NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1 are part of Class 10 Maths NCERT Solutions. Here we have given NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-1/

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 1
Chapter NameReal Numbers
ExerciseEx 1.1
Number of Questions Solved5
CategoryNCERT Solutions

Maths Class 10 NCERT Solutions will help you to score more marks in your CBSE board Examination.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

Ex 1.1 Class 10 Question 1.
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255.
Solutions:
(i) Given numbers are 135 and 225.
On applying Euclid’s division algorithm, we have
225 = 135 x 1 + 90
Since the remainder 90 ≠ 0, so again we apply Euclid’s division algorithm to 135 and 90, to get
135 = 90 x 1 + 45
Since the remainder 45 ≠ 0, so again we apply Euclid’s division algorithm to 90 and 45, to get
90 = 45 x 2 + 0
The remainder has now become zero, so we stop.
∵ At the last stage, the divisor is 45
∴ The HCF of 135 and 225 is 45.
Ex 1.1 Class 10 Maths Chapter 1 Real Numbers NCERT Solutions
(ii) Given numbers are 196 and 38220
On applying Euclid’s division algorithm, we have
38220 = 196 x 195 + 0
Since we get the remainder zero in the first step, so we stop.
∵ At the above stage, the divisor is 196
∴ The HCF of 196 and 38220 is 196.
Exercise 1.1 Class 10 Maths NCERT Solutions Chapter 1 Real Numbers
(iii) Given numbers are 867 and 255
On applying Euclid’s division algorithm, we have
867 = 255 x 3 + 102
Since the remainder 102 ≠ 0, so again we apply Euclid’s division algorithm to 255 and 102. to get
255 = 102 x 2 + 51
Since the remainder 51 ≠ 0, so again we apply Euclid’s division algorithm to 102 and 51, to get
102 = 51 x 2 + 0
We find the remainder is 0 and the divisor is 51
∴ The HCF of 867 and 255 is 51.

Exercise 1.1 Class 10 Maths NCERT Solutions Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solutions:
Let ‘a’ be any positive integer and b = 6.
∴ By Euclid’s division algorithm, we have
a = bq + r, 0 ≤ r ≤ b
a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4,5
Now, ‘a’ may be of the form of 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5
If ‘a’ is of the form 6q, 6q + 2, 6q + 4 then ‘a’ is an even.
Class 10 Maths Chapter 1 Real Numbers Ex 1.1 NCERT
In above we can see clearly that the numbers of the form 6q, 6q + 2, 6q + 4 are having the factor 2.
∴ The numbers of the form 6q, 6q + 2, 6q + 4 are even.
If ‘a’ is of the form 6q +1,6q +3, 6q + 5 then ‘a’ is an odd.
As if
Class 10 Maths Ex 1.1 Real Numbers NCERT
∵ We know that the number of the form 2k + 1 is odd.
∴ The numbers of the form 6q + 1, 6q + 3, 6q + 5 are odd.

Class 10 Maths Chapter 1 Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solutions:
Maximum number of columns = HCF of (616, 32)
For finding the HCF we should apply Euclid’s division algorithm
Given numbers are 616 and 32
On applying Euclid’s division algorithm, we have
616 = 32 x 19 + 8
Since the remainder 8 ≠ 0, so again we apply Euclid’s division algorithm to 32 and 8, to get
32 = 8 x 4 + 0
Maths 1.1 Class 10 Chapter 1 Real Numbers NCERT
The remainder has now become zero, so we stop,
∵ At the last stage, the divisor is 8
∴ The HCF of 616 and 32 is 8.
Therefore, the maximum number of columns in which an army contingent of 616 members can march behind an army band of 32 members in a parade is 8.

Class 10 Maths Ex 1.1 Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solutions:
Let ‘a’ be any positive integer and b = 3.
∴ By Euclid’s division algorithm, we have a = 3q + r, 0 ≤ r < b
a = 3q + r, 0 ≤ r < 3 [ ∵ b = 3] where q ≥ 0 and r = 0,1, 2
∴ a = 3q or 3q + 1 or 3q + 2
Now
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 6
Thus, the square of any positive integer is either of the form 3m or 3m + 1.

Maths 1.1 Class 10 Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solutions:
Let ‘a’ be any positive integer and b = 3.
∴ By Euclid’s division algorithm, we have a = bq + r,0 ≤ r ≤ b
a = 3q + r,0 ≤ r < 3 [ ∵ b = 3] where q ≥ 0 and r = 0. 1, 2
∴ a = 3q or 3q + 1 or 3q + 2
Now
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 7
Thus, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

We hope the NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1 help you. If you have any query regarding NCERT Solutions for Class 10 Mathematics Chapter 1 Real Numbers Ex 1.1, drop a comment below and we will get back to you at the earliest.

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data,Cumulative Frequency Graph and O give Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data,Cumulative Frequency Graph and O give Test Yourself.

Other Exercises

Question 1.
Solution:
Correct option: (b)
The cumulative frequency table is useful in determining the median.

Question 2.
Solution:
Correct option: (c)
Mean = 27
Median = 33
Mode = 3Median – 2Mean
= 3 x 33 – 2 x 27
= 99 – 54
= 45

The Mean Median Calculator is used to calculate the count, sum, mean, median, mode and range of a set of numbers.

Question 3.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 1

Question 4.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 2

Question 5.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 3

Question 6.
Solution:
Number of athletes who completed the race in less than 14.6 seconds
= 2 + 4 + 15 + 54
= 75.

Question 7.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 4

Question 8.
Solution:
The frequency table is as follows:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 5
The frequency corresponding to the class 20 – 25 is 4.

Question 9.
Solution:

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 6

Question 10.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 7

Question 11.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 8

Question 12.
Solution:

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 9

Question 13.
Solution:
We plot the points (10, 3), (20, 11), (30, 28), (40, 48) and (50, 70) to get the cumulative frequency curve as follows:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 10

Question 14.
Solution:

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 11

Question 15.
Solution:
We plot the points (20, 2), (30, 7), (40, 17), (50, 40), (60, 60), (70, 82), (80, 85), (90, 90) and (100, 100) to get the cumulative frequency curve as follows:

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 12
Here, N = 100 ⇒ N/2 = 50
At y = 50, affix A.
Through A, draw a horizontal line meeting the curve at P.
Through P, a vertical line is drawn which meets OX at M.
OM = 56.
Hence, median = 56

Question 16.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 13

Question 17.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 14

Question 18.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 15

Question 19.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 16

Question 20.
Solution:
Less Than Series:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 17

We plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get less than ogive.

More Than Series:

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 18
We plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3) to get more than ogive.

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 19

The two curves intersect at L. Draw LM ⊥ OX.

Thus, median = OM = 16.

Question 21.
Solution:
Less Than Series:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 20

We plot the points (45, 1), (50, 10), (55, 25), (60, 43), (65, 83), (35, 27), (70, 109), (75, 125), (80, 139) and (85, 149) to get less than ogive.

More Than Series:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 21

We plot the points (40, 149), (45, 148), (50, 139), (55, 124), (60, 106), (65, 66), (70, 40), (75, 24) and (80, 10) to get more than ogive.

RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 22

Question 22.
Solution:
RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself 23

We hope the RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself help you. If you have any query regarding RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Test Yourself, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Hindi Kshitiz, Kritika, Sparsh, Sanchayan | Chapterwise Class 10 Hindi NCERT Book Solutions

NCERT Solutions for Class 10 Hindi

NCERT Solutions for Class 10 Hindi are the part of NCERT Solutions for Class 10. Here we have given Class 10 Hindi NCERT Solutions.

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Students will understand how to answer a particular question in the exam after preparing the concepts from NCERT Solutions for Grade 10 Hindi. You can cover Hindi Course-A Kshitij Bhag 2 (17 chapters), Kritika Bhag 2 (5 chapters), and Hindi Course-B Sparsh Bhag 2(17 chapters), Sanchayan Bhag 2 (3 chapters) in the NCERT Solutions PDF links available here. So download them and ace up your preparation.

Chapter Wise NCERT Solutions for Class 10 Hindi

Each and every chapter in the NCERT textbook for Class 10 Hindi are explained in a simple language for students to understand easily and prepare all answers properly. These NCERT 10th Hindi Textbook Solutions help you to practice all exercise questions and score well in the main examinations. Overall, Class 10 Hindi NCERT Solutions PDF for all chapters clear all your doubts and make you learn each and every point perfectly.

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NCERT Solutions for Class 10 Hindi – A

NCERT Solutions for Class 10 Hindi Kshitij Bhag 2 क्षितिज भाग 2

काव्य – खंड

गद्य – खंड

NCERT Solutions for Class 10 Hindi Kritika Bhag 2 कृतिका भाग 2

NCERT Solutions for Class 10 Hindi – B

NCERT Solutions for Class 10 Hindi Sparsh Bhag 2 स्पर्श भाग 2

काव्य – खंड

गद्य – खंड

NCERT Solutions for Class 10 Hindi Sanchayan Bhag संचयन भाग 2

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NCERT Solutions for Class 10 Hindi Kshitij क्षितिज भाग 2 | Chapterwise Class 10 Hindi Kshitij Updated for 2021-22

NCERT Solutions for Class 10 Hindi Kshitij

NCERT Solutions for Class 10 Hindi Kshitij क्षितिज भाग 2 are the part of NCERT Solutions for Class 10 Hindi. Here we have given NCERT Solutions for Class 10 Hindi Kshitij Bhag 2.

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Chapter Wise NCERT Solutions for Class 10 Hindi Kshitij क्षितिज भाग 2 Free

Kshitij is the most important textbook for Hindi course A. This book comprises compositions of seventeen creators just like eight prose works. The poems portray several trends of Ritikal, Bhaktikal, and modern times, thus students get familiar with the development journey. You can check all your subject knowledge by answering the textbook questions. Check your answers by using the NCERT Solutions for Class 10 Hindi Kshitij (क्षितिज भाग 2) and identify your weak areas. Also, download NCERT Class 10 Hindi Kshitij (क्षितिज भाग 2) Solutions in PDF format by accessing the links below:

NCERT Solutions for Class 10 Hindi Kshitij (क्षितिज) भाग 2

NCERT Solutions for Class 10 Hindi Kshitij क्षितिज भाग 2

काव्य – खंड

गद्य – खंड

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NCERT Solutions for Class 10 English Footprints Without Feet Chapter 8 The Hack Driver

NCERT Solutions for Class 10 English Footprints Without Feet Chapter 8 The Hack Driver are part of NCERT Solutions for Class 10 English. Here we have given NCERT Solutions for Class 10 English Footprints Without Feet Chapter 8 The Hack Driver.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectEnglish Footprints Without Feet
ChapterChapter 8
Chapter NameThe Hack Driver
CategoryNCERT Solutions

The Hack Driver Class 10 English Footprints Without Feet Chapter 8 Questions and Answers

The Hack Driver Class 10 English Textual Exercises Questions and Answers

Read and Find Out {Pages 47 & 50)

Question 1.
Why is the lawyer sent to New Mullion ? What does he first think about the place ? (CBSE 2011)
Or
Why is the lawyer sent to New Mullion ? (CBSE 2015)
Answer:
The lawyer is sent to New Mullion to serve summons on Oliver Lutkins. He first thinks about the place to be a sweet and simple country village. But he finds it dirty and unclean.

Question 2.
Who befriends him ? Where does he take him ?
Answer:
A delivery man (Oliver Lutkins himself) befriends him. He takes him to the Fritz’s, Gustaffs barber shop and Wade’s Hill. He finally takes him to Lutkins’ house and his mother’s.

Question 3.
What does he say about Lutkins ?
Answer:
He says that Lutkins never pays anybody a cent. He is not really bad. But it is hard to make him part with his money. Then he escapes meeting.

Question 4.
What more does Bill say about Lutkins and his family ?
Answer:
Bill says that Lutkins has a mother. They have a farm three miles north. But her mother is a real terror. She is about nine feet tall and four feet thick. She is as quick as a cat.

Question 5.
Does the narrator serve the summons that day ? (CBSE 2011)
Answer:
The narrator does not serve the summons that day. He comes again to New Mullion. He serves it on Lutkins when he is recognised by another official. This official is with him.

Question 6.
Who is Lutkins ?
Answer:
Lutkins is a cheat. He borrows money from different persons. But he never returns that money. He escapes meeting also. He has a cheerful and friendly manner. He is a hack driver in New Mullion.

Think About It (Page 53)

Question 1.
When the lawyer reached New Mullion, did ‘Bill’ know that he was looking for Lutkins ? When do you think Bill came up with his plan for fooling the lawyer ?
Answer:
Bill (Lutkins himself) at once knew that the lawyer was looking for him. The lawyer’s first sentence to him is : ‘I want to find a man named Oliver Lutkins’.

Hearing him and knowing about his ways Bill (Lutkins himself) comes up with his plan. This plan is for fooling the lawyer at once.

Question 2.
Lutkins openly takes the lawyer all over the village. How is it that no one lets out the secret ?

(Hint: Notice that the hack driver asks the lawyer to keep out of sight behind him when they go into Fritz’s.) Can you find other such subtle ways in which Lutkins manipulates the tour ?

Answer:

Lutkins acts very cleverly. Wherever he takes the lawyer to find ‘Lutkins’, he gives out hints. These are to keep everything in secret. For example, Fritz looks at the lawyer hiding behind Bill and ‘hesitates’. Then he tells a lie.

Yes, he manipulates the tour cleverly. When he goes to Gustaffs barber shop, he first enters the shop. But the lawyer remains outside. In fact, he talks very confidently to the lawyer. His honesty and helpful nature impress him very much. But he plays a trick before it.

Question 3.
Why do you think Lutkins’ neighbours were anxious to meet the lawyer ? (CBSE 2011)
Answer:
The neighbours were anxious to meet the lawyer because he was a lawyer. A lawyer is supposed to be intelligent and clever. But Lutkins deceived such a man. This was really superb of him. So they were anxious to see such a lawyer who had been befooled.

Question 4.
After his first day’s experience with the hack driver the lawyer thinks of returning to New Mullion to practise law. Do you think he would have reconsidered this idea after his second visit ?
Answer:
No, he really changed his mind on his second visit. In his first visit he thought of the villagers as simple-hearted. But it was not so. They were crafty and deceptive as he learnt later. They also told lies as Lutkins proved so.

Yes, he would surely have reconsidered this idea after his second visit.

Question 5.
Do you think the lawyer was gullible ? How could he have avoided being taken for a ride ?
Answer:
Yes, the lawyer was surely gullible. If he had had a worldly experience, he would not have been befooled by Lutkins himself.

Talk About It

Question 1.
Do we come across persons like Lutkins only in fiction or do we encounter them in real life as well ? You can give examples from fiction, or narrate an incident that you have read in the newspaper, or an incident from real life.
Answer:
We do come across persons like Lutkins in our real life also. They are successful in deceiving people like Lutkins. They can easily be identified after their first or second tricks or deceptions.

The best example of such man is that of Mr. Natwarlal. He landed in jail after deceiving lots of people.

Yes, I remember one such incident. A young man known to me got his first salary. He was very excited. He wanted to give a little gift of gold to his mother. He went to the bullion market. A con man met. He told him that he could get him one real gold chain cheaply. The young man fell into his trap. He bought the chain. That was found to be fake one.

Question 2.
Who is a ‘con man’, or a confidence trickster ?
Answer:
A ‘con man’ is a trickster. He is a master at deceiving people like the young lawyer being deceived by Lutkins himself.

We hope the NCERT Solutions for Class 10 English Footprints Without Feet Chapter 8 The Hack Driver help you. If you have any query regarding NCERT Solutions for Class 10 English Footprints Without Feet Chapter 8 The Hack Driver drop a comment below and we will get back to you at the earliest.