# Chapter 5let Us C Solutions

Chapter 5 Solution P5.2- 2, 3, 6. P5.3-3, 5, 8, 15. P5.4-3, 6, 8, 16. P5.5-2, 4, 6, 11. P 5.2-2 Consider the circuit of Figure P 5.2-2. Find i a by simplifying the circuit (using source transformations) to a single-loop circuit so that you need to write only one KVL equation to find i a. The NCERT Solutions For Class 11 Maths Chapter 15 Statistics are available as a pdf on this page. The NCERT Solutions are authored by the most experienced educators in the teaching industry making the solution of every problem straightforward and justifiable. Every solution written in the pdf given underneath empowers the students to get ready for the test and accomplish it. (c) If the marks obtained by a student in five different subjects are input through the keyboard, find out the aggregate marks and percentage marks obtained by the student. Assume that the maximum marks that can be obtained by a student in each subject are 100.

**NCERT Solutions for Class 9 English Moments Chapter 10 The Beggar **are part of NCERT Solutions for Class 9 English. Here we have given CBSE Class 9 English Moments Chapter 10 The Beggar.

## NCERT Solutions For Class 9 English Moments Chapter 10 The Beggar

**NCERT Textbook Questions**

**Think about it ****(Page 67)**

Question 1.

Has Lushkoff become a beggar by circumstance or by choice?

Answer:

Lushkoff was a middle-aged man and belonged to the Russian choir. He was not a bom beggar. He was sacked from the choir because of his drinking habits. He became a beggar by choice as he did not like to work hard at that stage.

Question 2.

What reasons does he give to Sergei for his telling lies?

Answer:

Lushkoff told the truth and accepted that he was a liar. Formerly he used to be a singer. He told lies to seek the favour of others. If he had spoken the truth none would have helped him. So, he decided to tell lies.

**More Resources for CBSE Class 9**

## Let Us C Solutions 5th Edition Chapter 5 Slideshare

Question 3.

Is Lushkoff a willing worker? Why, then, does he agree to chop wood for Sergei?

Answer:

No, he was not a willing worker. He was too weak to work. He had lost his strength and stamina due to his habit of drinking and secondly he was a middle aged man. He agreed to chop wood because of pride and shame and he had been trapped by his own words. So he had no other way but to accept Sergei’s offer.

Question 4.

Sergei says, “I am happy that my words have taken effect.” Why does he say so? Is he right in saying so?

Answer:

Sergei looked satisfied with the performance of the beggar and felt happy. When Lushkoff’s job of packing and hauling of the furniture was over, he praised him while handing him a rouble. But he was not right in saying so because Lushkoff had not developed the habit of working hard. He was still an idle fellow.

Question 5.

Lushkoff is earning thirty five roubles a month. How is he obliged to Sergei for this?

Answer:

Sergei played very important role in improving the condition of the beggar. It was because of Sergei that Lushkoff could earn thirty five roubles a month. He offered him the job to chop wood at his home. Later on he sent him to one of his friends to do the job of copying. Lushkoff was highly obliged to Sergei because now he was a notary because of him.

Question 6.

During their conversation Lushkoff reveals that Sergei’s cook, Olga, is responsible for the positive change in him. How has Olga saved Lushkoff?

Answer:

Olga was a social and sympathetic lady. She loved humanity. She realised the condition of Lushkoff and tried her best to improve the same. She did the work of his part as she wanted to change him. She kept criticizing him in order to improve him. He realised her feelings towards him and a change took place in his heart. So, Olga was responsible for the positive change in him.

**Talk about it ****(Page 68)**

Question 1.

How can we help beggars/abolish begging?

Answer:

The whole world is facing the nuisance of begging. The number of beggars is increasing day by day. Beggars can be seen at all public places. Some of the beggars have made it a business. It has become a serious problem. Our society and the government should take necessary steps to solve this problem. Global spread of education is required. Our government should pass strict laws against begging. Beggars should be given an opportunity to work. Financial support can be provided to them in order to set up some work. Beggars may be turned into skilled labourers. The government should set up beggar’s home only for the handicapped. Begging is a bad practice and is an impediment in the way of progress. So, we should discourage begging and beggars.

Vmix crack key. We hope the NCERT Solutions for Class 9 English Moments Chapter 10 The Beggar help you. If you have any query regarding CBSE Class 9 English Moments Chapter 10 The Beggar, drop a comment below and we will get back to you at the earliest.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.5

Choose the correct or the most suitable answer.

Question 1.

If A = {(x, y) : y = e^{x} ; x ∈ R } and B = {(x, y) : y = e^{-x}, x ∈ R } then n(A ∩ B)

(a) Infinity

(b) 0

(c) 1

(d) 2

Solution:

(c) 1

Hint.

A∩B = (0, 1)

n(A∩B) = 1

Question 2.

IfA {(x, y) : y = sin x, x ∈ R) and 8= (x, y) : y = cos x, x ∈ R) then A∩B contains …….

(a) no element

(b) infinitely many elements

(c) only one element

(d) cannot be determined.

Solution:

(b) infinitely many elements

Question 3.

The relation R defined on a set A = {0, -1, 1, 2} by xRy if x^{2} +y^{2} ≤ 2, then which one of the following is true?

(a) R = {(0, 0), (0, -1), (0, 1), (-1, 0), (-1, 1), (1, 2), (1, 0)}

(b) R = {(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)

(c) Domain of R is {0, -1, 1, 2}

Solution:

(a) Range of R is {0, -1, 1}

Hint.

A= {0, -1, 1, 2}

x^{2} + y^{2} ≤ 2

The values of x and y can be 0, -1 or 1

So range = {0, -1, 1}

Question 4.

If f(x) = x – 2 + x + 2 , x ∈ R, then

Solution:

Hint.

f(x) = x – 2 + x + 2

Question 5.

Let R be the set of all real numbers. Consider the following subsets of the plane R x R: S = {(x, y): y = x + 1 and 0 < x < 2} and T = {(x, y) : x – y is an integer} Then which of the following is true?

(a) T is an equivalence relation but S is not an equivalence relation.

(b) Neither S nor T is an equivalence relation

(c) Both S and T are equivalence relation

(d) S is an equivalence relation but T is not an equivalence relation.

Solution:

(a) T is an equivalence relation but S is not an equivalence relation.

Hint.

(0, 1), (1, 2) it is not an equivalence relation

T is an equivalence relation

Question 6.

Let A and B be subsets of the universal set N, the set of natural numbers. Then

A’ ∪ [(A ∩ B) ∪ B’] is ………

(a) A

(b) A’

(c) B

(d) N

Solution:

(d) N

Hint.

Question 7.

The number of students who take both the subjects Mathematics and Chemistry is 70. This represents 10% of the enrollment in Mathematics and 14% of the enrollment in Chemistry. How many students take at least one of these two subjects?

(a) 1120

(b) 1130

(c) 1100

(d) insufficient data

Solution:

(b) 1130

Hint.

n(M ∪ C) = n(M) + n(C) – n(M ∩ C)

= 700 + 500 – 70 = 1130

Question 8.

If n[(A × B) ∩ (A × C)] = 8 and n(B ∩ C) = 2 , then n(A) is

(a) 6

(b) 4

(c) 8

(d) 16

Solution:

(b) 4

Question 9.

If n(A) = 2 and n(B ∪ C) = 3, then n[(A × B) ∪ (A × C)] is …….

(a) 2^{3}

(b) 3^{2}

(c) 6

(d) 5

Solution:

(c) 6

Hint.

n[(A × B) ∪ (A × C)] = n[ A × (B ∪ C)] = n(A) × n(B ∪ C) = 2 × 3 = 6

Question 10.

If two sets A and B have 17 elements in common, then the number of elements common to the set A × B and B × A is

(a) 2^{17}

(b) 17^{2}

(c) 34

(d) insufficient data

Solution:

(b) 17^{2}

Hint.

n (A ∩ B) = 17

So n [(A × B) ∩ (B × A)]

= n(A ∩ B) × n(B ∩ A) = 17 × 17 = 17^{2}

Question 11.

For non-empty sets A and B, if A ⊂ B then (A × B) ∩ (B × A) is equal to ……….

(a) A ∩ B

(b) A × A

(c) B × B

(d) None of these

Solution:

(b) A × A

Hint.

When A ⊂ B, (A × B) ∩ (B × A) = A × A

Question 12.

The number of relations on a set containing 3 elements is

(a) 9

(b) 81

(c) 512

(d) 1024

Solution:

(c) 512

Hint.

Number of relations = 2^{n2} = 2^{32} = 2^{9} = 512

Question 13.

Let R be the universal relation on a set X with more than one element. Then R is

(a) Not reflexive

(b) Not symmetric

(c) Transitive

(d) None of the above

Solution:

(c) Transitive

Question 14.

Let X = {1, 2, 3, 4} and R = {(1, 1), (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1,4), (4, 1)}. Then R is …….

(a) Reflexive

(b) Symmetric

(c) Transitive

(d) Equivalence

Solution:

(b) Symmetric

Hint.

(4, 4} ∉ R ⇒ R is not reflexive

(1, 4), (4, 1) ∈ R ⇒ R is symmetric

(1, 4), (4, 1) ∈ R but (4, 4) ∉ R

So R is not transitive

Question 15.

Solution:

Question 16.

Solution:

(c) [0, 1)

Question 17.

The rule f(x) = x^{2} is a bijection if the domain and the co-domain are given by ….

(a) R,R

(b) R, (0, ∞)

(c) (0, ∞), R

(d) [0, ∞), [0, ∞)

Solution:

(d) [0, ∞), [0, ∞)

## Chapter 5let Us C Solutions Collection Agency

Question 18.

The number of constant functions from a set containing m elements to a set containing n elements is

(a) mn

(b) m

(c) n

(d) m + n

Solution:

(c) n

## Chapter 5let Us C Solutions Llc

Question 19.

The function f: [0, 2π] ➝ [-1, 1] defined by f(x) = sin x is

(a) One to one

(b) Onto

(c) Bijection

(d) Cannot be defined

Solution:

(b) Onto

So it is not one-to-one

So it is an onto function

## Chapter 5let Us C Solutions Corp

Question 20.

If the function f : [-3, 3] ➝ S defined by f(x) = x^{2} is onto, then S is ………

(a)[-9, 9]

(b) R

(c) [-3, 3]

(d) [0, 9]

Solution:

(d) [0, 9]

Question 21.

Let X = {1, 2, 3, 4}, Y = {a, b, c, d) and f = {(1, a), (4, b), (2, c), (3, d) (2, d)}. Then f is ………

(a) An one-to-one function

(b) An onto function

(c) A function which is not one-to-one

(d) Not a function

Solution:

(d) Not a function

Hint.

Since the element 2 has two images, it is not a function

Question 22.

Solution:

## Chapter 5let Us C Solutions Inc

Question 23.

Let f : R ➝ R be defined by f(x) = 1 – x . Then the range of f is ………

(a) R

(b) (1, ∞)

(c) (-1, ∞)

(d) (-∞, 1]

Solution:

(d) (-∞, 1]

Hint.

f: R ➝ R defined by

f(x) = 1 – x

For example,

f(1) = 1 – 1 = 0

f(8) = 1 – 8 = -1

f(-9) = 1 – 9 = -8

f(-0.2) = 1 – 0.2 = 0.8

so range = (-∞, 1]

## Let Us C Chapter 5 Solutions

Question 24.

The function f : R ➝ R is defined by f(x) = sin x + cos x is ……

(a) An odd function

(b) Neither an odd function nor an even function

(c) An even function

(d) Both odd function and even function

Solution:

(b) Neither an odd function nor an even function

Question 25.

(a) An odd function

(b) Neither an odd function nor an even function

(c) An even function

(d) Both odd function arid even function

Solution:

(c) An even function