The mean of six positive integers is 15. The median is 18, and the onl

Question

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

MathRevolution · Answer

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

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

Let the six positive integers be A1 <= A2<= A3 <= A4 <= A5 <= A6. Since the median is 18, A3+A4=2*18=36. By the condition of “maximum possible value of the largest” A6 should be as large as possible, while others be as small as possible. Since mode of six integers is less than 18, A1 should be equal to A2 thereby they are mode and they should be 1. Moreover since the mode is unique, A4 < A5 < A6 and A4, A5 should be as small as possible. If A4=18, A3 should be 18 since the mode is 18(it’s impossible since the only mode of integers is less than 18). So A4 should be 19, A3 be 17 and A5 be 20. As a result A1= A2 =1, A3=17, A4=19, A5=20. Since the mean of six positive integers is 15, A1 + A2 + A3 + A4 + A5 + A6 = 6*15 = 90. So A6= 90-58=32. The answer is, therefore, (d)

sharma123 · Answer

MathRevolution  i'm getting 34 as the answer, as i used A1=A2=0. why cant we use 0 in the set??

dominicraj · Answer

"six  positive  integers" dude..

Regards,
Dom.

KarishmaB · Answer

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)

srvkmr88 · Answer

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.

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The mean of six positive integers is 15. The median is 18, and the onl

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The mean of six positive integers is 15. The median is 18, and the onl

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