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The mean of twenty-five consecutive positive integers numbers is what

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The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A) 4%

(B) 5%

(C) 20%

(D) 25%

(E) Cannot be determined by the information provided.
[Reveal] Spoiler: OA

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The mean of twenty-five consecutive positive integers numbers is what [#permalink]

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Bunuel wrote:
The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A) 4%

(B) 5%

(C) 20%

(D) 25%

(E) Cannot be determined by the information provided.


let the consecutive numbers be from 1 to 25
Sum = \(\frac{n}{2}*(first term + last term) =\) \(\frac{25}{2}*(1+25)\)
or Sum = 25*13
Therefore Mean of 25 terms will be = \(\frac{25*13}{25}\) = 13
so Mean as % of Sum = \(\frac{13}{(25*13)}*100\) = 4%
Option A

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I assume when the question says "total" it means to say "sum". By definition, mean = sum/n, where n is the number of terms.

We have n = 25, so:

mean = sum / 25

so the mean is 1/25 of the sum, or 4% of the sum.

It's not important that the set consists of consecutive integers.
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Bunuel wrote:
The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A) 4%

(B) 5%

(C) 20%

(D) 25%

(E) Cannot be determined by the information provided.


Solution



    • Let the numbers be \(n-12,n-11,…………. , n, ,………….n+11,n+12\)

      o The mean of the above would be \(= n\)

      o The sum of 25 numbers will be \(= 25n\)

      o Thus, the percentage \(= n*\frac{100}{25n} = 4\)%


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Re: The mean of twenty-five consecutive positive integers numbers is what [#permalink]

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New post 29 Jul 2017, 09:41
IanStewart wrote:
I assume when the question says "total" it means to say "sum". By definition, mean = sum/n, where n is the number of terms.

We have n = 25, so:

mean = sum / 25

so the mean is 1/25 of the sum or 4% of the sum.

It's not important that the set consists of consecutive integers.



I guess this would be the best way to solve it. If we focus on the definition, we can indeed come to conclusion that the relation between Mean and Sum is -


Mean/Sum = 1/N, where N is the total number of terms.

With this inference in mind, it would hardly take any time to solve this question. :)



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given, mean = sum /25

1/25= 4/100...
hence mean = 4% of sum..

ans A

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Re: The mean of twenty-five consecutive positive integers numbers is what [#permalink]

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Bunuel wrote:
The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A) 4%

(B) 5%

(C) 20%

(D) 25%

(E) Cannot be determined by the information provided.


----------ASIDE----------------------
There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median."
For example, in each of the following sets, the mean and median are equal:
{7, 9, 11, 13, 15}
{-1, 4, 9, 14}
{3, 4, 5, 6}
------------------
For this question, let x = first value
So, x+1 = second value
x+2 = third value
.
.
.
x+24 = last value

This means x+12 is the median AND the mode

Since x+12 = the mean of the 25 numbers, we can conclude that (25)(x+12 ) = the SUM of the 25 numbers

The mean of twenty-five consecutive positive integers numbers is what percent of the total?
We must convert (x+12)/(25)(x+12) to a PERCENT
(x+12)/(25)(x+12) = 1/25
= 4/100
= 4%
Answer: A

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Re: The mean of twenty-five consecutive positive integers numbers is what   [#permalink] 10 Nov 2017, 17:05
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