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

Complete solutions of Ex 1.1 Class 10 Maths Chapter 1 with additional questions and answers from new NCERT syllabus textbook Class 10 Maths.

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

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

You can also Download NCERT Solutions for class 10 Maths in Hindi to help you to revise complete Syllabus and score more marks in your examinations.

Ex 1.1 Class 10 Maths 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.

Ex 1.1 Class 10 Maths 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.

Ex 1.1 Class 10 Maths 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.

Ex 1.1 Class 10 Maths 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.

Ex 1.1 Class 10 Maths 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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