#### class ten maths chapter 1 real number NCERT Solutions

## Chapter 1: Real Number

**TEXTBOOK’S 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. **(i) In 135 and 225, 225 is a larger integer.

Using Euclid’s division algorithm,

225 = 135 × 1 + 90

[where 135 is the divisor, 90 is remainder]

Since, remainder 90 ≠ 0, we apply Euclid’s division algorithm to 135 and 90

∴ 135 = 90 × 1 + 45

Again since, remainder 45 ≠ 0, we apply Euclid’s division algorithm to 90 and 45

∴ 90 = 45 × 2 + 0

Now, the remainder is zero so, our procedure stops.

Hence, HCF of 135 and 225 is 45.

(ii) In 196 and 38220, 38220 is a larger integer.

Using Euclid’s division algorithm,

38220 = 196 × 195 + 0

The remainder is zero, so our procedure stops.

Hence, HCF of 196 and 38220 is 196.

(iii) In 867 and 255, 867 is a larger integer.

Using Euclid’s division algorithm

867 = 255 × 3 + 102

[where 255 is the divisor, 102 is remainder]

Since, remainder 102 ≠ 0, we apply Euclid’s division algorithm to 255 and 102

∴ 255 = 102 × 2 + 51

Since, remainder 51 ≠ 0, we apply Euclid’s division algorithm to 102 and 51

So, 102 = 51 × 2 + 0

Remainder = 0

The remainder is zero, so our procedure stops.

Hence, HCF of 867 and 255 and 51.

**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 be any positive odd integer. We
apply Euclid’s division algorithm with a
and b = 6.

a = 6q + r, 0 ≤ r < 6

Since 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.

i.e., a can be 6q or 6q + 1 or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is the quotient. But, a is odd, so a cannot be equal to 6q, 6q + 2, 6q + 4, which are even numbers.

⇒ Any odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5. **Proved. **

**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. **Total members in the army are 616 and 32.

616 is larger integer i.e., 616 > 32

Using Euclid’s division algorithm,

616 = 32 × 19 + 8 [where 32 is divisor, 8 is remainder]

Since, remainder 8 ≠ 0, by applying Euclid’s division algorithm, we get

32 = 8 × 4 + 0

The remainder is zero,

**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.

**Sol.
**Let x be any
positive integer, then it is of the form 3q, 3q + 1 or 3q + 2.

If x = 3q

⇒ (x)^{2} =
(3q)^{2} [Squaring both
sides]

⇒ x^{2} = 9q^{2} = 3(3q^{2}) = 3m, where m = 3q^{2 }

Hence, x^{2} = 3m …(i)

If x = 3q + 1

⇒ x^{2} =
(3q + 1)^{2} [Squaring
both sides]

⇒ x^{2} = 9q^{2} + 1 + 2 × 3q × 1

⇒ x^{2} =
3(3q^{2} + 2q) + 1 = 3m + 1,

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

⇒ x^{2} = 3m + 1 …(ii)

If x = 3q + 2

⇒ x^{2} =
(3q + 2)^{2} = 9q^{2} + 4 + 12q

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

⇒ x2 = 3m + 1 …(iii)

From (i) , (ii), and (iii), we get

x2 = 3m, 3m + 1

∴ Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
**Proved. **

**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.

**Sol.
**Let a be any
positive integer, then it is of the form 3q, 3q + 1 or 3q + 2.

If a = 3q,

Then a^{3} = 27q^{3} [Cubing both sides]

⇒ a^{3} = 9
× 3q^{3} = 9m …(i)

where, m = 3q^{3}

If a = (3q + 1)

⇒ a^{3} =
(3q + 1)^{3} [Cubing
both sides]

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

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

⇒ a^{3}
= 27q^{3} + 1 + 27q^{2} + 9q

⇒ a^{3}
= 27q^{3} + 27q^{2} + 9q + 1

⇒ a^{3}
= 9(3q^{3} + 3q^{2} + q) + 1

⇒ a^{3}
= 9m + 1 …(ii)

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

If a = 3q + 2

⇒ a^{3}
= (3q + 2)^{3} [Cubing
both sides]

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

⇒ a^{3}
= 27q^{3} + 8 + 54q^{2} + 36q

⇒ a^{3}
= 27q^{3} + 54q^{2} + 36q + 8

⇒ a3
= 9(3q^{3} + 6q^{2} + 4q) + 8

⇒ a^{3}
= 9m + 8 …(iii)

where, m = 3q^{3} + 6q^{2} + 4q

Hence,
from (i), (ii) and (iii) we can say that the cube of any integer is of the form 9m, 9m + 1 or 9m + 8.
**Proved.**

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