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07 Dec 2020, 04:45
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D
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Difficulty:
35%
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Question Stats:
67% (01:42) correct
33%
(01:53)
wrong
based on 154
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In the range of integers from 1 to 200, how many of the numbers will contain 3 as at least one of its digits?
A. 19
B. 20
C. 30
D. 38
E. 40
A. 19
B. 20
C. 30
D. 38
E. 40
07 Dec 2020, 05:35
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Bunuel
A fact that will help you solve these type of questions in 10 to 20 seconds..
00 to 99 are 100 * 2-digit numbers, so total of 200 digits are used for integers from 00 to 99.
Here we write single digit also as 2-digit to ensure equal participation of 0.
All the digits have to be equally present, so each is used 200/10 or 20 times. However, one of the eintegers uses the digit twice for example 22, 33 and so on.
S0 the number of integers containing a specific digit ( except 0) is 20-1=19.
Once you know this we can tackle any question.
from 0-99.... 19 will contain digit 3.
from 100-199...another 19 will contain digit 3.
Total 38
D
17 Dec 2020, 23:17
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Bunuel
In the range single digits containing 3
only one - 3
So total single digits have 3 = 1
Double digits containing 3:
a) If unit digit = 3
8*1 - one option for unit digit i.e. 3 and 8 options for tens digit, for tens digit we are leaving 0 and 3 hence 8 options
b) If tens digit is 3
1*9 - one option for tens digit i.e. 3 and 9 options for tens digit, for tens digit we are leaving only 3.
c)Both unit and tens digit are 3 :
1*1 = 1 One option for units digit and tens digit i.e. 3
So total 2 digits have at least one three - 8+9+1= 18
3 digits containing 3:
1*9*1 : unit digit can only be 1 , i.e. 3. For tens digit we are leaving 3 hence 9 options. For hundred digit one option i.e. 1
1*1*9: Tens digit one options i.e. 3. Hundred's digit 1 option i.e. 1 and units digit 9 options because we are leaving out 3.
1*1*1: Units digit one option i.e. 3. Tens digit one option i.e. 3 and hundred digit one option i.e. 1 .
So among 3 digits total number of 3's = 9 + 9 +1 = 19
Hence total three's -> 1+18+19= 38
Ans- D
Hope it's clear.
