The average points accumulated by 6 children is 87.5. They have all attained different points which are whole numbers. The lowest point is 59 while the highest point is 100.
- Find the average points achieved by the 4 children whose marks lie between the highest and the lowest.
- Find the smallest possible second lowest mark among these 6 children.
- Find the largest possible second lowest mark among these 6 children.
(a)
Total points of 6 children
= 6 x 87.5
= 525
Total points of the rest of the 4 children
= 525 - 100 - 59
= 366
Average points of 4 children
= 366 ÷ 4
= 91.5
(b)
To have the smallest possible second lowest mark among the 6 children, the 3rd child to the 6th child must have as high a score as possible and the 2nd child must have the second lowest score among the 6 children.
So the 6th child will have the highest score. The 5th child will have 1 mark will less than 6th child and this pattern continues to the 3rd child.
1st child's score = 59
2nd child's score = ?
3rd child's score = 100 - 3 = 97
4th child's score = 100 - 2 = 98
5th child's score = 100 - 1 = 99
6th child's score = 100
The smallest possible second lowest mark among these 6 children
= 366 - 99 - 98 - 97
= 72
(c)
To have the largest possible second lowest mark among these 6 children, the 2nd child to the 5th child must have as high a score as possible but lower than the 6th child.
So the 2nd child 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 child's score = 59
2nd child's score = 1 u
3rd child's score = 1 u + 1
4th child's score = 1 u + 2
5th child's score = 1 u + 3
6th child's score = 100
1 u + (1 u + 1) + (1 u + 2) + (1 u + 3) = 366
4 u + 6 = 366
4 u = 366 - 6
4 u = 360
1 u = 360 ÷ 4 = 90
90 x 4 + 6 = 366 (Equal to 366)
Largest possible second lowest mark among these 6 children = 90
Answer(s): (a) 91.5; (b) 72; (c) 90
