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

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

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Exercise 1.1 class 10 maths Chapter 1. Real Numbers

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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 6q + 1, or 6q + 3, or 6q + 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 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.]

 

Solution :

To Show :

a² = 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 = 3q + r for some integer q ≥ 0

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

Now we have;

⇒ a² = (3q + 0)² or (3q + 1)² or (3q +2)²

⇒ a² = 9q² or 9q² + 6q + 1 or 9q² + 12q + 4

⇒ a² = 9q² or 9q² + 6q + 1 or 9q² + 12q + 3 + 1

⇒ a² = 3(3q²) or 3(3q² + 2q) + 1 or 3(3q² + 4q + 1) + 1

Let m = (3q²) or (3q² + 2q)  or (3q² + 4q + 1)

Then we get;

a² = 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 9m, 9m + 1 or 9m + 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

a³  = (9q)³ = 9(81q³) or 9m where m = 81q³

when r = 1

a = 9q + 1 

a³  = (9q + 1)³ = 9(81q³ + 27q² + 3q) + 1

      = 9m + 1  where m = 81q³ + 27q² + 3q

when r = 2

a = 9q + 2 

a³  = (9q + 2)³ = 9(81q³ + 54q² + 12q) + 8

      = 9m + 2  where m = 81q³ + 54q² + 12q

⇒ The End