NCERT Textbook Exercise 1.1 Solutions: Chapter 1- Real Numbers, Class 10, Mathematics PDF Download

NCERT Mathematics Solutions

Maths Textbook Exercise 1.1 Solutions

(Page 7)

Q.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) We have,

a = bq + r

Applying division lemma to 225 and 135 we obtain,

225 = 135 x 1 + 90

and 135 = 90 x 1 + 45

and  90 = 45 x 2 + 0

Therefore, HCF of 225, 135 = 45 

(ii) We have,

a = bq + r

Applying division lemma to 196 and 38220 we obtain,

38220 = 196 x 195 + 0

Therefore, HCF of 196 and 38220 = 196

(iii) We have,

a = bq + r

Applying division lemma to 867 and 255 we obtain,

867 = 255 x 3 + 102

255 = 102 x 2 + 51

102 = 51 x 2 + 0

Therefore, HCF of 867 and 255 is 51.

Q.2: Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or6q + 5, where q is some integer.

Solution:

Let a be any positive number and a = 6. Then, by Euclid’s algorithm,

a = 6q + r (0 ≤ r < 6)

say, r = 0, 1, 2, 3, 4, 5

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

Also, 6q + 1 = 2 × 3q + 1 = 2k₁ + 1, where k₁ is a positive integer,

Similarly, 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k₂ + 1, where k₂ is a positive integer,

and, 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k₃ + 1, where k₃ is a positive integer.

From these we observe that 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1. So, these numbers are not divisible by 2 and hence, are odd positive integers.

Q.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:

To find the maximum number of columns, we have to find HCF of 616 and 32.

Applying Euclid’s algorithm to find the HCF we get,

616 = 32 × 19 + 8

32 = 8 × 4 + 0

Or, the HCF (616, 32) = 8.

Therefore, maximum number of column is 8.

Q.4: Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Solution:

a = bq + r;

Let a be any positive integer, b = 3 and r = 0, 1, 2 because 0 ≤ r < 3

Then a = 3q + r for some integer q ≥ 0

Therefore, a = 3q + 0 or 3q + 1 or 3q + 2

From the above we can say that the square of any positive integer is either of the form 3m or 3m + 1.

Q.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.

[Taking hint from the above do it yourselves]