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.

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

(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.

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

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.

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

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.

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

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 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

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