Latest TCS Aptitude Question SOLUTION: Two consecutive numbers are removed from progression 1,2,3,4,......n.The arithmetic mean of remaining numbers is 26 1/4.The value of n is

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Two consecutive numbers are removed from progression 1,2,3,4,......n.The arithmetic mean of remaining numbers is 26 1/4.The value of n is

Asked In TCS mallumounika (9 years ago)

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Read Solution (11) :

sayanee nandi  (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 (22) | No (3)  

Eshwar  (9 years ago)

n(n+1)/2-n/2-n/2 +1 = 26.25(n) -52.5 solving this u will get n=50

Is this solution Helpfull? Yes (9) | No (1)  

mahesh  (9 years ago)

answer may be 136

Is this solution Helpfull? Yes | No (7)  

ANKIT SHARMA  (9 years ago)

1+2+3+.............+23=286
Hence ans is 23

Is this solution Helpfull? Yes | No (4)  

ANKIT SHARMA  (9 years ago)

answer is n=54
becoz
26.25*52=1365
1+2+3.........+54=1431
and 1+2+3...............+53=1378
condition satisified in 54 ans

Is this solution Helpfull? Yes | No (4)  

raviteja  (9 years ago)

n(n+1)/2=26.5
n(n+1)=53
n2+n=53
n^2+n-53=0
n^
n

Is this solution Helpfull? Yes | No (2)  

mounika  (9 years ago)

wt's d crct ans?

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bhanu shakya  (9 years ago)

can u explain little more saynee

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shailendra mittal  (9 years ago)

if something missing digit is x,y
so avg. of this no. is x+y/2
airthmtic mean of 1,2 ,3,4.......n
n(n+1)/2n-x+y/2=52.5/2
(n+1)/2-x+y/2=52.5/2 53/2-.5/2
(n+1)/2=53/2 n=52

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swati richhariya  (9 years ago)

i dont get it

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ARCHIE GUPTA  (9 years ago)

average (n(n+1)/2)/n=109/4
n=53.5 or 54

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