Video: AQA GCSE Mathematics Higher Tier Pack 3 β€’ Paper 1 β€’ Question 11

The table contains six positive integers arranged in ascending order. π‘Ž, 𝑏, 8, 𝑐, 𝑐 + 1, 𝑑. Find the values of π‘Ž, 𝑏, 𝑐, and 𝑑 such that (1) the range is as large as possible, (2) the median of the 6 integers is 9, (3) the mode is also the smallest value, and (4) the mean is 11. Show all your working.

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Video Transcript

The table contains six positive integers arranged in ascending order. π‘Ž, 𝑏, eight, 𝑐, 𝑐 plus one, and 𝑑. Find the values of π‘Ž, 𝑏, 𝑐, and 𝑑 such that the range is as large as possible, the median of the six integers is nine, the mode is also the smallest value, and the mean is 11. Show all your working.

So we were told that the table contains six positive integers. Positive integers begin at one and go up from there: one, two, three, four, five, and so on, excluding all of the fractions and decimals in between. We are also told that they are arranged in ascending order, meaning left to right; they go smallest to largest. And out of all six positive integers, we’re only given one β€” eight.

So π‘Ž and 𝑏 are to the left of eight. So they should be smaller than eight. And then 𝑐, 𝑐 plus one, and 𝑑 are to the right of eight. So they should be larger than eight. So let’s look through the list of four things that we must follow to find the values of π‘Ž, 𝑏, 𝑐, and 𝑑.

Number one says, β€œthe range is as large as possible.” The range is the difference between the smallest number and the largest number or we can think of the distance between the smallest and the largest number, so π‘Ž and 𝑑. So we want π‘Ž to be as small as it can and 𝑑 to be as large as it can be.

Well, since we know that we have six positive integers, the smallest value that π‘Ž could take would be one. So let’s allow the π‘Ž to equal one. Now, 𝑑, we’re not sure yet because positive integers go all the way up to infinity. So we’re not exactly sure where 𝑑 can land just yet.

Let’s go to point two: β€œthe median of the six integers is nine.” The median would be considered the middle number, the number found in the middle of this list. Well, we have a total of one, two, three, four, five, six numbers.

So the one that would be in the middle would be located where the star is because we want the same amount of numbers to the left of that number and to the right of that number. That’s what the median is; it’s right in the middle of this set.

So we want to know what number is between eight and 𝑐. The way that we would do that would be to take eight and 𝑐 and add them together and divide by two. And we know what the median is. They told us it’s nine. So we set this equal to nine.

So this gives us an opportunity to solve for 𝑐. We need to multiply both sides of the equation by two. This way, the twos cancel on the left. And we have that eight plus 𝑐 is equal to nine times two which is 18. So we can solve for 𝑐 by subtracting eight from both sides and we find that 𝑐 is equal to 10. And this makes sense.

Let’s go ahead and plug the value of 𝑐 into our table. We said 𝑐 was 10 and π‘Ž was one. Now, we were told that the median, the middle number, was nine. And we know it would be between the third and the fourth number. So what number is between eight and 10? Nine, just like what was supposed to be.

Now, point three says, β€œthe mode is also the smallest value.” The mode is the number that occurs the most often. Well, right now, we have one, 𝑏, eight, 10, 𝑐 plus one, and 𝑑. We can actually fill in the fifth number because if we know that 𝑐 is 10, 𝑐 plus one will be this value. So 10 plus one means that this must be 11.

So we are told that the mode is also the smallest value. So our smallest value that we have is one. So the number that must occur the most often must be one. And we’re only missing 𝑏 and 𝑑. So the only value that can be one would be 𝑏. 𝑑 must be larger than 11 because it’s all the way to the right. So 𝑏 is equal to one.

Now, before we start the fourth point that the mean is 11 and we solve for 𝑑, we were asked to show all of our working. So let’s go ahead and make little notes for each point that we’ve done so far.

One says the range is as large as possible. So we can say that we minimised π‘Ž. So π‘Ž must equal one because it was a positive integer and it needed to be the smallest, minimising it.

Next, the median of the six integers was nine. So we would find the median by finding the number in between eight and 𝑐. So we needed to add together eight and 𝑐 and divide it by two. And we knew that the median was supposed to be nine. So we set it equal to nine, multiplied both sides by two to get eight plus 𝑐 equals 18, and then subtract eight from both sides. And we found that 𝑐 was equal to 10. And that made sense because the number in between eight and 10 is nine.

Next, the mode β€” that’s the number that occurs the most often. And we were told that it was also the smallest value. So we decided that π‘Ž must be equal to 𝑏, making π‘Ž and 𝑏 both one. So lastly, we’re told that the mean is 11. The mean is the average. So to find the mean or the average of these numbers, we need to add all of the numbers together and then divide by the number of integers that we have.

So again, we add all of our positive integers together. However, we don’t know 𝑑 and we’ll solve here. So we add them all up. And how many numbers did we have? They’re six. So we divide by six. And this is how we would find the mean. And we are told that the mean is 11. So we set this equal to 11.

So to solve for 𝑑, let’s multiply both sides of the equation by six to get rid of our denominator. So we have one plus one plus eight plus 10 plus 11 plus 𝑑 equals 66 because 11 times six gave us 66.

Now, let’s add the numbers that are on the left. One plus one plus eight plus 10 plus 11 is equal to 31. So to solve for 𝑑, we subtract 31 from both sides, making 𝑑 equal to 35.

So this means that our values are π‘Ž equals one, 𝑏 equals one, 𝑐 equals 10, and 𝑑 equals 35.

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