The average points accumulated by 6 students is 79.5. They have all attained different points which are whole numbers. The lowest point is 55 while the highest point is 92.
- Find the average points achieved by the 4 students whose marks lie between the highest and the lowest.
- Find the smallest possible second lowest mark among these 6 students.
- Find the largest possible second lowest mark among these 6 students.
(a)
Total points of 6 students
= 6 x 79.5
= 477
Total points of the rest of the 4 students
= 477 - 92 - 55
= 330
Average points of 4 students
= 330 ÷ 4
= 82.5
(b)
To have the smallest possible second lowest mark among the 6 students, the 3rd student to the 6th student must have as high a score as possible and the 2nd student must have the second lowest score among the 6 students.
So the 6th student will have the highest score. The 5th student will have 1 mark will less than 6th student and this pattern continues to the 3rd student.
1st student's score = 55
2nd student's score = ?
3rd student's score = 92 - 3 = 89
4th student's score = 92 - 2 = 90
5th student's score = 92 - 1 = 91
6th student's score = 92
The smallest possible second lowest mark among these 6 students
= 330 - 91 - 90 - 89
= 60
(c)
To have the largest possible second lowest mark among these 6 students, the 2nd student to the 5th student must have as high a score as possible but lower than the 6th student.
So the 2nd student will have the highest possible second lowest score. The 3rd child must have at least 1 mark higher than the 2nd child and the pattern continues to the 5th child.
1st student's score = 55
2nd student's score = 1 u
3rd student's score = 1 u + 1
4th student's score = 1 u + 2
5th student's score = 1 u + 3
6th student's score = 92
1 u + (1 u + 1) + (1 u + 2) + (1 u + 3) = 330
4 u + 6 = 330
4 u = 330 - 6
4 u = 324
1 u = 324 ÷ 4 = 81
81 x 4 + 6 = 330 (Equal to 330)
Largest possible second lowest mark among these 6 students = 81
Answer(s): (a) 82.5; (b) 60; (c) 81