On writing first 252 positive integers in a straight line, how many ti

04,14,24,34, 54,...,94 = 9 nos of 4.
40,41,42,43,44,...,49 = 11 nos of 4.
104,114,124,134, 154,...,194 = 9 nos of 4.
140,141,142,143,144,...,149 = 11 nos of 4.
204,214,224,234 = 4 nos of 4.
240,241,242,243,244,...,249 = 11 nos of 4.

Total 4s = 9+11+9+11+4+11=55 nos

FINAL ANSWER IS (D)

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To solve the question, I just counted all possible 4's, like from 1 to 100 there are ten 4's. likewise from 100 to 200 there are ten 4's. then there are five 4's between 200 and 252. However, there are the numbers beginning with 4, such as 40, 41, 42,... and 140, 141, 142,... as well as 240, 241, 242,.... These all amount for thirty 4's. Finally, we need to sum all 4's up.
10+10+5+10+10+10=55

4,14,24,34,40-49,54,64,74,84,94
There is an extra 4 in 44

So total 20 in (1-100), Similarly 20 in (101-200) and 15 in 201-252 => 55

Answer - D

On writing first 252 positive integers in a straight line, how many times digit 4 appears?

A. 50
B. 52
C. 54
D. 55
E. 56

Hundred's place can be taken by three digits i.e. 0,1 and 2.
For 0's and 1's at hundred place possibilities are
2 * 10 * 1 = 20 where 2 is for 0 and 1 and 10 is for 0 to 9
For 2's at hundred place, possibilities are
1 * 5 * 1 = 5

For 4 at tens place possibilities are
3 * 1 * 10 = 30 where 3 is for 0, 1 and 2 at hundred's place and 10 is for 0 to 9 for unit's place

Hence total appearances = 20 + 5 + 30 = 55

Answer D.

On writing first 252 positive integers in a straight line, how many times digit 4 appears?

A. 50
B. 52
C. 54
D. 55
E. 56
From 1 to 10 ----------1 times 4
From 11 to 20……….....1 times 4
From 21 to 30 ---------1 times 4
From 31 to 39 ---------1 times 4
From 40 to 49 ---------10 times 4
From 50 to 99 ---------5 times 4

Total 20 times 4 from 1 to 100

In the same way, 20 times 4 from 101 to 200 and
14 times 4 from 201 to 252.
Total 54 times 4 on writing 252 positive integers in a straight line.

Answer: C

1 to 252;
4 appears?
04,44,94=10+([4]4)=11
104,144,194=10+(1[4]4)=11
204,244=5+(2[4]4)=6
40,49=10-(44)=9
140,149=10-(144)=9
240,249=10-(244)=9
total: 11+11+6+9(3)=55

Ans (D)

001, 002, 003, 004, ... 009, 010.. ... 250, 251, 252

————252 ————
From 1 to 100 —> 20 times digit 4 is used
From 101 to 200 —> 20 times digit 4 is used
From 201 to 252:
—> from 240 to 249 — 11 times
—> 204, 214, 224, 234 —4 times

In total, 20+20+11+4= 55

The answer is D


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I tried one although i can't say that that is a definite one.
See the combinations i did if you are okay with it.

Thank you lnm87 for explanation but I have not understood Might be your way is correct but P&C are my weaknesses

A year ago i was in a similar position on P&C but Ron's 'Thursday's with Ron' videos helped.You might have seen them,if not, you must.

I used to run away the moment i saw P&C problem; today the difference is not much - I try to walk away but not without giving a try.
The catch is, first, more important, the P&C problems are worded in that manner, second counting is involved. If counting is not correct there's no way they can be solved. Its still not that that i solve every &C problem.

In this one, let me know if i can he you.

If you are looking for a method that does not entail looking for each number separately, we can use combination..
Now digit 4 can be at any of the places - tens, units but not hundred as we are counting till 252.

Tens place
Place 4 at tens place --- a4b
Now a or hundreds place can be taken by 0, 1 or 2, so 3 ways
b or units place can take any of the 0-9, so 10 ways
Total - 3*1*10=30 times

units place
Place 4 at units place --- ab4
Now a or hundreds place can be taken by 0, 1 or 2, so 3 ways
b or units place can take any of the 0-9, so 10 ways
Total - 3*1*10=30 times
BUT we have counted till 299 while we had to count only till 252, so minus 254, 264, 274, 284, 294 or 5 numbers, so 30-5=25

Total 30+25=55