# 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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