The average points accumulated by 6 pupils is 84.5. They have all attained different points which are whole numbers. The lowest point is 75 while the highest point is 90.
- Find the average points achieved by the 4 pupils whose marks lie between the highest and the lowest.
- Find the smallest possible second lowest mark among these 6 pupils.
- Find the largest possible second lowest mark among these 6 pupils.
(a)
Total points of 6 pupils
= 6 x 84.5
= 507
Total points of the rest of the 4 pupils
= 507 - 90 - 75
= 342
Average points of 4 pupils
= 342 ÷ 4
= 85.5
(b)
To have the smallest possible second lowest mark among the 6 pupils, the 3rd pupil to the 6th pupil must have as high a score as possible and the 2nd pupil must have the second lowest score among the 6 pupils.
So the 6th pupil will have the highest score. The 5th pupil will have 1 mark will less than 6th pupil and this pattern continues to the 3rd pupil.
1st pupil's score = 75
2nd pupil's score = ?
3rd pupil's score = 90 - 3 = 87
4th pupil's score = 90 - 2 = 88
5th pupil's score = 90 - 1 = 89
6th pupil's score = 90
The smallest possible second lowest mark among these 6 pupils
= 342 - 89 - 88 - 87
= 78
(c)
To have the largest possible second lowest mark among these 6 pupils, the 2nd pupil to the 5th pupil must have as high a score as possible but lower than the 6th pupil.
So the 2nd pupil 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 pupil's score = 75
2nd pupil's score = 1 u
3rd pupil's score = 1 u + 1
4th pupil's score = 1 u + 2
5th pupil's score = 1 u + 3
6th pupil's score = 90
1 u + (1 u + 1) + (1 u + 2) + (1 u + 3) = 342
4 u + 6 = 342
4 u = 342 - 6
4 u = 336
1 u = 336 ÷ 4 = 84
84 x 4 + 6 = 342 (Equal to 342)
Largest possible second lowest mark among these 6 pupils = 84
Answer(s): (a) 85.5; (b) 78; (c) 84