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

# NCERT Solutions for Exercise 1.6 Class 11 Maths Chapter 1 - Sets

Edited By Vishal kumar | Updated on Nov 02, 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

## 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 ).

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.

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

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.

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.

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.

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.

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.

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.

## More About NCERT Solutions for Class 11 Maths Chapter 1 Exercise 1.6:-

In the NCERT syllabus Class 11th Maths chapter 1 exercise 1.6, you will get eight questions based on the practical applications of union, the intersection of the set, and Venn's diagram. Before Class 11 Maths ch 1 ex 1.6, there are some solved examples given in the textbook which you can solve. Solving these examples will help you to understand the concept clearly. The last concept of this class 11 ex 1.6 chapter is useful in solving relations and function problems as well as in probability.

Also Read| Sets Class 11th Notes

## 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.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%

## Key Features of 11th Class Maths Exercise 1.6 Answers

1. 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.

2. Clarity and accuracy: The class 11 ex 1.6 answers are clear and accurate, ensuring that students can confidently prepare for their exams.

3. Additional resources: The availability of a downloadable PDF format makes it convenient for students to access the answers for study and revision.

4. 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-

## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Can I get NCERT exemplar solutions for Class 11 Maths ?

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

2. 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.

3. Can I get chemistry Class 11 NCERT exemplar solutions ?
4. 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

5. What is the weightage of Algebra in the Class 11 Maths ?

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

6. What is the syllabus for CBSE Class 11 Maths ?

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

Exam Date:19 September,2024 - 19 September,2024

Exam Date:20 September,2024 - 20 September,2024

#### National Institute of Open Schooling 12th Examination

Exam Date:20 September,2024 - 07 October,2024

Get answers from students and experts

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) Option 2) Option 3) Option 4)

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 3.8×107 J 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

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) Option 2) Option 3) Option 4)

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

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, 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

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.

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.

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

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

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9