# NCERT Solutions for Class 11 Mathematics english Medium 1. Sets

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##### NCERT Solutions⇒Class 11th ⇒ Mathematics ⇒ Chapter 1. Sets:

## 1. Sets

#### Exercise 1.6

**Exercise 1.6**

**Q1. If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38, find n ( X ∩ Y ).**

**Solution:** Given that

n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38

n ( X ∩ Y ) = ?

n(X ∪ Y ) = n(X) + n(Y) - n(X ∩ Y)

=> 38 = 17 + 23 - n(X ∩ Y)

=> 38 = 40 - n(X ∩ Y)

=> n(X ∩ Y) = 40 - 38

=> n(X ∩ Y) = 2

**Q2. If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements ; how many elements does X ∩ Y have?**

**Solution:** Given that;

n ( X ) = 8, n ( Y ) = 15 and n ( X ∪ Y ) = 18

n ( X ∩ Y ) = ?

n(X ∪ Y ) = n(X) + n(Y) - n(X ∩ Y)

=> 18 = 8 + 15 - n(X ∩ Y)

=> 18 = 23 - n(X ∩ Y)

=> n(X ∩ Y) = 23 - 18

=> n(X ∩ Y) = 5

**Q3. 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 the people can speak hindi be n(H) = 250,

The people can speak English be n(E) = 200 and

and Total people be n(H ∪ E) = 400

and people can speak both Hindi and English be n(H ∩ E) = ?

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

∴ 400 = 250 + 200 – n(H ∩ E)

⇒ 400 = 450 – n(H ∩ E)

⇒ n(H ∩ E) = 450 – 400

∴ n(H ∩ E) = 50

Thus, 50 people can speak both Hindi and English.

**Q4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?**

**Solution:** Given that: n(S) = 21, n(T) = 32, n(S ∩ T) = 11

We know that: n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

∴ n (S ∪ T) = 21 + 32 – 11 = 42

Thus, the set (S ∪ T) has 42 elements.

**Q5. If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, how many elements does Y have?**

**Solution: **

Given that: n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10

We know that:

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

∴ 60 = 40 + n(Y) – 10

∴ n(Y) = 60 – (40 – 10)

= 60 - 30

= 30

Thus, the set Y has 30 elements.

**Q6. 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 then

n(C ∪ T) = 70, n(C) = 37, n(T) = 52

We know that: n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

∴ 70 = 37 + 52 – n(C ∩ T)

⇒ 70 = 89 – n(C ∩ T)

⇒ n(C ∩ T) = 89 – 70 = 19

Thus, 19 people like both coffee and tea.

**Q7. 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?**

**Solutions: **

Let C denote the set of people who like cricket, and T denote the set of people who like tennis then-

∴ n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10

We know that: n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

∴ 65 = 40 + n(T) – 10

⇒ 65 = 30 + n(T)

⇒ n(T) = 65 – 30 = 35

Therefore, 35 people like tennis.

Now,

(T – C) ∪ (T ∩ C) = T

Also,

(T – C) ∩ (T ∩ C) = Φ

∴ n (T) = n (T – C) + n (T ∩ C)

⇒ 35 = n (T – C) + 10

⇒ n (T – C) = 35 – 10 = 25

Thus, 25 people like only tennis.

**Q8. 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 then-

∴ n(F) = 50, n(S) = 20, n(S ∩ F) = 10

We know that: n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

= 20 + 50 – 10

= 70 – 10 = 60

Thus, 60 people in the committee speak at least one of the two languages.

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