Gujarat Board Solutions Class 10 Maths Chapter 1 Real Numbers
Gujarat Board Textbook Solutions Class 10 Maths Chapter 1 Real Numbers
Ex 1.1
Question 1.
Use Euclid’s division algorithm to find the HCF of
- 135 and 225
- 196 and 38220
- 867 and 255
Solution:
1. Let a = 225 and b = 135
Since a > b
Applying Euclid division Algorithm
a = bq + r (where 0 ≤ b < r)
225 = 135 × 1 + 90
Dividend = 225
Divisor = 135
Since 90 is the remainder but 90 ≠ 0. So we apply Euclid division lemma to 135 and 90
a = bq + r (Here a = 135 and b = 90)
135 = 90 × 1 + 45
Since remainder 45 ≠ 0
So again, we apply Euclid division lemma to 90 and 45
a = bq + r (Here a = 90 and 6 = 45)
90 = 45 × 2 + 0
Since remainder is zero, the process stops. Since the divisor at this stage is 45, therefore HCF of 135 and 225 is 45.
2. Let a = 38220 and b = 196
Since a > b
Applying Euclid division lemma,
a = bq + r (where 0 ≤ r < b)
38220 = 196 × 195 + 0
Since remainder is zero, at this stage our process stops and divisor is 196,
Therefore 196 is the HCF.
3. Let a = 867 and b = 255
Since a > b
Applying Euclid division lemma
a = bq + r (where 0 ≤ b < r)
867 = 255 × 3 + 102
Since 102 ≠ 0, we apply Euclid division lemma to a = 255 and b = 102 we get
a = bq + r (where 0 ≤ b <r)
255 = 102 × 2 + 51
Since 51 ≠ 0 we apply Euclid division lemma to a = 102 and 6 = 51
a = bq + r
102 = 51 × 2 + 0
Since remainder is zero, our process stops. At this stage divisor is 51, therefore HCF 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.(CBSE)
Solution:
Let a be any positive integer and 6 = 6. Then by Euclid’s division algorithm,
a = 6q + r (0 ≤ r < 6) for some integer q > 0, the values of remainder must be less than 6, i.e.,
r = 0, 1, 2, 3, 4, 5
We take only odd numbers like 1, 3 and 5.
For r = 0, a = 6q + r = 6q + 0
⇒ a = 6q
⇒ a = 2(3q) = 3m
where m is any positive integer
∴ a = 6q (not odd)
Similarly,
for r = 2, a = 6q + 2 = 2 (3q + 1) (not odd)
for r = 4, a = 6q + 4 = 2 (3q + 2) (not odd)
for r = 6, a = 6q + 6 = 2 (3q + 3) (not odd)
But, for r = 1
a = 6q + 1
for r = 3
a = 6q + 3
and, for r = 5
a = 6q + 5
Thus, for r = 1, 3 and 5, a is odd.
So we take only odd value. Therefore any positive integer of the form of 617 + 1, 6q + 3 and 6q + 5 is odd.
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:
HCF of 616 and 32 will give the maximum number of columns in which they can march. Now we use Euclid’s Algorithm to find HCF.
Let a = 616 and b = 32
a = bq + r
616 = 32 × 19 + 8
Since 8 ≠ 0 we again apply Euclid’s Algorithm
a = bq + r
32 = 8 × 4 + 0
Since remainder is zero.
Hence the HCF of 616 and 32 is 8.
Therefore an army contingent can march in 8 columns each.
Question 4.
Use Euclid division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.(CBSE)
[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 of (3m or 3m + 1)]
Solution:
Let a be any positive integer and b = 3
By applying Euclid’s Algorithm
a = bq + r (0 ≤ r < 6)
then a = 3q + r (0 ≤ r < 3)
Values of r = 0, 1 and 2
when r = 0
then a = 3q + 0
a = 3q
when r = 1
then a = 3q + 1
when r = 2
then a = 3q + 2
Squaring of these positive integers as:
1. a = 3q
a2 = (3q)2
a2 = 9q2
a2 = 3 × (3q2)
a2 = 3m
Thus, 9q2 = 3 (3q2) is divisible by 3
2. a = 3q + 1
a2 = (3q + 1)2
a2 = 9q2 + 6q + 1
a2 = 3 (3q2 + 2q) + 1
(where, m = 3q2 + 2q)
= 3m + 1
Thus, a2 is written in the form of 3m + 1.
a = 3q + 2
a2 = (3q + 2)2
a2 = 9q2 + 12q + 4
a2 = 9q2 + 12q + 3 + 1
a2 = 3 (3q2 + 4q + 1) + 1
a2 = 3m + 1
where m = 3q2 + 4q + 1
Thus, a2 is written in the form of 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.(CBSE)
Solution:
Let a be any positive integer and b = 3
By Applying Euclid’s division lemma
a = bq + r (0 ≤ r < b)
then a = 3q + r (0 ≤ r < 3)
Values of r = 0, 1, 2
when r = 0
then a = 3q + 0
a = 3q
Cubing both sides
a3 = (3q)3
a3 = 27q3
a3 = 9 (3q3) [27q3 is divisible by 9]
a3 = 9m
when r = 1
a = 3q + 1
Cubing both sides
a3 = (3 q + 1)3
= 27q3 + 9q2 + 9q + 1
= 9 (3q3 + q2 + q) + 1
[27q3 + 9q2 + 9q is divisible by 9]
a3 = 9m + 1
when r = 2
a = 3q + 2
Cubing both sides
a3 = (3q + 2)3
= 27q3 + 54q2 + 36q + 8
= 9 (3q3 + 6q2 + 4q) + 8
[27q3 + 54q2 + 36q is divisible by 9]
a3 = 9m + 8
Therefore the cube of any positive integer is 9m, 9m + 1 or 9m + 8.
Ex 1.2
Question 1.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
- 12, 15 and 21
- 17, 23 and 29
- 8, 9 and 25
Solution:
1. Prime factorisation of 12 = 2 × 2 × 3
Prime factorisation of 15 = 3 × 5
Prime factorisation of 21 = 3 × 7
HCF of 12, 15 and 21 = 3
LCM of 12, 15 and 21
= 2 × 2 × 3 × 5 × 7 = 420
2. Prime factorisation of 17 = 17 × 1
Prime factorisation of 23 = 23 × 1
Prime factorisation of 29 = 29 × 1
HCF of 12, 15, 29 = 1
LCM of 17, 23, 29 = 17 × 23 × 29
= 11339
3. Prime factorisation of 8 = 2 × 2 × 2
Prime factorisation of 9 = 3 × 3
Prime factorisation of 23 = 23 × 1
HCF of 8, 9 and 23 = 1
LCM of 8, 9, 23 = 2 × 2 × 2 × 3 × 3 × 23
= 1656
Question 2.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
Given that HCF (306, 657) = 9
LCM × HCF = Product of two numbers
LCM × 9 = 306 × 657
LCM = 306×657/9
LCM = 34 × 657
LCM = 22338
∴ LCM of (306, 657) = 22338.
Question 3.
Check whether 6n can end with the digit 0 for any natural number n.(CBSE)
Solution:
6n will end with the digit zero if 6n is divisible by 2 and 5.
But 6n = (2 × 3)n = 2n × 3n
i.e. in the factorisation of 6n, no factor is of 5. Therefore by the fundamental theorem of arithmetic every composite number can be expressed a product of primes and this factorisation is unique apart from the order in which the prime factors occur.
Therefore our assumption is wrong that 6n ends in zero, thus there does not exist any natural number n for which 6n ends with zero.
Question 4.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Given that 7 × 11 × 13 + 13
= (7 × 11 × 1 + 1) × 13
= (77 + 1) × 13
= 78 × 13
Composite number because it is product of more than two factors.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= (7 × 6 × 4 × 3 × 2 × 2 + 1) × 5
= (1008 + 1) × 5
= 1009 × 5
Product of more than two factors, which is a composite number.
Ex 1.3
Question 1.
Question 3.
Prove that the following are irrationals:
Ex 1.4
Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
Question 2.
Write down decimal expansions of the rational numbers in question 1 above which have terminating decimal expansions.
Solution:
Question 3.
The following real numbers have decimal expansions as given below. In each case decide whether they are rational or not. If they are rational and of the form p/q , what can you say about the prime factor of q?
2. 0.120 1200 12000 120000 …
The decimal expansion is neither terminating nor recurring, therefore, the given number is an irrational number.
