How many times the digit 5 will come in counting from 1 to 99 excludin

1.	How many times the digit 5 will come in counting from 1 to 99 excluding those which are divisible by 3?1. 192. 203. 144. 13
Answer» Correct Answer - Option 3 : 14 Formula Used: The n^th term of an AP is given by: a_n = a + (n – 1) × d ----(i) where a_n = n^th term of the AP, a = first term of the AP, n = number of terms within the AP, d = difference between successive terms in the AP. Calculation: The numbers between 1 and 99, which are divisible by 3, are: 3, 6, 9,......., 99. Using the equation (i), we get: 99 = 3 + (n – 1) × 3 ⇒ n = 33 So, the remaining terms, not divisible by 3 = 99 – 33 = 66 Among these 66 terms, 5 appears once each in the following terms: 5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 56, 57, 58, 59, 65, 75, 85, and 95. But, 15, 45, 51, 54, 57 and 75 are divisible by 3, and hence excluded. So, the total number of remaining terms with 5 appearing only once = 18 – 6 = 12. And appears twice in the term 55. So, the total number of times the digit 5 appears = 12 + (1 × 2) = 14 ∴ The total number of times 5 appears between 1 to 99, when excluding the numbers divisible by 3, will be 14

How many times the digit 5 will come in counting from 1 to 99 excluding those which are divisible by 3?1. 192. 203. 144. 13

Answer» Correct Answer - Option 3 : 14

Formula Used:

The n^th term of an AP is given by:

a_n = a + (n – 1) × d ----(i)

where a_n = n^th term of the AP,

a = first term of the AP,

n = number of terms within the AP,

d = difference between successive terms in the AP.

Calculation:

The numbers between 1 and 99, which are divisible by 3, are:

3, 6, 9,......., 99.

Using the equation (i), we get:

99 = 3 + (n – 1) × 3

⇒ n = 33

So, the remaining terms, not divisible by 3 = 99 – 33 = 66

Among these 66 terms, 5 appears once each in the following terms:

5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 56, 57, 58, 59, 65, 75, 85, and 95.

But, 15, 45, 51, 54, 57 and 75 are divisible by 3, and hence excluded.

So, the total number of remaining terms with 5 appearing only once = 18 – 6 = 12.

And appears twice in the term 55.

So, the total number of times the digit 5 appears = 12 + (1 × 2) = 14

∴ The total number of times 5 appears between 1 to 99, when excluding the numbers divisible by 3, will be 14