Video: AQA GCSE Mathematics Foundation Tier Pack 4 • Paper 3 • Question 8

A group of 20 students sat four exams for one module. The graph shows some information about how many students passed and failed these exams. a) Which exam was the mode for students failing the exam? Circle your answer. [A] 1 [B] 2 [C] 3 [D] 4 b) Calculate the mean pass rate for the four exams. c) Over the four exams, how many more passes were there than fails? d) In order to pass this module, a student needs to pass at least two of the exams. Juliet says, “Over half the students must have passed the module because there were more passes in the exams than fails.” Is she correct? You must justify your answer.

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Video Transcript

A group of 20 students sat four exams for one module. The graph shows some information about how many students passed and failed these exams. There are four parts to the question. Part a), which exam was the mode for students failing the exam? Circle your answer. Was it one, two, three, or four.

Before starting this question, it is important to understand what is shown on the graph. The number of students is on the 𝑦-axis. This goes from zero to 16. Each square represents one student. There were four exams numbered one to four. The grey block represents those students who passed the exam. And the white block represents those who failed the exam. This can be seen from the key below the graph. 12 students passed the first exam, as the bar goes up to 12. Eight students failed the first exam, as the white bar goes up to eight. We know that 20 students sat each of the exams. So it is worth checking that 12 plus eight equals 20, which it does.

Nine students passed the second exam, and 11 students failed it. In the third exam, 15 students passed and only five failed. Finally, in the fourth exam, seven students passed and 13 failed. The pairs of numbers nine and 11, 15 and five, and seven and 13 all add up to 20.

In this first part of the question, we are asked to work out the mode for students who failed the exam. The mode is the most popular or most common occurrence. We know that eight students failed the first exam. 11 failed the second exam. Five failed the third exam, and 13 failed the fourth exam. The highest frequency or number of students here is 13. This corresponds to the fourth exam. Therefore, the mode is four. More students failed the fourth exam than any of the others. The second part of the question says the following.

b) Calculate the mean pass rate for the four exams.

A phrase that is often used to help us calculate the mean is add them up and then divide by how many there are. In this case, the mean will be calculated by dividing the total number of passes by the total number of exams. 12 students passed the first exam, nine passed the second, 15 the third, and seven the fourth. Therefore, the total number of passes is equal to 12 plus nine plus 15 plus seven. This is equal to 43. As 20 students sat each of the four exams, the total number of exams sat can be calculated by multiplying four by 20. This is equal to 80. We could also have worked this out by adding all eight numbers from the bar chart. The mean pass rate for the four exams is therefore 43 out of 80.

This can be converted into a decimal by dividing 43 by 80. Therefore, the mean pass rate as a decimal is 0.5375. We could also convert this into a percentage by multiplying by 100, as percent means out of 100. 0.5375 multiplied by 100 is equal to 53.75. We can therefore say that the mean pass rate for the four exams was 43 out of 80 or 53.75 percent. The third part of the question says the following.

c) Over the four exams, how many more passes were there than fails?

In order to do this, we firstly need to calculate the total number of passes. As we’ve already worked out in part b), this is equal to 12 plus nine plus 15 plus seven. This tells us that there were 43 total passes. The total number of fails can be calculated by adding eight, 11, five, and 13. This is equal to 37. We know that 43 and 37 must add or sum to 80, as there are 80 exams sat all together. To calculate how many more passes there were than fails, we need to subtract 37 from 43. This is equal to six. Therefore, there were six more passes than there were fails. The final part of the question says the following.

d) In order to pass this module, a student needs to pass at least two of the exams. Juliet says, “Over half the students must have passed the module because there were more passes in the exams than fails.” Is she correct? You must justify your answer.

We’re told in the question that in order to pass the module, a student needs to pass at least two of the exams. We have also worked out previously that over half of the exams were passed. In fact, 53.75 percent of the exams were passed. This tends to suggest that over half of the students must have passed, and Juliet is correct. However, let’s look at this in a bit more detail.

Let’s consider the total of 43 exams that were passed. The lowest pass rate was on the fourth exam, where only seven students passed. It is possible that all seven of these students passed all four exams. Seven multiplied by four is equal to 28. So if this was the case, they would’ve passed 28 of the 43 exams between them. The next lowest bar was nine. And this is two greater than seven. It is therefore also possible that these two extra students passed exams one, two, and three. Two multiplied by three is equal to six. Therefore, these two students take up another six of the passes. As 28 plus six is equal to 34, we have now covered 34 of the 43 passes in the exams.

If we now consider the next highest bar, 12, this is three higher than nine, as 12 minus nine is equal to three. The three students that passed the first exam do not need to be the same as any of the six other students that passed the third exam. It is therefore possible that nine different students passed one exam, either exam one or exam three. Nine multiplied by one is equal to nine. 28 plus six plus nine is equal to 43. Therefore, we have now covered the 43 exams that were passed. This would result in two students passing none of the exams, as seven plus two plus nine plus two is equal to 20. And we know that there were 20 students in total.

This situation means that nine out of the 20 students passed more than two exams. Seven of them passed all four, and two of them passed three exams. This is less than half of the students. Therefore, Juliet in this circumstance is incorrect. To justify our answer, we only need to find one example for which Juliet is wrong. In this case, it was when seven students passed all four exams, two students passed three exams, nine students passed one exam, and two students passed no exams. In this situation, only nine out of 20 of the students passed the module which is less than half.

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