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In a set of 9 numbers, both the median and the mean are 15. If the

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In a set of 9 numbers, both the median and the mean are 15. If the [#permalink]

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In a set of 9 numbers, both the median and the mean are 15. If the largest number in the set is 3 less than twice the smallest number, what is the largest possible number in the set?

A 23
B. 24
C. 25
D. 27
E. 29
[Reveal] Spoiler: OA

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Re: In a set of 9 numbers, both the median and the mean are 15. If the [#permalink]

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Since the smallest no. Will be less than mean, therefore comparing the options, only A and C fall under conditions, in which is C is bigger no.

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Re: In a set of 9 numbers, both the median and the mean are 15. If the [#permalink]

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New post 13 Sep 2017, 21:14
It would be A.
To have the largest possible number, we need to have all the four terms before median equal to the smallest . And 3 terms after median should be equal to median. Adding all the numbers and equalling the equation to 9*15, we can find the value of largest number.

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In a set of 9 numbers, both the median and the mean are 15. If the [#permalink]

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In a set of 9 numbers, the mean and median is the 5th number.
Since the mean is 15, the sum of all the elements in the set is 15*9(135)
If the smallest number(S) be x and the largest number(L) is 2x - 3!

If we need the largest possible number for the set, the list must be S,S,S,S,M,M,M,M, and L

4x + 60 + 2x - 3 = 135
6x + 57 = 135
6x = 78
x = 13

The largest number is 2x - 3 = 23(Option A)
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Re: In a set of 9 numbers, both the median and the mean are 15. If the [#permalink]

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Must be A.

4 numbers must be less than or equal to 15 and 4 numbers must be greater than or equal to 15 (because median = 15)
Now, the mean is 15. Hence, the numbers must add up to 135.

Option E is eliminated because it does not satisfy the condition of median. We get the smallest number as 16, which is greater than median. Therefore, it is impossible.

Option D: If 27 is the largest number, 15 is the smallest numbers; then all other numbers upto median must be 15. That would add up to 75. Summing that up with 27 gives us 102. Thus, we need to fit in the next three numbers which are greater than the median (15), however, their total must not be more than 135-102 = 33. This is impossible.

Similar calculations show that Option B and Option C also don't satisfy the condition.

Option A: If 23 is the largest number, the smallest number is (23+3)/2=13. If all numbers upto the median are 13, then they would add up to 52, plus the median i.e 15 would give 67. The largest number is 23. Therefore, the sum is 67+23=90. 135 - 90 = 45. Fitting three numbers greater than or equal to 15 and adding up to 45 is possible. Therefore, A is the answer.

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Re: In a set of 9 numbers, both the median and the mean are 15. If the [#permalink]

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New post 18 Sep 2017, 04:51
Bunuel wrote:
In a set of 9 numbers, both the median and the mean are 15. If the largest number in the set is 3 less than twice the smallest number, what is the largest possible number in the set?

A 23
B. 24
C. 25
D. 27
E. 29


The sum of the numbers is 9 x 15 = 135.

We can let the smallest number = x, so the largest is 2x - 3. To make the largest number in the set as large as possible, we will minimize (as much as possible) all the other values in the set. In ascending order, the first 4 numbers in the set will all be x. Because the median is 15, the 5th number in the set must be 15, and the 6th, 7th, and 8th numbers will also be 15. We can now use the following equation to determine the largest number in the set:

x + x + x + x + 15 + 15 + 15 + 15 + 2x - 3 = 135

6x = 78

x = 13

Thus, the largest possible number is 2(13) - 3 = 23

Answer: A
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Re: In a set of 9 numbers, both the median and the mean are 15. If the   [#permalink] 18 Sep 2017, 04:51
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