
How many times does the digit $1$appear in numbers from $1$ to $100$?
(A)$18$ (B)$19$ (C)$20$ (D)$21$
Answer
593.4k+ views
Hint: First we need to divide the given set of numbers into some subsets. We can divide the set into subsets depending upon our convenience. Because of this particular process the chances of making mistakes become zero.
Complete step by step solution-
From $1$ to $100$ there are three types of numbers-
4. 1 digit numbers- $1 - 9$
5. 2 digit numbers- $10 - 99$
6. 3 digit numbers- $100$
So we are having total $100$ numbers.
In first type there are $9$ numbers which are
$\left\{ {1,2,3,4,5,6,7,8,9} \right\}$
Therefore we can conclude that there is only $1$ number where the digit ‘$1$’ appears. $N = 1$
In second type there are $90$numbers which are
o $\left\{ {10,11,12,13,14,15,16,17,18,19,20} \right\}$
We need to check each number in this group since number ‘$11$’ has to times the digit $1$,
there are more chances to make a mistake. So there are a total $11$ number of digits ‘$1$’.
o $21 - 30$. This range has only $1$ digit ‘$1$’ in the number ‘$21$’.
o $31 - 40$. This range has only $1$ digit ‘$1$’ in the number ‘$31$’.
o $41 - 50$. This range has only $1$ digit ‘$1$’ in the number ‘$41$’.
o $51 - 60$. This range has only $1$ digit ‘$1$’ in the number ‘$51$’.
o $61 - 70$. This range has only $1$ digit ‘$1$’ in the number ‘$61$’.
o $71 - 80$. This range has only $1$ digit ‘$1$’ in the number ‘$71$’.
o $81 - 90$. This range has only $1$ digit ‘$1$’ in the number ‘$81$’.
o $91 - 99$. This range has only $1$ digit ‘$1$’ in the number ‘$91$’.
Therefore in the second type digit ‘$1$’ appears $19$ times. $N = 19$
In third type there is $1$ number which is
100
Therefore we can conclude that there is only $1$ number where digit 1 appears. $N = 1$
Hence the number of times the digit ‘$1$’ appears in numbers from $1$ to $100$ is $21$.
Answer- (D)
Note: This type of question seems easy but there are more chances of making mistakes whenever there will be repeating the required digit (for example in our question there is that number $11$) in a number. So you have to be careful while solving the question.
Complete step by step solution-
From $1$ to $100$ there are three types of numbers-
4. 1 digit numbers- $1 - 9$
5. 2 digit numbers- $10 - 99$
6. 3 digit numbers- $100$
So we are having total $100$ numbers.
In first type there are $9$ numbers which are
$\left\{ {1,2,3,4,5,6,7,8,9} \right\}$
Therefore we can conclude that there is only $1$ number where the digit ‘$1$’ appears. $N = 1$
In second type there are $90$numbers which are
o $\left\{ {10,11,12,13,14,15,16,17,18,19,20} \right\}$
We need to check each number in this group since number ‘$11$’ has to times the digit $1$,
there are more chances to make a mistake. So there are a total $11$ number of digits ‘$1$’.
o $21 - 30$. This range has only $1$ digit ‘$1$’ in the number ‘$21$’.
o $31 - 40$. This range has only $1$ digit ‘$1$’ in the number ‘$31$’.
o $41 - 50$. This range has only $1$ digit ‘$1$’ in the number ‘$41$’.
o $51 - 60$. This range has only $1$ digit ‘$1$’ in the number ‘$51$’.
o $61 - 70$. This range has only $1$ digit ‘$1$’ in the number ‘$61$’.
o $71 - 80$. This range has only $1$ digit ‘$1$’ in the number ‘$71$’.
o $81 - 90$. This range has only $1$ digit ‘$1$’ in the number ‘$81$’.
o $91 - 99$. This range has only $1$ digit ‘$1$’ in the number ‘$91$’.
Therefore in the second type digit ‘$1$’ appears $19$ times. $N = 19$
In third type there is $1$ number which is
100
Therefore we can conclude that there is only $1$ number where digit 1 appears. $N = 1$
Hence the number of times the digit ‘$1$’ appears in numbers from $1$ to $100$ is $21$.
Answer- (D)
Note: This type of question seems easy but there are more chances of making mistakes whenever there will be repeating the required digit (for example in our question there is that number $11$) in a number. So you have to be careful while solving the question.
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