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What is the average (arithmetic mean) of all the multiples of ten from

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Bunuel
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What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   24 May 2016, 12:18
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What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

A. 90
B. 95
C. 100
D. 105
E. 110
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C
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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   24 May 2016, 17:09
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10 and 190 inclusive would mean there are 19 terms.

No need to add up all the terms and further divide by 19 since the terms are evenly spaced in ascending order from 10, 20, 30........190

The middle term is the average which is the tenth term = 100

C is the answer.


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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   24 May 2016, 20:54
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Bunuel wrote:
What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

A. 90
B. 95
C. 100
D. 105
E. 110


The multiples of ten from 10 to 190 inclusive would be an evenly spaced set with 19 terms - 10, 20, 30,.......190
So average = (First term + Last Term)/2 = (10 + 190)/2 = 200/2 = 100
Hence, the correct answer is C.
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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   27 Oct 2016, 15:30
190/10 = 19 --> there are 19 multiples of 10 (i.e. there are 19 numbers in the set)

19/2 = 9... --> so we're looking for the 10th number in the set because it's an odd number of numbers in the set.

100 is the 10th number in the set.

Thus, C is correct. :)
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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   28 Oct 2016, 09:37
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Bunuel wrote:
What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

A. 90
B. 95
C. 100
D. 105
E. 110


All the multiples of ten from 10 to 190 inclusive = 10 + 20 + 30 + 40........190

10 + 20 + 30 + 40........190 = 10 ( 1 + 2 + 3 ....... 19 )

Now, ( 1 + 2 + 3 ....... 19 ) = 19*20/2 => 190

So, The sum of the series is 10*190

Average of the series will be 10*190/19 = 100

Hence answer will be (C) 100
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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   10 May 2017, 18:55
Bunnel,

I have a question regarding this question: Can we say since this data set (the multiples of 10) is equally distributed and the two lower and upper limits included in the data set, bringing the total data points to an odd number, therefore the average will be equal to the middle number: 100?
IE: 12345 what is the average: 3
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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   10 May 2017, 21:52
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Heseraj wrote:
What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

A. 90
B. 95
C. 100
D. 105
E. 110

Bunnel,

I have a question regarding this question: Can we say since this data set (the multiples of 10) is equally distributed and the two lower and upper limits included in the data set, bringing the total data points to an odd number, therefore the average will be equal to the middle number: 100?
IE: 12345 what is the average: 3


Multiples of 10 represent arithmetic progression (aka evenly spaced set). In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula \(mean=median=\frac{a_1+a_n}{2}\), where \(a_1\) is the first term and \(a_n\) is the last term.

First term is 10 and last term is 190. So \(mean=median=\frac{10+190}{2}=100\).

Answer: C.
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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   15 May 2017, 17:29
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Bunuel wrote:
What is the average (arithmetic mean) of all the multiples of ten from 10 to 190 inclusive?

A. 90
B. 95
C. 100
D. 105
E. 110


We can calculate the average using the following formula for evenly spaced sets:

average = (first multiple in the set + last multiple in the set)/2

average = (10 + 190)/2 = 200/2 = 100

Answer: C
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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   25 Jun 2018, 03:36
Sum of arithmetic series = n(a+l)/2 , where n= number of terms, a = first term, l= last term

Average = sum/n

Using both the equations, average = (a+l)/2
=> 200/2 = 100, option C

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Re: What is the average (arithmetic mean) of all the multiples of ten from [#permalink] New post   03 Feb 2023, 03:20
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