How many times the digit 9 occurs in the sequence? A sequence is given say : 100, 101, 102, 103, 104, 105………, 364, 365. Now, How many times the digit 9 occurs in the above sequence? 
How do I approach this kind of problems. I am preparing for my placements and one of the sections is Quantitative Aptitude. Are there good resources for such problems?
 A: Systematic enumeration seems best for such a small problem.
$ {109, 119, 129,...189} = 9 $ now replace the first digit with a 2 so there are 18 of those.
${190...199} = 11$ now replace the front digit with a 2 so there are 22 of them.
$ {309, 319, 329,...359} = 6 $
$18 + 22 + 6 = 46$
A: the one's digit will be 9 every 10 advancements in the sequence (109, 119, 129,...)
the ten's digit will be 9 every 100 advancements in the sequence, with the caveat that it will be 9 ten times in a row (190,191,192,... and 290,291,292...)
the hundred's digit will never be 9
so 10 (190->199) + 10 (290->299) + 10 (109->199) + 10 (209->299) + 6 (309->359) - 1 (199 double counted) - 1 (299 double counted)
=10+10+10+10+6-1-1=44
Depending on the amount of time you are given, I think brute forcing it like this is perfectly adequate.
A: starting from 101, 102..the first 9 we will get in digit 109, 2nd in 119..similarly in 129, 139, 149, 159,169,179,189 = total 9
now from 190 , 191... to 199 = total 11
So b/w 101 to 200 there will be 20 9's.
similarly b/w 201 to 300 there will be 20 other 9's.
Then from 301 to 365 the 9's will be 6 ( one each in 109,119,129,139,149 and 159)
So total number of 9 = 20+20+6 = 46
