WEBVTT
00:00:01.800 --> 00:00:06.360
Jenny sits nine exams over the course of a year, which are all out of 30 marks.
00:00:06.920 --> 00:00:11.000
After she received the results of her eighth exam, her average score was 27.
00:00:11.840 --> 00:00:15.200
After the final exam, her average mark decreased by two.
00:00:15.960 --> 00:00:18.560
Work out what score Jenny got on her last exam.
00:00:19.160 --> 00:00:21.120
You must show how you got to your answer.
00:00:23.240 --> 00:00:27.680
We’re told that Jenny’s average score was 27 after the first eight exams.
00:00:27.680 --> 00:00:32.840
And as we haven’t been told what type of average this is, we, therefore, assume that this is the mean average.
00:00:34.560 --> 00:00:40.520
To find the mean average of a set of numbers, we add all the numbers up and then we divide by how many there are.
00:00:42.080 --> 00:00:48.800
In this question, we know what the mean average is — it’s 27 — and we know how many data points they are — they’re eight.
00:00:49.000 --> 00:00:51.760
But we don’t know what the total of all the scores was.
00:00:53.000 --> 00:00:56.560
We can, however, work this out and we’ll do so by forming an equation.
00:00:57.080 --> 00:01:02.640
We’ll introduce the letter 𝑥 to represent the sum of all of Jenny’s marks for the first eight exams.
00:01:04.400 --> 00:01:12.320
As the average was 27 and the number of results was eight, we therefore have the equation 27 is equal to 𝑥 over eight.
00:01:13.760 --> 00:01:20.840
To solve this equation for 𝑥, we can multiply both sides by eight to give 27 multiplied by eight is equal to 𝑥.
00:01:22.360 --> 00:01:25.880
We can work out 27 multiplied by eight using a column method.
00:01:27.240 --> 00:01:29.560
Seven multiplied by eight is 56.
00:01:31.120 --> 00:01:33.000
Two multiplied by eight is 16.
00:01:33.240 --> 00:01:38.880
And adding the five that we’ve carried, we see that 27 multiplied by eight is 216.
00:01:40.440 --> 00:01:45.080
So we know that Jenny’s total score after eight exams was 216.
00:01:46.880 --> 00:01:51.440
We’ve been told that after the final exam, Jenny’s average mark has decreased by two.
00:01:51.760 --> 00:01:53.560
So her new average is 25.
00:01:55.440 --> 00:02:15.520
If we now let 𝑦 equal Jenny’s score on just the ninth exam, we can form another equation: the total across all nine papers — so that’s 𝑦 — plus the total across the previous eight papers 216 divided by how many papers there are which is nine is equal to the new average of 25.
00:02:16.320 --> 00:02:18.960
We can now solve this equation to find the value of 𝑦.
00:02:20.600 --> 00:02:27.200
We begin by multiplying both sides by nine giving 𝑦 plus 216 is equal to 225.
00:02:28.120 --> 00:02:31.000
25 multiplied by 10 is 250.
00:02:31.360 --> 00:02:35.080
So to find 25 multiplied by nine, we just subtract 25.
00:02:37.240 --> 00:02:42.640
Next, we subtract 216 from each side of the equation, giving 𝑦 is equal to nine.
00:02:44.280 --> 00:02:47.320
Remember that 𝑦 represents Jenny’s score on the ninth exam.
00:02:47.720 --> 00:02:50.120
So our answer to the question is nine marks.
00:02:51.880 --> 00:02:54.400
Now, remember all of these papers are out of 30.
00:02:54.400 --> 00:03:02.920
And you may find it a little odd that Jenny scored a very low mark on the ninth paper — only nine out of 30 — and her average only decreased by two.
00:03:04.440 --> 00:03:10.280
The reason for this is that Jenny’s previous high average of 27 was across a much larger number of papers.
00:03:10.520 --> 00:03:12.440
It was across eight papers.
00:03:12.440 --> 00:03:19.760
Therefore, this one low result of nine does affect the mean by bringing it down by two, but it doesn’t affect a huge amount.