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jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,302
Given Kudos: 355
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
20 Sep 2013, 17:04
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
53% (02:15) correct
47%
(02:06)
wrong
based on 153
sessions
History
Date
Time
Result
Not Attempted Yet
A machinist's salary at a factory increases by $2,000 at the end of each full year the machinist works. If the machinist's salary for the fifth year is $39,000, what is the machinist's average annual salary for his first 21 years at the factory?
(A) $12,000
(B) $33,000
(C) $37,000
(D) $51,000
(E) $54,000
(A) $12,000
(B) $33,000
(C) $37,000
(D) $51,000
(E) $54,000
20 Sep 2013, 17:16
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jlgdr
The average salary = the median salary = {11th year} = {5th year} + 6*2,000 = 39,000 + 12,000 = 51,000.
General Discussion
06 Sep 2021, 04:40
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jlgdr
Solution:
We are given that machinist's salary for the fifth year is \($39,000\). And the salary at a factory increases by \($2,000\) at the end of each year.
So, we know \(S_5=$39000\) and consequently:
\(S_4=$39000-2000=$37000\),
\(S_3=$37000-2000=$35000\),
\(S_2=$35000-2000=$33000\) and
\(S_1=$33000-2000=$31000\)
We can draw an inference that we are dealing with arithmetic progression (AP) with first term \(a=31000\) and common difference \(d=2000\)
We are asked the average of annual salary for his first 21 years. This means we are asked the \(\frac{sum of first 21 terms}{21}\)
We can use the formula of sum of AP to get sum of 21 terms. Sum of 21 terms \(= \frac{n}{2}[2a+(n-1)d]=\frac{21}{2}[2\times 31000+20\times 2000]= \frac{21}{2}[102000]=21\times 51000\)
Thus average \(= \frac{21\times 51000}{21}=$51000\)
Hence the right answer is \($51,000\)
