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Statistics Question

Expert replies
by coolhabhi » Sat Mar 09, 2013 10:58 am
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
Official Answer: D
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Source: — Problem Solving |

by Anju@Gurome » Sat Mar 09, 2013 11:19 am
coolhabhi 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
Let us assume the integers are in increasing order are a, b, c, d, e, and f.
Hence, (a + b + c + d + e + f) = 6*15 = 90
Now, median = (c + d)/2 = 18 ---> (c + d) = 36

Now, to maximize f we need to minimize all the others.
Minimum possible value of a and b is 1.
Now (c + d) = 36 but c and d cannot be equal to 18 as that would make 18 a mode of the integers. Hence, least possible value of d is 19.

Now, e must be as close as possible to d.
But, again e cannot be equal to d which will make e = d = a mode > 18.
Hence, e = 20

Hence, f = 90 - (a + b + c + d + e) = 90 - (1 + 1 + 36 + 20) = 90 - 58 = 32

The correct answer is D.
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by GMATGuruNY » Sat Mar 09, 2013 11:39 am
coolhabhi 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
Official Answer: D
Sum of the 6 integers = (number)(average) = 6*15 = 90.
Let the 6 integers be -- in ascending order -- a, b, c, d, e, f.

To MAXIMIZE the value of f, we must MINIMIZE the rest.

The MODE is the value that appears the MOST.
Since the mode must be less than 18, the smallest possible value for the mode is 1.
Let a=1 and b=1, yielding the following:
1, 1, c, d, e, f.

Since the median is 18, the average of c and d is 18:
(c+d)/2 = 18
c+d = 36.

To minimize the value of e, we must minimize the value of d.
Since there is only ONE MODE, there cannot be another integer that appears twice.
Thus, c and d must be different integers, implying that the smallest possible combination for c and d is c=17 and d=19:
1, 1, 17, 19, e, f.

Since no integer other than 1 can appear twice, the smallest possible value for e is 20, yielding the following:
1, 1, 17, 19, 20, f.

Thus:
f = 90-1-1-17-19-20 = 32.

The correct answer is D.
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