Latest TCS Aptitude Question SOLUTION: the consecutive numbers are removed from thr progression 1,2,3,....,n the arithmetic mean of the remaining number is 261/4 the value of n is

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(#M40038749) TCS Numerical Ability question finding n Keep an EYE

the consecutive numbers are removed from thr progression 1,2,3,....,n the arithmetic mean of the remaining number is 261/4 the value of n is

Asked In TCS sukanya T R (9 years ago)

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SUMANTA SAMANTA  (9 years ago)

We know that ,
sum of natural number or consecutive = n*(n+1)/2
And the mean of this consecutive no is
= (n*(n+1)/2)/n
= (n*(n+1)/(2*n))
= (n+1)/2
Therefore,
(n+1)/2 = 261/4
(n+1)= 261/2
So, n = (261/2-1)=129.5
which near equals to 130

Is this solution Helpfull? Yes (9) | No (25)  

rajesh kumar  (9 years ago)

any two consecutive numbers you eliminate,but the average always must be in the range from (n+1/2)+1 to (n+1/2)-1
here the average after removing consecutive nos is 26.25
now we have to check options.say the option is given 50.
then,average of (1+2+....+50)50=51/2=25.5
it covers the range 24.5 to 26.5
so 26.25 belongs to that range.so ans=50

Is this solution Helpfull? Yes (8) | No   

shikha panwar  (9 years ago)

mean of remaining numbers is 26 1/4 and not 261/4.....kindly change ur ans sumanta

Is this solution Helpfull? Yes (6) | No (1)  

nandita shrivastava  (9 years ago)

but the condition given is that 2 consecutive numbers are removed.so this z not the exact correct answer ryt

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

ALLAM ANUSHA  (9 years ago)

mean of n-2 numbers (i.e.,after removing two consecutive numbers from n natural numbers) is=n/2(2*a+(n-1)*d)/n
here n=n-2
((n-2)/2)((2+n-2-1)*1)/(n-2)=105/4
after solving n=53.5

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

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