
If A = $\{ x \in {\rm N}:{\text{x is a multiple of }}3\} $ and B = $\{ x \in {\rm N}:{\text{x is a multiple of 6}}\} $ then A-B is equals to
\[
A)\{ 6,12,18,.....\} \\
B)\{ 3,6,9,12,......\} \\
C)\{ 3,9,15,21,.....\} \\
D){\text{none of the above}} \\
\]
Answer
629.7k+ views
Hint: Here to proceed the solution we need to know the multiples of 3 and 6. Make set A and set B as per given condition.
Here we are given with two sets where with the condition, Where
A = $\{ x \in {\rm N}:{\text{x is a multiple of }}3\} $
Here set A is a multiple of 3 which are natural numbers.
Then $A = \{ 3,6,9,12,15,18,.......\} $
$B = \{ x \in {\rm N}:{\text{x is a multiple of 6}}\} $
Here set B is a multiple of 6 which are natural numbers.
Then $B = \{ 6,12,18,24,.....\} $
Now we got both the values of set A and set B
Then
$
A - B = \{ 3,6,9,12,15,18,.....\} - \{ 6,12,18,24,30,....\} \\
A - B = \{ 3,9,12,15,21,.....\} \\
$
Here A-B means we have removed the element of set B from set A.
Option C is the correct answer.
NOTE: In these problems first we have to find the value of set A by given condition and in the same way we have to find the values of set B .Later we have to find A-B which means we have to subtract values of set B from set A. In this type of problem mainly we have to focus on the conditions given to the sets because different sets have different conditions.
Here we are given with two sets where with the condition, Where
A = $\{ x \in {\rm N}:{\text{x is a multiple of }}3\} $
Here set A is a multiple of 3 which are natural numbers.
Then $A = \{ 3,6,9,12,15,18,.......\} $
$B = \{ x \in {\rm N}:{\text{x is a multiple of 6}}\} $
Here set B is a multiple of 6 which are natural numbers.
Then $B = \{ 6,12,18,24,.....\} $
Now we got both the values of set A and set B
Then
$
A - B = \{ 3,6,9,12,15,18,.....\} - \{ 6,12,18,24,30,....\} \\
A - B = \{ 3,9,12,15,21,.....\} \\
$
Here A-B means we have removed the element of set B from set A.
Option C is the correct answer.
NOTE: In these problems first we have to find the value of set A by given condition and in the same way we have to find the values of set B .Later we have to find A-B which means we have to subtract values of set B from set A. In this type of problem mainly we have to focus on the conditions given to the sets because different sets have different conditions.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Class 10 Question and Answer - Your Ultimate Solutions Guide

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
Who is known as the "Little Master" in Indian cricket history?

A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

State and explain Ohms law class 10 physics CBSE

Distinguish between soap and detergent class 10 chemistry CBSE

a Why did Mendel choose pea plants for his experiments class 10 biology CBSE

Draw the diagram of the sectional view of the human class 10 biology CBSE

