Question

How many numbers between $1$ and $200$ are multiples of $5$ and divisible by $3$?
A. $100$
B. $50$
C. $95$
D. $93$

Answer 1

Indirectly we need to calculate the terms that are multiples of 5 plus the terms that divisible by 3 minus the terms that are divisible by ${\text{15}}$, as there will be few common terms in both the series. Here we use the concept of AP to get the number of terms.

Complete step by step answer:

As per the given data, we have to find the between $1$ and $200$ that are multiples of $5$ and should also be divisible by $3$
The number that is multiple of $5$ between $1$ and $200$ is $5,10,15...200$ and number divisible by $3$ are $3,6,9...198$
The number satisfying both the conditions are $15,30,...195$
In the above-obtained series, we can observe that the first term is $15$ and the difference is also $15$ while the last term is $195$.
Now, substituting the values in the formula of A.P, i.e.
${{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}$ we get,

$${\text{195 = 15 + }}\left( {{\text{n - 1}}} \right){\text{15}} \\\ \Rightarrow {\text{180 = (n - 1)15}} \\\$$

Simplifying it further and calculating the value of n, we get,
$\Rightarrow \dfrac{{180}}{{15}}{\text{ = n - 1}}$
On simplifying further we get,
$\Rightarrow {\text{12 = n - 1}}$
On rearranging we get,
$\Rightarrow {\text{12 + 1 = n}}$
Hence , ${\text{n = 13}}$
Hence , In the above obtained series we can observe that first term is ${\text{5}}$ and difference is also ${\text{5}}$ while the last term is $195$
Now, substituting the values in the formula of A.P, i.e.
${{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}$ we get,

$$\Rightarrow {\text{195 = 5 + }}\left( {{\text{n - 1}}} \right){\text{5}} \\\ \Rightarrow {\text{190 = (n - 1)5}} \\\$$

On calculating the value of n we get,
$\Rightarrow \dfrac{{190}}{5}{\text{ = n - 1}}$
On simplifying we get,
$\Rightarrow {\text{38 = n - 1}}$
$\Rightarrow {\text{ n = 39}}$
Now, calculating the total number of terms that are divisible by $3$.
Hence, In the above-obtained series we can observe that first term is ${\text{3}}$ and difference is also ${\text{3}}$ while the last term is $198$
Now, substituting the values in the formula of A.P, i.e.
${{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}$ we get,

$$\Rightarrow {\text{198 = 3 + }}\left( {{\text{n - 1}}} \right)3 \\\ \Rightarrow {\text{195 = (n - 1)3}} \\\$$

Hence calculating the value of n,
$\Rightarrow \dfrac{{195}}{3}{\text{ = n - 1}}$
On further simplification we get,
$\Rightarrow {\text{66 = n - 1}}$
$\Rightarrow {\text{n = 67}}$
Hence , we need to add all the terms satisfying the condition of divisible by $3$ and multiples of $5$ and subtract the terms common in both of them
So, it will be
${\text{67 + 39 - 13 = 93}}$
Hence, option (d) is our correct answer.

Note: Multiples of a number like here we take 5, are always divisible by 5. So we can say that if a number is multiple of 5 it is divisible by 5. Also here we subtract the numbers divisible by 15 from the list, as we have already considered them in the list of numbers divisible by 3 and in the list of numbers which are multiples of 5, so to count them only once we subtract the total sum, and hence our required answer is obtained.