NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Class 10 Maths Exercise 1.1

1. Express each number as product of its prime factors:
(i) 140            ​​(ii) 156 ​​              (iii) 3825 ​​            (iv) 5005 ​​             (v) 7429
See Answer(i) 140 = 2 × 2 × 5 × 7  = 22 × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91        ​
See Answer(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Product of the two numbers ​= 26 × 91 ​= 2366
HCF × LCM ​​​= 13 × 182 ​= 2366
Hence, product of two numbers = HCF × LCM
(ii) 510 and 92
See Answer(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of the two numbers ​= 510 × 92 ​= 46920
HCF × LCM ​= 2 × 23460 ​= 46920
Hence, product of two numbers = HCF × LCM
(iii) 336 and 54
See Answer(iii) 336 and 52
336 = 2 × 2 × 2 × 2 × 3 × 7 ​= 24 × 3 × 7
54 = 2 × 3 × 3 × 3 ​​= 2 × 33
HCF = 2 × 3 = 6
LCM = 24 × 33 × 7 ​​= 3024
Product of the two numbers​= 336 × 54 = 18144
HCF × LCM ​= 6 × 3024​ = 18144
Hence, product of two numbers = HCF × LCM

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21 ​​      ​​
See Answer(i) 12, 15 and 21
12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5​
21 = 3 × 7
HCF = 3 and LCM = 22 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29
See Answer(ii) 17, 23 and 29
17 = 1 × 17
23 = 1 × 23​
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25
See Answer(iii) 8, 9 and 25
8 = 2 × 2 × 2 = 23
9 = 3 × 3 = 32​
25 = 5 × 5 = 52
HCF = 1
LCM = 23 × 32 × 52 = 8 × 9 × 25 = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).
See AnswerHCF (306, 657) = 9
We know that,
LCM × HCF = Product of two numbers
Therefore, LCM = Product of two numbers/HCF
So, LCM = (306 × 657)/9 = 22338
Hence, LCM (306, 657) = 22338

5. Check whether 6ⁿ can end with the digit 0 for any natural number n.
See AnswerIf any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5.
Prime factorisation of 6ⁿ = (2 × 3)ⁿ
It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6ⁿ will not be divisible by 5.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
See AnswerNumbers are of two types – prime and composite.
Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
It can be observed that:
7 × 11 × 13 + 13
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
Now consider the next number:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 ×1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

7. 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?
See AnswerIt can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path.
As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia.
The total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
18 = 2 × 3 × 3 and
12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Class 10 Maths Chapter 1 Board Questions
10th Maths Chapter 1 Assertion Reason Questions
Class 10 Maths Chapter 1 Case Study

Class 10 Maths Exercise 1.2

1. Prove that √5 is irrational.
See AnswerLet √5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that √5 = a/b. Let a and b have a common factor other than 1. Then we can divide them by the common factor and assume that a and b are co-prime.
a = √5b
⇒ a² = 5b²
Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer.
⇒ (5k)² = 5b²
⇒ 25k² = 5b²
⇒ 5k² = b²
This means that b² is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
This is a contradiction to the fact that a and b are co-prime.
Hence, √5 cannot be expressed as p/q or it can be said that √5 is irrational.

2. Prove that is 3 + 2√5 irrational.
See AnswerLet 3 + 2√5 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
3 + 2√5 = a/b
⇒ 2√5 = a/b – 3
⇒ √5 = 1/2(a/b – 3)
Since a and b are integers, 1/2(a/b – 3) will also be rational.
Therefore, √5 is rational.
This contradicts the fact that √5 is irrational.
Hence, our assumption that 3 + 2√5 is rational is false.
Therefore, 3 + 2√5 is irrational.

3. Prove that the following are irrationals:
(i) 1/√2
See Answer(i) 1/√2
Let 1/√2 is rational.
Therefore, we can assume two co-prime integers a, b (b ≠ 0) such that
1/√2 = a/b
⇒ √2 = b/a
b/a is rational as a and b are integers.
Therefore, √2 is also rational, which contradicts to the fact that √2 is irrational.
Hence, our assumption is false and 1/√2 is irrational.

(ii) ​​​7√5
See Answer(ii) 7√5
Let 7√5 is rational.
Therefore, we assume find two co-prime integers a, b (b ≠ 0) such that
7√5 = a/b
⇒ √5 = a/7b
a/7b is rational as a and b are integers.
Therefore, √5 should be rational.
This contradicts the fact that √5 is irrational.
Therefore, our assumption that 7√5 is rational is false.
Hence, 7√5 is irrational.

(iii) 6 + √2

See Answer(iii) 6 + √2
Let 6 + √2 be rational.
Therefore, we can assume two co-prime integers a, b (b ≠ 0) such that
6 + √2 = a/b
⇒ √2 = a/b – 6
Since a and b are integers, a/b – 6 is also rational and hence, √2 should be rational.
This contradicts the fact that √2 is irrational.
Therefore, our assumption is false and hence, 6 + √2 is irrational.

Class 10 Maths Chapter 1 MCQ

Board Paper Questions From 10th Maths Chapter 1
Class 10 Maths Chapter 1 Board Questions and their answers with explanation are given here. These questions are asked in CBSE Board Exams in previous years.
1. Prove that √2 is an irrational number. [CBSE 2025]
See AnswerWe’ll use proof by contradiction.
Assume that √2 is rational. Then it can be written as √2 = p/q where p and q are integers with no common factors (in lowest form) and q ≠ 0.
This means: p² = 2q²
This equation shows that p² is even (as it equals 2q²), which means p must be even (since odd numbers squared remain odd).
So we can write p = 2k for some integer k.
Substituting: (2k)² = 2q²
4k² = 2q²
q² = 2k²
Now this shows that q² is even, which means q must be even.
But this contradicts our initial assumption that p and q have no common factors (since both would be divisible by 2).
Therefore, our initial assumption must be false and √2 cannot be expressed as a ratio of integers, making it irrational.

2. Show that 6ⁿ cannot end with digit 0 for any natural number n. [CBSE 2023]
See SolutionFor a number to end in 0, it must be divisible by 10.
10 = 2 × 5
Prime factorization of 6 = 2 × 3
Any power of 6ⁿ will only have 2 and 3 as prime factors.
Hence, it can never be divisible by 5.
Therefore, 6ⁿ cannot end in 0.

3. Find the HCF and LCM of 72 and 120. [CBSE 2023]
See SolutionPrime factorization:
72 = 2³ × 3²
120 = 2³ × 3 × 5
HCF = 2³ × 3 = 24
LCM = 2³ × 3² × 5 = 360

4. Using prime factorisation, find HCF and LCM of 96 and 120. [CBSE 2023]
See Solution96 = 2⁵ × 3
120 = 2³ × 3 × 5
HCF = 2³ × 3 = 24
LCM = 2⁵ × 3 × 5 = 480

5. Find the greatest 3-digit number which is divisible by 18, 24 and 36. [CBSE 2023]
See SolutionFind LCM of 18, 24, 36
= 2³ × 3² = 72
Greatest 3-digit number = 999
Divide: 999 ÷ 72 = 13.875
→ Take floor = 13
13 × 72 = 936

6. If √2 is an irrational number, prove (5 – 2√2) is also an irrational number. [CBSE 2023 Compt]
See SolutionAssume 5 – 2√2 is rational.
Then, 2√2 = 5 – (rational number) = rational
⇒ √2 = rational
⇒ Contradiction.
So, 5 – 2√2 is an irrational number.

7. Prove that √2 is an irrational number. [CBSE 2023, 2020 Compt]
See SolutionLet √2 = a/b in lowest terms.
⇒ 2 = a²/b²
⇒ a² = 2b²
⇒ a is even
⇒ a = 2k
⇒ (2k)² = 4k² = 2b²
⇒ b² = 2k²
⇒ b is even
⇒ Both a and b even
⇒ Contradiction.
So, √2 is irrational.

8. If p = ab³ and q = a²b, a and b are prime, find LCM(p, q). [CBSE 2020 Compt]
See Solutionp = ab³
q = a²b
LCM = a²b³

9. Prove that √3 is an irrational number. [CBSE 2020]
See SolutionSame method as √2 proof.
Assume √3 = a/b
⇒ 3 = a²/b²
⇒ a² = 3b²
⇒ a is divisible by 3, let a = 3k
⇒ (3k)² = 9k² = 3b²
⇒ b² = 3k²
⇒ b divisible by 3
⇒ Both divisible by 3
⇒ Contradiction
So, √3 is an irrational number.

10. The traffic lights at three different road crossings change after every 48s, 72s, and 138s. If they change simultaneously at 7am, at what time will they change together next? [CBSE 2020]
See SolutionLCM(48, 72, 138) = LCM = 432 sec
Therefore, after 432 s, they will change simultaneously. We know that 60s = 1 min
= 1 s = 1/60 min
= 432 s = 432/60 min = 7 min 12 s
Hence, the lights will change simultaneously at 7 : 07 : 12 am.

11. Find the HCF and LCM of 26, 65, 117 using prime factorisation. [CBSE 2020]
See Solution26 = 2 × 13
65 = 5 × 13
117 = 3 × 3 × 13
HCF = 13
LCM = 2 × 3² × 5 × 13 = 1170.

12. Prove that √5 is an irrational number. [CBSE 2020]
See SolutionSame process as √2 proof.
Let √5 = a/b
⇒ 5 = a²/b²
⇒ a² = 5b²
Then a and b both divisible by 5
⇒ contradiction
⇒ √5 is irrational number.

13. Prove 2 + 5√3 is an irrational number, given √3 is irrational. [CBSE 2019]
See SolutionAssume 2 + 5√3 is rational
⇒ Subtract 2: 5√3 = rational
⇒ √3 = rational, which is contradiction.
So, 2 + 5√3 is irrational.

14. Prove (2 + √3)/5 is an irrational, given √3 is an irrational number. [CBSE 2019]
See SolutionAssume (2 + √3)/5 is rational
⇒ 2 + √3 = rational
⇒ √3 = rational
⇒ Contradiction
So, (2 + √3)/5 is irrational.

15. Write the smallest number divisible by both 306 and 657. [CBSE 2019]
See SolutionFind LCM(306, 657)
Prime factors:
306 = 2 × 3² × 17
657 = 3 × 219 = 3 × 3 × 73 = 3² × 73
LCM = 2 × 3² × 17 × 73 = 22338

16. a = x³y² and b = xy³, where x and y are primes. Find LCM(a, b). [CBSE 2019]
See SolutionGiven two positive integers
a = x³y² = x × x × x × y × y 
and b = xy³ = x × y × y × y
LCM (a,b) = Product of the greatest power of each prime factor involved in the numbers
 = x³ × y³ 
LCM = x³ y³

17. If HCF(336, 54) = 6, find LCM(336, 54). [CBSE 2019]
See SolutionProduct of numbers = HCF × LCM
⇒ 336 × 54 = 6 × LCM
⇒ LCM = (336 × 54)/6 = 3024

18. LCM is 9 times HCF. LCM + HCF = 500. Find HCF.  [CBSE 2019 Compt]
See SolutionLet HCF = x ⇒ LCM = 9x
x + 9x = 500
⇒ 10x = 500
⇒ x = 50

19. Show that (3 + √7)/5 is an irrational number, given √7 is irrational. [CBSE 2019 Compt]
See SolutionAssume (3 + √7)/5 is rational
⇒ Multiply both sides by 5: 3 + √7 = rational
⇒ √7 = rational
⇒ Contradiction
So, (3 + √7)/5 is irrational

Assertion-Reason Based Questions

(A) Both Assertion and Reason are true and Reason is the correct Explanation of the Assertion.
(B) Both Assertion and Reason are true but Reason is not correct explanation of the Assertion.
(C) Assertion is true but Reason is false.
(D) Assertion is false but Reason is true.
1. Assertion (A): The product of two consecutive positive integers is divisible by 2.
Reason (R): 132, 343, 565 is a composite number.
See Answer(B) Both Assertion and Reason are true but Reason is not correct explanation of the Assertion.

2. Assertion (A): (7 × 13 × 11) + 11 and (7 × 6 × 5 × 4 × 3 × 2 × 1) + 3 have exactly composite numbers.
Reason (R): (3 × 12 × 101) + 4 is not a composite number.
See Answer(C) Assertion is true but Reason is false.

3. Assertion (A) The number 5ⁿ cannot end with the digit 0, where n is a natural number.
Reason (R) Prime factorisation of 5 has only two factors, 1 and 5.
See Answer(A) Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.

4. Assertion (A) 15ⁿ ends with the digit zero, where n is natural number.
Reason (R) Any number ends with digit zero, if its prime factor is of the form 2ᵐ × 5ⁿ, where m and n are a natural numbers.
See Answer(D) Assertion is false but Reason is true.

5. Assertion (A) The HCF of two numbers is 5 and their product is 150. Then, their LCM is 40.
Reason (R) For any two positive integers a and b, HCF(a,b) × LCM(a,b) = a × b
See Answer(D) Assertion is false but Reason is true.

6. Assertion (A) The HCF of two numbers is 18 and their product is 3042, then their LCM is 169.
Reason (R) For any two positive integers a and b, HCF(a,b) + LCM(a,b) = a × b.
See Answer(C) Assertion is true but Reason is false.

7. Assertion (A) If HCF(336, 54) = 6, then LCM(336, 54) = 2043.
Reason (R) The sum of exponents of prime factors in the prime factorisation of 196 is 4.
See Answer(D) Assertion is false but Reason is true.

8. Assertion (A) The HCF of 24, 30 and 42 is 6.
Reason (R) HCF of two or more numbers = Product of the smallest power of each common prime factor, involved in the numbers.
See Answer(A) Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.

9. Assertion (A) √x is an irrational number, where x is a prime number.
Reason (R) Square root of any prime number is an irrational number.
See Answer(A) Both Assertion and Reason are true and Reason is the correct explanation of the Assertion.

10. Assertion (A) Product of HCF and LCM of three numbers is equal to the product of those numbers.
Reason (R) Product of HCF and LCM of two numbers is equal to the product of those numbers.
See Answer(D) Assertion is false but Reason is true.

Case Based Questions

Case Study 1

Teaching Mathematics through activities is a powerful approach that enhances students’ understanding and engagement. Keeping this in mind, Ms. Mukta planned a prime number game f o r class 5 students. She announces the number 2 in her class and asked the first student to multiply it by a prime number and then pass it  to second student. Second student also multiplied it by a prime number and passed it to third student. In this way by multiplying to a prime number, the last student got 173250. Now, Mukta asked some questions as given below t o t h e students. [CBSE 2024]
See SolutionWe are given:
The game starts with number 2.
Each student multiplies the current number by a prime number.
Final number = 173250
Each multiplication step represents a student.
So, number of students = number of prime factors multiplied (excluding the initial 2)
Let’s factorize 173250 completely to answer the questions.
Step 1: Prime factorization of 173250
173250 ÷ 2 = 86625
86625 ÷ 3 = 28875
28875 ÷ 3 = 9625
9625 ÷ 5 = 1925
1925 ÷ 5 = 385
385 ÷ 5 = 77
77 ÷ 7 = 11
11 ÷ 11 = 1
So, prime factorization:
173250 = 2 × 3² × 5³ × 7 × 11

Now we can answer the questions:
(i) What is the least prime number used by students?
See SolutionThe prime factors used: 2, 3, 5, 7, 11
Least prime number = 2

(ii) (a) How many students are in the class?
See SolutionEach student multiplies by one prime number.
Since starting number is 2, and the final product is:
2 × 3² × 5³ × 7 × 11 = total of 1 + 2 + 3 + 1 + 1 = 8 prime multiplications
So, 8 students

(b) What is the highest prime number used by students?
See SolutionPrime factors used: 2, 3, 5, 7, 11
Highest = 11

(iii) Which prime number has been used maximum times?
See SolutionCount of each prime:
2 → once
3 → 2 times
5 → 3 times
7 → once
11 → once
5 is used 3 times, which is the maximum.

Case Study 2

February 14 is celebrated a s International Book Giving Day and many Countries in the world celebrate this day. Some people in India also started celebrating this day and donated the following number of books of various subjects to a public library : History = 96, Science = 240 , Mathematics = 336 These books have to be arranged in minimum number of stacks such that each stack contains books of only one subject a n d the number of books o n each stack is the same. [CBSE 2023 Compt.]
See SolutionWe are given:
History books = 96
Science books = 240
Mathematics books = 336
We need to arrange the books in minimum number of stacks, with:
Books of only one subject per stack
Each stack containing the same number of books
This means: We need to find the HCF (Highest Common Factor) of 96, 240 and 336.
This gives the number of books per stack.
Prime Factorization:
96 = 2⁵ × 3
240 = 2⁴ × 3 × 5
336 = 2⁴ × 3 × 7
Now take the common factors with lowest powers:
Common: 2⁴ × 3 = 16 × 3 = 48

Based on the above information, answer the following questions.
(i) How many books are arranged in each stack?
See Solution48 books per stack

(ii) How many stacks are used to arrange all the Mathematics books?
See SolutionTotal Math books = 336
Books per stack = 48
Number of stacks = 336 ÷ 48 = 7 stacks

(iii) (a) Determine the total number of stacks that will be used for arranging all the books.
See SolutionHistory: 96 ÷ 48 = 2 stacks
Science: 240 ÷ 48 = 5 stacks
Maths: 336 ÷ 48 = 7 stacks
Total stacks = 2 + 5 + 7 = 14 stacks

(b) If the thickness of each book of History, Science and Mathematics is 1.8 cm , 2.2 cm and 25 cm respectively, then find the height of each stack of History, Science and Mathematics books.
See SolutionIf the thickness of each book is:
History = 1.8 cm
Science = 2.2 cm
Maths = 2.5 cm (likely meant 2.5, not 25 — assuming typo)
We calculate the height of each stack:
History stack: 48 × 1.8 = 86.4 cm
Science stack: 48 × 2.2 = 105.6 cm
Maths stack: 48 × 2.5 = 120 cm

Case Study 3

In a classroom activity on real numbers, the students have to pick a number from a pile and frame question on it if it is not a rational number for the rest of the class. The number cards picked up by first 3 students and their questions on the numbers for the rest of the class are as shown below. Answer them.
i) Suraj picked up J8 and his question was — which of the following is true about V8?
(a) It is a natural number
(b) It is a n irrational number
(c) It is a rational number
(d) None of the above
See Solution(i) Suraj picked up √8 and asked:
Which of the following is true about √8?
√8 = √(4 × 2) = √4 × √2 = 2√2
→ √2 is irrational ⇒ 2√2 is also irrational
Correct answer: (b) It is an irrational number

(ii) Neha picked up ‘bonus’ and her question was—which of the following is not irrational?
(a) 3 – 4√5
(b) 2 + 2√9
(c) √7 – 6
(d) 4√11 – 6
See Solution(ii) Neha picked up ‘bonus’ and asked:
Which of the following is not irrational?
Let’s check each option:
(a) 3 − 4√5
→ √5 is irrational ⇒ 4√5 is irrational ⇒ 3 − 4√5 is irrational
(b) 2 + 2√9
→ √9 = 3 ⇒ 2 + 2 × 3 = 2 + 6 = 8 (rational)← this one is not irrational
(c) √7 − 6
→ √7 is irrational ⇒ irrational − 6 = irrational
(d) 4√11 − 6
→ √11 is irrational ⇒ 4√11 is irrational ⇒ irrational − 6 = irrational
Correct answer: (b) 2 + 2√9

(iii) Anu pick up √15 – √10 and her question was 115 – 110 is.. number.
(a) a natural
(b) an irrational
(c) a whole
(d) a rational
See Solution(iii) Anu picked up √15 − √10 and asked:
√15 − √10 is… number?
→ Both √15 and √10 are irrational numbers
→ Their difference is not a rational number, as subtracting two irrationals doesn’t necessarily make it rational
→ In this case, √15 − √10 is irrational
Correct answer: (b) an irrational

Case Study 4

Sakshara International School organised a combined exhibition for grade 10 students of its three branches. 345 students from Mumbai branch, 405 students from Pune branch and 270 students from Nagpur branch participated in it. The following were planned for the exhibition:
• Group projects The students of each of the three branches were divided in groups for making various group projects such that each group had equal number of students and the number of groups was minimum.
• Individual project Each of total 1020 students had to submit an individual project. A fixed number of topics were allotted such that each topic had equal number of students.
• Inter-state model making competition A few equal number of students were selected from each branch t o participate in the competition. Each branch was supposed to submit between 3 to 7 models.
See Solution
Given:
– Mumbai: 345 students
– Pune: 405 students
– Nagpur: 270 students
– Total students = 345 + 405 + 270 = 1020

(i) A Maths teacher asked his student to solve the below puzzle regarding the individual project.
The number of students who got the same topic can be represented as (2ⁿ x 5), where n is a positive integer having the maximum possible value. Find n and the number of topics allotted.
See Solution(i) Individual Project
Puzzle:
Each of the 1020 students submits an individual project.
They are divided equally among a number of topics.
Each topic has number of students = (2ⁿ × 5), where n is the maximum possible.
Step 1: Factorize 1020
1020 = 2² × 3 × 5 × 17
We want the number of students per topic = 2ⁿ × 5
→ That means we must divide 1020 by a number of the form (2ⁿ × 5), such that n is maximum and result is an integer
Try increasing values of n:
– n = 1 → 2¹ × 5 = 10 → 1020 ÷ 10 = 102
– n = 2 → 2² × 5 = 20 → 1020 ÷ 20 = 51
– n = 3 → 2³ × 5 = 40 → 1020 ÷ 40 = 25.5
– n = 4 → 2⁴ × 5 = 80 → 1020 ÷ 80 = 12.75
Maximum valid n = 2, number of students per topic = 20, total topics = 1020 ÷ 20 = 51
Answer:
– n = 2
– Number of topics = 51

(ii) Mumbai branch divided the students selected for inter-state model making competition into the groups of 12 students, Pune into the groups of 10 students and Nagpur into the groups of 15 students.
(a) How many students were selected from each branch?
(b) How many models were submitted by individual branches and all the branches together?
See Solution(ii) Inter-state Model Making Competition
– Mumbai: Groups of 12 students
– Pune: Groups of 10 students
– Nagpur: Groups of 15 students
– Each branch to submit between 3 and 7 models
– All groups must have equal number of students in each branch
Let’s find common multiples between 3 and 7 for each group size.
(a) How many students selected from each branch?
Find a number divisible by:
– 12 (Mumbai)
– 10 (Pune)
– 15 (Nagpur)
And gives number of groups between 3 and 7
Let’s list first few multiples of LCM(12, 10, 15)
Step 1: LCM(12, 10, 15)
– 12 = 2² × 3
– 10 = 2 × 5
– 15 = 3 × 5
LCM = 2² × 3 × 5 = 60
Try multiples of 60 that give between 3 and 7 groups for each branch:
– 60 × 3 = 180
– 60 × 4 = 240
– 60 × 5 = 300
– 60 × 6 = 360
– 60 × 7 = 420
We need a number divisible by:
– 12 for Mumbai
– 10 for Pune
– 15 for Nagpur
Try 60:
– 60 ÷ 12 = 5 → 5 groups
– 60 ÷ 10 = 6 → 6 groups
– 60 ÷ 15 = 4 → 4 groups
→ Number of students per branch = 60
→ Number of models per branch = 5 (Mumbai), 6 (Pune), 4 (Nagpur)
Each branch submits models = Number of groups
Total models = 5 + 6 + 4 = 15 models
Answer:
(a) 60 students selected from each branch
(b) Mumbai: 5 models, Pune: 6 models, Nagpur: 4 models
→ Total models = 15

Class 10 Maths Chapter 1 Solutions
Class 10 Maths Chapter 1 Main Topics
Class 10 Maths Chapter 1 Practice Tests
10th Maths Chapter 1 Question wise Solutions
Class 10 Maths Chapter 1 Case Study
Class 10 Maths Chapter 1 Assignments

Preparing for mathematics exams becomes simpler with NCERT Class 10 Maths Chapter 1 Important Questions and CBSE Previous Year Questions. These resources help students focus on frequently asked problems, enhancing their problem-solving skills. For instance, understanding Real Numbers in depth requires familiarity with proofs of irrational numbers and applications of the Fundamental Theorem of Arithmetic. Textbook Class 10 Maths Chapter 1 Notes and Summary act as quick references, saving time while revising key concepts.

Solving 10th Math Exercise 1.1 to Exercise 1.4 further aids in solidifying these concepts. For students looking for additional support, online video lectures and study materials provide step-by-step explanations of the chapter’s topics. Incorporating NCERT Complete Solutions for Real Numbers Class 10 into daily study routines can significantly boost confidence, ensuring students are well-prepared to perform in their board exams. Learning Class 10 Maths Chapter 1 Real Numbers effectively requires a systematic approach and a clear understanding of the concepts involved. Here are some steps and strategies to help you learn class 10 mathematics chapter 1.

Euclid’s Division Lemma states that for any two positive integers a and b, there exist unique integers q and r such that: a = bq + r, where 0 ≤ r < b.
The Fundamental Theorem of Arithmetic states that every positive integer greater than 1 can be expressed as a product of prime numbers in a unique way, up to the order of factors.

Key Highlights of Class 10 Maths Chapter 1 Real Numbers

Class 10 Maths Chapter 1, Real Numbers, focuses on Euclid’s Division Lemma, the Fundamental Theorem of Arithmetic, proving the irrationality of numbers, terminating and non-terminating decimals and revisiting HCF and LCM. These concepts form the basis of board exam questions, often involving direct proofs, applications and numerical problems. Mastery ensures strong foundational skills.

Class 10 Maths NCERT Book Chapter 1 Real Numbers is not only useful for CBSE board exams but also lays the groundwork for advanced mathematics. Tools like NCERT Class 10 Maths Chapter 1 Exercises Formula List and Practice Problems simplify understanding and application of theorems. Teachers often recommend using NCERT Simplified Solutions for Real Numbers Class 10 to enhance topics clarity. Students can also test their knowledge with online tests, MCQs and sample papers prepared for Chapter 1. For comprehensive preparation, the Class 10 Maths Chapter 1 Exercise 1.1, Exercise 1.2 and Exercise 1.3 Solutions are particularly helpful in mastering the topic step by step. Resources such as Class 10th NCERT Maths Textbook Chapter 1 Study Material ensure that learners have everything they need in one place. With consistent practice and a focus on important questions, students can confidently approach any problem from this critical chapter.

Methods for proving the irrationality of certain numbers, such as the square root of a prime number, are discussed. Understanding decimal expansions of real numbers. Terminating decimals and recurring decimals. Rationalization is the process of converting an irrational number into a rational one by multiplying or dividing by a suitable number.

According to new syllabus for academic session, only two exercises are in chapter 1. The exercise 1.1 deleted and exercise 1.2 become the exercise 1.1. Similarly exercise 1.3 become exercise 1.2 and exercises 1.4 is also deleted.
The course structure of class 10 mathematics chapter 1 is given below:
Fundamental Theorem of Arithmetic – statements after reviewing work done earlier and
after illustrating and motivating through examples, Proofs of irrationality of √2, √3, √5.

>> Total Number of Exercises: 2
>> Total Marks in Board Exam: 6
>> Number of Periods needed: 15
Class 10 Maths Chapter 1 Real Numbers, is an important chapter that lays the foundation for understanding various mathematical concepts. The main points covered in this chapter include introduction to Real Numbers. Real numbers are a set of numbers that includes all rational and irrational numbers. Real numbers can be represented on the number line.

Utilize online resources like video tutorials, educational websites like Tiwari Academy and interactive quizzes to reinforce your understanding of the chapter. As you progress, practice solving sample papers or past year question papers. This will help you become familiar with the exam pattern and time management. Periodically test your knowledge by attempting self-assessment quizzes or problems from the chapter. This will help you gauge your progress.

In class 10 Maths Chapter we will learn about understanding square numbers and their properties. Finding the square root of a number using prime factorization. Real numbers are used in various mathematical applications, such as finding the distance between two points in coordinate geometry.

Class 10 Maths Chapter Real Numbers
The chapter 1 provides numerous examples and practice problems to reinforce the concepts discussed. These are the main points covered in Class 10 Maths Chapter 1 Real Numbers. It serves as the building block for further mathematical concepts and is essential for a solid foundation in mathematics. Get here topic wise solutions for 10th Mathematics chapter 1 all exercises in Hindi and English medium. Students can take help of videos for solving questions of each exercise.

Class 10 Maths Chapter 1 Exercise 1.1 Question 1

Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
The step by step solutions of each part is given here with practice questions>. You can discuss with us Part (i), Part (ii), Part (iii), Part (iv), and Part (v) to learn more about the solution.

Class 10 Maths Chapter 1 Exercise 1.1 Question 2
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54.
Get the solution of question 2 and discuss Part (i), Part (ii) and Part (iii) with us along with assignments and other doubts for more clarification.

Class 10 Maths Chapter 1 Exercise 1.1 Question 3
Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25.
Discuss with part one, two and three in detail and get worksheets to understand the solution in simple way.

Class 10 Maths Chapter 1 Exercise 1.1 Question 4
Given that HCF (306, 657) = 9, find LCM (306, 657).
The solution of this question is very simple and usually asked in MCQ type questions. For more revision, please write to us.

Class 10 Maths Chapter 1 Exercise 1.1 Question 5
Check whether 6ⁿ can end with the digit 0 for any natural number n.
The solution of question 5 is similar to the example given in textbook. Discuss with us and Practice here more questions based on question 5.

Class 10 Maths Chapter 1 Exercise 1.1 Question 6
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Extra questions for practice based on the solution of question 6 are given here to download. Put your view to discuss with us.

Class 10 Maths Chapter 1 Exercise 1.1 Question 7
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?
Question number 7 is important for exams. Get here the solution and practice assignments with answers for revision.

Class 10 Maths Chapter 1 Exercise 1.2 Question 1
Prove that √5 is irrational.
To understand the solution of question 1, student need to practice assignments and discuss the explanation in depth.

Class 10 Maths Chapter 1 Exercise 1.2 Question 2
Prove that 3 + 2√5 is irrational.
Get here the assignments and practice questions based on the solution of question 2. If still there is any doubt, please visit to discussion forum.

Class 10 Maths Chapter 1 Exercise 1.2 Question 3
Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2
The solution of question 3 is similar to question 1 and 2. Get more practice to learn the concepts in better way. Discuss in detail here.

Use number lines and diagrams to visualize concepts like decimal expansions and the placement of real numbers on the number line. Visual aids can help make abstract concepts more concrete. Don’t hesitate to ask your teacher or a classmate for help if you encounter difficulties. Sometimes, discussing problems with others can provide valuable insights. Maintain a study schedule and allocate specific time slots for learning and practicing this chapter.

The complete NCERT chapter 1 of 10th Maths 1 is divided in four main parts. The explanation of each part with video explanation is given below. Start by reading the chapter thoroughly and grasp the basic definitions and concepts. Ensure that you understand the definitions of real numbers, Euclid’s Division Lemma and the Fundamental Theorem of Arithmetic. Take concise notes while reading the chapter. Write down key formulas, definitions and important points. These notes will be helpful for quick revision.

Learn here how to find the prime factorization of numbers. This skill is essential for various aspects of number theory. Practice finding the prime factorization of different numbers. Practice working with terminating and recurring decimals. Solve NCERT textbook problems that involve decimal expansions of real numbers. Familiarize yourself with methods for rationalizing irrational numbers, especially square roots. Practice these techniques with different examples. Remember that learning mathematics often requires patience and persistence. Take your time to understand each concept thoroughly before moving on to the next. By following these steps and strategies, you can effectively learn Class 10 Maths NCERT Chapter 1 Real Numbers.

Class 10 Maths Chapter 1 Main Topics
Mathematics is best learned through practice. Solve a variety of problems related to real numbers, including examples from your textbook and additional practice questions. Pay attention to different types of questions, such as finding HCF, LCM, prime factorization and working with decimal expansions.

Euclid’s Division Lemma is a fundamental concept in this chapter. Make sure you understand how it works and can apply it to solve problems related to divisibility and remainders. There are four exercises in chapter 1 covering all the four topics on Real Numbers. Students can practice well using the 6 question papers given below. It will help to prepare the chapter for CBSE exams also.

The word algorithm come from the name of the name of the 9th century Persian mathematician al-Khwarizmi. The word ‘algebra’ is derived from a book, he wrote, called Hisab al-jabr w’al-muqabala.

An equivalent version of Fundamental theorem of Arithmetic was probably first recorded as Proposition 14 of Book IX in Euclid’s Elements, before it came to be known as the Fundamental Theorem of Arithmetic.

However, the first correct proof was given by Carl Friedrich Gauss in his al-Khwarizmi.
Carl Friedrich Gauss is often referred to as the ‘Prince of Mathematicians’ and is considered one of the three greatest mathematicians of all time, along with Archimedes and Newton. He has made fundamental contributions to both mathematics and science. Learn here how to make NCERT class 10 Maths chapter 1 easy to learn.