Jharkhand Board JAC Class 10 Maths Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.1

Question 1.

Use Euclid’s division algorithm to find the of (1) 135 and 225 (2) 196 and 38220 (3) 867 and 255.

Solution:

1. 135 and 225

Here, 225 > 135

∴ 225 = 135 × 1 +90

Since remainder ≠ 0, we apply division lemma to 135 and 90.

∴ 135 = 90 × 1 + 45

Since remainder ≠ 0, we apply division lemma to 90 and 45.

∴ 90 = 45 × 2 + 0

Since remainder = 0, the divisor 45 is the HCF.

Hence, HCF (135, 225) = 45.

2. 196 and 38220

Here, 38220 > 196

∴ 38220 = 196 × 195 +0

Since remainder = 0, the divisor 196 is the HCF.

Hence, HCF (196, 38220) = 196.

3. 867 and 255

Here, 867 > 255

∴ 867 = 255 × 3 + 102

Since remainder ≠ 0, we apply division lemma to 255 and 102.

∴ 255 = 102 × 2 + 51

Since remainder ≠ 0. we apply division lemma to 102 and 51.

∴ 102 = 51 × 2 + 0

Since remainder = 0, the divisor 51 is the HCF.

Hence, HCF (867, 255) = 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.

Solution:

Let a be any positive integer and b = 6. Then. by Euclid’s division lemma, a = 6q + r, for some integer q ≥ 0 and r = 0, 1, 2, 3, 4 or 5. because 0 ≤ r < 6.

So, a = 6q or a = 6q + 1 or

a = 6q + 2 = 2(3q + 1) or a = 6q + 3 or

a = 6q + 4 = 2(3q + 2) or a = 6q + 5.

Since a is an odd integer, a cannot be 6q or 6q + 2 or 6q + 4 as they are all divisible by 2.

Therefore, any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.

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:

To arrive at the answer, we have to find the HCF of 616 and 32.

Here, 616 > 32

∴ 616 = 32 × 19 + 8

∴ 32 = 8 × 4 + 0

Thus, HCF (616, 32)=8

Hence, the maximum number of columns in which they can march is 8 columns.

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 a 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:

Let a be any positive integer and b = 3. Then, by Euclid’s division lemma, a = 3q or a = 3q + 1 or a = 3q + 2; where q is a non- negative integer.

1. If a = 3q, then a^{2} = (39)^{2} = 9q^{2} = 3(3q^{2}) = 3m, where m = 3q^{2} is some integer.

2. If a = 3q + 1, then a^{2} = (3q + 1)^{2} = 9q2 + 6q + 1 = 3(3q^{2} + 2q) + 1 = 3m+ 1. where m = 3q^{2} + 2q is some integer.

3. If a = 3q + 2, then a^{2} = (3q + 2)^{2} = 9q^{2} + 12q + 4 = 9q^{2} + 12q + 3 + 1 = 3 (3q^{2} + 4q + 1) + 1 = 3m + 1, where m = 3q^{2} + 4q + 1 is some integer.

Thus, in either case, the square of any positive integer is of the form 3m or 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.

Solution:

Let a be any positive integer and b = 3. Then, by Euclid’s division lemma, a = 3q or a = 3q + 1 or a = 3q + 2; where q is negative integer.

1. If a = 3q, then

a^{3} = (39)^{3} = 27q^{3} = 9(3q^{3}) = 9m,

where m = 3q^{3} is some integer.

2. If a = 3q + 1, then

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

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

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

= 9m + 1.

where m = 3q^{3} + 3q^{2} + q is some integer.

3. If a = 3q + 2, then

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

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

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

= 9m + 8,

where m = 3q^{3} + 6q^{2} + 4q is some integer.

Thus, in either case, the cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.