Study Materials

NCERT Solutions for Class 10th Mathematics

Page 1 of 4

Chapter 1. Real Numbers

Exercise 1.1

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

Page 1 of 4

Chapter Contents: