Plus One Math's Solution Ex 1.6 Chapter 1 Sets


Keralanotes brings to you the CBSE solved questions of Ex 1.6 Chapter1 Sets. This NCERT solution is specially prepared for the students of class 11 Maths. In this set, we will give you 1.6 and all its questions with their solutions. In these solutions, the working of every step is clearly illustrated to help you better understand it and to improve your understanding of the topics.

This solution is for exercise 1.6 Chapter1 of NCERT Solution for Class 11 Maths textbook solution for the CBSE students. You can free download it on our websites.

Board SCERT, Kerala
Text Book NCERT Based
Class Plus One 
Subject Math's Textbook Solution
Chapter Chapter 1
Exercise Ex 1.6
Chapter Name Sets
Category Plus One Kerala

Kerala Syllabus Plus One Math's Textbook Solution Chapter  1 Sets Exercises 1.6

Chapter  1  Sets Textbook Solution

Kerala plus One maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

Chapter  1  Sets Exercise   1.6

    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?



    n(S ∩ T)=11

    n(S ∪ T)=n(T)+n(S)-n(S ∩ T)


    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?

    n(X)= 40

    n(XUY)= 60

    n(X∩Y)= 10

    n(XUY)= n(X) + n(Y) -n(X∩Y)

    60= 40 + n(Y)- 10

    n(Y) = 60 - 40 + 10 = 30

    Thus, the set Y has 30 elements.

    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?

    C-coffee T-tea







    since each person likes at least one of the two drinks,n(U)=n(CUT)

    number of peoples like both tea and coffee=19

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

    It is given that:

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

    n(X ∩ Y) = ?

    We know that:

    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?

    It is given that:

    n(X ∩ Y) = ?

    We know that:

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

    number of people speak hindi, n(H)=250 

    number of people speak english, n(E)=200

    n (HUE) = 400

    n(HUE)= n(H) +n(E) -n(H∩E)

    400 = 250+200 - n(H∩E)


    n(H∩E)=450-400= 50

    Thus, 50 people can speak both Hindi and English.

    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?

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

    T denote the set of people who like tennis

    ∴ 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.


    (T – C) ∪ (T ∩ C) = T


    (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.

    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?

    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

    ∴ 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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Chapter 1: Sets EX 1.6 Solution

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