This same concept is applied in mathematics. There is a limit in which a measurement or value cannot go beyond and above. In this article, we will learn about lower and upper-bound limits of accuracy, their definition, rules, and formulas, and see examples of their applications.

## Lower and Upper bounds definition

The **lower bound** (LB) refers to the lowest number that can be rounded to get an estimated value.

The **upper bound** (UB) refers to the highest number that can be rounded to get an estimated value.

Another term that you'll come across in this topic is **error interval.**

**Error intervals** show the range of numbers that are within the limits of accuracy. They are written in the form of inequalities.

The lower and upper bounds can also be called the** limits of accuracy**.

Consider a number 50 rounded to the nearest 10.

Many numbers can be rounded to get 50, but the lowest is 45. This means that the lower bound is 45 because it is the lowest number that can be rounded to get 50.

The upper bound is 54 because it is the highest number that can be rounded to get 50.

As explained earlier, the lower and upper bound can be found by just figuring out the lowest and highest number that can be rounded to get the estimated value, but there is a simple procedure that you can follow to achieve this. The steps are below.

1. You should first know the degree of accuracy, DA.

The **degree of accuracy** is the measure to which a value is rounded.

2. Divide the degree of accuracy by 2,

$\frac{DA}{2}$.

3. Add what you got to the value to get the upper bound, and subtract to get the lower bound.

$Lowerbound=Value-\frac{DA}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Upperbound=Value+\frac{DA}{2}$

## Rules and formulas for upper and lower bounds

You may come across questions involving formulas, and you will have to work with multiplication, division, addition, and subtraction. In cases like this, you have to follow some rules to get the correct answers.

### For Addition.

This usually happens when we have a value that undergoes an increase. We then have an original value and its range of increase.

When you have a question involving addition, do the following:

1. Find the upper and lower bounds of the original value, UB_{value}, and of its range of increase, UB_{range}.

2. Use the following formulas to find the upper and lower bounds of the answer.

$U{B}_{new}=U{B}_{value}+U{B}_{range}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}L{B}_{new}=L{B}_{value}+L{B}_{range}$

3. Considering the bounds, decide on a suitable degree of accuracy for your answer.

### For Subtraction.

This usually happens when we have a value that undergoes a decrease. We then have an original value and its range of decrease.

When you have a question involving subtraction, do the following.

1. Find the upper and lower bounds of the original value, UB_{value}, and of its range of increase, UB_{range}.

2. Use the following formulas to find the upper and lower bounds of the answer.

$U{B}_{new}=U{B}_{value}-U{B}_{range}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}L{B}_{new}=L{B}_{value}-L{B}_{range}$

3. Considering the bounds, decide on a suitable degree of accuracy for your answer.

### For Multiplication.

This usually happens when we have quantities that involve the multiplication of other quantities, such as areas, volumes, and forces.

When you have a question involving multiplication, do the following.

1. Find the upper and lower bounds of the numbers involved. Let them be quantity 1, q1, and quantity 2, q2.

2. Use the following formulas to find the upper and lower bounds of the answer.

$U{B}_{new}=U{B}_{q1}\times U{B}_{q2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}L{B}_{new}=L{B}_{q1}\times L{B}_{q2}$

3. Considering the bounds, decide on a suitable degree of accuracy for your answer.

### For Division.

Similarly to the multiplication, this usually happens when we have a quantity that involves the division of other quantities, such as velocity, and density.

When you have a question involving division, do the following.

1. Find the upper and lower bounds of the numbers involved. Let's denote them quantity 1, q1, and quantity 2, q2.

2. Use the following formulas to find the upper and lower bounds of the answer.

$U{B}_{new}=\frac{U{B}_{q1}}{L{B}_{q2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}L{B}_{new}=\frac{L{B}_{q1}}{U{B}_{q2}}$

3. Considering the bounds, decide on a suitable degree of accuracy for your answer.

## Upper and Lower bounds examples

Let's take some examples.

Find the upper and lower bound of the number 40 rounded to the nearest 10.

**Solution.**

There are lots of values that could be rounded to 40 to the nearest 10. It can be 37, 39, 42.5, 43, 44.9, 44.9999, and so on.

But the lowest number which will be the lower bound is 35 and the highest number is 44.4444, so we will say the upper bound is 44.

Let's call the number that we start with, 40, $x$. The error interval will be:

$35\le x<45$This means x can be equal to or more than 35, but less than 44.

Let's take another example, now following the steps we've mentioned earlier.

The length of an object y is 250 cm long, rounded to the nearest 10 cm. What is the error interval for y?

**Solution.**

To know the error interval, you have to first find the upper and lower bound. Let's use the steps we mentioned earlier to get this.

**Step 1:** First, we have to know the degree of accuracy, DA. From the question, the degree of accuracy is DA = 10 cm.

**Step 2:** The next step is to divide it by 2.

$\frac{DA}{2}=\frac{10}{2}=5$

**Step 3:** We will now subtract and add 5 to 250 to get the lower and upper bound.

$Upperbound=value+\frac{Da}{2}=250+5=255\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Lowerbound=value+\frac{Da}{2}=250-5=245$

The error interval will be:

$245\le y<255$

This means that the length of the object can be equal to or more than 245 cm, but less than 255 cm.

Let's take an example involving addition.

The length of a rope x is 33.7 cm. The length is to be increased by 15.5 cm. Considering the bounds, what will be the new length of the rope?

**Solution.**

This is a case of addition. So, following the steps for addition above, the first thing is to find the upper and lower bounds for the values involved.

**Step 1:** Let's start with the original length of the rope.

The lowest number that can be rounded to 33.7 is 33.65, meaning that 33.65 is the lower bound, LB_{value}.

The highest number is 33.74, but we will use 33.75 which can be rounded down to 33.7, UB_{value}.

So, we can write the error interval as:

$33.65\le x<33.75$

We will do the same for 15.5 cm, let's denote it y.

The lowest number that can be rounded to 15.5 is 15.45 meaning that 15.45 is the lower bound, LB_{range}.

The highest number is 15.54, but we will use 15.55 which can be rounded down to 15.5, UB_{range}.

So, we can write the error interval as:

$15.45\le y\le 15.55$

**Step 2:** We will use the formulas for finding upper and lower bounds for addition.

$U{B}_{new}=U{B}_{value}+U{B}_{range}$

We are to add both upper bounds together.

$U{B}_{new}=33.75+15.55\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=49.3cm$

The lower bound is:

$L{B}_{new}=L{B}_{value}+L{B}_{range}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=33.65+15.45\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=49.1cm$

**Step 3:** We now have to decide what the new length will be using the upper and lower bound we just calculated.

The question we should be asking ourselves is to what degree of accuracy does the upper and lower bound round to the same number? That will be the new length.

Well, we have 49.3 and 49.1 and they both round to 49 at 1 decimal place. Therefore, the new length is 49 cm.

Let's take another example involving multiplication.

The length L of a rectangle is 5.74 cm and the breadth B is 3.3 cm. What is the upper bound of the area of the rectangle to 2 decimal places?

**Solution.**

**Step 1:** First thing is to get the error interval for the length and breadth of the rectangle.

The lowest number that can be rounded to the length of 5.74 is 5.735 meaning that 5.735 is the lower bound, LB_{value}.

The highest number is 5.744, but we will use 5.745 which can be rounded down to 5.74, UB_{value}.

So, we can write the error interval as:

$5.735\le L\le 5.745$

The lowest number that can be rounded to the breadth of 3.3 is 3.25 meaning that 3.25 is the lower bound.

The highest number is 3.34, but we will use 3.35, so we can write the error interval as:

$3.25\le B\le 3.35$

The area of a rectangle is: $Length\times Breadth$

**Step 2:** So to get the upper bound, we will use the upper bound formula for multiplication.

$U{B}_{new}=U{B}_{value}\times U{B}_{range}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=5.745\times 3.35\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=19.24575cm$

**Step 3:** The question says to get the answer in 2 decimal places. Therefore, the upper bound is:

$U{B}_{new}=19.25cm$

Let's take another example involving division.

A man runs 14.8 km in 4.25 hrs. Find the upper and lower bounds of the man's speed. Give your answer in 2 decimal places.

**Solution**

We are asked to find the speed, and the formula for finding speed is:

$Speed=\frac{Dis\mathrm{tan}ce}{Time}=\frac{d}{t}$

**Step 1:** We will first find the upper and lower bounds of the numbers involved.

The distance is 14.8 and the lowest number that can be rounded to 14.8 is 14.75 meaning that 14.75 is the lower bound, LB_{d}.

The highest number is 14.84, but we will use 14.85 which can be rounded down to 14.8, UB_{d}.

So, we can write the error interval as:

$14.75\le d<14.85$

The speed is 4.25 and the lowest number that can be rounded to 4.25 is 4.245 meaning that 4.245 is the lower bound, LB_{t}.

The highest number is 4.254, but we will use 4.255 (which can be rounded down to 4.25), UB_{t}, so we can write the error interval as:

$4.245\le t<4.255$

**Step 2:** We are dealing with division here. So, we will use the division formula for calculating the upper and lower bound.

$U{B}_{new}=\frac{U{B}_{d}}{L{B}_{t}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{14.85}{4.245}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=3.4982\approx 3.50(2d.p.)$

The lower bound of the man's speed is:

$L{B}_{new}=\frac{L{B}_{d}}{U{B}_{t}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{14.75}{4.255}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=0.4665\approx 0.47(2d.p.)$

$\approx $ is the symbol for approximation.

**Step 3:** The answers for the upper and lower bound are approximated because we are to give our answer in 2 decimal places.

Therefore, the upper and lower bound for the man's speed are 3.50 km/hr and 0.47 km/hr respectively.

Let's take one more example.

The height of a door is 93 cm to the nearest centimetre. Find the upper and lower bounds of the height.

**Solution.**

The first step is to determine the degree of accuracy. The degree of accuracy is to the nearest 1 cm.

Knowing that the next step is to divide by 2.

$\frac{1}{2}=0.5$To find the upper and lower bound, we will add and subtract 0,5 from 93 cm.

The Upper bound is:

$UB=93+0.5\phantom{\rule{0ex}{0ex}}=93.5cm$

The Lower bound is:

$LB=93-0.5\phantom{\rule{0ex}{0ex}}=92.5cm$

## Lower and Upper bound limits of accuracy - Key takeaways

- The lower bound refers to the lowest number that can be rounded to get an estimated value.
- The upper bound refers to the highest number that can be rounded to get an estimated value.
- Error intervals show the range of numbers that are within the limits of accuracy. They are written in the form of inequalities.
- The lower and upper bounds can also be called the
**limits of accuracy**.

###### Learn with 5 Lower and Upper Bounds flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Lower and Upper Bounds

What are upper and lower bounds?

Upper bound refers to the highest number that can be rounded to get an estimated value.

Lower bound refers to the lowest number that can be rounded to get an estimated value.

How do you find upper and lower bounds?

The following steps can be used to find upper and lower bounds.

- You should first know the degree of accuracy is. The degree of accuracy is the measure to which a value is rounded.
- Divide the degree of accuracy by 2.
- Add what you got to the value to get the upper bound and subtract to get the lower bound.

What are lower and upper bounds example?

Consider a number 50 rounded to the nearest 10. There are many numbers that can be rounded to get 50, but the lowest is 45. This means that the lower bound is 45 because it is the lowest number that can be rounded to get 50. The upper bound is 54 because it is the highest number that can be rounded to get 50.

What does bounds mean in maths?

Bounds in maths refers to limits. It shows the highest and lowest point a value cannot go beyond.

Why use upper and lower bounds?

Upper and lower bounds are used to determine accuracy.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more