NCERT Solutions for Exercise 1.6 Class 11 Maths Chapter 1

Q: What is the syllabus for CBSE Class 11 Maths ?

Here you will get S yllabus for CBSE Class 11 Maths .

Vishal kumarUpdated on 02 Nov 2023, 07:03 AM IST

NCERT Solutions for Class 11 Maths Chapter 1 - Sets Exercise 1.6- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 1: Sets Exercise 1.6- In NCERT Solutions for Class 11 Maths Chapter 1 exercise 1.6 you will learn about the practical problems on union and intersection of two sets. Class 11th Maths Chapter 1 Exercise 1.6 is all about the applications of the union of two sets, the intersection of two sets, and applications of Venn's diagram. You should try to solve all the problems given in the Class 11 Maths ch 1 ex 1.6 on your own. These problems are related to our daily life problems which you must solve to get clarity. You may not be able to solve these NCERT problems at first, Class 11 Maths Chapter 1 exercise 1.6 solutions are here to help you. This NCERT book ex 1.6 class 11 is not only important for this chapter but formulae derived in this exercise will be used in the upcoming chapter probability. If you are looking for NCERT Solutions, click on the given link where you will find the detailed NCERT Solutions for Classes 6 to 12 in one place.

The provided class 11 maths ex 1.6 solution are designed to facilitate students in grasping fundamental concepts effectively. With the detailed, step-by-step solutions offered by Careers360, achieving success in final exams becomes a straightforward endeavour. You can access the NCERT Class 11 Maths Solutions Chapter 1 Exercise 1.6 in PDF format via the link provided. Additionally, the PDF version is readily accessible for your convenience.

Also, see

NCERT Solutions for Class 11 Maths Chapter 1 – Sets Exercise 1.6

Download PDF

Access Sets Class 11 Chapter 1 Exercise: 1.6

Question:1 If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X $\cup$ Y ) = 38, find n ( X $\cap$ Y ).

Answer:

n( X ) = 17, n( Y ) = 23 and n( X $\cup$ Y ) = 38

n( X $\cup$ Y ) = n( X ) + n( Y ) - n ( X $\cap$ Y )

38 = 17 + 23 - n ( X $\cap$Y )

n ( X $\cap$Y ) = 40 - 38

n ( X $\cap$Y ) = 2.

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?

Answer:

n( X $\cup$ Y ) = 18

n(X) = 8

n(Y) = 15

n( X $\cup$ Y ) = n(X) + n(Y) - n(X $\cap$ Y)

n(X $\cap$ Y) = 8 + 15 - 18

n(X $\cap$ Y) = 5

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?

Answer:

Hindi speaking = 250 = n(A)

English speakind = 200 = n(B)

Group of people = 400 = n(A $\cup$ B)

People speaking both Hindi and English = n(A $\cap$ B)

n(A $\cup$ B) = n(A) + n(B) - n(A $\cap$ B)

n(A $\cap$ B) = 250 + 200 - 400

n(A $\cap$ B) = 450 - 400

n(A $\cap$ B) = 50

Thus,50 people speak hindi and english both.

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?

Answer:

n(S) = 21

n(T) = 32

n(S $\cap$ T) = 11

n(S $\cup$ T) = n(S) + n(T) - n(S $\cap$ T)

n(S $\cup$T) = 21 + 32 - 11

n(S $\cup$T) = 53 - 11

n(S $\cup$ T) = 42

Hence, the set (S $\cup$ T) has 42 elements.

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?

Answer:

n( X $\cup$ Y) = 60

n( X ∩ Y) = 10

n(X) = 40

n( X $\cup$ Y) = n(X) + n(Y) - n( X ∩ Y)

n(Y) = 60 - 40 + 10

n(Y) = 30

Hence,set Y has 30 elements.

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?

Answer:

n(people like coffee) = 37

n(people like tea ) = 52

Total number of people = 70

Total number of people = n(people like coffee) + n(people like tea ) - n(people like both tea and coffee)

70 = 37 + 52 - n(people like both tea and coffee)

n(people like both tea and coffee) = 89 - 70

n(people like both tea and coffee) = 19

Hence,19 people like both coffee and tea.

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?

Answer:

Total people = 65

n(like cricket) = 40

n(like both cricket and tennis) = 10

n(like tennis) = ?

Total people = n(like cricket) + n(like tennis) - n(like both cricket and tennis)

n(like tennis) = 65 - 40 + 10

n(like tennis) = 35

Hence,35 people like tennis.

n (people like only tennis) = n(like tennis) - n(like both cricket and tennis)

n (people like only tennis) = 35 - 10

n (people like only tennis) = 25

Hence,25 people like only tennis.

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?

Answer:

n(french) = 50

n(spanish) = 20

n(speak both french and spanish) = 10

n(speak at least one of these two languages) = n(french) + n(spanish) - n(speak both french and spanish)

n(speak at least one of these two languages) = 50 + 20 - 10

n(speak at least one of these two languages) = 70 -10

n(speak at least one of these two languages) = 60

Hence,60 people speak at least one of these two languages.

Benefits of NCERT Solutions for Class 11 Maths Chapter 1 Exercise 1.6:-

NCERT Solutions for 11th class maths exercise 1.6 answers is not only useful for this chapter but also very important to understand other chapters like probability.
Exercise 1.6 Class 11 Maths can be solved using Venn's diagram, but the formulas derived from it will be useful in other chapters also.
Class 11th Maths chapter 1 exercise 1.6 solutions can be used as a reference.

Key Features of 11th Class Maths Exercise 1.6 Answers

Step-by-step solutions: The exercise 1.6 class 11 maths solution provides a detailed, step-by-step explanation of each problem, making it easier for students to understand the concepts.
Clarity and accuracy: The class 11 ex 1.6 answers are clear and accurate, ensuring that students can confidently prepare for their exams.
Additional resources: The availability of a downloadable PDF format makes it convenient for students to access the answers for study and revision.
Alignment with NCERT curriculum: The class 11 maths chapter 1 exercise 1.6 answers are designed to align with the NCERT curriculum, ensuring that students are well-prepared for their examinations.

Also see-

NCERT Solutions of Class 11 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

Q: Can I get NCERT exemplar solutions for Class 11 Maths ?

Yes, click on the link to get NCERT Exemplar Solutions for Class 11 Maths.

Q: Do I need to buy NCERT exempler solution book for Class 11 Maths ?

No, you don't need to buy NCERT exemplar solutions book for class 11 maths. It is available online on Careers360.

Q: Can I get chemistry Class 11 NCERT exemplar solutions ?

Here you will get the NCERT Exemplar Solutions for Class 11 Chemistry.

Q: If X and Y are two sets such that n ( X ) = 10, n ( Y ) = 13 and n ( X ∪ Y ) = 20, find n ( X ∩ Y ).

n ( X ∩ Y ) = n ( X ) + n ( Y ) - n ( X ∪ Y ) = 10 + 13 - 20 = 3

Q: What is the weightage of Algebra in the Class 11 Maths ?

The unit Algebra has 30 marks weightage in the CBSE Class 11 Maths.

Q: What is the syllabus for CBSE Class 11 Maths ?

Here you will get S yllabus for CBSE Class 11 Maths.

Articles

DU BA LLB Eligibility Criteria 2026 - Age Limit, Qualifications, Marks

Sep 22, 2025

DU LLB Counselling 2026 – Seat Allotment, Documents, Admission Process

Sep 21, 2025

CGPA Calculator - Know How to Calculate CGPA

Sep 19, 2025

SOF IHO Exam 2026 – Dates, Registration, Syllabus & Pattern

Sep 19, 2025

Upcoming School Exams

National Means Cum-Merit Scholarship

Ongoing Dates

NMMS Application Date

25 Jul'25 - 30 Sep'25 (Online)

Uttar Pradesh Board 12th Examination

Ongoing Dates

UP Board 12th Others

10 Aug'25 - 27 Sep'25 (Online)

Mock Test

Exam Pattern

Result

Uttar Pradesh Board 10th examination

Ongoing Dates

UP Board 10th Others

10 Aug'25 - 27 Sep'25 (Online)

Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Read Complete Answer

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

Read Complete Answer

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Read Complete Answer

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Read Complete Answer

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Read Complete Answer

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Read Complete Answer

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Read Complete Answer

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Read Complete Answer

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Read Complete Answer

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is