Latest Elitmus Aptitude Question SOLUTION: one boy has all the 2 digit even n0. and other has all the 2 digit odd n0.Both square their no.and add up .what is the difference of these two addition??? a) 0 b) 4905 c) other tw

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(#M40020988) ELITMUS question Keep an EYE

one boy has all the 2 digit even n0. and other has all the 2 digit odd n0.Both square their no.and add up .what is the difference of these two addition???
a) 0
b) 4905
c) other two options i forgot...

Asked In Elitmus mayank chaudhary (11 years ago)

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Anmol Modi  (11 years ago)

The total 2 digits no.=90
As per questn ,we have 2 find
((11^2)+(13^2)+(15^2)+(17^2)+...+(99^2))-((10^2)+(12^2)+(14^2)+(16^2)+...+(98^2))
now it can be written as
(11^2)-(10^2)+(13^2)-(12^2)+(15^2)-(14^2).........+(99^2)-(98^2)
=(121-100)+(169-144)+(225-196)+......+(9801-9604)
=21+25+29+.......+197
now these terms are in A.P. where a=21, d=4,
n=45(because 1 odd and 1 even term is used to make 1 A.P. terms)
so total difference =
(45/2)*(2*21+(45-1)*4)
=4905

Is this solution Helpfull? Yes (67) | No   

komal  (11 years ago)

for AP series with same common difference,sum of the two series=difference of sum of squares of those 2 series.
for example,
take 2 AP series as follows:
odd series 1=1+3+5+7
even series 2=2+4+6+8
sum1={n/2(a+l)}={4/2(1+7)}=16 //l=last term,a=first term,n=no of terms
sum2={n/2(a+l)}={4/2(2+8)}=20
sum of squares 1=1+9+25+49=84
sum of squares 2=4+16+36+64=120
difference of these additions=120-84=36
sum of these two series=20+16=36
hence to find answer to the given question,
its simply addition of these 2 series with
series 1 in AP with a=10,n=45,and l=98
series 2 in AP with a=11,n=45 and l=99
sum1={45/2(10+98)}=2430
sum2={45/2(11+99)}=2475
as proved.
difference=sum of above 2=2430+2475=4905
hence ,ans=(b)

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rahul singh  (11 years ago)

other option are 1none of these 2)11...

Is this solution Helpfull? Yes (2) | No (1)  

rahul singh  (11 years ago)

3rd OPTION 1105

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ABDUL KHALAQUE  (11 years ago)

series 1 in AP with a=10,n=45,and l=98
series 2 in AP with a=11,n=45 and l=99
sum1={45/2(10+98)}=2430
sum2={45/2(11+99)}=2475
as proved.
difference=sum of above 2=2430+2475=4905
hence ,ans=(b)

Is this solution Helpfull? Yes (1) | No   

sonali randave  (11 years ago)

there is no option like 11...

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SACHIN AGGARWAL  (11 years ago)

Its 4905.just try to solve it. the difference will become the addition of all two digit no.s.

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Ankit Goyal  (11 years ago)

it can be written as
(11^2 - 10^2) + (13^2 - 12^2) + ........ + (99^2 - 98^2)
= 21 + 25 + 29 ....... upto 45 terms
= (45/2)[2*21 + (45 - 1)*4] = 4905

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