NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

Updated on Jan 25, 2021 - 10:27 a.m. IST by Saumya.Srivastava #CBSE Class 10th

NCERT solutions for class 10 Maths chapter 1 Real Numbers - In mathematics, there are two kinds of numbers - Real number and Imaginary number. Real numbers are those numbers that can be represented on the number line. NCERT solutions for Class 10 Maths Chapter 1 Real Numbers will help the students to prepare for the exams in a better way. Real numbers include the in-depth classification of numbers, its applications, and properties related to various kinds of numbers. In this particular chapter, we are going to talk about real numbers. The NCERT solutions for class 10 maths chapter 1 real numbers carry detailed explanations for each and every question. NCERT solutions for other subjects and classes can be downloaded by clicking on the above link. Here you will get NCERT solutions for class 10 also. Students must complete the NCERT Class 10 syllabus to the earliest to score well in the exams.

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NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.1

Q1 (1) Use Euclid’s division algorithm to find the HCF of 135 and 225

Answer:

225 > 135. Applying Euclid's Division algorithm we get

$225=135\times 1+90$

since remainder $\neq$ 0 we again apply the algorithm

$135=90\times 1+45$

since remainder $\neq$ 0 we again apply the algorithm

$90=45\times 2$

since remainder = 0 we conclude the HCF of 135 and 225 is 45.

Q1 (2) Use Euclid’s division algorithm to find the HCF of 196 and 38220

Answer:

38220 > 196. Applying Euclid's Division algorithm we get

$38220=196\times 195+0$

since remainder = 0 we conclude the HCF of 38220 and 196 is 196.

Q1 (3) Use Euclid’s division algorithm to find the HCF of 867 and 255

Answer:

867 > 225. Applying Euclid's Division algorithm we get

$867=255\times 3+102$

since remainder $\neq$ 0 we apply the algorithm again.

since 255 > 102

$255=102\times 2+51$

since remainder $\neq$ 0 we apply the algorithm again.

since 102 > 51

$102=51\times 2+0$

since remainder = 0 we conclude the HCF of 867 and 255 is 51.

Q2 Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer:

Let p be any positive integer. It can be expressed as

p = 6q + r

where $q\geq 0$ and $0\leq r< 6$

but for r = 0, 2 or 4 p will be an even number therefore all odd positive integers can be written in the form 6q + 1, 6q + 3 or 6q + 5.

Q3 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?

Answer:

The maximum number of columns in which they can march = HCF (32, 616)

Since 616 > 32, applying Euclid's Division Algorithm we have

$616=32\times 19+8$

Since remainder $\neq$ 0 we again apply Euclid's Division Algorithm

Since 32 > 8

$32=8\times 4+0$

Since remainder = 0 we conclude, 8 is the HCF of 616 and 32.

The maximum number of columns in which they can march is 8.

Q4 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.

[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1]

Answer:

Let x be any positive integer.

It can be written in the form 3q + r where $q\geq 0$ and r = 0, 1 or 2

Case 1:

For r = 0 we have

x ² = (3q) ²

x ² = 9q ²

x ² = 3(3q ² )

x ² = 3m

Case 2:

For r = 1 we have

x ² = (3q+1) ²

x ² = 9q ² + 6q +1

x ² = 3(3q ² + 2q) + 1

x ² = 3m + 1

Case 3:

For r = 2 we have

x ² = (3q+2) ²

x ² = 9q ² + 12q +4

x ² = 3(3q ² + 4q + 1) + 1

x ² = 3m + 1

Hence proved.

Q5 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer:

Let x be any positive integer.

It can be written in the form 3q + r where $q\geq 0$ and r = 0, 1 or 2

Case 1:

For r = 0 we have

x ³ = (3q) ³

x ³ = 27q ³

x ³ = 9(3q ³ )

x ³ = 9m

Case 2:

For r = 1 we have

x ³ = (3q+1) ³

x ³ = 27q ³ + 27q ² + 9q + 1

x ³ = 9(3q ³ + 3q ² +q) + 1

x ³ = 3m + 1

Case 3:

For r = 2 we have

x ³ = (3q+2) ³

x ³ = 27q ³ + 54q ² + 36q + 8

x ³ = 9(3q ³ + 6q ² +4q) + 8

x ³ = 3m + 8

Hence proved.

NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.2

Q1 (1) Express each number as a product of its prime factors: 140

Answer:

The number can be as a product of its prime factors as follows

$\\140=2\times 2\times 5\times 7\\ 140=2^{2}\times 5\times 7$

Q 1 (2) Express each number as a product of its prime factors: 156

Answer:

The given number can be expressed as follows

$\\156=2\times 2\times 3\times 13\\ 156=2^{2}\times 3\times 13$

Q1 (3) Express each number as a product of its prime factors: 3825

Answer:

The number is expressed as the product of the prime factors as follows

$\\3825=3\times 3\times 5\times 5\times 17\\ 3825=3^{2}\times 5^{2}\times 17$

Q1 (4) Express each number as a product of its prime factors: 5005

Answer:

The given number can be expressed as the product of its prime factors as follows.

$5005=5\times 7\times 11\times 13$

Q1 (5) Express each number as a product of its prime factors: 7429

Answer:

The given number can be expressed as the product of their prime factors as follows

$7429=17\times 19\times 23$

Q2 (1) Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers: 26 and 91

Answer:

26 = 2 x 13

91 = 7 x 13

HCF(26,91) = 13

LCM(26,91) = 2 x 7 x 13 = 182

HCF x LCM = 13 x 182 = 2366

26 x 91 = 2366

26 x 91 = HCF x LCM

Hence Verified

Q2 (2) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 510 and 92

Answer:

The number can be expressed as the product of prime factors as

510 = 2 x 3 x 5 x 17

92 = 2 ² x 23

HCF(510,92) = 2

LCM(510,92) = 2 ² x 3 x 5 x 17 x 23 = 23460

HCF x LCM = 2 x 23460 = 46920

510 x 92 = 46920

510 x 92 = HCF x LCM

Hence Verified

Q2 (3) Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. 336 and 54

Answer:

336 is expressed as the product of its prime factor as

336 = 2 ⁴ x 3 x 7

54 is expressed as the product of its prime factor as

54 = 2 x 3 ³

HCF(336,54) = 2 x 3 = 6

LCM(336,54) = 2 ⁴ x 3 ³ x 7 = 3024

HCF x LCM = 6 x 3024 = 18144

336 x 54 = 18144

336 x 54 = HCF x LCM

Hence Verified

Q3 (1) Find the LCM and HCF of the following integers by applying the prime factorization method. 12, 15 and 21

Answer:

The numbers can be written as the product of their prime factors as follows

12 = 2 ² x 3

15 = 3 x 5

21 = 3 x 7

HCF = 3

LCM = 2 ² x 3 x 5 x 7 = 420

Q3 (2) Find the LCM and HCF of the following integers by applying the prime factorisation method. 17, 23 and 29

Answer:

The given numbers are written as the product of their prime factors as follows

17 = 1 x 17

23 = 1 x 23

29 = 1 x 29

HCF = 1

LCM = 17 x 23 x 29 = 11339

Q3 (3) Find the LCM and HCF of the following integers by applying the prime factorization method. 8, 9 and 25

Answer:

The given numbers are written as the product of their prime factors as follows

8 = 2 ³

9 = 3 ²

25 = 5 ²

HCF = 1

LCM = 2 ³ x 3 ² x 5 ² = 1800

Q4 Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer:

As we know the product of HCF and LCM of two numbers is equal to the product of the two numbers we have

HCF (306, 657) x LCM (306, 657) = 306 x 657

$\\ LCM (306, 657) = \frac{306\times 657}{HCF (306, 657) }\\ LCM (306, 657) = \frac{306\times 657}{9}\\ LCM (306, 657) =22338$

Q5 Check whether $6 ^n$ can end with the digit 0 for any natural number n.

Answer:

By prime factorizing we have

6 ⁿ = 2 ⁿ x 3 ⁿ

A number will end with 0 if it has at least 1 as the power of both 2 and 5 in its prime factorization. Since the power of 5 is 0 in the prime factorization of 6 ⁿ we can conclude that for no value of n 6 ⁿ will end with the digit 0.

Q6 Explain why 7 $\times$ 11 $\times$ 13 + 13 and 7 $\times$ 6 $\times$ 5 $\times$ 4 $\times$ 3 $\times$ 2 $\times$ 1 + 5 are composite numbers.

Answer:

7 x 11 x 13 + 13

= (7 x 11 + 1) x 13

= 78 x 13

= 2 x 3 x 13 ²

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

= (7 x 6 x 4 x 3 x 2 x 1 + 1) x 5

= 5 x 1008

After Solving we observed that both the number are even numbers and the number rule says that we can take atleast two common out of two numbers. So that the number is composite number.

Q7 There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Answer:

The time after which they meet again at the starting point will be equal to the LCM of the times they individually take to complete one round.

Time taken by Sonia = 18 = 2 x 3 ²

Time taken by Ravi = 12 = 2 ² x 3

LCM(18,12) = 2 ² x 3 ² = 36

Therefore they would again meet at the starting point after 36 minutes.

NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.3

Q1 Prove that $\sqrt 5$ is irrational.

Answer:

Let us assume $\sqrt{5}$ is rational.

It means $\sqrt{5}$ can be written in the form $\frac{p}{q}$ where p and q are co-primes and $q\neq 0$

$\\\sqrt{5}=\frac{p}{q}$

Squaring both sides we obtain

$\\\left ( \sqrt{5} \right )^{2}=\left (\frac{p}{q} \right )^{2}\\ 5=\frac{p^{2}}{q^{2}}\\ p^{2}=5q^{2}$

From the above equation, we can see that p ² is divisible by 5, Therefore p will also be divisible by 5 as 5 is a prime number. $(i)$

Therefore p can be written as 5r

p = 5r

p ² = (5r) ²

5q ² = 25r ²

q ² = 5r ²

From the above equation, we can see that q ² is divisible by 5, Therefore q will also be divisible by 5 as 5 is a prime number. $(ii)$

From (i) and (ii) we can see that both p and q are divisible by 5. This implies that p and q are not co-primes. This contradiction arises because our initial assumption that $\sqrt{5}$ is rational was wrong. Hence proved that $\sqrt{5}$ is irrational.

Q2 Prove that $3 + 2 \sqrt 5$ is irrational.

Answer:

Let us assume $3 + 2 \sqrt 5$ is rational.

This means $3 + 2 \sqrt 5$ can be wriiten in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\3+2\sqrt{5}=\frac{p}{q}\\ 2\sqrt{5}=\frac{p}{q}-3\\ \sqrt{5}=\frac{p-3q}{2q}\\$

As p and q are integers $\frac{p-3q}{2q}\\$ would be rational, this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $3 + 2 \sqrt 5$ is rational was wrong. Therefore $3 + 2 \sqrt 5$ is irrational.

Q3 Prove that the following are irrationals :

(i) $1/ \sqrt 2$

Answer:

Let us assume $\frac{1}{\sqrt{2}}$ is rational.

This means $\frac{1}{\sqrt{2}}$ can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\\frac{1}{\sqrt{2}}=\frac{p}{q}\\ \sqrt{2}=\frac{q}{p}$

Since p and q are co-prime integers $\frac{q}{p}$ will be rational, this contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $\frac{1}{\sqrt{2}}$ is rational was wrong. Therefore $\frac{1}{\sqrt{2}}$ is irrational.

Q3 (2) Prove that the following are irrationals :

(ii) $7 \sqrt 5$

Answer:

Let us assume $7 \sqrt 5$ is rational.

This means $7 \sqrt 5$ can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\7\sqrt{5}=\frac{p}{q}\\ \sqrt{5}=\frac{p}{7q}$

As p and q are integers $\frac{p}{7q}\\$ would be rational, this contradicts the fact that $\sqrt{5}$ is irrational. This contradiction arises because our initial assumption that $7 \sqrt 5$ is rational was wrong. Therefore $7 \sqrt 5$ is irrational.

Q3 (3) Prove that the following are irrationals : $6 + \sqrt 2$

Answer:

Let us assume $6 + \sqrt 2$ is rational.

This means $6 + \sqrt 2$ can be written in the form $\frac{p}{q}$ where p and q are co-prime integers.

$\\6+\sqrt{2}=\frac{p}{q}\\ \sqrt{2}=\frac{p}{q}-6\\ \sqrt{2}=\frac{p-6q}{q}$

As p and q are integers $\frac{p-6q}{q}$ would be rational, this contradicts the fact that $\sqrt{2}$ is irrational. This contradiction arises because our initial assumption that $6 + \sqrt 2$ is rational was wrong. Therefore $6 + \sqrt 2$ is irrational.

NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.4

Q1 (1) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:13/3125

Answer:

$\frac{13}{3125}=\frac{13}{5^{5}}$

The denominator is of the form 2 ^a x 5 ^b where a = 0 and b = 5. Therefore the given rational number will have a terminating decimal expansion.

Q1 (2) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:17 / 8

Answer:

$\frac{17}{8}=\frac{17}{2^{3}}$

The denominator is of the form 2 ^a x 5 ^b where a = 3 and b = 0. Therefore the given rational number will have a terminating decimal expansion.

Q1 (3) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 64 / 455

Answer:

$\frac{64}{455}=\frac{64}{5\times 7\times 13}$

The denominator is not of the form 2 ^a x 5 ^b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

Q 1 (4) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 15 / 1600

Answer:

$\\\frac{15}{1600}=\frac{3\times 5}{2^{6}\times 5^{2}}\\ \frac{15}{1600}=\frac{3}{2^{6}\times 5}$

The denominator is of the form 2 ^a x 5 ^b where a = 6 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

Q 1 (5) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 29 / 343

Answer:

$\frac{29}{343}=\frac{29}{7^{3}}$

The denominator is not of the form 2 ^a x 5 ^b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

Q1 (6) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: $\frac{23}{2 ^3 5 ^ 2 }$

Answer:

$\frac{23}{2 ^3 \times 5 ^ 2 }$

The denominator is of the form 2 ^a x 5 ^b where a = 3 and b = 2. Therefore the given rational number will have a terminating decimal expansion.

Q1 (7) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: $\frac{129 }{2 ^ 2 7^5 5 ^ 7 }$

Answer:

$\frac{129 }{2 ^ 2\times 7^5 \times 5 ^ 7 }$

The denominator is not of the form 2 ^a x 5 ^b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

Q1 (8) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 6 / 15

Answer:

$\\\frac{6}{15}=\frac{2\times 3}{3\times 5}\\ \frac{6}{15}=\frac{2}{5}$

The denominator is of the form 2 ^a x 5 ^b where a = 0 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

Q1 (9) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 35 / 50

Answer:

$\\\frac{35}{50}=\frac{5\times 7}{2\times 5^{2}}\\\frac{35}{50}=\frac{7}{2\times 5}$

The denominator is of the form 2 ^a x 5 ^b where a = 1 and b = 1. Therefore the given rational number will have a terminating decimal expansion.

Q1 (10) Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: 77 / 210

Answer:

$\\\frac{77}{210}=\frac{7\times 11}{2\times 3\times5\times 7 }\\ \frac{77}{210}=\frac{11}{2\times 3\times5}$

The denominator is not of the form 2 ^a x 5 ^b . Therefore the given rational number will have a non-terminating repeating decimal expansion.

Q2 Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Answer:

decimal expansions of rational numbers are

(i) $\frac{13}{3125}=0.00416$

(ii)&nbsnbsp; $\frac{17}{8}=2.125$

(iv) $\frac{15}{1600}=0.009375$

(vI) $\frac{23}{2^{3}\times 5^{2}}=\frac{23}{200}=0.115$

(viii) $\frac{6}{15}=\frac{2}{5}=0.4$

(ix) $\frac{35}{50}=\frac{7}{2\times 5}=0.7$

Q3 (1) The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p / q what can you say about the prime factors of q? 43.123456789

Answer:

$\\43.123456789=\frac{43123456789}{10^{9}}\\ 43.123456789=\frac{43123456789}{2^{9}\times 5^{9}}$

The denominator is of the form 2 ^a x 5 ^b where a = 9 and b = 9. Therefore the given number is rational and has a terminating decimal expansion.

Q3 (2) The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p / q what can you say about the prime factors of q? 0.120120012000120000

Answer:

Since the decimal part of the given number is non-terminating and non-repeating we can conclude that the given number is irrational and cannot be written in the form $\frac{p}{q}$ where p and q are integers.

Q3 (3) The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p/q what can you say about the prime factors of q?

$43. \overline{123456789}$

Answer:

As the decimal part of the given number is non-terminating and repeating, the number is rational but its denominator will have factors other than 2 and 5.

Types of questions asked from Class 10 maths Chapter 1 Real Numbers

CBSE Class 10 maths board exam will have the following types of questions from real numbers:

Euclid's division algorithm
Prime factorization
Highest common factor
Prime factorization
Prime & composite numbers
Rational & irrational numbers

NCERT solutions for class 10 maths chapter wise

Chapter No.	Chapter Name
Chapter 1	NCERT solutions for class 10 maths chapter 1 Real Numbers
Chapter 2	NCERT solutions for class 10 maths chapter 2 Polynomials
Chapter 3	NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables
Chapter 4	NCERT solutions for class 10 maths chapter 4 Quadratic Equations
Chapter 5	NCERT solutions for class 10 chapter 5 Arithmetic Progressions
Chapter 6	NCERT solutions for class 10 maths chapter 6 Triangles
Chapter 7	NCERT solutions for class 10 maths chapter 7 Coordinate Geometry
Chapter 8	NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry
Chapter 9	NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry
Chapter 10	NCERT solutions class 10 maths chapter 10 Circles
Chapter 11	NCERT solutions for class 10 maths chapter 11 Constructions
Chapter 12	NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles
Chapter 13	NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes
Chapter 14	NCERT solutions for class 10 maths chapter 14 Statistics
Chapter 15	NCERT solutions for class 10 maths chapter 15 Probability

NCERT solutions of class 10 subject wise

How to use NCERT Solutions for Class 10 maths chapter 1 Real Numbers?

Now you have gone through the NCERT solutions for class 10 maths chapter 1 real numbers and have a good knowledge of answer writing in a structural manner. It's time to practice various kinds of problems based on real numbers.
After the completion of the NCERT syllabus, you can check past 5-year papers of board exams. Previous year papers will increase your dealing ability with a variety of questions.
NCERT syllabus coverage and previous year papers are enough tools to get a good score in the board examination. After covering NCERTs and the previous year papers, you can jump to the next chapters.

Benefits of NCERT solutions for Class 10 Maths Chapter 1 Real Numbers

Using the NCERT Class 10 Maths solutions chapter 1, students will be able to confirm the right answers once they are done solving the questions themselves.
Class 10 Maths chapter 1 NCERT solutions are solved by the subject matter experts. Therefore, the answers to all the questions are reliable.
NCERT Class 10 solutions for Class 1 Maths gives the step by step solution to all the questions which makes it easy for the students to understand.

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

Question: What is the weightage of the chapter real numbers for CBSE board exam?

Answer:

As per the CBSE the total weightage of real number is 4 marks in the final board exam.

Question: Where can I find the complete solutions of NCERT class 10 maths?

Answer:

Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link.

Question: What are the important topics of class 10 maths chapter 1 real numbers.

Answer:

Euclid's division algorithm, Euclid's division algorithm to find the highest common factor, prime & composite numbers, rational & irrational numbers are the important topics from the chapter real numbers.

Question: How many chapters are there in the class 10 maths ?

Answer:

There are 15 chapters in the class 10 Maths NCERT.

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Questions related to CBSE Class 10th

Showing 31 out of 31 Questions

371 Views

Board cbse class 10 Is cancelled and promote all student please sir because of corona do not study

Aadarsh Gupta 9th Apr, 2021

Hello aspirant,

No CBSE and ICSE board didn't not cancelled the 10th and 12th exam.

As they had declared that 10th and 12th class exam will not be cancelled and students will have to give it in offline mode in this pandemic also.

Hope this helps you

All the best for your future

52 Views

CBSE class 10th malayalam can someone please tell me a good one

Ashish Kumar Verma 8th Apr, 2021

Hello Dear,

Malayalam is the mother tongue of those who have their roots in Kerala. It is one of the Dravidian languages that has similarities in words with its other contemporary languages such as Tamil and Telugu. People who speak Malayalam are referred to as Mallus. You can download the previous year papers from the link below:

https://school.careers360.com/download/sample-papers/cbse-10th-class-malayalam-solved-sample-paper-2021

Good Luck

62 Views

When will the Karnataka NTSE Stage 1 Merit list(i.e people who have cleared stage-1) and the cut off come out?

Souhrid Sarkar 23rd Mar, 2021

Hi Shreyas,

DSERT, Karnataka has released the NTSE Karnataka 2021 Result stage 1 on March 9, 2021 on their official website ( dsert.kar.nic.in ). It is available in the form of district-wise marks lists , consisting of registration numbers and marks of all students.

Along with the NTSE stage 1 result, cutoff and merit list of selected candidates is also released.

The students those who have cleared the stage 1 exam are eligible to seat for the stage 2 exam which is going to be held on June 13, 2021 . The scholarship will be given to the students who will pass the stage 2 examination.

To know more about NTSE Karnataka Important Dates, Cut-off, Scholarship visit :

https://www.google.com/amp/s/school.careers360.com/articles/ntse-karnataka-result/amp

Hope it helps
Thank You !!!

638 Views

Age criteria to appear in class 10 CBSE exam

Manisha 31st Jan, 2021

Hello Bishal

According to guidelines of CBSE, minimum age to appear for class x must be 14 years. There is no upper limit to appear for class x cbse board.

A candidate can appear for maximum three attempts.

Some candidates give private exams or sometimes students fail in standard ix then they privately appear for class x then their age must be more than 14 years. Sometimes students appear for x class after one year gap of passing class ix then also their age would be 15 or 16 as there is no upper limit age.

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My child is supposed to make another attempt at compartment exam of 10th cbse Missed last date for applying. since the boards are postponed to May 4th can I pay her fees and apply now.

Yash Dev Student Expert 15th Jan, 2021

Hello sir I'm sorry to inform you but now you're not eligible to fill the registration form as last date to fill the registration Form was 9th December 2020. Yes examination gets postponed to May but portal to fill the registration form is not open , in case if the registration form portal will open again you can fill the registration form and make the payment.

Feel free to comment if you've any doubt

Good luck

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NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers

NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.1

NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.2

NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.3

NCERT solutions for class 10 maths chapter 1 Real Numbers Excercise: 1.4

Types of questions asked from Class 10 maths Chapter 1 Real Numbers

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