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sunita123
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Set S contains 100 consecutive integers. If the range of the Updated on: 06 Feb 2018, 01:40
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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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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Set S contains 100 consecutive integers. If the range of the 06 Apr 2014, 15:51
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sunita123
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?
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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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Set S contains 100 consecutive integers. If the range of the 06 Apr 2014, 16:07
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Thanks Bunuel:) ,now i get it.
Bunuel
sunita123
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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Set S contains 100 consecutive integers. If the range of the 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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Set S contains 100 consecutive integers. If the range of the 20 Apr 2022, 12:24
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How are we able to solve these within 2 mins ?
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Set S contains 100 consecutive integers. If the range of the 25 Feb 2026, 00:35
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hii Bunuel
can there be other 100 consecutive integers as -200 to -101 etc. I am not getting how -1 must be the largest integer , feeling difficulty to visualise it properly .Please elaborate little bit
Bunuel


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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Set S contains 100 consecutive integers. If the range of the 25 Feb 2026, 03:22
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shaliny
hii Bunuel
can there be other 100 consecutive integers as -200 to -101 etc. I am not getting how -1 must be the largest integer , feeling difficulty to visualise it properly .Please elaborate little bit

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Please read the question and solution:

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

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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