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nitin

· started a discussion

· 1 Months ago

Not satisfied with the solution (Question: 107565.0 )

I think there is one and only one case this can happen when average of first 30 is being calculated. As you are giving 15 in your question.

average of n natural number is (n+1)/2.

(n+1)/2=15 => n=29 but he has missed one number hence ans will be 30

Question:

A teacher gave sum to his class to find the average of first n numbers viz, 1, 2, 3, 4, 5, 6... etc. But when the teacher checked the solution, he has found that during the calculation a student just missed a number for the addition thus his average of the n numbers was 15. The value of n is: 

Options:
A)

30

B) 26  
C) 31
D) Not unique
Solution:
Ans: (d) Let Number leave by student be k, then

\(\cfrac{\left ( \cfrac {n(n+1)}{2}-k \right )}{n}\)   = 15

n (n+1) - 2k = 30n \(\Rightarrow\) n2 - 29n - 2k = 0

n (n-29) = 2k

Hence from option 

∵ k is non 0 and non negative no., then least possible value of n be 30 then n = 30 then k = 15

When n = 31 then k = 31

Hence (d) will be are answer.

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