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Bunuel
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Bunuel
Is x an integer?

(1) The average (arithmetic mean) of x – 1 and x + 1 is an integer.
(2) The average (arithmetic mean) of x and –x is an integer.

Stmnt 1: Let X = 1.. we get average as Integer, let x = 1/2 we get average as Integer.. So not suffiecient.
Stmn 2: X = 1, Average Integer... X = 1/2, Average = Integer

So E.
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Average of x-1 and x+1 is X which is given to be an integer so X is an integer so sufficient. Statement 2 the average is 0 which does not give any idea abt X. So not sufficient. So answer is A.

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rahul16singh28
Bunuel
Is x an integer?

(1) The average (arithmetic mean) of x – 1 and x + 1 is an integer.
(2) The average (arithmetic mean) of x and –x is an integer.

Stmnt 1: Let X = 1.. we get average as Integer, let x = 1/2 we get average as Integer.. So not suffiecient.
Stmn 2: X = 1, Average Integer... X = 1/2, Average = Integer

So E.

I
Average of x-1 and x+1 is nothing but x itself and it is given as integer, so x is integer..
average of x-1 and x+1 is \(\frac{x-1+x+1}{2}=\frac{2x}{2}=x\)
so statement I is suff..

II
Whatever be value of x, the average of x and -x should be 0
so statement II does not tell anything new
insuff

A
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Average is 0 in statement 2 and not the value of x. X can be integer or non integer the average of X and -x is always 0. So it doesnt say the type of X

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akhilbatra06
Zero is an integer


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Hi

Yes, zero is an integer. So average of x and -x will be 0, which is an integer. If x=2, then also average of x and -x is zero. If x=1.5, then also average of x and -x is zero.
So zero being an integer doesnt tell us anything whether x will be an integer or not. '

So answer should be A.
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