Latest Elitmus Aptitude Question SOLUTION: The sum of first n numbers where n is very large is calculated. What is the probability thai it is divisible by n? a) 1 b) 3/4 c) 1/2 d) can't determine

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(#M40037474) ELITMUS Numerical Ability question Keep an EYE

The sum of first n numbers where n is very large is calculated. What is the probability thai it is divisible by n?

a) 1
b) 3/4
c) 1/2
d) can't determine

Asked In Elitmus Saket Kumar Agrawal (10 years ago)

Unsolved Read Solution (11)

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NITIN RAJ G  (10 years ago)

sum of n nos.= n(n+1)/2
dividing it by n =n(n+1)/2n = (n+1)/2
it is divisible by n if n+1 is even..
so probability =1/2

Is this solution Helpfull? Yes (28) | No (3)  

Saket Kumar Agrawal  (10 years ago)

sum of 1st n numbers = n(n+1)/2

as this is divisible by any value of n so the probability of it to be divisible by n would be 1.

so coreect ans is (a)1

Is this solution Helpfull? Yes (10) | No (12)  

Deepesh  (10 years ago)

Sum of 1st n numbers = n(n+1)/2

This means that either n or (n+1) is divisible by 2

(1)If n is divisible by 2,then S(n) will not be divisible by n because original value of n is reduced to half.

But when n is not divisible by 2, this means that (n+1) is divisible by 2, and the originality of n is maintained

i.e.

let us assume (n+1)/2 = a

S(n) = n(n+1)/2
=> n*a
S(n) = n*a

=> n*a is divisible by n

Hence Total probability = 1/2

Is this solution Helpfull? Yes (7) | No (6)  

Gaurav Malhotra  (10 years ago)

yuuppieee its 1/2

Is this solution Helpfull? Yes (4) | No (1)  

deepak  (10 years ago)

rakesh bhai kaha ho
solution do bhai

Is this solution Helpfull? Yes (4) | No (2)  

ritesh kumar raushan  (10 years ago)

Sum of 1st n numbers = n(n+1)/2
now, n(n+1)/2n = (n+1)/2(dividing sum by n)
Since, (n+1)/2 is divisible only if n is odd i.e. (n+1) is even.
so probability must be 1/2.

Is this solution Helpfull? Yes (3) | No   

Ajay Lamoria  (9 years ago)

ans is (c) becz
sum of 1st n numbers = n(n+1)/2
We have to check when n(n+1)/2 is divisible by n
Case 1: When n is even
n(n+1) will be divisible by n resulting in (n+1) and since n is even (n+1) being odd will not be divisible by 2.
Case 2: When n is odd
n(n+1) will be divisible by n resulting in (n+1) and since n is odd (n+1) being even will be divisible by 2.
So when n is very large, we have almost equal no of even and odd numbers.
So the probability is 1/2.

Is this solution Helpfull? Yes (2) | No   

Shreesh Tripathi  (10 years ago)

As it is said n is very large thus n+1~n so n*(n+1)~n*n what in that case n dividea or not

Is this solution Helpfull? Yes | No (1)  

AVNISH RANJAN JHA  (10 years ago)

since n is very large
sum can be written as s = n(n+1)/2=(n^2)/2
when devided by n q =n/2
so probability will be 1

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vandana  (10 years ago)

sum of n numbrs= n(n+1)/2
dividing it by n =n(n+1)/2=>n = (n+1)/2
it is divisible by n if n+1 is even.............
so probability is 1/2

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will  (9 years ago)

its obviously 1 becoz question ask it is divisible by n or not ...
hw mny times u gonna divide the sum ..
its surely 1

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