Latest Elitmus Aptitude Question SOLUTION: 3. Soldier were practicing to shoot. Coincidence occurred, first soldier shoot the target in his first chance, second shoot the target in his 3rd chance (i.e. 1 +2), third shoot in

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(#M40038369) ELITMUS Logical Reasoning question Keep an EYE

3. Soldier were practicing to shoot. Coincidence occurred, first soldier shoot the target in his first chance, second shoot the target in his 3rd chance (i.e. 1 +2), third shoot in his 6th chances (i.e. 1+2+3), and so on.. Total 9880 bullet were fired. Find the number of soldier.

Asked In Elitmus Ritesh Anand (9 years ago)

Unsolved Read Solution (6)

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DIMPLE SINGH  (9 years ago)

series formed will be
1 3 6 .....
it's in the form of n(n+1)/2

sum of ( n(n+1)/2) = 9880
1/2 sum of (n^2+n)=9880
1/2 [ n(n+1)(2n+1)/6 + n(n+1)] = 9880
n(n+1)(n+2) = 59280
38*39*40 = 59280

therefore n= 38 = no. of soldier

Is this solution Helpfull? Yes (17) | No (2)  

Deepak Yadav  (9 years ago)

1st soldier shoot = 1 chance
2nd soldier shoot = 1+2 chance
3rd soldier shoot = 1+2+3 chance
4
.
.
.
nth soldier shoot = 1+2+3+4+.......+n chance
so,
1+2+3+4+.......+n= n*(n+1)/2
Total 9880 bullet were fired, so
n*(n+1)/2 =< 9880
solve hit and trail method
when n=140 then total bullet fired 9800
the 141 men are firing other(80) bullet but not get shoot.
so,
141 soldier........ ANS.

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amits  (9 years ago)

the series formed is- 1+3+6+10+15....
it is in the form of 1/2*(n(n+1))
summation of 1/2*(n(n+1))= 9880
i.e summation of 1/2*(n^2+n)=9880
1/2[n(n+1)(2n+1)/6+n(n+1)/2]=9880
n*(n+1)*(2n+4)=118560
on checking, 38*39*80=118560
so, n=38

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priyanka  (9 years ago)

plz xpln 5th step how is it??

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aquib aslam  (9 years ago)

take 1+3+6+10+15+21 nw this will be the series and sum upto
1+3+6+10+15+21...=9880
n/2*(n+1) =9880
now take summation of above equation by solving it we get a quadtatic equation solve it the ans will be 40

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Deepak Yadav  (9 years ago)

sry frnds for mis typing
when n=140 then total bullet fired 9800 (wrong)
....................................9870

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