# NCERT Solutions For Class 11 Maths Chapter 1 Exercise 1.6

Sets

## 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.