# NCERT Solutions for CBSE Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

Here is the solution of NCERT Class 10 Maths Chapter 1 Real Number Exercise 1.1. There are 5 Questions in Class 10 NCERT Maths Chapter 1 Real Numbers Exercise 1.1. All 5 Questions under Exercise 1.1 are based on Euclid's Division Lemma. You can go through each one of them one by one.

Question 1: Use Euclid’s division algorithm to find the HCF of:

(i)135 and 225

Solution 1 (i): Since 225 > 135,

225 = 135 × 1 + 90

Also,

135 = 90 × 1 + 45

90 = 2 × 45 + 0

Therefore, the HCF of 135 and 225 is 45.

(ii)196 and 38220

Solution 1 (ii): Since 38220 > 196,

38220 = 196 × 195 + 0

Therefore, HCF of 196 and 38220 is 196.

(iii)867 and 255

Solution 1 (iii): Since 867 > 255,

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0

Therefore, HCF of 867 and 255 is 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 2: By Euclid’s algorithm,

a = 6q + r, and r = 0, 1, 2, 3, 4, 5

Hence, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.

Hence, these numbers are odd numbers.

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 3: Euclid’s algorithm

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF of (616, 32) is 8.

Therefore, the maximum number of columns in which they can march 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 4: a = 3q + r, r = 0, 1, 2

Therefore, a= 3q or 3q + 1 or 3q + 2

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.

Question 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 5: a = 3q + r, 0 ≤ r < 3

a = 3q, 3q + 1, 3q + 2

There are three cases.

Case 1: When a = 3q,

a3 = (3q)3 = 27q3 = 9(3q3) = 9m

Case 2: When a = 3q + 1,

a3 = (3q + 1)3

a3 =27q3 + 27q2 + 9q + 1

a3 = 9(3q3 + 3q2 + q) + 1

a3 = 9m+ 1

Case 3: When a = 3q + 2,

a= (3q + 2)3

a= 27q+ 54q+ 36q + 8

a3= 9(3q+ 6q+ 4q) + 8

a3=9m+ 8

Therefore, the cube of any positive integer is of the form 9m, 9m + 1,or 9m + 8.