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Set S contains 100 consecutive integers. If the range of the

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sunita123
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Set S contains 100 consecutive integers. If the range of the [#permalink] New post   Updated on: 06 Feb 2018, 01:40
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Set S contains 100 consecutive integers. If the range of the negative elements of Set S equals 80, what is the average (arithmetic mean) of the positive numbers in the set?

A. 8
B. 8.5
C. 9
D. 9.5
E. 10
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Originally posted by sunita123 on 06 Apr 2014, 15:40.
Last edited by Bunuel on 06 Feb 2018, 01:40, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Set S contains 100 consecutive integers. If the range of the [#permalink] New post   06 Apr 2014, 15:51
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sunita123 wrote:
Set S contains 100 consecutive integers. If the range of the negative elements of Set S equals 80, what is the average (arithmetic mean) of the positive numbers in the set?


So, we have a set of 100 consecutive integers. Some of them must be negative and some not (if all of them were negative then the range of 100 consecutive negative integers would be 99, not 80). This implies that the greatest negative integer must be -1.

Now, we are told that the range of the negative elements of Set S equals 80: {range} = {largest} - {smallest} --> 80 = -1 - {smallest} --> {smallest} = -81.

So, the set consists of 81 negative elements (from -81 to -1), 0 and 18 positive integers (from 1 to 18), the total of 81+1+18=100 consecutive integers.

The average of integers from 1 to 18 is (first + last)/2 = (1 + 18)/2 = 9.5.

Hope it's clear.
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Re: Set S contains 100 consecutive integers. If the range of the [#permalink] New post   06 Apr 2014, 16:07
Thanks Bunuel:) ,now i get it.
Bunuel wrote:
sunita123 wrote:
Set S contains 100 consecutive integers. If the range of the negative elements of Set S equals 80, what is the average (arithmetic mean) of the positive numbers in the set?


So, we have a set of 100 consecutive integers. Some of them must be negative and some not (if all of them were negative then the range of 100 consecutive negative integers would be 99, not 80). This implies that the greatest negative integer must be -1.

Now, we are told that the range of the negative elements of Set S equals 80: {range} = {largest} - {smallest} --> 80 = -1 - {smallest} --> {smallest} = -81.

So, the set consists of 81 negative elements (from -81 to -1), 0 and 18 positive integers (from 1 to 18), the total of 81+1+18=100 consecutive integers.

The average of integers from 1 to 18 is (first + last)/2 = (1 + 18)/2 = 9.5.

Hope it's clear.
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Re: Set S contains 100 consecutive integers. If the range of the [#permalink] New post   13 Dec 2016, 10:25
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This is a great Question.
Here is my solution o this one -->

Range of negatives = 80
Hence -1-p=80=> p=-81
Hence,the series will be like => -81,-80,....0,1,...18
Hence the average of positive integers will be 1+18/2 => 9.5

Hence Required Mean = 9.5
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Re: Set S contains 100 consecutive integers. If the range of the [#permalink] New post   20 Apr 2022, 12:24
How are we able to solve these within 2 mins ?
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Re: Set S contains 100 consecutive integers. If the range of the [#permalink]
20 Apr 2022, 12:24
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