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
_________________
The only thing more dangerous than ignorance is arrogance.
-- Albert Einstein
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80% (01:52) correct 
