A survey was conducted to know the hobbies of 220 students of class 9th. Out of which 130 students informed about their hobby as rock climbing and 180 students informed about their hobby as sky watching. There are 110 students who follow both hobbies. Then, how many students do not have any of the two hobbies? How many of them follow the hobby of rock climbing only? How many students follow the hobby of sky watching only?
Answer
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Hint- The intersection of two sets A and B is the set of all those elements which belong to both A and B.
$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
In this question, we need to determine the total number of students who do not have any of these hobbies of rock climbing as well as sky watching, who have the hobby of rock climbing only and who have the hobby of sky watching only for which we need to evaluate the intersection of both the hobbies.
Complete step by step solution:
Let $n\left( A \right)$ = The number of students whose hobby is rock climbing.
$n\left( B \right) = $ The number of students whose hobby is sky watching.
$n\left( {A \cap B} \right) = $ The number of students follows both hobbies.
We have,
$n\left( {A \cup B} \right) = $ Number of students whose hobby is either rock climbing or sky watching
$
n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \\
= 130 + 180 - 110 \\
= 200 \\
$
Here, the total number of students = 220
So, the number of students who do not have any of these two hobbies is$220 - 200 = 20$
Also, the number of students who follow the hobby of rock climbing only is given as:
$
n\left( {A - B} \right) = n\left( A \right) - n\left( {A \cap B} \right) \\
= 130 - 110 \\
= 20 \\
$
Again, the number of students who follow the hobby of sky watching only is given as:
$
n\left( {B - A} \right) = n\left( B \right) - n\left( {A \cap B} \right) \\
= 180 - 110 \\
= 70 \\
$
Hence,
The numbers of students who do not have any of these hobbies are 20.
The number of students who have the hobby of rock climbing are 20.
The numbers of students who have the hobby of sky watching are 70.
Note: For representing the intersection of subsets A and B of U, we draw two circles, and their intersecting regions will represent the intersection of both the sets. To draw two disjoint sets, we normally draw two circles which will not intersect each other within a rectangle.
$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
In this question, we need to determine the total number of students who do not have any of these hobbies of rock climbing as well as sky watching, who have the hobby of rock climbing only and who have the hobby of sky watching only for which we need to evaluate the intersection of both the hobbies.
Complete step by step solution:
Let $n\left( A \right)$ = The number of students whose hobby is rock climbing.
$n\left( B \right) = $ The number of students whose hobby is sky watching.
$n\left( {A \cap B} \right) = $ The number of students follows both hobbies.
We have,
$n\left( {A \cup B} \right) = $ Number of students whose hobby is either rock climbing or sky watching
$
n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \\
= 130 + 180 - 110 \\
= 200 \\
$
Here, the total number of students = 220
So, the number of students who do not have any of these two hobbies is$220 - 200 = 20$
Also, the number of students who follow the hobby of rock climbing only is given as:
$
n\left( {A - B} \right) = n\left( A \right) - n\left( {A \cap B} \right) \\
= 130 - 110 \\
= 20 \\
$
Again, the number of students who follow the hobby of sky watching only is given as:
$
n\left( {B - A} \right) = n\left( B \right) - n\left( {A \cap B} \right) \\
= 180 - 110 \\
= 70 \\
$
Hence,
The numbers of students who do not have any of these hobbies are 20.
The number of students who have the hobby of rock climbing are 20.
The numbers of students who have the hobby of sky watching are 70.
Note: For representing the intersection of subsets A and B of U, we draw two circles, and their intersecting regions will represent the intersection of both the sets. To draw two disjoint sets, we normally draw two circles which will not intersect each other within a rectangle.
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