In this post we are providing NCERT Solutions Class 10 Maths Chapter 1 Real Numbers also we are added class 10 maths chapter 1 pdf. These solutions are prepared by our team faculty to help students with board exam preparations. They solve and provide the NCERT Soultion for Maths to aid the students to solve the problems easily. Our expert also focuses on preparing the solutions in such a way that it is under easy to understand and simple way for the students.
Answers to the questions present in Real Numbers are given in the first chapter of Maths solution for class 10. Here, students are introduced on various important concepts that will be useful for those who wish to pursue mathematics as a subject in their Class 11 and Class 12. Let’s start our Exercises.
Class 10 Maths Chapter 1 Exercise 1.1 Solutions
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. 135 and 225
Let’s apply this algorithm to find the HCF of 135 and 225.
First, we divide the larger number (225) by the smaller number (135) and find the remainder:
225 = 135 × 1 + 90
The remainder is 90, so we continue by dividing 135 by 90 and finding the remainder:
135 = 90 × 1 + 45
The remainder is 45, so we continue by dividing 90 by 45 and finding the remainder:\
90 = 45 × 2 + 0
Since the remainder is 0, we stop. The last non-zero remainder was 45, so the HCF of 135 and 225 is 45.
Therefore, the HCF of 135 and 225 is 45.
ii. 196 and 38220
Let’s apply this algorithm to find the HCF of 196 and 38220.
First, we divide the larger number (38220) by the smaller number (196) and find the remainder:
38220 = 196 × 194 + 76
The remainder is 76, so we continue by dividing 196 by 76 and finding the remainder:
196 = 76 × 2 + 44
The remainder is 44, so we continue by dividing 76 by 44 and finding the remainder:
76 = 44 × 1 + 32
The remainder is 32, so we continue by dividing 44 by 32 and finding the remainder:
44 = 32 × 1 + 12
The remainder is 12, so we continue by dividing 32 by 12 and finding the remainder:
32 = 12 × 2 + 8
The remainder is 8, so we continue by dividing 12 by 8 and finding the remainder:
12 = 8 × 1 + 4
The remainder is 4, so we continue by dividing 8 by 4 and finding the remainder:
8 = 4 × 2 + 0
Since the remainder is 0, we stop. The last non-zero remainder was 4, so the HCF of 196 and 38220 is 4.
Therefore, the HCF of 196 and 38220 is 4.
iii. 867 and 255
Let’s use Euclid’s division algorithm to find the HCF (highest common factor) of 867 and 255.
First, we divide the larger number (867) by the smaller number (255) and find the remainder:
867 = 255 × 3 + 102
The remainder is 102, so we continue by dividing 255 by 102 and finding the remainder:
255 = 102 × 2 + 51
The remainder is 51, so we continue by dividing 102 by 51 and finding the remainder:
102 = 51 × 2 + 0
Since the remainder is 0, we stop. The last non-zero remainder was 51, so the HCF of 867 and 255 is 51.
Therefore, the HCF of 867 and 255 is 51.
Also Check : All Maths Formulas For Class 10
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let n be any positive odd integer. Then n can be written in the form of 2k+1 where k is some integer.
Now, we consider the remainders when k is divided by 3.
Case 1: k ≡ 0 (mod 3) In this case, we can write k as 3q for some integer q. Therefore, n = 2k + 1 = 2(3q) + 1 = 6q + 1.
Case 2: k ≡ 1 (mod 3) In this case, we can write k as 3q + 1 for some integer q. Therefore, n = 2k + 1 = 2(3q + 1) + 1 = 6q + 3.
Case 3: k ≡ 2 (mod 3) In this case, we can write k as 3q + 2 for some integer q. Therefore, n = 2k + 1 = 2(3q + 2) + 1 = 6q + 5.
Thus, in all three cases, we have expressed n in the form of 6q + 1, 6q + 3, or 6q + 5, where q is some integer. Therefore, any positive odd integer can be expressed in one of these three forms.
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?
Solution:
Given,
The Number of army contingent members is = 616
The Number of army band members is = 32
By using Euclid’s algorithm method to find their HCF,
We get, Since, 616>32, therefore we get,
616 = 32 × 19 + 8
Since, 8 ≠ 0, then we need to taking 32 as a new divisor,
32 = 8 × 4 + 0
Then, we have get remainder as 0, therefore, HCF (616, 32) = 8.
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 x = positive integer
y = 3.
By Euclid’s division algorithm, then,
x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.
Therefore, x = 3q, 3q+1 and 3q+2
According to given the question , by squaring both the sides, we get,
x2 = (3q)2 = 9q2 = 3 × 3q2
Let 3q2 = m
Therefore, x2= 3m ……………………..(1)
x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1
Substitute, 3q2+2q = m, to get,
x2= 3m + 1 ……………………………. (2)
x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1
Again, substitute, 3q2+4q+1 = m, to get,
x2= 3m + 1…………………………… (3)
Hence, from equation 1, 2 and 3, we can say that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
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.
Solution:
Let x = positive integer and y = 3.
By Euclid’s division algorithm, then,
x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.
Therefore, putting the value of r, we get,
x = 3q
or
x = 3q + 1
or
x = 3q + 2
Now, by taking the cube of all three above expressions, we get,
Case (i): When r = 0, then,
x2= (3q)3 = 27q3= 9(3q3)= 9m; where m = 3q3
Case (ii): When r = 1, then,
x3 = (3q+1)3 = (3q)3 +13+3×3q×1(3q+1) = 27q3+1+27q2+9q
Taking 9 as common factor,
x3 = 9(3q3+3q2+q)+1
Putting = m, we get,
Putting (3q3+3q2+q) = m, we get ,
x3 = 9m+1
Case (iii): When r = 2, then,
x3 = (3q+2)3= (3q)3+23+3×3q×2(3q+2) = 27q3+54q2+36q+8
Taking 9 as common factor,
x3=9(3q3+6q2+4q)+8
Putting (3q3+6q2+4q) = m
x3 = 9m+8
Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Also Check : Important Maths Formulas
Class 10 Maths Chapter 1 Exercise 1.2 Solutions
1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solutions:
(i) 140
To express 140 as a product of its prime factors, we can use the following steps:
First, we can divide 140 by 2, which gives 70. So we can write:
140 = 2 × 70
Next, we can divide 70 by 2 again, which gives 35. So we can write:
140 = 2 × 2 × 35
We cannot divide 35 by 2, but we can divide it by 5, which gives 7. So we can write:
140 = 2 × 2 × 5 × 7
Therefore, the prime factorization of 140 is 2 × 2 × 5 × 7.
(ii) 156
To express 156 as a product of its prime factors, we can use the following steps:
First, we can divide 156 by 2, which gives 78. So we can write:
156 = 2 × 78
Next, we can divide 78 by 2 again, which gives 39. So we can write:
156 = 2 × 2 × 39
We cannot divide 39 by 2, but we can divide it by 3, which gives 13. So we can write:
156 = 2 × 2 × 3 × 13
Therefore, the prime factorization of 156 is 2 × 2 × 3 × 13.
(iii) 3825
To express 3825 as a product of its prime factors, we can use the following steps:
First, we can divide 3825 by 3, which gives 1275. So we can write:
3825 = 3 × 1275
Next, we can divide 1275 by 3 again, which gives 425. So we can write:
3825 = 3 × 3 × 425
We cannot divide 425 by 3, but we can divide it by 5, which gives 85. So we can write:
3825 = 3 × 3 × 5 × 17
Therefore, the prime factorization of 3825 is 3 × 3 × 5 × 17.
(iv) 5005
To express 5005 as a product of its prime factors, we can use the following steps:
First, we can divide 5005 by 5, which gives 1001. So we can write:
5005 = 5 × 1001
Next, we can divide 1001 by 7, which gives 143. So we can write:
5005 = 5 × 7 × 143
We cannot divide 143 by 5 or 7, but we can divide it by 11, which gives 13. So we can write:
5005 = 5 × 7 × 11 × 13
Therefore, the prime factorization of 5005 is 5 × 7 × 11 × 13.
(v) 7429
To express 7429 as a product of its prime factors, we can use the following steps:
We can check for divisibility by 2, 3, 5, and 7, but we find that none of these primes divide 7429 exactly. So we need to try other primes.
We can try dividing 7429 by 11, which gives 673. So we can write:
7429 = 11 × 673
We cannot divide 673 by any of the primes we have tried so far, but we can check for divisibility by primes up to the square root of 673, which is approximately 26. So we can try dividing by 13, which gives a remainder of 4. We can try dividing by 17.
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
(i) 26 and 91
For the pair of numbers 26 and 91, we can find their prime factorizations as:
26 = 2 × 13 91 = 7 × 13
The common factors are 13, so HCF(26, 91) = 13.
To find the LCM, we can list the multiples of each number until we find a common multiple:
Multiples of 26: 26, 52, 78, 104, 130, 156, 182, 208, 234, 260, 286, 312, 338, 364, 390, 416, 442, 468, 494, 520, 546, 572, 598, 624, 650, 676, 702, 728, 754, 780, 806, 832, 858, 884, 910, …
Multiples of 91: 91, 182, 273, 364, 455, 546, 637, 728, 819, 910, …
The first common multiple we find is 182, so LCM(26, 91) = 182.
We can verify that LCM(26, 91) × HCF(26, 91) = 182 × 13 = 2366, which is equal to the product of the two numbers: 26 × 91 = 2366.
(ii) 510 and 92
For the pair of numbers 510 and 92, we can find their prime factorizations as:
510 = 2 × 3 × 5 × 17 92 = 2^2 × 23
The common factors are 2, so HCF(510, 92) = 2. To find the LCM, we can use the highest powers of each prime factor:
LCM(510, 92) = 2^2 × 3 × 5 × 17 × 23 = 197820
We can verify that LCM(510, 92) × HCF(510, 92) = 197820 × 2 = 98160, which is equal to the product of the two numbers: 510 × 92 = 46920.
(iii) 336 and 54
For the pair of numbers 336 and 54, we can find their prime factorizations as:
336 = 2^4 × 3 × 7 54 = 2 × 3^3
The common factors are 2 and 3, so HCF(336, 54) = 2 × 3 = 6. To find the LCM, we can use the highest powers of each prime factor:
LCM(336, 54) = 2^4 × 3^3 × 7 = 9072
We can verify that LCM(336, 54) × HCF(336, 54) = 9072 × 6 = 54432, which is equal to the product of the two numbers: 336 × 54 = 18144.
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
Solution:
(i) 12, 15 and 21
Prime factorisation of 12 = 2 × 2 × 3 Prime factorisation of 15 = 3 × 5 Prime factorisation of 21 = 3 × 7
HCF of 12, 15 and 21 = 3
LCM of 12, 15 and 21 = 2 × 2 × 3 × 5 × 7 = 420
Verification: HCF × LCM = 3 × 420 = 1260 Product of the three numbers = 12 × 15 × 21 = 3780 HCF × LCM = Product of the three numbers
(ii) 17, 23 and 29
Prime factorisation of 17 = 17 Prime factorisation of 23 = 23 Prime factorisation of 29 = 29
HCF of 17, 23 and 29 = 1
LCM of 17, 23 and 29 = 17 × 23 × 29 = 11339
Verification: HCF × LCM = 1 × 11339 = 11339 Product of the three numbers = 17 × 23 × 29 = 11339 HCF × LCM = Product of the three numbers
(iii) 8, 9 and 25
Prime factorisation of 8 = 2 × 2 × 2 Prime factorisation of 9 = 3 × 3 Prime factorisation of 25 = 5 × 5
HCF of 8, 9 and 25 = 1
LCM of 8, 9 and 25 = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 900
Verification: HCF × LCM = 1 × 900 = 900 Product of the three numbers = 8 × 9 × 25 = 1800 HCF × LCM ≠ Product of the three numbers
Therefore, there must be a mistake in either the HCF or LCM calculation. We can check that the HCF is actually 1 by observing that the only common factor of the three numbers is 1. Therefore, the correct HCF is 1, and the correct LCM is 2 × 2 × 2 × 3 × 3 × 5 × 5 = 900.
Verification: HCF × LCM = 1 × 900 = 900 Product of the three numbers = 8 × 9 × 25 = 1800 HCF × LCM ≠ Product of the three numbers.
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
As we know that,
HCF×LCM=Product of the two given numbers
Therefore,
9 × LCM = 306 × 657
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.
Solution:
No, 6n cannot end with the digit 0 for any natural number n.
To see why, consider the prime factorization of 6:
6 = 2 x 3
Any multiple of 6 will have at least one factor of 2 and one factor of 3.
For a number to end with the digit 0, it must be divisible by 10, which means it must have a factor of 5 and a factor of 2.
However, since 6 has only one factor of 2, any multiple of 6 will have at most one factor of 2. Thus, it cannot end with the digit 0.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Let’s consider the two given expressions:
7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
To prove that they are composite numbers, we need to show that they are not prime numbers.
For the first expression, we can factor out 13 as a common factor:
7 × 11 × 13 + 13 = 13(7 × 11 + 1)
We can see that both 7 and 11 are factors of the expression inside the parentheses, which means that 13(7 × 11 + 1) is not a prime number. Therefore, it is a composite number.
For the second expression, we can simplify it using factorials:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 7 × 6! + 5
We can see that 5 is a factor of the expression, which means that 7 × 6! + 5 is not a prime number. Therefore, it is a composite number.
Hence, we have shown that both expressions are composite numbers, as required.
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?
Since the circular path is of a fixed length, we can assume the length to be the LCM of the time taken by Sonia and Ravi to complete one round of the field.
LCM of 18 and 12
LCM(18,12) = 2×3×3×2×1=36
So they will meet again at the starting point after 36 minutes.
Class 10 Maths Chapter 1 Exercise 1.3 Solutions
1. Prove that √5 is irrational.
Solutions:
To prove that √5 is irrational, we will assume the opposite, that √5 is a rational number. This means that we can write:
√5 = a/b
where a and b are co-prime integers.
Squaring both sides, we get:
5 = (a/b)^2
Multiplying both sides by b^2, we get:
5b^2 = a^2
This implies that a^2 is divisible by 5, and hence a must be a multiple of 5. Let a = 5k, where k is an integer.
Substituting this value of a in the equation 5b^2 = a^2, we get:
5b^2 = (5k)^2
Simplifying this equation, we get:
b^2 = 5k^2
This means that b^2 is divisible by 5, and hence b must also be a multiple of 5. But this contradicts our assumption that a and b are co-prime.
Therefore, our initial assumption that √5 is rational must be false, and we can conclude that √5 is irrational.
2. Prove that 3 + 2√5 + is irrational.
Let us assume 3 + 2√5 is rational.
Then we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y
Rearranging, we get,
Since, x and y are integers, thus,
is a rational number.
Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational.
So, we conclude that 3 + 2√5 is irrational.
3. Prove that the following are irrationals:
(i) 1/√2
(ii) 7√5
(iii) 6 + √2
Solutions:
(i) 1/√2
To prove that 1/√2 is irrational, we will assume the opposite, that 1/√2 is a rational number. This means that we can write:
1/√2 = a/b
where a and b are co-prime integers.
Squaring both sides, we get:
1/2 = a^2/b^2
Multiplying both sides by 2b^2, we get:
b^2 = 2a^2
This means that b^2 is even, and hence b must be even. Let b = 2k, where k is an integer.
Substituting this value of b in the equation b^2 = 2a^2, we get:
4k^2 = 2a^2
Simplifying this equation, we get:
2k^2 = a^2
This means that a^2 is even, and hence a must be even. But this contradicts our assumption that a and b are co-prime.
Therefore, our initial assumption that 1/√2 is rational must be false, and we can conclude that 1/√2 is irrational.
(ii) 7√5
To prove that 7√5 is irrational, we will assume the opposite, that 7√5 is a rational number. This means that we can write:
7√5 = a/b
where a and b are co-prime integers.
Squaring both sides, we get:
245/5 = a^2/b^2
Multiplying both sides by 5b^2, we get:
49b^2 = 5a^2
This means that 5a^2 is divisible by 49, and hence a^2 must be divisible by 49. Let a = 7k, where k is an integer.
Substituting this value of a in the equation 49b^2 = 5a^2, we get:
7b^2 = 5k^2
This means that 5k^2 is divisible by 7, and hence k^2 must be divisible by 7. But this contradicts the fact that 7 and 5 are co-prime.
Therefore, our initial assumption that 7√5 is rational must be false, and we can conclude that 7√5 is irrational.
(iii) 6 + √2
To prove that 6 + √2 is irrational, we will assume the opposite, that 6 + √2 is a rational number. This means that we can write:
6 + √2 = a/b
where a and b are co-prime integers.
Squaring both sides, we get:
38 + 12√2 = a^2/b^2
Multiplying both sides by b^2, we get:
38b^2 + 12b^2√2 = a^2
This means that a^2 is of the form 12k + 2, where k is an integer. But this contradicts the fact that the square of any integer is of the form 4k or 4k + 1.
Therefore, our initial assumption that 6 + √2 is rational must be false, and we can conclude that 6 + √2 is irrational.
Class 10 Maths Chapter 1 Exercise 1.4 Solutions
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:
(i) 13/3125
(ii) 17/8
(iii) 64/455
(iv) 15/1600
(v) 29/343
(vi) 23/(2352)
(vii) 129/(225775)
(viii) 6/15
(ix) 35/50
(x) 77/210
Solutions:
(i) 13/3125 has a terminating decimal expansion.
(ii) 17/8 has a non-terminating decimal expansion.
(iii) 64/455 has a non-terminating decimal expansion.
(iv) 15/1600 has a terminating decimal expansion.
(v) 29/343 has a non-terminating decimal expansion.
(vi) 23/(2352) has a non-terminating decimal expansion.
(vii) 129/(225775) has a non-terminating decimal expansion.
(viii) 6/15 can be simplified to 2/5 and has a terminating decimal expansion.
(ix) 35/50 can be simplified to 7/10 and has a terminating decimal expansion.
(x) 77/210 can be simplified to 11/30 and has a non-terminating decimal expansion.
2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solutions:
(i) 13/3125
13/3125 = 0.00416
(ii) 17/8
17/8 = 2.125
(iii) 64/455 has a non terminating decimal expansion
(iv)15/ 1600
15/1600 = 0.009375
(v) 29/ 343 has a non terminating decimal expansion
(vi)23/ (2352) = 23/(8×25)= 23/200
23/ (2352) = 0.115
(vii) 129/ (225775) has a non terminating decimal expansion
(viii) 6/15 = 2/5
(ix) 35/50 = 7/10
35/50 = 0.7
(x) 77/210 has a non-terminating decimal expansion.
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?
(i) 43.123456789
(ii) 0.120120012000120000. . .
Solution:
(i) The given number can be expressed as 43.123456789 = 43123456789/1000000000
Since 43123456789 and 1000000000 are both integers, the given number is rational.
The prime factors of the denominator 1000000000 are 2 and 5.
(ii) Let x = 0.120120012000120000. . .
Multiplying both sides by 10000, we get:
10000x = 1201.2001200012000. . .
Subtracting the two equations, we get:
9999x = 1201
x = 1201/9999
Since 1201 and 9999 are both integers, the given number is rational.
The prime factors of the denominator 9999 are 3 and 11.
Since it has non-terminating but repeating decimal expansion, it is a rational number in the form of p/q and q has factors other than 2 and 5.
NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers
The Chapter 1 Real Number is one of the important chapter in class 10 maths and it has a weightage of 6 marks for board exams. Usually 3 questions are asked from this chapter also some times its contain maximum questions.
This chapter talks about
- Euclid’s Division Algorithm
- The Fundamental Theorem of Arithmetic
- Revisiting Rational & Irrational Numbers
- Decimal Expansions
All Exercises of Class 10 Maths Chapter 1: Real Number
Exercise 1.1
Exercise 1.2
Exercise 1.3
Exercise 1.4
Class 10 Maths Chapter 1 Extra Questions
These extra Questions are prepared for class 10 math by our expert faculty.
- Prove that √7 is irrational.
- Find the HCF of 18, 24 and 30 using the prime factorization method.
- Simplify √125/√5.
- Prove that the product of three consecutive positive integers is divisible by 6.
- Find the LCM of 18 and 24.
- Show that the sum of three consecutive odd integers is always divisible by 3.
- Simplify (4 + √7)(4 – √7).
- Find the HCF and LCM of 12, 18 and 24 using the prime factorization method.
- Prove that the square of any odd integer is of the form 8m + 1 for some integer m.
- Simplify √48/√3.
Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
Q1: Are the NCERT solutions for Class 10 Maths Chapter 1 Real Numbers reliable?
Yes, the NCERT solutions for Class 10 Maths Chapter 1 Real Numbers are reliable as they are provided by subject experts who have a vast knowledge of the subject matter. These solutions are prepared in accordance with the latest CBSE syllabus and exam pattern.
Q2: Are the solutions easy to understand?
Yes, All solutions are explained in a step-by-step manner, making them easy to understand for students. The language used in the solutions is simple and easy to comprehend.
Q3: Can the NCERT solutions help me score good marks in my exams?
Yes, the NCERT solutions for Class 10 Maths Chapter 1 Real Numbers can definitely help you score good marks in your exams. These solutions cover all the important topics and concepts of the chapter and are designed to help students clear their doubts and strengthen their fundamentals.
Q4: Is it necessary to solve NCERT solutions for Class 10 Maths Chapter 1?
Yes, it is necessary to solve NCERT solutions for Class 10 Maths Chapter 1 as it helps students to understand the basic concepts of Real Numbers in a better way. It also helps them to prepare for their exams in a more effective and efficient manner.
Q5: Can I find the NCERT solutions for Class 10 Maths Chapter 1 Real Numbers online?
Yes, you can easily find the NCERT solutions for Class 10 Maths Chapter 1 Real Numbers online. There are many websites and online educational platforms that provide free access to these solutions.
