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Re: The average (arithmetic mean) of five positive integers a, b, c, d, e
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29 Jan 2020, 05:06
The average (arithmetic mean) of five positive integers a, b, c, d, e is 22 and a < b < c < d < e. If e is 40, what is the least possible value of the median of the five integers?
A. 10
B. 11
C. 12
D. 13
E. 14
Soln
Given- a,b,c,d,e== Positive Integers & a < b < c < d < e
a+b+c+d+e/5=22
so a+b+c+d+e=110
we are given e=40 so now we know a+b+c+d+40=110
Sum of remaining four integers (a+b+c+d)=70 , also we already know e is largest Integer among all 5 , so remaining four integers have to be less than 40 and Positive.
To find least value of median ( median in odd number of terms is the middle value) .So Median here is c,
Question has asked us to find least possible value of c.
So far we know sum of a,b,c,d=70 ,
We want least value of c, so lets maximize the d , condition is d<e so as e=40 , d can max be 39
So now we are left with a+b+c= 70-39=31
sum of remaining 3 integers has to be 31, now think of ways of distributing 31 among a,b,c such that a<b<c
Take clue from ans options Ans choice A : if c=10, b=9, a=8 Not equal to 31 Eliminate A
Ans choice B: if c=11, b=10, a=9 Not equal to 31 Eliminate B
Ans choice C: if c=12, b=10, a=9 Yes now its sum =31 CORRECT
Ans choice D: if c=13, b =10 a=7 also equals 31 but we are asked least value of c
Ans choice E: if c=14, b=10, a=6 also equals 31 but we are asked least value of c
Ans is B
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