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If the average (arithmetic mean) of 18 consecutive odd integers is 534

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If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 30 May 2015, 17:47
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If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525
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Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 30 May 2015, 19:04
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A very helpful rule to know in arithmetic is the rule that in evenly spaced sets, average = median. By evenly spaced sets I mean any set where the numbers are evenly spaced from one another (they do not have to necessarily be even numbers, or spaced 2 apart, or adhere to any other rule). The sets (-3,-2,-1,0,1,2), (15,20,25), (12,14,16,18,20,22,24), and (-35,-28,-21,-14) are all examples of evenly spaced sets. Because the average will equal the median in these sets, then we quickly know that the median of this set of consecutive odd integer numbers is 534.

There are 18 numbers in the set, and in a set with an even number of terms the median is just the average of the two most median terms (here the 9th and 10th numbers in the set). This means that numbers 9 and 10 in this set are 533 and 535. Because we know that number 9 is 533, we know that the smallest number is 8 odd numbers below this, which means that it is 8*2 = 16 below this (every odd number is every other number). Therefore 533-16 = 517, answer choice A.

I hope this helps!
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Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 30 May 2015, 22:59
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Hello BuggerinOn

I would be most comfortable solving it in this fashion

\(S = \frac{n}{2} [ 2a + (n-1)d ]\)
(you would like to remember this formula which is the sum of an arthimetic progression. also notice number of terms multiplied by average of those numbers is the sum of the set)


\(534*18 = \frac{18}{2} [ 2a + (18-1)2 ]\)

\(1068 = (2a + 34)\)

\(a = 517\)

:)
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Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 27 Oct 2016, 10:53
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BuggerinOn wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525


Odd numbers can be represented in the form : n , ( n + 2 )............

So , n + ( n + 2 ) + ( n + 4 ) + ( n + 6 )............+ ( n + 34 ) = 18*534

Or, 18n + 306 = 18*534

Or, 18 ( n + 17 ) = 18*534

Or, n + 17 = 534

So, n = 517

Hence, answer will be (A) 517...

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Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 12 Dec 2016, 11:01
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Here is my solution to this one ->
Let n be the first term
Last term => \(n+17*2=> n+34\)

Now as the series will be in AP.
Mean = Average of the first and the last term
Hence 534 =\(\frac{n+n+34}{2}=> n+17\)
Hence n+17=534=> n=517

Hence A

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If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 10 Oct 2018, 00:57
Bunuel wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

(A) 517
(B) 518
(C) 519
(D) 521
(E) 525

Sum = n/2[2a+(n-1)d] , a = is the first term , d = is the common difference n=18 as mentioned in the question
Over here
Sum = 18*534 = 9612 , N=18 , A = a ,D=2 (because consecutive odd)
9612 = 18/2(2a+(18-1)2)
9612 = 9(2a + 34)
9612/9 = 2a + 34
1068 = 2a + 34
1068 - 34 = 2a
1034 = 2a
517 = a !

I will go with A
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Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 10 Oct 2018, 01:08
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Bunuel wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

(A) 517
(B) 518
(C) 519
(D) 521
(E) 525



Note: we have total 18 consecutive integers. It means we got average by dividing the sum of 2 integers . Question is which 2 integers?


533 + 535 / 2 = 534.........................a bit common sense is required here.

If total number of integers were 19 , the average would be 10th integers. So, in this case we take care both 9th and 10th integer of the serious.


9th = 533

17-19-21-23-25-27-29-31


517 is the least among the odd integers.


The best answer is A.
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Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 10 Oct 2018, 07:34
BuggerinOn wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525



\(\frac{x}{18}\)= \(534\)

x = 9,532

if each of the numbers were 501 than total would be 501*18 = 9,018

so if i deduct 1 from each of the odd numbers i get 534-18 = 516

so smallest odd number is 517
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Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

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New post 13 Oct 2018, 18:15
BuggerinOn wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525


Since the average of 18 consecutive odd integers is 534, the average of the middle two integers is also 534. Therefore, the middle two integers are 533 and 535. There will be 8 odd integers less than 533 (and 8 more greater than 535). So the smallest of these integers is 533 - 8(2) = 533 - 16 = 517.

Alternate Solution:

If the smallest of these integers is x, then the largest of these integers will be x + 34. Since consecutive odd integers form an evenly spaced set of integers, the average is equal to the average of the smallest and the largest integers; thus:

(x + x + 34)/2 = 534

2x + 32 = 1068

2x = 1034

x = 517

Answer: A
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Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534   [#permalink] 13 Oct 2018, 18:15
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