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
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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

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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In the depths of winter, I finally learned
that within me there lay an invincible summer.
-- Albert Camus, "Return to Tipasa"

When you remove the largest element of the consecutive integers list, the mean moves down by 0.5. So mean of 5 smallest integers will be 18.

This post discusses this and other related concepts: https://www.veritasprep.com/blog/2015/0 ... -the-gmat/
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Contact: bansal.karishma@gmail.com