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# Exercise 1.1 Class 10 maths 1. Real Numbers - ncert solutions - Toppers Study

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## Exercise 1.1 Class 10 maths 1. Real Numbers - ncert solutions - Toppers Study

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### Exercise 1.1 class 10 Mathematics Chapter 1. Real Numbers

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- Exercise 1.1 Class 10 Maths 1. Real Numbers - Ncert Solutions - Toppers Study
- Class 10 Ncert Solutions
- Solution Chapter 1. Real Numbers Class 10
- Solutions Class 10
- Chapter 1. Real Numbers Exercise 1.1 Class 10

## 1. Real Numbers

### | Exercise 1.1 |

## Exercise 1.1 Class 10 maths 1. Real Numbers - ncert solutions - Toppers Study

### Exercise 1.1 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

**Sol: **

**(1) **135 and 225

*a = 225, b = 135 {Greatest number is ‘a’ and smallest number is ‘b’}*

* Using Euclid’s division algorithm*

* a = bq + r (then)*

* 225 = 135 ×1 + 90*

* 135 = 90 ×1 + 45*

* 90 = 45 × 2 + 0 {when we get r=0, our computing get stopped}*

* b = 45 {b is HCF}*

*Hence: HCF = 45*

**Sol:**

(ii) 196 and 38220

* a = 38220, b = 196 {Greatest number is ‘a’ and smallest number is ‘b’}*

* Using Euclid’s division algorithm*

* a = bq + r (then)*

* 38220= 196 ×195 + 0 {when we get r=0, our computing get stopped} *

* b = 196 {b is HCF}*

*Hence: HCF = 196*

**Sol:**

(iii) 867 and 255

*a = 867, b = 255 {Greatest number is ‘a’ and smallest number is ‘b’}*

*Using Euclid’s division algorithm*

*a = bq + r (then)*

*38220= 196 ×195 + 0 {when we get r=0, our computing get stopped} *

*b = 196 {b is HCF}*

*Hence: HCF = 196*

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

Sol:

Let *a* is the positive odd integer

Where b = 6,

When we divide *a* by 6 we get reminder 0, 1, 2, 3, 4 and 5, {r < b}

Here *a *is odd number then reminder will be also odd one.

We get reminders 1, 3, 5

*Using Euclid’s division algorithm*

*So we get *

*a = 6q + 1, 6q+3 and 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?

Sol:

Maximum number of columns = HCF (616, 32)

* a = 616, b = 32 {Greatest number is ‘a’ and smallest number is ‘b’}*

* Using Euclid’s division algorithm*

* a = bq + r (then)*

* 616 = 32 ×19 + 8 {when we get r=0, our computing get stopped} *

* 32 = 8 × 4 + 0*

* b = 8 {b is HCF}*

* HCF = 8*

* Hence:* Maximum number of columns = 8

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

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

**Solution : **

To Show :

a^{2} = 3m or 3m + 1

*a = bq + r *

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

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

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

Now we have;

⇒ a^{2} = (3q + 0)^{2} or (3q + 1)^{2} or (3q +2)^{2}

⇒ a^{2} = 9q^{2} or 9q^{2} + 6q + 1 or 9q^{2} + 12q + 4

⇒ a^{2} = 9q^{2} or 9q^{2} + 6q + 1 or 9q^{2} + 12q + 3 + 1

⇒ a^{2} = 3(3q^{2}) or 3(3q^{2} + 2q) + 1 or 3(3q^{2} + 4q + 1) + 1

Let m = (3q^{2}) or (3q^{2} + 2q) or (3q^{2} + 4q + 1)

Then we get;

*a ^{2} = 3m or 3m + 1 or 3m + 1 *

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

**Solution: **

Let , a is any positive integer

By using Euclid’s division lemma;

*a = bq + r where; *0 ≤ *r* < b

Putting b = 9

*a = 9q + r where; *0 ≤ *r* < 9

when r = 0

*a = 9q + 0 = 9q *

*a ^{3} = (9q)^{3} = 9(81q^{3}) or 9m where m = 81q^{3}*

when r = 1

*a = 9q + 1 *

*a ^{3} = (9q + 1)^{3} = 9(81q^{3} + 27q^{2} + 3q) + 1*

* = 9m + 1 where m = 81q ^{3} + 27q^{2} + 3q*

when r = 2

*a = 9q + 2 *

*a ^{3} = (9q + 2)^{3} = 9(81q^{3} + 54q^{2} + 12q) + 8*

* = 9m + 2 where m = 81q ^{3} + 54q^{2} + 12q*

⇒ The End

##### Other Pages of this Chapter: 1. Real Numbers

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