
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Answer
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Hint: In this question, we are given the total people who like cricket or tennis, the number of people who like cricket, and the number of people who like both cricket and tennis. We will solve this sum using the union and intersection of sets. Here union $\left( \cup \right)$ will be used for 'or' statement and intersection $\left( \cap \right)$ will be used for 'and' statement. For finding number of people who like tennis we will use the formula given as: $n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)$ where $n\left( A\cup B \right)$ represent number of elements in A or B, n(A) represent number of elements in A, n(B) represent number of elements in B and $n\left( A\cap B \right)$ represent number of elements in A and B. To find the number of people who like tennis only and not cricket we will subtract the number of people who like both tennis and cricket from the number of people who like tennis.
Complete step-by-step solution:
Now, let us find a number of people who like tennis. For this, let C and T denote a set of people who like cricket and tennis respectively.
We are given a number of people = 65.
Therefore, the number of people who like cricket or tennis = 65.
Thus, \[n\left( C\cup T \right)=65\cdots \cdots \cdots \cdots \cdots \left( 1 \right)\]
Also, we are given that, the number of people who like cricket is 40.
Thus, \[n\left( C \right)=40\cdots \cdots \cdots \cdots \cdots \left( 2 \right)\]
Also, the number of people who like both tennis and cricket is 10.
Thus, \[n\left( C\cap T \right)=10\cdots \cdots \cdots \cdots \cdots \left( 3 \right)\]
As we know, for any two set
\[n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)\]
So, let's use it for set C and T, we get:
\[n\left( C\cup T \right)=n\left( C \right)+n\left( T \right)-n\left( C\cap T \right)\]
Putting value from (1), (2) and (3) we get:
\[\begin{align}
& 65=40+n\left( T \right)-10 \\
& \Rightarrow 65=30+n\left( T \right) \\
& \Rightarrow n\left( T \right)=35 \\
\end{align}\]
n(T) represents the number of people who like tennis.
Therefore, the number of people who like tennis = 35.
Now, we have to find a number of people who like only tennis and not cricket. Since n(T) can have people who like cricket also, so let us subtract the number of people who like both cricket and tennis from the number of people who like tennis to get the number of people who only like tennis but not cricket, we get:
The number of people who like only tennis but not cricket = number of people who like tennis - the number of people who like both cricket and tennis.
\[\begin{align}
& \Rightarrow n\left( T \right)-n\left( T\cap C \right) \\
& \Rightarrow 35-10 \\
& \Rightarrow 25 \\
\end{align}\]
Hence, 25 people like only tennis and not cricket.
Note: Students should not get confused between the number of people who like tennis and the number of people who like tennis only, as some of the people who like tennis can also like cricket. Students should note that we use union $n\left( A\cup B \right)$ when values can be taken from set A or set B and we use intersection $n\left( A\cap B \right)$ when values are taken from common in both A and B.
Complete step-by-step solution:
Now, let us find a number of people who like tennis. For this, let C and T denote a set of people who like cricket and tennis respectively.
We are given a number of people = 65.
Therefore, the number of people who like cricket or tennis = 65.
Thus, \[n\left( C\cup T \right)=65\cdots \cdots \cdots \cdots \cdots \left( 1 \right)\]
Also, we are given that, the number of people who like cricket is 40.
Thus, \[n\left( C \right)=40\cdots \cdots \cdots \cdots \cdots \left( 2 \right)\]
Also, the number of people who like both tennis and cricket is 10.
Thus, \[n\left( C\cap T \right)=10\cdots \cdots \cdots \cdots \cdots \left( 3 \right)\]
As we know, for any two set
\[n\left( A\cup B \right)=n\left( A \right)+n\left( B \right)-n\left( A\cap B \right)\]
So, let's use it for set C and T, we get:
\[n\left( C\cup T \right)=n\left( C \right)+n\left( T \right)-n\left( C\cap T \right)\]
Putting value from (1), (2) and (3) we get:
\[\begin{align}
& 65=40+n\left( T \right)-10 \\
& \Rightarrow 65=30+n\left( T \right) \\
& \Rightarrow n\left( T \right)=35 \\
\end{align}\]
n(T) represents the number of people who like tennis.
Therefore, the number of people who like tennis = 35.
Now, we have to find a number of people who like only tennis and not cricket. Since n(T) can have people who like cricket also, so let us subtract the number of people who like both cricket and tennis from the number of people who like tennis to get the number of people who only like tennis but not cricket, we get:
The number of people who like only tennis but not cricket = number of people who like tennis - the number of people who like both cricket and tennis.
\[\begin{align}
& \Rightarrow n\left( T \right)-n\left( T\cap C \right) \\
& \Rightarrow 35-10 \\
& \Rightarrow 25 \\
\end{align}\]
Hence, 25 people like only tennis and not cricket.
Note: Students should not get confused between the number of people who like tennis and the number of people who like tennis only, as some of the people who like tennis can also like cricket. Students should note that we use union $n\left( A\cup B \right)$ when values can be taken from set A or set B and we use intersection $n\left( A\cap B \right)$ when values are taken from common in both A and B.
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