## What you need to know

Upper and Lower Bounds and Truncating

This topic is about the consequences of two processes – **rounding **and **truncating**. If you’re not sure about rounding, more information can be found here (https://mathsmadeeasy.co.uk/gcse-maths-revision/rounding-numbers-gcse-math-revision/). Truncating is a fancy term for making something shorter, and it’s actually easier than rounding – rather than worrying about whether we need to round up or round down, when you truncate a number it is effectively always rounded down. Truncating is just chopping part of the number off.

**Example: **Truncate 35.7289 to 1dp and 3dp.

To truncate, we just count along 1 decimal place and then cut off the rest of the number. So, 35.7289 truncated to 1dp is 35.7.

Same goes for 3 decimal places, so 35.7289 truncated to 3dp is 35.728.

Rounding and truncating are useful tools, but we must be aware that it comes with consequences. When we round/truncate a number, we lose some information. In the example above, if I only gave the number 57.7 to someone and told them it had been rounded to 1 decimal place, they couldn’t possibly know whether the original number was 57.653 or 57.72 or 57.6834 or 57.7468 or…

As you can tell, there’s a whole bunch of different things that could’ve ended up being rounded to 57.7. Specifically, there is a range of values which we call the **error interval**. The error interval has two **error bounds**, the lower bound is the lower limit for numbers that will round *up* to the number in question, and the upper bound is the upper limit for numbers that will round *down* to the number in question.

**Example: **Let’s stick with 57.7 being the result of rounding some number (that we’re going to pretend we don’t know) to 1 decimal place, and find the range of values that starting number, let’s call it x, could’ve been.

Let’s first recall the rules of rounding. When rounding, we look to the digit after the one we wish to round up/down (i.e. if we’re rounding to 1 decimal place, we look at the second decimal place), and if it is 5 or above we round the relevant digit up, but if it’s below 5 we round it down. Note: this is all because 5 is the middle point between 0 and 10.

Firstly, the lower bound. We know by the rules of rounding that 5 is the cut-off point – specifically, having a 5 in the digit after the one we’re rounding to. So, we know that any number between 57.65 and 57.7 will round up to 57.7 just fine, but any number below 57.65 will round down. This makes 57.65 the lower bound.

Secondly, the upper bound. By the rules of rounding again, any number which is between 57.7 and 57.75 will round down to 57.7, but anything which is greater than (or equal to!) 57.75 will round up. This makes 57.75 the upper bound.

So, we express our error interval for x like

57.65 \leq x < 57.75

**Note:** whilst 57.65 is included in the interval because it does round to 57.7, 57.75 is not included because although it provides the upper limit for numbers which round down to 57.7, it does not itself round down to 57.7. This will always be the case, i.e. the inequality symbols will always be this way round when expressing an error interval due to rounding.

In practice, answering the questions is pretty quick task once you get the hang of it.

**Example: **The weight of a cat has been measured at 209 ounces to the nearest ounce. Work out the interval within which w, the weight of the cat, lies.

Firstly, the lower bound. In order to round up to 209 to the nearest whole number, the original weight of the cat must have been bigger or equal to 208.5. Any lower, and we’ll be rounding down to 208.

Next, the upper bound. In order to round down to 209, the weight of the cat must have been strictly less than 209.5. Any value which was equal to 209.5 or higher and it would’ve rounded up to 210.

So, the error bounds are 208.5 and 209.5, and we express the interval like

208.5 \leq w < 209.5

Bear in mind, when we **truncate**, the rules for finding error bounds are different.

**Example: **The weight of a dog has been truncated to 402.3 ounces to 1dp. Work out the interval within which w, the weight of the dog, lies.

In order for some number to be truncated to 402.3, it must have begun 402.3 followed by some other digits. Maybe 402.31, 402.395, 402.35467, and so on. As long as the number begins “402.3” then we’re fine.

In other words, anything that is greater than or equal to 402.3 but less than 402.4 (at which point it would just truncate to 402.4). We express this error interval like

402.3 \leq w < 402.4

This will always be the case for truncations – the first part will be the number itself, and the second part will be that number with the last digit increased by 1.

A common type of question you may be asked is to find the upper or lower limit for some quantity (such as the area of a shape or the speed of a car) given that some of the measurements have been rounded. Let’s take a look.

**Example: **The base and height of a triangle have been measured to 2 significant figures to be 12cm and 5.0cm respectively. Find the lower limit for area of the triangle.

The formula for the area of a triangle is \dfrac{1}{2}bh, so for the minimum area of the triangle, we will need the smallest possible values of both b and h.

The lower bound for b when rounding to 12cm to 2 significant figures is 11.5cm.

The lower bound for h when rounding to 5.0cm to 2 significant figures is 4.95cm.

Therefore, the lower bound for the area of the triangle is

\dfrac{1}{2} \times 11.5 \times 4.95 = 28.4625 \text{cm}^2.

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### Example Questions

1) The capacity of a jug has been measured to be 5.43 litres to 2dp. Work out the interval within which C, the capacity of the jug, lies.

Firstly, the lower bound. In order to round up to 5.43 to 2dp, the capacity of the jug must have been bigger or equal to 5.425. Any lower and it would’ve rounded down to 5.42.

Next, the upper bound. In order to round down to 5.43, the capacity of the jug must have been strictly less than 5.435. Any value which was equal to 5.435 or higher and it would’ve rounded up to 5.44.

Our lower bound is 5.425 and our upper bound is 5.435, so we express our answer like

5.425 \leq C < 5.435

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2) A formula describing the driving force, F, in Newtons of a vehicle is given by

F = \dfrac{P}{v}

Where P is the power in Watts, and v is the velocity in \text{ms}^{-1}.

P = 145 to 3sf,

v = 23.4 to 1dp.

Work out the lower bound for the driving force of the car. Give your answer to a suitable degree of accuracy.

Because our formula involves a fraction, we can’t just choose the smallest value for both quantities in order to determine the lower bound of the force. In order to make a fraction as small as possible, we want the numerator to be as a small as possible but the denominator to be as big as possible (the bigger the number you divide by is, the smaller your answer).

The lower bound for P is 144.5.

The upper bound for v is 23.45.

Therefore, the lower bound for the driving force of the car is

F = \dfrac{144.5}{23.45} = 6.16\text{ Newtons, 3sf}

This is a suitable degree of accuracy as it is precisely how accurately both of the initial values are given.

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3) The distance of a car journey is measured at 80.3 miles to 1dp, whilst the time taken for the journey is 1.87 hours to 2dp. By considering both upper and lower bounds, work out the average speed of the car, in miles per hour, over the course of the journey to a suitable degree of accuracy.

This one is a little different. Firstly, recall that the formula for speed is

\text{speed} = \dfrac{\text{distance}}{\text{time}}

The question tells us to consider upper and lower bounds, so we’ll do just that.

To find the upper bound for the average speed, we will need the upper bound for the distance and the lower bound for the time (a smaller denominator means a bigger number). On the contrary, to find the lower bound for the average speed we will need the lower bound for the distance and the upper bound for the time (a bigger denominator means a smaller number).

The upper bound for distance is 80.35, whilst the lower bound is 80.25

The upper bound for time is 1.875, whilst the lower bound is 1.865.

Therefore, we get:

\text{Upper bound for speed } = \dfrac{\text{distance upper bound}}{\text{time lower bound}} = \dfrac{80.35}{1.865} = 43.0831…

\text{Lower bound for speed } = \dfrac{\text{distance lower bound}}{\text{time upper bound}} = \dfrac{80.25}{1.875} = 42.8

The actual value of the average speed must fall between these two numbers. They’re obviously different, but when rounded to 2 significant figures they both come to 43. Any more significant figures and the two outcomes are not the same, so the most accurate statement we can make is that the average speed of the car must be 43 miles per hour, to 2sf.

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### Worksheets and Exam Questions

### Videos

#### Bounds and Truncation Q1

GCSE MATHS#### Bounds and Truncation Q2

GCSE MATHS#### Bounds and Truncation Q3

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