Gujarat Board GSEB Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.1

Question 1.

Use Euclid’s division algorithm to find the HCF of

- 135 and 225
- 196 and 38220
- 867 and 255

Solution:

1. Let a = 225 and b = 135

Since a > b

Applying Euclid division Algorithm

a = bq + r (where 0 ≤ b < r)

225 = 135 × 1 + 90

Dividend = 225

Divisor = 135

Since 90 is the remainder but 90 ≠ 0. So we apply Euclid division lemma to 135 and 90

a = bq + r (Here a = 135 and b = 90)

135 = 90 × 1 + 45

Since remainder 45 ≠ 0

So again, we apply Euclid division lemma to 90 and 45

a = bq + r (Here a = 90 and 6 = 45)

90 = 45 × 2 + 0

Since remainder is zero, the process stops. Since the divisor at this stage is 45, therefore HCF of 135 and 225 is 45.

2. Let a = 38220 and b = 196

Since a > b

Applying Euclid division lemma,

a = bq + r (where 0 ≤ r < b)

38220 = 196 × 195 + 0

Since remainder is zero, at this stage our process stops and divisor is 196,

Therefore 196 is the HCF.

3. Let a = 867 and b = 255

Since a > b

Applying Euclid division lemma

a = bq + r (where 0 ≤ b < r)

867 = 255 × 3 + 102

Since 102 ≠ 0, we apply Euclid division lemma to a = 255 and b = 102 we get

a = bq + r (where 0 ≤ b <r)

255 = 102 × 2 + 51

Since 51 ≠ 0 we apply Euclid division lemma to a = 102 and 6 = 51

a = bq + r

102 = 51 × 2 + 0

Since remainder is zero, our process stops. At this stage divisor is 51, therefore HCF is 51.

Question 2.

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

Solution:

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

a = 6q + r (0 ≤ r < 6) for some integer q > 0, the values of remainder must be less than 6, i.e.,

r = 0, 1, 2, 3, 4, 5

We take only odd numbers like 1, 3 and 5.

For r = 0, a = 6q + r = 6q + 0

⇒ a = 6q

⇒ a = 2(3q) = 3m

where m is any positive integer

∴ a = 6q (not odd)

Similarly,

for r = 2, a = 6q + 2 = 2 (3q + 1) (not odd)

for r = 4, a = 6q + 4 = 2 (3q + 2) (not odd)

for r = 6, a = 6q + 6 = 2 (3q + 3) (not odd)

But, for r = 1

a = 6q + 1

for r = 3

a = 6q + 3

and, for r = 5

a = 6q + 5

Thus, for r = 1, 3 and 5, a is odd.

So we take only odd value. Therefore any positive integer of the form of 617 + 1, 6q + 3 and 6q + 5 is odd.

Question 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?

Solution:

HCF of 616 and 32 will give the maximum number of columns in which they can march. Now we use Euclid’s Algorithm to find HCF.

Let a = 616 and b = 32

a = bq + r

616 = 32 × 19 + 8

Since 8 ≠ 0 we again apply Euclid’s Algorithm

a = bq + r

32 = 8 × 4 + 0

Since remainder is zero.

Hence the HCF of 616 and 32 is 8.

Therefore an army contingent can march in 8 columns each.

Question 4.

Use Euclid division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.(CBSE)

[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 of (3m or 3m + 1)]

Solution:

Let a be any positive integer and b = 3

By applying Euclid’s Algorithm

a = bq + r (0 ≤ r < 6)

then a = 3q + r (0 ≤ r < 3)

Values of r = 0, 1 and 2

when r = 0

then a = 3q + 0

a = 3q

when r = 1

then a = 3q + 1

when r = 2

then a = 3q + 2

Squaring of these positive integers as:

1. a = 3q

a^{2} = (3q)^{2}

a^{2} = 9q^{2}

a^{2} = 3 × (3q^{2})

a^{2} = 3m

Thus, 9q^{2} = 3 (3q^{2}) is divisible by 3

2. a = 3q + 1

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

a^{2} = 9q^{2} + 6q + 1

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

(where, m = 3q^{2} + 2q)

= 3m + 1

Thus, a^{2} is written in the form of 3m + 1.

a = 3q + 2

a^{2} = (3q + 2)^{2}

a^{2} = 9q^{2} + 12q + 4

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

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

a^{2} = 3m + 1

where m = 3q^{2} + 4q + 1

Thus, a^{2} is written in the form of 3m + 1.

Question 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.(CBSE)

Solution:

Let a be any positive integer and b = 3

By Applying Euclid’s division lemma

a = bq + r (0 ≤ r < b)

then a = 3q + r (0 ≤ r < 3)

Values of r = 0, 1, 2

when r = 0

then a = 3q + 0

a = 3q

Cubing both sides

a^{3} = (3q)^{3}

a^{3} = 27q^{3}

a^{3} = 9 (3q^{3}) [27q^{3} is divisible by 9]

a^{3} = 9m

when r = 1

a = 3q + 1

Cubing both sides

a^{3} = (3 q + 1)^{3}

= 27q^{3} + 9q^{2} + 9q + 1

= 9 (3q^{3} + q^{2} + q) + 1

[27q^{3} + 9q^{2} + 9q is divisible by 9]

a^{3} = 9m + 1

when r = 2

a = 3q + 2

Cubing both sides

a^{3} = (3q + 2)^{3}

= 27q^{3} + 54q^{2} + 36q + 8

= 9 (3q^{3} + 6q^{2} + 4q) + 8

[27q^{3} + 54q^{2} + 36q is divisible by 9]

a^{3} = 9m + 8

Therefore the cube of any positive integer is 9m, 9m + 1 or 9m + 8.