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Abhii46
There will be 10 numbers from 200 to 299 that will end with 3 ( 203, 213, 223........ 293 ).
There will be 100 numbers from 300 to 400 that will begin with 3, we will not count the numbers that will end with because they will be repeated.
Total numbers = 10 + 100 = 110
Answer is D.

Kindly give a kudo if you like my explanation.

I would have fallen into a trap on that one...I was thinking units and tens positions.
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Abhii46
There will be 10 numbers from 200 to 299 that will end with 3 ( 203, 213, 223........ 293 ).
There will be 100 numbers from 300 to 400 that will begin with 3, we will not count the numbers that will end with because they will be repeated.
Total numbers = 10 + 100 = 110
Answer is D.

I would have fallen into a trap on that one...I was thinking units and tens positions.

I agree, it's pretty easy to see that all the 300s will count, so the answer needs to be 100 or more, but then a lot of people will fall into the trap of picking 120 instead of 110. The 230s don't count (except for 233), but there could have easily been a question asking how many of these numbers contain at least one 3. Always important to reread the question before answering in order to avoid falling into some of the traps the exam has laid out for you.

Hope this helps!
-Ron
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The first number is 203, thereafter only possible case is numbers ending with 3. 3 repeats itself after every 10 digits
so between 200 and 299, periodicity of numbers ending with 3 is 10, so there are 10 such numbers 303,313......393

All numbers between 300-399, are starting with 3, so a total of 99-0+1 = 100 numbers.
So total numbers is 110.
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Dear all, isn't the question asking for all the numbers that either begin OR end with 3? In this case, shouldn't we be subtracting the numbers which do both, begin AND end with 3? 10 (ten numbers between 200 and 299)+100 (numbers between 300 and 399)-10 (ten numbers between 300 and 399, which begin AND end with 3)?
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- 10 Numbers from 200 to 299 (203, 213, 223, 233, 243, 253, 263, 273, 283, 293)
- 100 numbers from 300 to 400 that will begin with 3, (300+(n-1)1 = 399 (counting the numbers))
- (number ending with 3, i.e.323 got already counted in number starting from 3 we won't count them again)
- hence Total numbers = 10 + 100 = 110
- Answer is D.
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