NCERT Solutions For Class 11 Maths Chapter 1 Exercise 1.6

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Chapter 1 Ex.1.6 Question 1

If \(X\) and \(Y\) are two sets such that \(n\left( X \right) = 17,\;n\left( Y \right) = 23\) and \(n\left( {X \cup Y} \right) = 38\), find \(n\left( {X \cap Y} \right)\).

Solution

It is given that:

\(n\left( X \right) = 17,\;n\left( Y \right) = 23\) and \(n\left( {X \cup Y} \right) = 38\)

We know that:

\[\begin{align}n\left( {X \cup Y} \right) &= n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right)\\n\left( {X \cap Y} \right) &= n\left( X \right) + n\left( Y \right) - n\left( {X \cup Y} \right)\\ &= 17 + 23 - 38\\& = 40 - 38\\& = 2\end{align}\]

Therefore, \(n\left( {X \cap Y} \right) = 2\)

Chapter 1 Ex.1.6 Question 2

If \(X\) and \(Y\) are two sets such that \(X \cup Y\) has \(18\) elements, \(X\) has \(8\) elements and \(Y\) has \(15\) elements; how many elements does \(X \cap Y\) have?

Solution

It is given that:

\(n\left( X \right) = 8,\;n\left( Y \right) = 15\) and \(n\left( {X \cup Y} \right) = 18\)

We know that:

\[\begin{align}n\left( {X \cup Y} \right) &= n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right)\\n\left( {X \cap Y} \right)& = n\left( X \right) + n\left( Y \right) - n\left( {X \cup Y} \right)\\ &= 8 + 15 - 18\\ &= 23 - 18\\ &= 5\end{align}\]

Therefore, \(n\left( {X \cap Y} \right) = 5\)

Chapter 1 Ex.1.6 Question 3

In a group of \(400\) people, \(250\) can speak Hindi and \(200\) can speak English. How many people can speak both Hindi and English?

Solution

Let \(H\) be the set of people who speak Hindi, and

\(E\) be the set of people who speak English

Therefore, \(n\left( H \right) = 250,\;n\left( E \right) = 200\) and \(n\left( {H \cup E} \right) = 400\)

We know that:

\[\begin{align}n\left( {H \cup E} \right)& = n\left( H \right) + n\left( E \right) - n\left( {H \cap E} \right)\\n\left( {H \cap E} \right) &= n\left( H \right) + n\left( E \right) - n\left( {H \cup E} \right)\\& = 250 + 200 - 400\\ &= 450 - 400\\ &= 50\end{align}\]

Therefore, \(50\) people can speak both Hindi and English.

Chapter 1 Ex.1.6 Question 4

If \(S\) and \(T\) are two sets such that \(S\) has \(21\) elements, \(T\) has \(32\) elements, and \(S \cap T\) has \(11\) elements; how many elements does \(S \cup T\) have?

Solution

It is given that:

\(n\left( S \right)=21,\ n\left( T \right)=32\) and \(n\left( {S \cap T} \right) = 11\)

We know that:

\[\begin{align}n\left( S\cup T \right)&=n\left( S \right)+n\left( T \right)-n\left( S\cap T \right) \\ & =21+32-11 \\& =53-11 \\& =42\end{align}\]

Therefore, \(S \cup T\) has \(42\) elements.

Chapter 1 Ex.1.6 Question 5

If \(X\) and \(Y\) are two sets such that \(X\) has \(40\) elements, \(X \cup Y\) has \(60\) elements and \(X \cap Y\) has \(10\) elements, how many elements does \(Y\) have?

Solution

It is given that:

\(n\left( X \right) = 40,\;n\left( {X \cup Y} \right) = 60\) and \(n\left( {X \cap Y} \right) = 10\)

We know that:

\[\begin{align}n\left( {X \cup Y} \right)&= n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right)\\n\left( Y \right)&= n\left( {X \cap Y} \right) + n\left( {X \cup Y} \right) - n\left( X \right)\\&= 60 + 10 - 40\\&= 70 - 40\\&= 30\end{align}\]

Therefore, \(n\left( Y \right) = 30\)

Thus, the set \(Y\) has \(30\) elements.

Chapter 1 Ex.1.6 Question 6

In a group of \(70\) people, \(37\) like coffee, \(52\) like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?

Solution

Let \(C\) denote the set of people who like coffee, and

\(T\) denote the set of people who like tea

Therefore, \(n\left( C \right) = 37,\;n\left( T \right) = 52\) and \(n\left( {C \cup T} \right) = 70\)

We know that:

\[\begin{align}n\left( {C \cup T} \right) &= n\left( C \right) + n\left( T \right) - n\left( {C \cap T} \right)\\n\left( {C \cap T} \right) &= n\left( C \right) + n\left( T \right) - n\left( {C \cup T} \right)\\&= 37 + 52 - 70\\&= 89 - 70\\&= 19\end{align}\]

Therefore, \(19\) people like both coffee and tea.

Chapter 1 Ex.1.6 Question 7

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?

Solution

Let \(C\) denote the set of people who like cricket, and

\(T\) denote the set of people who like tennis

Therefore, \(n\left( C \right) = 40,\;n\left( {C \cup T} \right) = 65\) and \(n\left( {C \cap T} \right) = 10\)

We know that:

\[\begin{align}n\left( {C \cup T} \right)&= n\left( C \right) + n\left( T \right) - n\left( {C \cap T} \right)\\n\left( T \right)&= n\left( {C \cup T} \right) + n\left( {C \cap T} \right) - n\left( C \right)\\&= 65 + 10 - 40\\&= 75 - 40\\&= 35\end{align}\]

Hence, number of people who like tennis, \(n\left( T \right) = 35\)

Now, number of people like tennis only and not cricket \( = n\left( {T - C} \right)\)

As we know

\[\begin{align}n\left( T \right)& = n\left( {T - C} \right) + n\left( {C \cap T} \right)\\n\left( {T - C} \right) &= n\left( T \right) - n\left( {C \cap T} \right)\\&= 35 - 10\\&= 25\end{align}\]

Therefore, \(35\) people like tennis and \(25\) people like tennis only and not cricket.

Chapter 1 Ex.1.6 Question 8

In a committee, \(50\) people speak French, \(20\) speak Spanish and \(10\) speak both Spanish and French. How many speak at least one of these two languages?

Solution

Let \(F\) be the set of people in the committee who speak French, and

\(S\) be the set of people in the committee who speak Spanish

Therefore, \(n\left( F \right) = 50,\;n\left( S \right) = 20\) and \(n\left( {S \cap F} \right) = 10\)

We know that:

\[\begin{align}n\left( {S \cup F} \right) &= n\left( S \right) + n\left( F \right) - n\left( {S \cap F} \right)\\ &= 50 + 20 - 10\\ &= 70 - 10\\ &= 60\end{align}\]

Hence, \(60\) people in the committee speak at least one of the two languages.

  
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