Students can Download Class 10 Maths Chapter 8 Real Numbers Ex 8.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 8 Real Numbers Ex 8.1

Question 1.

Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Answer:

i) Step 1: Consider a = 225 and , b = 135

By Euclid’s lemma.

a = (b × q) + r

225 = (135 × 1) + 90

Step 2: Consider a = 135 and b = 90

Here q = 1, r = 45

By Euclid’s lemm

a = (b × q) + r

135 = (90 × 1) + 45

Step 3: Consider a = 90 and b = 45

Here q = 2, r = 0

By Euclid’s lemma

a = (b × q) + r

90 = (45 × 2) + 0

Since the remainder is zero,

∴ Last divisor is HCF

∴ HCF of 135 and 225 is 45.

ii) 196 and 38220

Step 1: Consider a = 38220 and b = 196. Here q = 195, r = 0.

By Euclid’s lemma

a = (b × q) + r

38220 = (196 × 195) + 0

∴ HCF of 196 and 38220 is 196

iii) 867 and 255

Step 1: Consider a = 867 and b = 255

Here q = 3, r = 102

By Euclid’s lemma

a = (b × q) + r

867 = (255 × 3) + 102

Step 2: Consider a = 255 and b = 102

Here q = 2 and r = 51

By Euclid’s lemma

a = (b × q) + r 255 = (102 × 2) + 51

Step 3 : Consider a = 102 and b = 51

Here q = 2, r = 0

By Euclid’s lemma

a = (b × q) + r

102 = (51 × 2) + 0

∴ HCF of 867 and 225 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.

Answer:

Let a be any positive odd integer and b = 6.

Then, by Euclid’s algorithm, a = bq + r, For some integer q ≥ 0 and 0 ≤ r < 6

∴ a = 6q + r [remainders are 0, 1, 2, 3, 4, 5].

Thus, a can be of the form 6q (or) 6q + 1 (or) 6q + 2 (or) 6q + 3 (or) 6q + 4 (or) 6q + 5 where q is some Quotient.

Since a is odd integer, so a cannot be of the form 6q, (or) 6q + 2 (or) 6q + 4. (since they are even)

Thus a is of form 6q + 1 (or) 6q + 3 (or) 6q + 5 where q is some integer.

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

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?

Answer:

HCF of 616 and 32.

Step 1: Euclid’s division lemma

a = 616 and b = 32

616 = 32 × 19 + 8

616 = 616.

Step 2:

Euclid’s division lemma,

a = 32 & b = 8.

a = bq + r

32 = 8 × 4 + 0

32 = 32

Here, Remainder is zero. So, HCF of (616, 32) = 8.

Hence, maximum number of columns is 8.

Question 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.]

Answer:

i) When n = 3m,

n^{2} = 9m^{2}

= 3 × 3m^{2}

= 3q (∵ q = 3m^{2})

ii) If n = 3m + 1

n^{2} = (3m + 1)^{2}

= 9m^{2} + 6m + 1

= 3 (3m^{2} + 2m) + 1

= 3q + 1 (∵ 3m^{2} + 2m = q)

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

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.

Answer:

Let ‘a’ and ‘b’ are positive integers,

a = b × q + r

Let b = 3. ∴ a = 3q + r

i) If r = 0 then a = 3q

ii) If r = l then a = 3q + 1

iii) If r = 2 then a = 3q + 2

If we consider cubes of these,

i) If a = 3q, then

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

= 27q^{3} = 9(3q)^{3} = 9m (∵ m = 3q^{2})

ii) If a = 3q + 1, then

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

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

= 9m + 1

(∵ m = 3q^{3}+ 3q^{2} + q)

iii) If a = 3q + 2, then

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

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

= 9m + 8

∴ Cube of any positive integer is of the form 9m, 9m + 1, 9m + 8.