PSEB Solutions for Class 10 Maths Chapter 1 Real Numbers
PSEB 10th Class Maths Solutions 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
Ex 1.2
Question 1.
Express each number as a product of its prime factors :
(i) 140 (Pb, 2019)
(ii) 156
(iii) 3825
(iv) 5005 (Pb. 2019, Set-1, II, III)
(v) 7429.
Solution:
(i) Prime factorisation of 140 = (2)2 (35) = (2)2 (5) (7)
(ii) Prime factorisation of 156 = (2)2 (39) = (2)2 (3) (13)
(iii) Prime factorisation of 3825 = (3)2 (425)
= (3)2 (5) (85)
= (3)2 (5)2 (17)
(iv) Prime factorisation of 5005
= (5) (1001)
= (5) (7) (143)
= (5) (7) (11) (13)
(v) Prime factorisation of 7429
= (17) (437)
= (17) (19) (23)
Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the two numbers.
(i) 26 and 91 [Pb. 2017 Set-C]
(ii) 510 and 92
(iii) 336 and 54.
Solution:
(i) Given numbers are 26 and 91 Prime factorisation of 26 and 91 are
26 = (2) (13) and
91 = (7) (13)
HCF (26, 91)
Product of least powers of common factors
∴ HCF (26, 91) = 13
and LCM (26, 91) = Product of highest powers of all the factors
= (2) (7) (13) = 182
Verification :
LCM (26, 91) × HCF (26, 91)
= (13) × (182) = (13) × (2) × (91)
= (26) × (91)
= Product of given numbers.
(ii) Given numbers are 510 and 92 Prime factorisation of 510 and 92 are
510 = (2) (255)
= (2) (3) (85)
= (2) (3) (5) (17)
and 92 = (2) (46) = (2)2 (23)
HCF (510, 92) = Product of least powers of common factors = 2
LCM (510, 92) = Product of highest Powers of all the factors
= (2)2 (3) (5) (17) (23) = 23460
Verification:
LCM (510, 92) × HCF (510, 92)
= (2) (23460)
= (2) × (2)2 (3) (5) (17) (23)
= (2) (3) (5) (17) × (2)2 (23)
= 510 × 92 = Product of given numbers.
(iii) Given numbers are 336 and 54
Prime factorisation of 336 and 54 are
336 = (2) (168)
= (2) (2) (84)
= (2) (2) (2) (42)
= (2) (2) (2) (2) (21)
= (2)4 (3) (7)
and 54 = (2) (27)
= (2) (3) (9)
= (2) (3) (3) (3)
= (2) (3)3
HCF (336, 54) = Product of least powers of common factors = (2) (3) = (6)
LCM (336, 54) = Product of highest powers of all the factors
= (2)4 (3)3 (7) = 3024
Verification :
LCM (336, 54) × H.C.F. (336, 54)
= 6 × 3024
= (2) (3) × (2)4 (3)3 (7)
= (2)4 (3) (7) × (2) (3)3
= 336 × 54
= Product of given numbers.
Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21 [MQP 2015, Pb. 2015 Set A, Set B, Set C]
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) Given numbers are 12, 15 and 21
Prime factorisation of 12, 15 and 21 are
12 = (2) (6)
= (2) (2) (3)
= (2)2 (3)
15 = (3) (5)
21 = (3) (7)
HCF (12, 15 and 21) = 3
LCM (12, 15 and 21) = (2)2 (3) (5) (7) = 420
(ii) Given numbers are 17, 23 and 29
Prime factorisation of 17, 23 and 29 are
17 = (17) (1)
23 = (23) (1)
29 = (29) (1)
HCF (17, 23 and 29) = 1
LCM (17, 23 and 29)
= 17 × 23 × 29 = 11339
(iii) Given numbers are 8, 9 and 25
Prime factorisation of 8, 9 and 25 are
8 = (2) (4)
= (2) (2) (2)
= (2)3 (1)
9 = (3) (3) = (3)2 (1)
25 = (5) (5) = (5)2 (1)
HCF (8, 9 and 25) = 1
LCM (8, 9 and 25) = (2)3 (3)2 (5)2 = 1800
Question 4.
Prove that HCF (306,657) = 9, find LCM (306,657). [Pb. 2016 Set-B, 2017 Set-B]
Solution:
Given numbers are 306 and 657 Prime factorisation of 306 and 657 are
306 = (2) (153)
= (2) (3) (51)
= (2) (3) (3) (17)
= (2) (3)2 (17)
657 = (3) (219)
= (3) (3) (73)
= (3)2 (73)
HCF (306, 657) = (3)2 = 9
∵HCF × LCM = Product of given number
∵9 × LCM (306, 657) = 306 × 657
= 34 × 657 = 22338
Question 5.
Check whether 6n can end with the digit 0 for any natural number n:
Solution:
Let us suppose that 6n ends with the digit 0 for some n ∈ N.
6n is divisible by 5.
But, prime factor of 6 are 2 and 3 Prime factor of (6)n are (2 × 3)n
⇒ It is clear that in prime factorisation of 6n there is no place for 5.
∵ By Fundamental Theorem of Arithmetic, Every composite number can be expressed as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur.
∴ Our supposition is wrong.
Hence, there exists no natural number n for which 6n ends with the digit zero.
Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Consider, 7 × 11 × 13 + 13 = 13 [7 × 11 + 1]
which is not a prime number because it has a factor 13. So, it is a composite number.
Also, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5[7 × 6 × 5 × 4 × 3 × 2 × 1 + 1], which is not a prime number because it has a factor 5. So it is a composite number.
Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for die same. Suppose they both start at the same point and at file same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Time taken by Sonia to drive one round of the field =18 minutes
Time taken by Ravi to drive one round of same field = 12 minutes
They meet again at the starting point = LCM (18, 12)
Now, Prime factorisation of 18 and 12 are 18 = (2) (9)
= (2) (3) (3)
= (2) (3)2
12 = (2) (6)
= (2) (2) (3)
= (2)2 (3)
LCM (18, 12) = (2)2 (3)2 = 4 × 9 = 36
Hence, After 36 minutes Sonia and Ravi will meet again at the starting point.
Ex 1.3
Question 1.
Prove that √5 is irrational.
Solution:
Let us suppose that √5 is rational so we can find integers r and s where s ≠ 0
such that √5 = r/s
Suppose r and s have some common factor other than 1, then divide r and s by the common factor to get :
√5 = a/b where a and b are coprime and b ≠ 0
b√5 = a
Squaring both sides,
⇒ (b√5)2 = a2
⇒ b2 (√5)2 = a2
⇒ 5b2 = a2 …………..(1)
5 divides a2.
By the theorem, if a prime number ‘p’ divides a2 then ‘p’ divides a where a is positive integer
⇒ 5 divides a …………(2)
So a = 5c for some integer c.
Put the value of a in (1),
5b2 = (5c)2
5b2 = 25c2
b2 = 5c2
or 5c2 = b2
⇒ 5 divides b2
∵ if a prime number ‘p’ divides b2, then p divides b ; where b is positive integer.
⇒ 5 divides b ………… (3)
From (2) and (3), a and b have at least 5 as common factor.
But this contradicts the fact that a and b are coprime i.e. no common factor other than 1.
∴ our supposition that √5 is rational wrong.
Hence √5 is irrational.
Question 2.
Prove that 3 + 2 √5 is irrational.
Solution:
Let us suppose that 3 + 2√5 is rational.
∴ we can find Co-Prime a and b, where a and b are integers and b ≠ 0
such that 3 + 2√5 = a/b
Since a and b both are integers,
Hence from (1), √5 is rational.
But this contradicts the fact that √5 is irrational.
∴ our supposition is wrong.
Hence 3 + 2√5 is irrational.
Question 3.
Prove that the following are irrationals :
(ii) Given that 7√5
Let us suppose that 7^5 is rational
∴ we can find coprime integers a and b where b ≠ 0
such that 7√5 = a/b
⇒ 7b√5 = a
⇒ √5 = a/7b ……………..(1)
Since a, 7 and b are integers, of two integers is a rational number.
i.e., a/7b = rational number
∴ from (1)
√5 = rational number
which contradicts the fact that √5 is irrational number.
∴ Our supposition is wrong.
Hence 7√5 is irrational.
(iii) Given that 6 + √2
Let us suppose that 6 + √2 is rational
∴ we can find coprime integers a and b where b ≠ 0
such that 6 + √2 = a/b
so from (1), √2 = rational number
which contradicts the fact that √2 is irrational number
∴ Our Supposition is wrong.
Hence 6 + √2 is irrational.
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 :
Solution:
(i)Let x = 13/3125 ………..(1)
Compare (1) with x = p/q
Here p = 13 and q = 3125
Prime factors of q = 3125 = 5 × 5 × 5 × 5 × 5 = 55 × 20
which are of the form 2n × 5m here n = 0, m = 5
which are non negative integers.
∴ x = 13/3125 have a terminating decimal expansion.
Question 2.
Write down the decimal expansions of those 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 f, what can you say about the prime factors of q?
(i) 43.123456789
(ii) O.120120012000120000……
(iii) 4.3.123456789
Solution:
(i) Let x= 43.123456789 ……….. (1)
It is clear from the number that x is rational number.
Now remove the decimal from the number
Where p = 43123456789 and q = 109
Now, Prime factors of q = 100 = (2 × 5)9
⇒ Prime factors of q are 29 × 59
(ii) Let x = 0.120120012000120000
It is clear from the number that it is an irrational number.
(iii) Let x = 43.123456789 …. (1)
It is clear that the given number is a rational number because it is non-terminating and repeating decimal.
To show that (i) is of the form p/q
Multiply (1) with 109 on both sides,
109 x = 43123456789.123456789 …………….(2)
Subtract (1) from (2), we get:
