Write down an inequality expressing the lower and upper bound for 23.071kg to three decimal places

Write down an inequality expressing the lower and upper bound for 0.3m given to 1 significant figure

Write down an inequality expressing the lower and upper bound for 0.023 given to 2 significant figure

The length of a football field is 500m to the nearest 10m. Jacob says the lower bound is 450m and the upper bound is 550m. Is he correct? Give a reason for your answer.

The mass of a loaf of bread is given as 1.7kg to the nearest 0.1kg. Write an inequality expressing the upper and lower bound of the mass.

A piece of string is measured at 1.57m to two decimal places. Write an inequality expressing the lower and upper bound within which l, the actual length of string lies.

Upper and Lower Bound Quiz

﻿Hello and welcome to this lesson on solving problems with rounding with me, Miss Oreyomi. As usual you will need a paper and your pen or something to write with or on. You may also need a calculator for this lesson. So if you do not have this equipment and you want to go and get them, pause this video right now, try to minimise distractions and press play when you're ready to begin the lesson. Okay, in today's lesson, you will be able to solve problems using upper and lower bounds. So your previous knowledge of upper and lower bounds will really come in handy. So try this says X and Y, rounded to the nearest 100, are both 100. Find the possible values of X? So if I'm rounding to the nearest 100 and they are both 100, I know that my value would range from 55 to 105 because if I round 55 to the nearest 100 it becomes 100. If I round say 104 to the nearest 100, it becomes 100. So I'm either going to put X or Y because they're both the same, would be greater than or equal to X and less than 105. I could possibly choose any value. First value I could take is 105 takeaway 55, you'd have 101. 101.25. takeaway 74. I could also have 98.2 take away 102. So any X or Y values between the range of 55 and 105. Okay, let's think about this question. Both sides are measured to the nearest centimetre, work out the smallest possible area. I want the smallest possible area. Let's start with the width eight centimetre, the lower bound that means, if I run 7.8 to the nearest centimetre, I would get eight centimetre. And that upper bound would be 8.5. Do the same for my width, five centimetre, the lower bound that when I round to the nearest centimetre would give me five is 4.5. And my upper bound would be 5.5 centimetre. Now let's think about it. If I want the smallest possible area, what two values should I be multiplying together that would give me the smallest possible area. Yes, it should be the lower bound for the length and the lower bound for the width. So let's try that now. 7.5 which is the lower bound for the length times 4.5 get your calculator out, what would you get? That gives me 33.75 centimetre squared. And that's the lowest possible value I could get using the lower bounds. Now, what if I want to work out the largest possible area? What if I want to work out the largest possible area? Again I'm going to do the same, my lower and upper bound for my length and my lower and upper bound for my width. If I want the largest possible area what bound should I be multiplying? You're correct, I should be multiply the upper bound for my length and the upper bound for my width. So again, using your calculator, it's going to be 8.5 multiplied by 5.5 that should give me 46.75. Okay, so the largest possible area I could possibly get would be multiplying my upper bounds together, and that's 46.75 centimetres squared. So just to recap, if I'm multiplying bounds, and I want the smallest possible value, I will multiply these lower bounds. If I am multiplying bounds, and I want the largest possible value, I will multiply both upper bounds. What of this one? Antoni is 152 centimetres tall to the nearest centimetre. And Yasmin is 143 centimetres tall to the nearest centimetre. Work out the greatest possible difference in their height. So in brackets, it's work out the upper bound for the difference in their height. So I want the largest value. Antoni is 152 centimetre, the lower bound would be 151.5 centimetre, and the upper bound would be 153.5 centimetre. The same for Yasmin, the lower bound for her height would be 142.5 centimetre, and the upper bound for her heights would be 143.5 centimetre. I'm going to give you five seconds to think about this. If I want the greatest possible value, what should I be taking away from what? Which bounds and which Antoni bound should I be using and which Yasmin bound should I be using, to find the greatest possible difference in their heights? Okay, you would see that it would be the upper bound of Antoni's height, which will be 152.5 take away the lower bound of Yasmin's height 142.5 that would give us the greatest value, because I am taking the smallest value away from the largest value, and that would give us the upper bound. So using your calculator, let's try this now. So I've got 152.5 take away 142.5, which is... don't need a calculator for that, do I? Which is 10. The difference in their heights would be 10. The upper bound difference in their heights would be 10. What if I want to work out the smallest difference? So again, think about this one. If I want to work out the smallest difference, what should I do here? I want to work out the smallest difference. Hopefully, you're able to see that it would make sense to take away the lowest bound of Antoni's height subtract the upper bound of Yasmin's height, as that would give me the smallest difference. So that would be 151.5 subtract that by 143. And that should give me eight centimetre. So hopefully it makes sense that when we're subtracting bounds, if I want to calculate the upper bound, which is here, I take the maximum upper bound for the first value, and I subtract it from the minimum lower bound of the second value to get the upper bound. And if I want to find the smallest difference or the lower bound, I take the lower bound mean of the first value and I subtract that from the upper bound max of the second value, as that gives me the smallest difference, that gives me the lower bound. Let's think about this question. A girl runs 200 metres measured the nearest 10 metre. She takes 13 seconds to the nearest second, what is her fastest possible average speed? Now we know that speed is equal to distance over time. But before we can work out how fast is average speed, we need to think about her upper or lower bound. So for the distance, her upper bound would be 250 metres, because run that down and we get 200 metres and the lower bound would be 195 metres to the nearest 10. For the time the lower bound to the nearest second will be 12.5 seconds and the upper bound would be 13.5 seconds. Now speed is equal to distance over time. If I want to work at the fastest possible average value, which value should I be taking for the distance and which value should I be using for the time. Hopefully you've said the upper bound and the lower bound, because if I travel a distance in a shorter time, that means I'm going at a greater speed. So to work out her fastest speed, I'm going to do her upper bound divided by the lower bound. Okay, the distance divided by the lower bound of time. In this case, it's 205 divided by 12.5 seconds, put that in your calculator. 12.5 seconds and the speed is 16.4. Okay, what if I want to work out her slowest possible average speed? The slowest possible average speed, which value for distance should I be using and which value for time should I be using? Yes, I should be using the lower bound for distance and the upper bound for time divided by 13.5 as that would give me the smallest possible value. So, again input the number in your calculator and what would you get? You get 14.44 recurring, so I'm just going to round that to one decimal place as 14.4 metres per second. Again, just to recap, if I want to find the upper bound or her fastest possible speed, I'm using the upper bound for the distance and the lower bound for the second value. And if I want to find the lower bound or the slowest possible time, I'll be using the lower bound for my first value divided by the upper bound of my second. Okay, let's think about this question. A bus can safely hold 940 kilogramme correct to two significant figures. An average human weigh 80 kilogrammes measured to the nearest 10 kilogramme. Can the bus hold 10 people? If I want to find out my lower bound and upper bound for my bus, so the lower bound correct two significant figures, the lower bound will be 935 and the upper bound will be 945. For my average human 80 kilogramme, the lower bound will be 75 kg, and the upper bound will be 85 kg. And I want to find out if the bus can hold 10 of my lower bound and 10 of my upper bound. So I'm going to multiply this by 10, I'm going to multiply both of these by 10. So my lower bound, if I've got 10 people that weigh approximately 75 kg, that would be a total of 750 kilogrammes and then multiply my upper bound by 10 people again and I would get 850 kilogrammes. Now, can 10 people fit in the bus where the upper bound and lower bound is 935 kilogrammes and 945 kilogrammes? Yes, they can because my upper bound and lower bound... my lower and upper bound for humans is less than my lower and upper bound for the bus. Okay, I want you to pause the screen now and go into your independent task. Complete as many questions as you can, and when you're done, come back and let's review the answers together. Okay, let's go for the answers very quickly. Find the least and greatest total length of eight hot dogs measuring nine centimetres to the nearest centimetre. The least would be 8.5 multiplied by eight is this round to the nearest centimetre would give me eight. And sorry, this rounded to the nearest centimetre will give me nine, and the greatest would be 9.5 times eight would be 76 centimetres. Now a playground has a width of 64 and a length of 113. And we want the upper bound for each of my value is the upper bound. And because it's the perimeter, I am adding all of it together, and my total answer is 356. Next one, 92.5 centimetres squared. I'm just going to check that there. Yeah. Okay, now we have six heights that has been given to us. And we're told to find the greatest possible mean, if we want the greatest possible mean, it's literally asking us for the upper bound of each of the values. And here you've been given a hint that to find the mean you're adding the values together that sum of the values divided by the number of values. So I want the upper bound of each one to one decimal place. Each height has been given to one decimal place, and I want the upper bound for each of these so it will be 2.95 plus 5.65 plus 7.85 plus 9.25 plus 6.55 plus 9.45, all of those values sum up together divided by six, and my greatest, my upper bound would be 6.95. Now I want the greatest possible range. Range means the biggest number takeaway the smallest number. In this case, to get the biggest possible range, I am going to have my upper bound for my biggest number, takeaway the lower bound for my smallest number, my smallest number is 2.9, the lower bound for 2.9 is 2.85. And the upper bound for my greatest number which is 9.4 is 9.45 subtract those two values and I get the biggest number the upper bound for this range. Okay. We want to work out the lower bound for the length correct to three significant figures. So if I round this up to three significant figures, I would get 125. This is the lower bound for my length of my rectangle. And then I want to work out the lower bound area so it would be 124.5 multiply by the lower bound for 60 centimetre would be 59.5 because if I round 59.5 up to two significant figures, I am going to get 60 centimetre and that should give you this area, the lower bound area right here. X and Y rounded to the nearest 100 are both 100. Find the possible values of X and Y, if X minus Y is equal to 25. I want you to pause the screen now and attempt this task there are more than one answer here, so you don't have to stop if you just found one, keep on going and try to find as many possible values of X and Y as you can. And if you're struggling with this, then this is when you keep on watching the video to see the help that I provided. However, if you're feeling quite confident then go for it, pause the video and attempt the work. Okay, if you need some support, we found out from the try this task that X and Y would be in the range of 50 and 105. So you want any two numbers between this range that if you subtract them from each other, you will get 25. A very easy one could be 75 subtract from 50 one possible answer, possible value for X is 75. Another possible value for Y is 50, so that would give you 25. Another possible value could be 90.5 subtract from 65.5 so those are two examples. Try to come up with your X and Y values, that when you subtract together in the range of 50 and 105 would give you 25. Okay, we have now reached the end of today's lesson, excellent job for sticking right through and getting stuck in and doing as much work as possible. Send your work to your teachers to show off, how excellent you've been this lesson and I will see next lesson.

A classroom is measured 5.4 x 3.4 metres, measured to 1 d.p. Calculate the lower bound of the area of the room.

James ran a distance of 400m in a time of 25.2 seconds. The distance of 400m was measured to the nearest 10 metres. The time of 25.2 was measured to the nearest tenth of a second. Work out the upper bound James' average speed.

James ran a distance of 400m in a time of 25.2 seconds. The distance of 400m was measured to the nearest 10 metres. The time of 25.2 was measured to the nearest tenth of a second. Work out the lower bound James' average speed.

Carl measured the length and width of a rectangle. He measured the length to be 19cm correct to the nearest centimetre. He measured the width to be 20cm correct to the nearest 10cm. Calculate the lower bound for the area of this rectangle.

6 heights are listed, given to one decimal place: 4.6cm, 6.3cm, 9.8cm, 3.2cm, 5.7cm, 3.6cm. Work out the greatest possible mean.

A circle has an area of 130cm^2 to the nearest 10cm^2. Work out the upper bound of the radius. Give your answer to 3 d.p.

Solving Problems with rounding

In this lesson, we will learn how to solve problems working with lower and upper bounds to find maximum and minimum values.

Maths

Y8 Maths

Key Stage 3

Year 8

Miss

Solving Problems with Rounding

Accuracy and estimation

Number

THEME

UNIT

Rounding numbers using a number line

Rounding from a calculator: Approximating

Rounding to significant figures (Part 1)

Rounding to significant figures (Part 2)

Rounding and Range

Upper and Lower Bound

Estimation

A teacher asked two students to round the number 5,469,704. Their answers are below. Which student(s) rounded the number correctly? Explain your answer.

Estimate the total cost of 32 books at £178 each and 57 markers at £22.30 each.

Student A estimated that 23.5 x 91 = 1800. Student B estimated that 23.5 x 91.4= 2160. What did student B do?

Work out an estimate for 9.97% of 3967

Estimate the value of: 6.2 + 7.9 x 3.2 - 9.8

I estimated that 7.85 x 29 = 240. Why is my estimated answer more than my exact answer?

Lesson overview:Solving Problems with Rounding

Core Content

Accuracy and estimation: