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NCERT Questions for Class 10 Maths Chapter 1 – Real Numbers
One of the more foundational chapters within Class 10 Mathematics, is the chapter known as Real Numbers. It discusses concepts that may be not just essential for one’s board exam but also beneficial in higher education. Mastering this chapter results in clear learning on topics that include Euclid’s division lemma, the Fundamental Theorem of Arithmetic, and applications of prime factorization. We have gathered critical questions on Class 10 Real Numbers to help you work on your practice and perform even better in examinations. These have been designed and presented in several formats, be it short answer and long answer or problem-based applications, and hence will bring about complete readiness.
Important Questions with Solutions of Class 10 Maths Chapter 1 – Real Numbers
1) Find the decimal expansion of the following rational numbers.
(i) 13/3125
(ii) 17/8
(iii) 15/1600
(iv) 23/23 x 52
(v) 6/15
(vi) 35/50
Ans: (i) The decimal expansion of
(ii) The decimal expansion of
(iii) The decimal expansion of
(iv) The decimal expansion of
(v) The decimal expansion of
(vi) The decimal expansion of
2) Prove that each positive odd number may be expressed in the manner 6q + 1, 6q + 3, or 6q + 5, where q is an integer.
Ans – Using Euclid’s algorithm, we can write a = 6q + r for some integer q ≥ 0 if an is any odd positive integer. …(1)
In formula (1), we divide a by 6, yielding a quotient q and a residual r, where r can be 1, 3, or 5, since 0 ≤ r < 6 and a is odd. …(2)
As a result, from (1) and (2), we derive that a = 6q + 1, 6q + 3, or 6q + 5.
Thus, any odd integer can be represented as 6q + 1, 6q + 3, or 6q + 5.
3) Prove that (n2 − 1) is divisible by 8, provided n is an odd positive integer.
Ans – Let us assume n = 4q + 1 be an odd positive integer. …..(1)
From (1) we get,
n2 − 1 = (4q + 1)2−1
⇒n2 − 1 = 16q2 + 8q …..(2)
By extracting 8 common from RHS of equation (2) we obtain,
n2 − 1 = 8(2q2 + q)
⇒n2 − 1 = 8m where m = 2q2 + q. …..(3)
Therefore, from (3), we can conclude that (n2 − 1) is divisible by 8 when n is an odd positive integer.
4) Show that any positive integer’s cube has the form 9m, 9m + 1, or 9m + 8 using Euclid’s division lemma.
Ans – Let a be any positive integer; by Euclid’s technique, we may express an as a = 3q + r, where q is a non-negative integer. …..(1)
In formula (1), we divide a by 3, yielding a quotient q and a residual r, where r can be 0, 1, or 2, as 0 ≤ r < 3. …..(2)
Thus, from (1) and (2), we may derive,
a = 3q, 3q + 1, or 3q + 2 … (3)
By cubing both sides of equation (3), we obtain,
Case 1: When a=3q,
a3 = 27q3
⇒ a3 = 9(3q3)
a3 = 9m
Let m be an integer defined as m = 3q3.
Case 2: When a = 3q + 1
a3 = (3q + 1)3
a3 = 9q3 + 9q2 + 9q + 1
⇒ a3 = 9(q3 + q2 + q) + 1
⇒ a3 = 9m + 1
Let m be an integer defined as m = (3q3 + 3q2 + q).
Case 3: When a = 3q + 2
a3 = (3q + 2)3
a3 = 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Let m be an integer defined as m = (3q3 + 6q2 + 4q).
Hence, given cases 1, 2, and 3, we conclude that the cube of every positive integer takes the form 9m, 9m + 1, or 9m + 8.
5) Determine the LCM and HCF of the specified pairs of integers and confirm that LCM × HCF equals the product of the two numbers.
(i) 26 and 91
Ans – First, let’s determine the HCF of the two provided numbers, 26 and 91.
The prime factorisation of the two numbers is as follows:
26 = 2 × 13 and 91 = 7 × 13 … (1)
HCF = 13 because 13 is a prime factor of both numbers.
From (1) we get, LCM = 2 × 7 × 13 = 182. …(3)
The product of the two numbers, 26 and 91, is 2366. …(4)
From (2) and (3), the product of HCF and LCM =13×182=2366. …..(5)
Therefore, given (4) and (5), the product of two numbers equals the HCF×LCM.
(ii) 510 and 92
Ans – First, let’s determine the HCF of the two provided numbers, 510 and 92
The prime factorisation of the two numbers is as follows: 510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23. …..(1)
HCF = 2 because both numbers have the same prime factorisation, which is 2.
From (1) we get, LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460. …..(3)
Product of two numbers, 510 × 92 = 46920. …..(4)
From (2) and (3), the product of HCF and LCM = 2 × 23460 = 46920.
Therefore, given (4) and (5), the product of two numbers equals the HCF × LCM.
(iii) 336 and 54
Ans – First, let’s determine the HCF of the two provided numbers, 336 and 54.
The prime factorisation of the numbers yields:
336 = 2 × 2 × 2 × 2 × 3 × 7 and 54 = 2 × 3 × 3 × 3 …..(1)
HCF = 2×3 = 6 because 2 and 3 are common prime factors.
From (1) we get, LCM = 24 × 33 × 7 = 3024. …..(3)
Product of two numbers = 336 × 54 = 18144 …..(4)
From (2) and (3), the product of the HCF and LCM = 6 × 3024 = 18144. …..(5)
Therefore, given (4) and (5), the product of two numbers equals the HCF × LCM.
6) Demonstrate that the square of any positive integer is either 4m or 4m + 1 for a given integer m.
Ans – Let a be any positive integer. Then, using Euclid’s algorithm, we can express a = 4q + r for some integer q ≥ 0. …..(1)
In equation (1), we divide a by 4, yielding a quotient q and a remainder r, where r can be 0, 1, 2, or 3, as 0 ≤ r < 4. …..(2)
As a result, from (1) and (2), we may derive,
a = 4q or 4q + 1 or 4q + 2 or 4q + 3 …..(3)
Case 1: If a = 4q, then squaring both sides gives,
a2 = 16q2
⇒a2 = 4(4q2) …..(4)
Equation (4) can be expressed as a2 = 4m for some integer m. ….. (5)
Case 2: If a = 4q+1, then squaring both sides gives,
a2 = 16q2 + 8q + 1
⇒a2 = 4(4q2 + 2q) + 1 …..(6)
Equation (6) can be expressed as a2 = 4m + 1 for some integer m ….. (7)
Case 3: If a = 4q + 2, then squaring both sides gives,
a2 = 16q2 + 16q + 4
⇒a2 = 4(4q2 + 4q + 1) …..(8)
Equation (8) can be expressed as a2 = 4m for some integer m ….. (9)
Case 4: If a = 4q + 3, then squaring both sides gives,
a2 = 16q2 + 24q + 9
a2 = 4(4q2 + 6q + 2) + 1 …..(10)
Equation (10) can be expressed as a2 = 4m + 1 for some integer m ….. (11)
As a result, it is possible to conclude that the square of any positive integer is either 4m or 4m + 1 for some integer m, as indicated by equations (5), (7), (9) and (11).
7) Demonstrate that the product of three consecutive positive integers is divisible by 6.
Ans – Let the three consecutive positive integers be x, (x + 1) and (x + 2).
Applying Euclid’s lemma, we can express x as x = 6q+r …..(1)
In equation (1), we divide a by 6, yielding a quotient q and a remainder r, where r can be 0, 1, 2, 3, 4, or 5, as 0 ≤ r < 6. …..(2)
Thus from (1) and (2) we can derive, x = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5
Case 1: If x = 6q then the product will be
x(x + 1)(x + 2) = 6q(6q + 1)(6q + 2), which is divisible by 6. …..(3)
Case 2: If x = 6q + 1 then product
x(x + 1)(x + 2) = (6q + 1)(6q + 2)(6q + 3)
⇒x(x + 1)(x + 2) = 2(3q + 1).3(2q + 1)(6q + 1)
⇒x(x + 1)(x + 2) = 6(3q+1).(2q + 1)(6q + 1)
which is divisible by 6. …..(4)
Case 3: If x = 6q + 2 then product
x(x + 1)(x + 2) = (6q + 2)(6q + 3)(6q + 4)
⇒x(x + 1)(x + 2) = 3(2q + 1).2(3q + 1)(6q + 4)
⇒x(x + 1)(x + 2) = 6(2q + 1)(3q + 1)(6q + 4)
which is divisible by 6. …..(5)
Case 4: If x = 6q + 3 then product
x(x + 1)(x + 2) = (6q + 3)(6q + 4)(6q + 5)
⇒x(x + 1)(x + 2) = 6(2q + 1)(3q + 2)(6q + 5)
which is divisible by 6. …..(6)
Case 5: If x = 6q + 4 then product
x(x + 1)(x + 2) = (6q + 4)(6q + 5)(6q + 6)
⇒x(x + 1)(x + 2) = 6(6q + 4)(6q + 5)(q + 1)
which is divisible by 6. …..(7)
Case 6: If x = 6q + 5 then product
x(x + 1)(x + 2) = (6q + 5)(6q + 6)(6q + 7)
⇒x(x + 1)(x + 2) = 6(6q + 5)(q + 1)(6q + 7)
which is divisible by 6. …..(8)
Equations (3) to (8) show that the product of three consecutivepositive numbers is divisible by 6.