VeritasKarishma wrote:
Prajat wrote:
The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the largest of the six integers is
(a) 26
(b) 28
(c) 30
(d) 32
(e) 34
...rephrasing an existing mean-median-mode problem
The mean of six positive integers is 15. - means the total sum is 90.
The median is 18, - so the two middle numbers have average 18. Say the numbers are ___, ____, 18, 18, ____, ____
and the only mode of the integers is less than 18. - Note above that if the numbers are as given above, the mode has to be 18. At max, two numbers other than 18 can be same but then there will be multiple modes. Hence the numbers in the middle must be different and the mode must be a smaller number. Let's make it as small as possible i.e. 1.
1, 1, 17, 19, ___, ____
To maximise the largest number, we have kept the fourth number as small as possible. Keep the fifth number as small as possible too i.e. 20 (it cannot be 19 because then there will be 2 modes)
1, 1, 17, 19, 20, ___
The last number must be 90 - (1 + 1+ 17 + 19 + 20) = 32
Answer (D)
Why the fifth number has to be 20? It can be 18, 16, 15 etc. It can also be 1.
A3+A4 = 36.
A1+A2+A5+A6 = 90-36 = 54
If we consider A1=A2=1, then
1+1+A5+A6=54
A5+A6=52
There are many values, which can be selected for A5 and A6. By manipulation, one can keep A6 at 34 and at 32 as well.
Correct me if I am missing any important point to be considered.