Upper and Lower Bounds

Introduction

• In daily life, measurements are often written as rounded numbers instead of exact values.
• When a number is rounded, the real value lies within a range.
• To find this range, we use Upper and Lower Bounds.

Example:

If a length is written as 10 cm (rounded to the nearest whole number), the actual length is not exactly 10 cm. The real value could be a little smaller or a little larger, but it must still round to 10. This means the true value lies in the range:

Lower Bound = 9.5

Upper Bound = 10.5

What are Upper and Lower Bounds?

• When a number is rounded, its exact value is not known. The upper and lower bounds show the smallest and largest possible values the number could be before rounding.
• The degree of accuracy shows the place value a number is rounded to and helps us find its bounds using the error interval.

The general form is:

• We use “$\le$” for the lower bound because it still rounds to the given number (so it is included), and “$<$" for the upper bound because that exact value would round up to the next number (so it is not included).

Solved Example

Problem 1
Find the upper and lower bound to the number 74, rounded off to the nearest integer value.

SOLUTION

1

Draw a number line and find the intervals.

• Place 74 in the middle of the number line.
• Since we are given that the degree of accuracy is an integer, we will add and subtract 1 to find the next and previous interval.

$$74 + 1 = 75$$ $$74 – 1 = 73$$

2

Find the midpoints of these intervals.

To find the upper bound, we calculate:

$$\frac{74+75}{2} = 74.5$$

To find the lower bound, we calculate:

$$\frac{73+74}{2} = 73.5$$

Final Answer: So, we write the inequality:

$$73.5 \le x < 74.5$$

How do we find Upper and Lower Bounds of single values using error intervals?

• To find bounds, we use the degree of accuracy and apply the error interval method by subtracting and adding half of the rounding unit.

Solved Example

Problem 2
Find the upper and lower bound to the number 75.5, rounded off to the nearest 0.5.

SOLUTION

1

Draw the number line and find the intervals.

• Since it is rounded to the nearest 0.5, we add and subtract 0.5.

$$75.5 + 0.5 = 76$$ $$75.5 – 0.5 = 75$$

2

Find the midpoints of these intervals.

Upper Bound:

$$\frac{75.5+76}{2} = 75.75$$

Lower Bound:

$$\frac{75+75.5}{2} = 75.25$$

3

Write the range

Final Answer:

$$75.25 \le x < 75.75$$

How do we find Upper and Lower Bounds in mathematical operations?

• Upper and lower bounds are used in mathematical operations to find the maximum and minimum possible values of an expression involving rounded numbers.

Final Summary:

Solved Examples

Solved Example

Problem 3
A rectangle has a length of 12.4 cm and a width of 8.6 cm, both measured correct to the nearest 0.1 cm. Find the upper and lower bounds of the perimeter and the area.

SOLUTION

Part 1: Find the upper and lower bound of length and width:

For the length:

1

Draw a number line and find the intervals

$$12.4 + 0.1 = 12.5$$ $$12.4 – 0.1 = 12.3$$

2

Find the midpoints of these intervals

Lower Bound:

$$\frac{12.3+12.4}{2} = 12.35$$

Upper Bound:

$$\frac{12.4+12.5}{2} = 12.45$$

For the width:

1

Draw a number line and find the intervals

$$8.6 + 0.1 = 8.7$$ $$8.6 – 0.1 = 8.5$$

2

Find the midpoints of these intervals

Lower Bound:

$$\frac{8.5+8.6}{2} = 8.55$$

Upper Bound:

$$\frac{8.6+8.7}{2} = 8.65$$

Part 2: Find the upper and lower bound of perimeter:

$$Perimeter = 2(length + width)$$

1

Find the upper and lower bounds

Upper Bound:

$$2(12.45 + 8.65) = 2(21.1) = 42.2$$

Lower Bound:

$$2(12.35 + 8.55) = 2(20.9) = 41.8$$

2

Write the range

$$41.8 \le P < 42.2$$

Part 3: Find the upper and lower bound of area:

The area of a rectangle is:

$$Area = length \times width$$

1

Find the upper and lower bounds

$$Upper~Bound: 12.45 \times 8.65 = 107.6925$$ $$Lower~Bound: 12.35 \times 8.55 = 105.5925$$

2

Write the range

Final Answer:

$$105.5925 \le A < 107.6925$$

Solved Example

Problem 4
A resistor has a voltage of 12 volts, correct to the nearest volt, and a resistance of 4 ohms, correct to the nearest 0.1 ohm. Find the upper and lower bounds of the voltage and resistance and then find the upper and lower bounds of the current.

SOLUTION

Part 1: Find the upper and lower bound of voltage and resistance:

For the voltage:

1

Draw a number line and find the intervals

$$12 + 1 = 13$$ $$12 – 1 = 11$$

Intervals for 12

2

Find the midpoints of these intervals

$$Lower~Bound: \frac{11+12}{2} = 11.5$$ $$Upper~Bound: \frac{12+13}{2} = 12.5$$

$$11.5 \le V < 12.5$$

For the resistance:

1

Draw a number line and find the intervals

$$4 + 0.1 = 4.1$$ $$4 – 0.1 = 3.9$$

2

Find the midpoints of these intervals

$$Lower~Bound: \frac{3.9+4}{2} = 3.95$$ $$Upper~Bound: \frac{4+4.1}{2} = 4.05$$

$$3.95 \le R < 4.05$$

Part 2: Find the upper and lower bound of current:

For division:

$$Upper~Bound = \frac{Upper~V}{Lower~R}$$
$$Lower~Bound = \frac{Lower~V}{Upper~R}$$

1

Calculate Upper Bound of Current

$$Upper~Bound~of~current = \frac{12.5}{3.95} = 3.165$$

2

Calculate Lower Bound of Current

$$Lower~Bound~of~current = \frac{11.5}{4.05} = 2.840$$

Solved Example

Problem 5
Ebony makes some bracelets to sell. The materials to make all the bracelets cost £190, correct to the nearest £5. Ebony sells all the bracelets for a total of £875, correct to the nearest £5 The total time taken to make and sell all these bracelets was 72 hours, correct to the nearest hour. Ebony uses this method to calculate her hourly rate of pay:

$$Hourly~rate~of~pay = \frac{total~selling~price – total~cost~of~materials}{total~time~taken}$$

The minimum hourly rate of pay for someone of Ebony’s age is £8.20. By considering bounds, determine if Ebony’s hourly rate of pay was definitely more than £8.20. You must show all your working.

SOLUTION

Bounds of total cost of Materials (£190)

1

Draw a number line and find the intervals

$$190 + 5 = 195$$ $$190 – 5 = 185$$

2

Find the midpoints of these intervals

$$Lower~Bound: \frac{185+190}{2} = 187.5$$ $$Upper~Bound: \frac{190+195}{2} = 192.5$$

$$187.5 \le c < 192.5$$

Bounds of total selling Price (£875)

1

Draw a number line and find the intervals

$$875 + 5 = 880$$ $$875 – 5 = 870$$

2

Find the midpoints of these intervals

$$Lower~Bound: \frac{870+875}{2} = 872.5$$ $$Upper~Bound: \frac{875+880}{2} = 877.5$$

$$872.5 \le S < 877.5$$

Bounds of Time (72 hours)

1

Draw a number line and find the intervals

$$72 + 1 = 73$$ $$72 – 1 = 71$$

2

Find the midpoints of these intervals

$$Lower~Bound: \frac{71+72}{2} = 71.5$$ $$Upper~Bound: \frac{72+73}{2} = 72.5$$

$$71.5 \le T < 72.5$$

Lower Bound of Hourly Pay

• We check the lower bound because we want to see if she definitely earns more than £8.20.

$$Lower~Bound = \frac{Lower~Numerator}{Upper~Denominator}$$

1

Find Lower Bound of Numerator

$$Numerator = Selling~price – Cost$$ $$Lower~numerator: 872.5 – 192.5 = 680$$

2

Use Upper Bound of Time

$$Upper~Bound~of~time = 72.5~hours$$

3

Calculate Lower Bound of Hourly Pay

$$\frac{680}{72.5} = 9.38$$

Final Answer:
Since $9.38 > 8.20$
Ebony is definitely earning more than £8.20 per hour.

Solved Example

Problem 6
A high-speed train travels a distance of 487km in 3 hours. The distance is measured correct to the nearest kilometre. The time is measured correct to the nearest minute. By considering bounds, work out the average speed, in $km/minute$, of the train to a suitable degree of accuracy. You must show all your working and give a reason for your answer.

SOLUTION

1

Convert Time into Minutes

$$3~hours = 3 \times 60 = 180~minutes$$

So, time = 180 minutes, correct to nearest minute.

2

Find Bounds of Distance

• Distance is correct to nearest km.
• Half of $1~km = 0.5~km$

Lower Bound:

$$487 – 0.5 = 486.5$$

Upper Bound:

$$487 + 0.5 = 487.5$$

$$486.5 \le D < 487.5$$

3

Find Bounds of Time

• Time is correct to nearest minute.
• Half of $1~minute = 0.5~minute$

Lower Bound:

$$180 – 0.5 = 179.5$$

Upper Bound:

$$180 + 0.5 = 180.5$$

$$179.5 \le T < 180.5$$

4

Find Upper and Lower Bounds of Speed

Lower Bound of Speed:

$$\frac{Lower~Distance}{Upper~Time}$$

$$= \frac{486.5}{180.5} = 2.695~km/min$$

Upper Bound of Speed:

$$\frac{Upper~Distance}{Lower~Time}$$

$$= \frac{487.5}{179.5}= 2.716~km/min$$

$$2.695 \le Speed < 2.716$$

Final Answer: Both the upper and lower bounds round to 2.7 km/min to 1 decimal place, so this is the suitable degree of accuracy.

Average speed = 2.7 km/min