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The average (arithmetic mean) of 6 consecutive integers is 18½. What

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The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]

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New post 27 Dec 2017, 08:36
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The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?

(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18
[Reveal] Spoiler: OA

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The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]

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New post 27 Dec 2017, 10:01
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?

(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18



The average of even number of consecutive integers is nothing but the
average of the middle two numbers - number 3(x) and number 4(y)
The average of odd number of consecutive integers is the middle integer.

Therefore, \(\frac{x+y}{2}= 18.5 => x+y = 37\) where x = 18 and y = 19
Hence, the third element must be the average of the 5 smallest integers, which is 18(Option E)

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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]

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New post 27 Dec 2017, 10:03
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?

(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18


Let the numbers be x-2, x-1, x, x+1, x+2 & x+3

Sum of six numbers \(= 6x+3= \frac{37}{2}*6\)

\(=>x=18\)

if largest no i.e x+3 is removed then median of the remaining 5 consecutive number, \(x=Average =18\)

Option E
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Re: The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]

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New post 27 Dec 2017, 10:06
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?

(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18


\(n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 18½*6 = 111\)

Or, \(n + (n +1) + (n +2) + (n +3) + (n +4) + (n +5) = 111\)

Or, \(6n + 15 = 111\)

Or, 6n = 96

So, n = 16

Average of the 5 smallest of these integers = \(\frac{16 + 17 + 18 + 19 + 20}{5}\) \(= \frac{90}{5} = 18\)

Thus, answer will be (E)

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The average (arithmetic mean) of 6 consecutive integers is 18½. What [#permalink]

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New post 27 Dec 2017, 11:15
Bunuel wrote:
The average (arithmetic mean) of 6 consecutive integers is 18½. What is the average of the 5 smallest of these integers?

(A) 12 ½
(B) 15
(C) 16
(D) 17 ½
(E) 18

Approach #1: (First Term + Last Term) ÷ 2

In an arithmetic progression, consecutive integers included, the mean equals

\(\frac{FirstTerm+LastTerm}{2}\)
Use that to find the smallest integer

The six consecutive integers are
x, (x+1), (x+2), (x+3), (x+4), (x+5)

Average (mean)= \(\frac{FirstTerm+LastTerm}{2}\)
\(\frac{x+(x+5)}{2}=18.5\)
\(2x + 5 = 37\)
\(2x = 32\), \(x = 16\)= first term

Average of first 5 terms: \(\frac{FirstTerm+LastTerm}{2}\)
A = \(\frac{16+20}{2}= 18\)

OR, in an AP {16, 17, 18, 19, 20}:
median = mean = 18

ANSWER E

Approach #2: Standard Average
\(\frac{S}{n}=A\)
Use the 6 terms defined in terms of x, above

\(\frac{6x + 15}{6} = 18.5\)
\(6x + 15 = 111\)
\(6x = 96\)
, \(x = 16\)= first term

First five terms: 16, 17, 18, 19, 20
Average: \(\frac{16+17+18+19+20}{5}=\frac{90}{5}=18\)

ANSWER E
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The average (arithmetic mean) of 6 consecutive integers is 18½. What   [#permalink] 27 Dec 2017, 11:15
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