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

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

• Real Numbers Class 10 Ex 1.1
• Real Numbers Class 10 Ex 1.2
• Real Numbers Class 10 Ex 1.3
• Real Numbers Class 10 Ex 1.4

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 1
Chapter Name	Real Numbers
Exercise	Ex 1.1
Number of Questions Solved	5
Category	NCERT Solutions

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

Question 1.
Use Euclid’s Division Algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) By Euclid’s Division Algorithm, we have
225 = 135 x 1 + 90 135
= 90 x 1 + 45 90
= 45 x 2 + 0
∴ HCF (135, 225) = 45.

(ii) By Euclid’s Division Algorithm, we have
38220 = 196 x 195 + 0
196 = 196 x 1 + 0
∴  HCF (38220, 196) = 196.

(iii) By Euclid’s Division Algorithm, we have
867 = 255 x 3 + 102
255 = 102 x 2 + 51
102 = 51 x 2 + 0
∴ HCF (867, 255) = 51.

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.
Solution:
Let a be a positive odd integer. Also, let q be the quotient and r the remainder after dividing a by 6.
Then, a = 6q + r, where 0 ≤ r < 6.
Putting r = 0, 1, 2, 3, 4, and 5, we get:
a = 6q, a = 6q + 1, a = 6q + 2, a = 6q + 3, a = 6q + 4, a = 6q + 5
But a = 6q, a = 6q + 2 and a = 6q + 4 are even.
Hence, when a is odd, it is of the form 6q + 1, 6q + 3, and 6q + 5 for some integer q.
Hence proved.

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?
Solution:
Let n be the number of columns such that the value of n be maximum and it must divide both the numbers 616 and 32.
Then, n = HCF (616, 32)
By Euclid’s Division Algorithm, we have:
616 = 32 x 19 + 8 32 = 8 x 4 + 0
∴ HCF (616, 32) = 8
i. e., n = 8
Hence, the maximum number of columns is 8.

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.
Solution:
Let a be a positive integer, q be the quotient and r be the remainder.
Dividing a by 3 using the Euclid’s Division Lemma,
we have:
a = 3q + r, where 0 ≤ r < 3
Putting r = 0, 1 and 2, we get:
a = 3q
⇒ a² = 9q²
= 3 x 3q²
= 3m (Assuming m = q²)
Then, a = 3q + 1
⇒  a² = (3q + l)² = 9q² + 6q + 1
= 3(3q ² + 2q) + 1
= 3m + 1 (Assuming m = 3q² + 2q)
Next, a = 3q + 2
⇒ a² = (3q + 2)² =9q² + 12q + 4
= 9q² + 12q + 3 + 1
= 3(3q² + 4q + 1) + 1
= 3m + 1.   (Assuming m = 3q² + 4q+l)
Therefore, the square of any positive integer (say, a²) is always of the form 3m or 3m + 1.
Hence, proved.

Question 5.
Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let a be a positive integer, q be the quotient and r be the remainder.
Dividing a by 3 using the Euclid’s Division Algorithm, we have,
a = 3q + r, where 0 ≤ r < 3
Putting r = 0, 1 and 2, we get:
a = 3q, a = 3q + 1 and a = 3q + 2
If a = 3q, then a³ = 27q³ = 9(3q³) = 9m. (Assuming m = 3q³.)
If a = 3q + 1, then
a³ = (3q + l)³ = 27q³ + 9q(3q + 1) + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1,  (Assuming m = 3q³ + 3q² + q)
If a = 3q + 2, then a³ = (3q + 2)³
= 27q³ + 18q(3q + 2) + (2)³
= 9(3q³ + 6q² + 4q) + 8
= 9m + 8, (Assuming m – 3q³ + 6q² + 4q)
Hence, a³ is of the form 9m, 9m + 1 or 9m + 8.

 

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