# PSEB 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.1

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 1 Real Numbers Ex 1.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 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) By Euclid’s division Algorithm

Step 1.
Since 225 > 135,
we apply the division Lemma to 225 and 135,
we get 225 = 135 × 1 + 90

Step 2.
Since the remainder 90 ≠ 0,
we apply the division Lemma to 135 and 90,
we get 135 = 90 × 1 + 45

Step 3. Since the remainder 45 ≠ 0,
we apply the division Lemma to 90 and 45,
we get 90 = 45 × 2 + 0

Since the remainder has now become zero, so we stop procedure.
∵ divisor in the step 3 is 45
∵ HCF of 90 and 45 is 45
Hence, HCF of 135 and 225 is 45. (ii) To find HCF of 196 and 38220
Step 1.
Since 38220 > 196,
we apply the division Lemma to 196 and 38220,
we get 38220 = 196 × 195 + 0
Since the remainder has now become zero so we stop the procedure.
∵ divisor in the step is 196
∵ HCF of 38220 and 196 is 196.
Hence, HCF of 38220 and 196 is 196.

(iii) To find HCF of 867 and 255
Step 1.
Since 867 > 255,
we apply the division Lemma to 867 and 255,
we get 867 = 255 × 3 + 102

Step 2.
Since remainder 102 ≠ 0,
we apply the divison Lemma to 255 and 102,
we get 255 = 102 × 2 + 51

Step 3.
Since remainder 51 ≠ 0,
we apply the division Lemma to 51 and 102, by taking 102 as division,
we get 102 = 51 × 2 + 0
Since the remainder has now become zero, so we stop the procedure.
∵ divisor in step 3 is 51.
HCF of 102 and 51 is 51.
Hence, 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 any positive odd integer, we apply the division algorithm with a and b = 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 quotient. However, since a is odd ∵ a cannot be equal to 6q, 6q + 2, 6q + 4 ∵ all are divisible by 2. Therefore, any 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:
Total number of members in army = 616 and 32 (A band of two groups)
Since two groups are to march in same number of columns and we are to find out the maximum number of columns.
∴ Maximum Number of columns = HCF of 616 and 32
Step 1.
Since 616 > 32, we apply the division Lemma to 616 and 32, to get
616 = 32 × 19 + 8

Step 2.
Since the remainder 8 ≠ 0, we apply the division Lemma to 32 and 8, to get
32 = 8 × 4 + 0.
Since the remainder has now become zero
∵ divisor in the step is 8
∵ HCF of 616 and 32 is 8.
Hence, maximum number of columns in which they can march 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 + L]
Solution:
Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2.
If x = 3 q
Squaring both sides,
(x)2 = (3q)2
– 9q2 = 3 (3q2) = 3m
where m = 3 q2
where m is also an integer
Hence x2 = 3m ………… (1)
If x = 3q + 1
Squaring both sides,
x2 = (3q + 1)2
x2 = 9q2 + 1 + 2 × 3q × 1
x2 = 3 (3 q2 + 2q) + 1
x2 = 3m + 1 …. (2)
where m = 3q2 + 2q where m is also an integer
From (1) and (2),
x2 = 3m, 3m + 1
Hence, 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.
Solution:
Let x be any positive integer and b = 3
x = 3 q + r where q is quotient and r is remainder
If 0 ≤ r < 3
If r = 0 then x = 3 q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2
x is of the form 3q or 3q + 1 or 3q + 2
If x = 3q
Cubing both sides,
x3 = (3q)3
x3 = 27q3 = 9 (3q3) = 9m
where m = 3q3 and is an integer .
x3 = 9m ……….. (1)
If x = 3q + 1 cubing both sides,
x3 = (3 q +1)3
x3 = 27q3 + 27q2 + 9q + 1
= 9 (3q3 + 3q2 + q) + 1
= 9m + 1
where m = 3q3 + 3q2 + q and is an integer
Again x3 = 9m + 1 …………. (2)
If x = 3q +2
Cubing both sides,
(x)3 = (3q + 2)2
= 27 q3 + 54 q2 + 36q + 8
x3 = 9 (3 q3 + 6q2 + 4q) + 8
x3 = 9m + 8 ………. (3)
where m = 3 q3 + 6q2 + 4q
Again x3 = 9m + 8
From (1) (2), & (3), we find that
x3 can be of the form 9m, 9m + 1, 9m + 8.
Hence, x3 of any positive integer can be of he form 9m, 9m + 1 or 9m + 8