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

Hint: 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.