## NCERT Solutions for Class 10 Maths Exercise 1.1 Chapter 1 Real Numbers â€“ 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 â€“ Real Numbers

1. Use Euclidâ€™s division algorithm to find the HCF of:

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

We have 225 > 135,

So, we apply the division lemma to 225 and 135 to obtain

Here remainder 90 â‰ 0, we apply the division lemma again to 135 and 90 to obtain

We consider the new divisor 90 and new remainder 45 â‰ 0, and apply the division lemma to obtain

Since at thisÂ time the remainder is zero, the process isÂ stopped.

The divisor at this stage is 45

Therefore, the HCF of 135 and 225 is 45.

(ii)Â 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)Â 867 and 255

We have 867 > 255,

So, we apply the division lemma to 867 and 255 to obtain

Here remainder 102 â‰ 0, we apply the division lemma again to 255 and 102 to obtain

Here remainder 51 â‰ 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.

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

**Ans.**Â (i)Â 135 and 225

We have 225 > 135,

So, we apply the division lemma to 225 and 135 to obtain

Here remainder 90 â‰ 0, we apply the division lemma again to 135 and 90 to obtain

We consider the new divisor 90 and new remainder 45 â‰ 0, and apply the division lemma to obtain

Since at thisÂ time the remainder is zero, the process isÂ stopped.

The divisor at this stage is 45

Therefore, the HCF of 135 and 225 is 45.

(ii)Â 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)Â 867 and 255

We have 867 > 255,

So, we apply the division lemma to 867 and 255 to obtain

Here remainder 102 â‰ 0, we apply the division lemma again to 255 and 102 to obtain

Here remainder 51 â‰ 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, whereÂ

Therefore,Â

Also, 6

6

6

Clearly, 6

Therefore, 6

And therefore, any odd integer can be expressed in the form 6

or 6

*q*Â is some integer.**Ans.**Â LetÂ

*a*Â be any positive integer andÂ

*b*Â = 6. Then, by Euclidâ€™s algorithm,

*a*Â = 6

*q*Â +Â

*rÂ*for some integerÂ

*q*Â â‰¥Â 0, andÂ

*r*Â = 0, 1, 2, 3, 4, 5 because 0Â â‰¤Â

*r*Â < 6.

Therefore,Â

*a*Â = 6

*q*Â or 6

*q*Â + 1 or 6

*q*Â + 2 or 6

*q +Â*3 or 6

*q*Â + 4 or 6

*q*Â + 5

Also, 6

*q*Â + 1 =Â = , where

*Â*Â is a positive integer

6

*q*Â + 3 = (6

*q*Â + 2) + 1 = 2 (3

*q*Â + 1) + 1 = , where

*Â*Â is an integer

6

*q*Â + 5 = (6

*q*Â + 4) + 1 = 2 (3

*q*Â + 2) + 1 = , where

*Â*Â is an integer

Clearly, 6

*q*Â + 1, 6

*q*Â + 3, 6

*q*Â + 5 are of the form 2

*k*Â + 1, whereÂ

*k*Â is an integer.

Therefore, 6

*q*Â + 1, 6

*q*Â + 3, 6

*q*Â + 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 6

*q*Â + 1, or 6

*q*Â + 3,

or 6

*q*Â + 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?

To find the HCF, we can use Euclidâ€™s algorithm.

Since, the last divisor is 8.

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

Therefore, they can march in 8 columns each.

**Ans.**Â 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 Euclidâ€™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 Euclidâ€™s division lemma to show that the square of any positive integer is either of form 3

ThenÂ

AndÂ

Therefore,Â

Or,

WhereÂ Â are some positive integers.

Hence, it can be said that the square of any positive integer is either of the form 3

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

*a*Â be any positive integer andÂ

*b*Â = 3.

ThenÂ

*a*Â = 3

*q*Â +Â

*r*Â for some integerÂ

*q*Â â‰¥ 0

AndÂ

*r*Â = 0, 1, 2 because 0 â‰¤Â

*r*Â < 3

Therefore,Â

*a*Â = 3

*q*Â or 3

*q*Â + 1 or 3

*q*Â + 2

Or,

WhereÂ Â are some positive integers.

Hence, it can be said that the square of any positive integer is either of the form 3

*m*Â or 3

*m*Â + 1.

5. Use Euclidâ€™s division lemma to show that the cube of any positive integer is of the form 9

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

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

We have three cases.

Case 1: WhenÂ

WhereÂ

Case 2: WhenÂ

Where

Case 3: WhenÂ

WhereÂ

Therefore, the cube of any positive integer is of the form 9

*m*, 9*mÂ*+ 1 or 9*m +Â*8.**Ans.**Â LetÂ

*a*Â be any positive integer andÂ

*b*Â = 3

*a =*Â 3

*q*Â +Â

*r*, whereÂ

*qÂ*â‰¥ 0 andÂ r=0,1,2r=0,1,2Â because 0 â‰¤Â

*r*Â < 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: WhenÂ

*a = 3q*,

WhereÂ

*m*Â is an integer such thatÂ

*m*Â =3q

^{3}

Case 2: WhenÂ

*a*Â = 3

*q*Â + 1,

Where

*Â m*Â is an integer such thatÂ

Case 3: WhenÂ

*a*Â = 3

*q*Â + 2,

WhereÂ

*m*Â is an integer such thatÂ

Therefore, the cube of any positive integer is of the form 9

*m*, 9

*mÂ*+ 1, or 9

*m +Â*8.