number rounded to the nearest whole number, the bounds are $\pm 0.5$. For a number rounded to 1 decimal place, the bounds are $\pm 0.05$. The upper bound is always < (strictly less than), not ≤.
Practice for Upper and Lower Bounds
- A number is 50 to the nearest 10. Find the bounds.
- A length is 8.7 cm to 1 decimal place. Find the error interval.
- A weight is 2.50 kg to 2 decimal places. Find the bounds.
- A distance is 12,000 m to 2 significant figures. Find the bounds.
- A mass is 0.060 kg to 2 significant figures. Find the bounds.
Bounds in Calculations
When performing calculations with rounded numbers, the result also has bounds. The maximum possible result comes from using the upper bounds appropriately, and the minimum possible result comes from using the lower bounds.
Rules for Bounds in Calculations:
For addition and subtraction: LB of sum = LB₁ + LB₂, UB of sum = UB₁ + UB₂
For multiplication and division:
- Maximum value: use UB for numerator and LB for denominator (or vice versa depending on operation)
- Minimum value: use LB for numerator and UB for denominator
For addition and subtraction: LB of sum = LB₁ + LB₂, UB of sum = UB₁ + UB₂
For multiplication and division:
- Maximum value: use UB for numerator and LB for denominator (or vice versa depending on operation)
- Minimum value: use LB for numerator and UB for denominator
Example 1: Bounds for Addition
Problem: Two lengths are 5.6 cm and 3.2 cm, each measured to 1 decimal place. Find the bounds for the total length.
Formula: LB_total = LB₁ + LB₂, UB_total = UB₁ + UB₂
Solution:
For 5.6: LB = 5.55, UB = 5.65
For 3.2: LB = 3.15, UB = 3.25
LB total = 5.55 + 3.15 = 8.70
UB total = 5.65 + 3.25 = 8.90
Answer: $8.70 \leq \text{total} < 8.90$ cm
Problem: Two lengths are 5.6 cm and 3.2 cm, each measured to 1 decimal place. Find the bounds for the total length.
Formula: LB_total = LB₁ + LB₂, UB_total = UB₁ + UB₂
Solution:
For 5.6: LB = 5.55, UB = 5.65
For 3.2: LB = 3.15, UB = 3.25
LB total = 5.55 + 3.15 = 8.70
UB total = 5.65 + 3.25 = 8.90
Answer: $8.70 \leq \text{total} < 8.90$ cm
Example 2: Bounds for Subtraction
Problem: Two weights are 12.5 kg and 4.7 kg, each to 1 decimal place. Find the bounds for the difference.
Formula: LB_diff = LB₁ - UB₂, UB_diff = UB₁ - LB₂
Solution:
For 12.5: LB = 12.45, UB = 12.55
For 4.7: LB = 4.65, UB = 4.75
LB difference = 12.45 - 4.75 = 7.70
UB difference = 12.55 - 4.65 = 7.90
Answer: $7.70 \leq \text{difference} < 7.90$ kg
Problem: Two weights are 12.5 kg and 4.7 kg, each to 1 decimal place. Find the bounds for the difference.
Formula: LB_diff = LB₁ - UB₂, UB_diff = UB₁ - LB₂
Solution:
For 12.5: LB = 12.45, UB = 12.55
For 4.7: LB = 4.65, UB = 4.75
LB difference = 12.45 - 4.75 = 7.70
UB difference = 12.55 - 4.65 = 7.90
Answer: $7.70 \leq \text{difference} < 7.90$ kg
Example 3: Bounds for Multiplication
Problem: A rectangle has length 8.5 cm and width 4.2 cm, each to 1 decimal place. Find the bounds for the area.
Formula: LB_area = LB_len × LB_wid, UB_area = UB_len × UB_wid
Solution:
For 8.5: LB = 8.45, UB = 8.55
For 4.2: LB = 4.15, UB = 4.25
LB area = 8.45 × 4.15 = 35.0675
UB area = 8.55 × 4.25 = 36.3375
Answer: $35.07 \leq \text{area} < 36.34$ cm² (rounded)
Problem: A rectangle has length 8.5 cm and width 4.2 cm, each to 1 decimal place. Find the bounds for the area.
Formula: LB_area = LB_len × LB_wid, UB_area = UB_len × UB_wid
Solution:
For 8.5: LB = 8.45, UB = 8.55
For 4.2: LB = 4.15, UB = 4.25
LB area = 8.45 × 4.15 = 35.0675
UB area = 8.55 × 4.25 = 36.3375
Answer: $35.07 \leq \text{area} < 36.34$ cm² (rounded)
Example 4: Bounds for Division
Problem: A car travels 250 km (to the nearest 10 km) in 4 hours (to the nearest hour). Find the bounds for the average speed.
Formula: Speed = distance ÷ time
Max speed = UB_distance ÷ LB_time
Min speed = LB_distance ÷ UB_time
Solution:
Distance 250 to nearest 10: LB = 245, UB = 255
Time 4 to nearest hour: LB = 3.5, UB = 4.5
Max speed = 255 ÷ 3.5 = 72.86 km/h
Min speed = 245 ÷ 4.5 = 54.44 km/h
Answer: $54.44 \leq \text{speed} < 72.86$ km/h
Problem: A car travels 250 km (to the nearest 10 km) in 4 hours (to the nearest hour). Find the bounds for the average speed.
Formula: Speed = distance ÷ time
Max speed = UB_distance ÷ LB_time
Min speed = LB_distance ÷ UB_time
Solution:
Distance 250 to nearest 10: LB = 245, UB = 255
Time 4 to nearest hour: LB = 3.5, UB = 4.5
Max speed = 255 ÷ 3.5 = 72.86 km/h
Min speed = 245 ÷ 4.5 = 54.44 km/h
Answer: $54.44 \leq \text{speed} < 72.86$ km/h
Watch Out!
For division, to get the maximum result, divide the largest possible numerator by the smallest possible denominator. To get the minimum result, divide the smallest possible numerator by the largest possible denominator.
For division, to get the maximum result, divide the largest possible numerator by the smallest possible denominator. To get the minimum result, divide the smallest possible numerator by the largest possible denominator.
Practice for Bounds in Calculations
- Add 15.6 and 8.3 (both to 1 d.p.). Find bounds for the sum.
- Subtract 23.4 from 56.7 (both to 1 d.p.). Find bounds for the difference.
- Multiply 6.5 by 3.2 (both to 1 d.p.). Find bounds for the product.
- Divide 120 (to nearest 10) by 5 (to nearest whole number). Find bounds for the quotient.
- A field is 45.0 m by 32.0 m (each to 1 d.p.). Find bounds for the area.
Methods & Techniques
Verification / Checking Strategy:
1. For rounding: Check that the rounded number is reasonable by comparing to the original.
2. For bounds: Verify that the original rounded number lies within the error interval.
3. For calculations with bounds: Check that the calculated bounds are consistent (LB ≤ exact value ≤ UB).
4. Use a number line: Visualising the bounds on a number line helps understand the interval.
1. For rounding: Check that the rounded number is reasonable by comparing to the original.
2. For bounds: Verify that the original rounded number lies within the error interval.
3. For calculations with bounds: Check that the calculated bounds are consistent (LB ≤ exact value ≤ UB).
4. Use a number line: Visualising the bounds on a number line helps understand the interval.
Summary: Rounding and Bounds
| Rounded to | Degree of Accuracy | Absolute Error | Lower Bound | Upper Bound |
|---|---|---|---|---|
| Nearest 10 | 10 | 5 | Value - 5 | Value + 5 |
| 1 decimal place | 0.1 | 0.05 | Value - 0.05 | Value + 0.05 |
| 2 decimal places | 0.01 | 0.005 | Value - 0.005 | Value + 0.005 |
| 1 significant figure | Varies | Half of place value of last s.f. | Value - half | Value + half |
Common Pitfalls & How to Avoid Them:
• Pitfall 1: Confusing decimal places with significant figures → Solution: Decimal places count digits after decimal point; significant figures count all meaningful digits.
• Pitfall 2: Forgetting that the upper bound is not inclusive (< not ≤) → Solution: Always write $LB \leq x < UB$.
• Pitfall 3: Using the wrong bounds in division (max = UB/UB) → Solution: Max = UB ÷ LB, Min = LB ÷ UB.
• Pitfall 4: Misidentifying significant figures in numbers with zeros → Solution: Apply the rules carefully: zeros between non-zero digits count; leading zeros don't.
• Pitfall 5: Rounding intermediate steps in bound calculations → Solution: Keep full precision until the final answer.
• Pitfall 1: Confusing decimal places with significant figures → Solution: Decimal places count digits after decimal point; significant figures count all meaningful digits.
• Pitfall 2: Forgetting that the upper bound is not inclusive (< not ≤) → Solution: Always write $LB \leq x < UB$.
• Pitfall 3: Using the wrong bounds in division (max = UB/UB) → Solution: Max = UB ÷ LB, Min = LB ÷ UB.
• Pitfall 4: Misidentifying significant figures in numbers with zeros → Solution: Apply the rules carefully: zeros between non-zero digits count; leading zeros don't.
• Pitfall 5: Rounding intermediate steps in bound calculations → Solution: Keep full precision until the final answer.
Technique Practice
- Verify that 4.567 rounded to 2 decimal places is 4.57.
- Check that the bounds for 3.4 (1 d.p.) are $3.35 \leq x < 3.45$.
- Identify the error: A student said the upper bound for 5.0 (1 d.p.) is 5.05. Is this correct?
- For a number 45 to 2 s.f., is the lower bound 44.5 or 45? Explain.
Real-World Applications
Application 1: Construction and Measurement
Scenario: A builder needs a beam of length 2.5 m to the nearest 0.1 m. What are the possible actual lengths?
Problem: Find the error interval.
Solution:
2.5 m to 1 d.p. → degree of accuracy = 0.1 m, absolute error = 0.05 m
LB = 2.45 m, UB = 2.55 m
Answer: The actual length is between 2.45 m and 2.55 m.
Scenario: A builder needs a beam of length 2.5 m to the nearest 0.1 m. What are the possible actual lengths?
Problem: Find the error interval.
Solution:
2.5 m to 1 d.p. → degree of accuracy = 0.1 m, absolute error = 0.05 m
LB = 2.45 m, UB = 2.55 m
Answer: The actual length is between 2.45 m and 2.55 m.
Application 2: Speed Estimation
Scenario: A car travels 240 km (to nearest 10 km) in 3 hours (to nearest hour). Estimate the maximum and minimum possible speed.
Problem: Find bounds for speed.
Solution:
Distance: LB = 235 km, UB = 245 km
Time: LB = 2.5 h, UB = 3.5 h
Max speed = 245 ÷ 2.5 = 98 km/h
Min speed = 235 ÷ 3.5 ≈ 67.1 km/h
Answer: Speed between 67.1 km/h and 98 km/h.
Scenario: A car travels 240 km (to nearest 10 km) in 3 hours (to nearest hour). Estimate the maximum and minimum possible speed.
Problem: Find bounds for speed.
Solution:
Distance: LB = 235 km, UB = 245 km
Time: LB = 2.5 h, UB = 3.5 h
Max speed = 245 ÷ 2.5 = 98 km/h
Min speed = 235 ÷ 3.5 ≈ 67.1 km/h
Answer: Speed between 67.1 km/h and 98 km/h.
Application 3: Shopping and Budgeting
Scenario: Items cost ₦12.50 and ₦7.80 (each to nearest 0.01). What is the maximum possible total?
Problem: Find the upper bound for the sum.
Solution:
UB for ₦12.50 = 12.505, UB for ₦7.80 = 7.805
UB total = 12.505 + 7.805 = 20.31
Answer: Maximum total is ₦20.31.
Scenario: Items cost ₦12.50 and ₦7.80 (each to nearest 0.01). What is the maximum possible total?
Problem: Find the upper bound for the sum.
Solution:
UB for ₦12.50 = 12.505, UB for ₦7.80 = 7.805
UB total = 12.505 + 7.805 = 20.31
Answer: Maximum total is ₦20.31.
Application 4: Population Estimates
Scenario: A city population is 250,000 to 2 significant figures. Find the error interval.
Problem: Find bounds.
Solution:
250,000 to 2 s.f. → the number is between 245,000 and 255,000
Answer: $245,000 \leq \text{population} < 255,000$
Scenario: A city population is 250,000 to 2 significant figures. Find the error interval.
Problem: Find bounds.
Solution:
250,000 to 2 s.f. → the number is between 245,000 and 255,000
Answer: $245,000 \leq \text{population} < 255,000$
Cross-Curricular Connections
- Science: Measurement uncertainty, experimental error, significant figures in data
- Engineering: Tolerances, safety margins, precision in manufacturing
- Economics: Rounding in financial statements, GDP estimates
- Geography: Population estimates, area measurements
Cumulative Practice Exercises
- Round 4.5678 to 2 decimal places.
- Round 0.003456 to 2 significant figures.
- Round 12,345 to 3 significant figures.
- A number is 80 to the nearest 10. Find the bounds.
- A measurement is 5.67 cm to 2 decimal places. Find the error interval.
- A distance is 4,500 m to 2 significant figures. Find the bounds.
- Add 12.3 and 7.8 (both to 1 d.p.). Find bounds for the sum.
- Subtract 45.6 from 89.2 (both to 1 d.p.). Find bounds for the difference.
- Multiply 8.4 by 3.6 (both to 1 d.p.). Find bounds for the product.
- Divide 200 (to nearest 10) by 5 (to nearest whole number). Find bounds for the quotient.
- A rectangle has length 15.0 cm and width 8.0 cm (each to 1 d.p.). Find bounds for the area.
- Error analysis: A student said 0.0056 rounded to 2 significant figures is 0.0056. Is this correct? Explain.
- A car travels 180 km (to nearest 10 km) in 2.5 hours (to 1 d.p.). Find the maximum possible speed.
- A population is 1,200,000 to 2 significant figures. Find the error interval.
- A weight is 75 kg to 2 significant figures. Find the bounds.
Answers to Cumulative Exercises
- Answer: 4.57
- Answer: 0.0035
- Answer: 12,300
- Answer: $75 \leq x < 85$
- Answer: $5.565 \leq x < 5.675$
- Answer: $4,450 \leq x < 4,550$
- Answer: $20.0 \leq \text{sum} < 20.2$
- Answer: $43.5 \leq \text{diff} < 43.7$
- Answer: $30.0 \leq \text{product} < 30.4$
- Answer: Min = 195 ÷ 5.5 ≈ 35.5, Max = 205 ÷ 4.5 ≈ 45.6 → $35.5 \leq \text{quotient} < 45.6$
- Answer: $119.0 \leq \text{area} < 121.0$ cm²
- Answer: Incorrect. 0.0056 to 2 s.f. is 0.0056 (already has 2 s.f. - the 5 and 6). The student is actually correct.
- Answer: Max speed = 185 ÷ 2.45 ≈ 75.5 km/h
- Answer: $1,150,000 \leq \text{population} < 1,250,000$
- Answer: $74 \leq x < 76$ (if 75 to 2 s.f., means between 74 and 76)
Conclusion & Summary
Approximation and bounds are essential for working with measured or rounded numbers. Rounding to decimal places and significant figures makes numbers easier to work with, while understanding upper and lower bounds allows us to know the possible range of true values. When performing calculations with rounded numbers, we can find the bounds of the result using the appropriate rules for addition, subtraction, multiplication, and division.
Key Takeaways:
1. Decimal places: Count digits after the decimal point.
2. Significant figures: Non-zero digits are always significant; zeros between are significant; leading zeros are not.
3. Bounds: LB = value - half of degree of accuracy, UB = value + half of degree of accuracy.
4. Error interval: $LB \leq x < UB$
5. Addition/Subtraction bounds: Use LB₁ ± LB₂ and UB₁ ± UB₂ appropriately.
6. Multiplication/Division bounds: For max, use UB/UB or UB/LB depending on operation; for min, use LB/LB or LB/UB.
Keep practising with different degrees of accuracy. Understanding bounds is crucial for science, engineering, and real-world measurements!
Video Resource
Watch this video for more examples of rounding, significant figures, and upper/lower bounds.
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