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QOTD # 15 The greatest of the 21 positive integers in a cert [#permalink]
10 Sep 2016, 02:04

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Question Stats:

78% (01:47) correct
22% (01:13) wrong based on 50 sessions

The greatest of the 21 positive integers in a certain list is 16. The median of the 21 integers is 10. What is the least possible average (arithmetic mean) of the 21 integers?

Re: QOTD # 15 The greatest of the 21 positive integers in a cert [#permalink]
19 Sep 2016, 17:00

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Carcass wrote:

The greatest of the 21 positive integers in a certain list is 16. The median of the 21 integers is 10. What is the least possible average (arithmetic mean) of the 21 integers?

A) 4 B) 5 C) 6 D) 7 E) 8

Let the following spaces represent the 21 integers listed in ASCENDING ORDER, with a red space indicating the median: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

The greatest of the 21 positive integers in a certain list is 16 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 16

The median of the 21 integers is 10 _ _ _ _ _ _ _ _ _ _ 10 _ _ _ _ _ _ _ _ _ 16

What is the least possible average (arithmetic mean) of the 21 integers? We must MINIMIZE the 19 remaining integers. So, the first 10 integers must be 1, since 1 is the smallest positive integer. 1 1 1 1 1 1 1 1 1 1 10 _ _ _ _ _ _ _ _ _ 16

The remaining integers must be greater than or equal to 10 and less than or equal to 16 Since, we're trying to MINIMIZE the remaining integers, we'll let each number be 10 to get: 1 1 1 1 1 1 1 1 1 1 10 10 10 10 10 10 10 10 10 10 16

Average = (1+1+1+1+1+1+1+1+1+1+10+10+10+10+10+10+10+10+10+10+16)/21 = 126/21 = 6

Re: QOTD # 15 The greatest of the 21 positive integers in a cert [#permalink]
04 May 2018, 09:21

2

This post received KUDOS

Expert's post

Carcass wrote:

The greatest of the 21 positive integers in a certain list is 16. The median of the 21 integers is 10. What is the least possible average (arithmetic mean) of the 21 integers?

A) 4 B) 5 C) 6 D) 7 E) 8

There are 21 numbers in the list, and, if we list the numbers in ascending order, the median is the 11th number.

Thus, we can let each of the first 10 numbers = 1, each of the next 10 numbers = 10, and the last number = 16.

Thus, the average is (10 x 1 + 10 x 10 + 16)/21 = 126/21 = 6.