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) ÷ 2In 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