RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.1

Question 1.
Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) 135 and 225
Using Euclid’s Division Algorithm
Step I : ∵ 225 > 135
Therefore according to Euclid’s Division Lemma
225 = 135 × 1 + 90
Step II : ∵ Remainder 90 ≠ 0, therefore now applying Euclid’s Division Lemma on 135 and 90
135 = 90 × 1 + 45
Step III : ∵ Remainder 45 ≠ 0, therefore now applying Euclid’s Division Lemma on 90 and 45
90 = 45 × 2 + 0
The remainder has now become zero, so this procedure stops. The divisor in this step is 45. Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220
Using Euclid’s Division Algorithm
Step I : ∵ 38220 > 196
Therefore according to Euclid’s Division Lemma
38220 = 196 × 195 + 0
Since the remainder obtained is zero. Therefore this procedure stops here. The divisor in this step is 196. Therefore, the HCF of 38220 and 196 is 196.

(iii) 867 and 255
Using Euclid’s Division Algorithm
Step I : ∵ 867 > 255
Therefore according to Euclid’s Division Lemma
867 = 255 × 3 + 102
Step II : ∵ Remainder 102 ≠ 0, therefore now applying Euclid’s Division Lemma on 255 and 102
225 = 102 × 2 + 51
Step III : Remainder 51 ≠ 0, therefore now applying Euclid’s Division Lemma on 102 and 51
102 = 51 × 2 + 0
The remainder has now become zero, so this procedure stops. The divisor in this step III is 51. Therefore, the HCF of 867 and 255 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.
Solution:
Let a be a positive odd integer. Now for a and b = 6, by the application of Euclid’s Division Algorithm 0 ≤ r < 6, i.e., the values of a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is some quotient. Now since a is odd positive integer. Therefore it cannot be of the form 6q, 6q + 2 or 6q + 4 since all these being divisible by 2 are even positive integers. Therefore, any positive odd 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?
Solution:
Group of members in parade = 616 and 32 According to the question, we are to find out the maximum number of columns, i.e., we have to find out the HCF of 616 and 32.
Step I : ∵ 616 > 32
Therefore according to Euclid’s Division Lemma
616 = 32 × 19 + 8
Step II : ∵ Remainder 8 ≠ 0, therefore now applying Euclid’s Division Lemma on 32 and 8
32 = 8 × 4 + 0
Since the remainder obtained now is zero. Therefore this procedure stops. The divisor in this step is 8. Therefore, the HCF of 616 and 32 is 8, i.e., the maximum number of columns in which they can march in parade 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].
Solution:
Let x be any positive integer. Then it can be of form 3q, 3q + 1 or 3q + 2.
Therefore if x = 3q
squaring both sides
x² = (3q)²
= 9q²
= 3(3q²)
= 3m
where m = 3q² and m is also an integer.
Hence x² = 3m ….(i)
Again if x = 3q + 1
squaring both sides
x² = (3q + 1)²
=> x² = 9q² + 2 × 3q × 1 + 1
=> x² = 3(3q² + 2q) + 1
=> x² = 3m + 1 …….(ii)
Where m = 3q² + 2q and m is also an integer.
In the last if x = 3q + 2
squaring both sides
x² = (3q + 2 )²
= 9q² + 12q + 4
= 3(3q² + 4q + 1) + 1
= 3m + 1 ……….(iii)
Where m = 3q² + 4q + 1 and m is also an integer.
Hence from (i), (ii) and (iii)
x² = 3m or 3m + 1
Therefore the square of any positive integer is 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.
Solution:
Let a be any positive integer and b = 3.
∴ a = 3q + r
Where q is the quotient and r is the remainder.
Here 0 ≤ r < 3
Therefore if r = 0 then a = 3q
if r = 1 then a = 3q + 1
if r = 2 then a = 3q + 2
i. e., a is of the form 3q, 3q + 1 or 3q + 2.
Now if a = 3q
Cubing both sides
a³ = (3q)³
a³ = 27q³ = 9(3q³) = 9m
Here m = 3q³ and m is also an integer
∴ a³ = 9m ….(i)
Again if a = 3q + 1 Cubing both sides
a³ = (3q + 1)³
⇒ a³ = 27q³ + 27q³ + 9q + 1
= 9 (3q³ + 3q³ + q) + 1
= 9m + 1
Here m = 3q³ + 3q² + q and m is also an integer.
Hence a³ = 9m + 1 ….(ii)
Now if a = 3q + 2
Cubing both sides
a³ = (3q + 2)³
= 27q³ + 54q² + 36q + 8
a³ = 9 (3q³ + 6q² + 4q) + 8
a³ = 9m + 8 …. (iii)
Here m = 3q³ + 6q² + 4q and m is also an integer.
Hence a³ = 9m + 8
Now from (i), (ii) and (iii) we find that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.