  # NCERT Solutions class 10 Maths Exercise 1.1 Ch 1 Real Numbers

## NCERT Solutions for Class 10 Maths Exercise 1.1 Chapter 1 Real Numbers – FREE PDF Download

NCERT Class 10 Maths Ch 1 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter 1 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Real Numbers solutions will help you understand the chapter thoroughly.

# NCERT Solutions for Class 10 Maths Chapter 1 – Real Numbers

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

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Ans. (i) 135 and 225

We have 225 > 135,

So, we apply the division lemma to 225 and 135 to obtain Here remainder 90 ≠ 0, we apply the division lemma again to 135 and 90 to obtain We consider the new divisor 90 and new remainder 45 ≠ 0, and apply the division lemma to obtain Since at this time the remainder is zero, the process is stopped.

The divisor at this stage is 45

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

We have 38220 > 196,

So, we apply the division lemma to 38220 and 196 to obtain As the remainder is zero, the process stops.

The divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

We have 867 > 255,

So, we apply the division lemma to 867 and 255 to obtain Here remainder 102 ≠ 0, we apply the division lemma again to 255 and 102 to obtain Here remainder 51 ≠ 0, we apply the division lemma again to 102 and 51 to obtain As the remainder is zero, the process stops.

The divisor at this stage is 51,

Therefore, HCF of 867 and 255 is 51.

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

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

a = 6q + for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

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

Also, 6q + 1 = = , where is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = , where is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = , where is an integer

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 expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,

or 6q + 5

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?

Ans. We have to find the HCF (616, 32) to find the maximum number of columns in which they can march.

To find the HCF, we can use Euclid’s algorithm.  Since, the last divisor is 8.

Therefore, the HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

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. [Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Ans. Let a be any positive integer and b = 3.

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

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or, Where are some positive integers.

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

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9+ 1 or 9m + 8.

Ans. Let a be any positive integer and b = 3

a = 3q + r, where ≥ 0 and r=0,1,2r=0,1,2 because 0 ≤ r < 3

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

Therefore, every number can be represented as these three forms.

We have three cases.

Case 1: When a = 3q, Where m is an integer such that m =3q3

Case 2: When a = 3q + 1,    Where m is an integer such that Case 3: When a = 3q + 2,    Where m is an integer such that Therefore, the cube of any positive integer is of the form 9m, 9+ 1, or 9m + 8.