Latest TCS Aptitude Question SOLUTION: What is the no of occurance of 5 in the given series121122123124125126.........356

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What is the no of occurance of 5 in the given series121122123124125126.........356

Asked In TCS pritish beura (10 years ago)

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ARPIT AGRAWAL  (10 years ago)

No. of 5 betw. 121-221= 20
No. of 5 betw. 222-322= 20
No. of 5 betw. 323-356= 11
so total no. of occurance of 5 will be 20+20+11=51

Is this solution Helpfull? Yes (13) | No (1)  

R NAGA HARIKANTH  (10 years ago)

51.First consider 125-200 the number of occurrences are 18. 200-300 are 20. and 13
..18+20+13=51

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Ann Mary Jolly  (10 years ago)

Ans: 51
On considering the individual terms of the series as 3-digit no.s
The series looks like.... 121 122 123 124 125........356
i.e. the series is just the no.s from 121 to 356.
On moving from 121 to 356,no. of occurrences of 5 = 51
(counting the 5s in units place and 10s places)

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k.v.v.varun  (10 years ago)

52 is the answer

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sravya  (10 years ago)

51 is the answer

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D Venkat Rami Reddy  (10 years ago)

from 121-200:125,135,145,150,151,...155,...159,165,175,185,195= 18 5's
from 200-300: 20 5's
from 300-356:305,315,325,335,345,350,351,352,353,354,355,356=12 5's
ans= 40

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sruthi  (10 years ago)

from 120 t0 200 18 5's are there,200 to 300 20 5's next 305,315,325,335,345,350,351,352,353,354,355,356 13 5's so total 20+18+13=51

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Shilajit Dutta  (10 years ago)

36 as it is from 121 then 122 then to 129 one 5 in each set so in this way there is a 5 in each of the sets.again in the set 151 to 159 there are all 5,s,so counting in total there are 36 5's in the set.

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poonam hemnani  (10 years ago)

It will be 47

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Santosh  (10 years ago)

5 in 1 to 100 = 20 (10in 05,15,25,...95(Unit digit) + 10 in 50,51,52,53....59(ten's place))
101 to 200 = 20
201 to 301 = 20
=60
+1(of 305)+1(315)+1(325)+1(335)+1(345)+2(355)+1(356)=8
=68

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sai arun kumar palwadi  (10 years ago)

121 to 200 - 18 5's
201 to 300 - 20 5's
301 to 356 - 13 5's
so total 51

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