NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

Here, you will find Chapter 1 Real Numbers Class 10 Maths NCERT Solutions which will enhance your knowledge and let you complete your homework on time. It will be helpful in knowing the way to solve questions effectively and easily. Class 10 Real Numbers Maths is also important for the purpose of board examination and help you in getting more marks.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers


Exercise 1.1

Page No. 7

1. Use Euclid’s division algorithm to find the HCF of:

i. 135 and 225

ii. 196 and 38220

iii. 867 and 225

Solution

(i) 225 > 135 we always divide greater number with smaller one.

225 = 135 × 1 + 90

Now, remainder 90 ≠ 0, thus again using Euclid's division lemma for 90, we get,

135 = 90 × 1 + 45

Again, 45 ≠ 0, repeating the above step for 45, we get,

90 = 45 × 2 + 0

As there are no remainder so divisor 45 is our HCF.

Hence, the HCF of 225 and 135 is 45.

(ii) 38220 > 196 we always divide greater number with smaller one.

38220 = 196 × 195 + 0

As there is no remainder so divisor 196 is our HCF.

Hence, the HCF of 196 and 38220 is 196.

(iii) 867 > 255 we always divide greater number with smaller one.

867 = 225 × 3 + 102

Remainder 102 ≠ 0, therefore taking 225 as divisor and applying the Euclid's division lemma method, we get,

225 = 102 × 2 + 51

Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,

102 = 51 × 2 + 0

As there is no remainder so divisor 51 is our HCF.

Hence, the HCF of 867 and 225 is 51.

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 integer and b = 6.
∴ By Euclid’s division algorithm, we have
a = bq + r, 0 ≤ r ≤ b
a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4,5
Now, ‘a’ may be of the form of 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5

6q + 0
6 is divisible by 2 so it is a even number 

6q + 1 
6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number 

6q + 2 
6 is divisible by 2 and 2 is also divisible by 2 so it is a even number 

6q  +3 
6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number 

6q + 4 
6 is divisible by 2 and 4 is also divisible by 2 it is a even number

6q + 5 
6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5.

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

Given,

Number of army contingent members=616

Number of army band members = 32

HCF (616, 32) will give the maximum number of columns in which they can march.

We can use Euclid's algorithm to find the HCF.

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

4. Use Euclid's division lemma to show that the square of any positive integer is either of 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 and b = 3
x = 3q + where q is quotient and r is remainder 0 ≤r<3
Ifr = 0 then x = 3q
If r = 1 then x = 3q +1
If r = 2 then x = 3q + 2
x is of the form 3 or 3q +1 or 3q + 2
If x = 3q
Squaring both sides,
(x)2 = (3q)2
= 9q2 = 3 (3q2) = 3m
m = 3q2
where m is also an integer
Hence, x2 = 3m  ....... (i)
If x = 3q +1
Squaring both sides,
x2 = (3q + 1)2
x2 = 9q2 +1 +2 × 3q × 1
x2 = 3 (3q2 +2q) +1
x2 = 3m +1  ........ (ii)
where, m = 3q2 + 2
where m is also an integer If
If x = 3q + 2
Squaring both sides,
x2 = (3q + 2)2
x2 = 9q2 + 6q +4
x2 = 9q2 + 6q +3+1
x22 = 3(3q22 + 2q +1)+1
x22 = 3m +1
where, m = 3q2 + 2q +1
where m is also an integer
From (i), (ii) and (iii),
x2 = 3m, 3m +1
Hence, square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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 ≥ 0 and 0 ≤ r < 3
∴ a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,
a3 = (3q)3 = 27q3 = 9(3q)3 = 9m,
Where m is an integer such that m = 3q3

Case 2: When = 3q + 1,
a3 = (3q +1)3
a3= 27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)

Case 3: When a = 3q + 2,
a3 = (3q +2)3
a3= 27q3 + 54q2 + 36q + 8
a3 = 9(3q3 + 6q2 + 4q) + 8
a3 = 9m + 8
Where m is an integer such that m = (3q3 + 6q2 + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Exercise 1.2

Page No: 11

1. Express each number as product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Solution

(i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7

(ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

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
(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)
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)× (3) × (5) × (17) × (23)
= (2) × (3) × (5) × (17) × (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) × HCF (336, 54)
= 6 × 3024 = (2) × (3) × (2)+ × (3)3 × (7)
= (2)4 × (3) (7) × (2) × (3)3 = 336 × 54
= Product of given numbers.

3. Find the LCM and HCF of the following integers by applying the prime factorization method.


(i) 12, 15 and 21 
(ii) 17, 23 and 29 
(iii) 8, 9 and 25

Solution

(i) 12 = 2 × 2 × 3
15 =3 × 5
21 =3 × 7
HCF = 3
LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 1 × 17 × 19 × 23 = 11339

(iii) 8 =1 × 2 × 2 × 2
9 =1 × 3 × 3
25 =1 × 5 × 5
HCF =1
LCM = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution

We have the formula that
Product of LCM and HCF = product of number
LCM × 9 = 306 × 657
Divide both side by 9 we get
LCM = (306 × 657) / 9 = 22338

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
∴ 6"nn is divisible by 5.
But, prime factors of 6 are 2 and 3.
∴ Prime factors of (6)n are (2 × 3)n
⇒ It is clear that in prime factorisation of 6 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.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution

7 × 11 × 13 + 13
Taking 13 common, we get
13 (7 x 11 +1 )
13(77 + 1 )
13 (78)
It is product of two numbers and both numbers are more than 1 so it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 +1)
5(1008 + 1)
5(1009)
It is product of two numbers and both numbers are more than 1 so it is a composite number.

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 the same. Suppose they both start at the same point and at the 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)× (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.

Exercise 1.3

Page No. 14

8. Let take √5 as rational number

Solution

If a and b are two co prime number and b is not equal to 0.
We can write √5 = a/b
Multiply by b both side we get
b5 = a
To remove root, Squaring on both sides, we get
5b2 = a2 …  (i)

Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also.
That means 5 will divide a. So we can write
a = 5c
Putting value of a in equation (i) we get
5b2 = (5c)2
5b2 = 25c2
Divide by 25 we get
b2/5 = c2
Similarly, we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it contradicts.
Hence √5 is not a rational number, it is irrational.

2. Prove that 3 + 2√5 is irrational.
Solution

Let take that 3 + 2√5 is a rational number.
So we can write this number as
3 + 2√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 3 both sides we get
2√5 = a/b – 3
2√5 = (a-3b)/b
Now divide by 2, we get
5 = (a-3b)/2b
Here a and b are integer so (a-3b)/2b is a rational number so √5 should be a rational number But √5 is a irrational number so it contradicts.
Hence, 3 + 2√5 is a irrational number.

3. Prove that the following are irrationals:

(i) 1/√2 (ii) 7√5 (iii) 6 + √2

Answer

(i) Let take that 1/2 is a rational number.
So we can write this number as
1/√2 = a/b
Here and b are two co prime number and b is not equal to 0
Multiply by √2 both sides we get
1 = (a2)/b
Now multiply by b
b = a2
divide by a we get
b/a = √2
Here a and b are integer so b/a is a rational number so √2 should be a rational number But √2 is a irrational number so it contradicts.
Hence, 1/√2 is a irrational number

(ii) Let take that 7√5 is a rational number.
So we can write this number as
7√5 = a/b
Here a and b are two co prime number and b is not equal to 0
Divide by 7 we get
5 = a/(7b)
Here a and b are integer so a/7b is a rational number so √5 should be a rational number but √5 is a irrational number so it contradicts.
Hence, 7√5 is a irrational number.

(iii) Let take that 6 + √2 is a rational number.
So we can write this number as
6 + √2 = a/b
Here a and b are two co prime number and b is not equal to 0
Subtract 6 both side we get
2 = a/b – 6
2 = (a-6b)/b
Here a and b are integer so (a-6b)/b is a rational number so √2 should be a rational number.
But √2 is a irrational number so it contradicts.
Hence, 6 + √2 is a irrational number.

Page No: 17

Exercise 1.4

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:
(i) 13/3125
(ii) 17/
(iii) 64/455 
(iv) 15/1600 
(v) 29/343 
(vi) 23/2× 52 
(vii) 129/2× 5× 75 
(viii) 6/15 
(ix) 35/50 
(x) 77210


Solution

(i) 13/3125
Factorize the denominator we get
3125 =5 × 5 × 5 × 5 × 5 = 55
So denominator is in form of 5m so it is terminating .

(ii) 17/8
Factorize the denominator we get
8 =2 × 2 × 2 = 23
So denominator is in form of 2m so it is terminating .

(iii) 64/455
Factorize the denominator we get
455 =5 × 7 × 13
There are 7 and 13 also in denominator so denominator is not in form of 2m × 5n . so it is not terminating.

(iv) 15/1600
Factorize the denominator we get
1600 =2 × 2 × 2 ×2 × 2 × 2 × 5 × 5 = 26 × 52
so denominator is in form of 2m × 5n
Hence it is terminating.

(v) 29/343
Factorize the denominator we get
343 = 7 × 7 × 7 = 73
There are 7 also in denominator so denominator is not in form of 2m × 5n
Hence it is non-terminating.

(vi) 23/(23 × 52)
Denominator is in form of 2m × 5n
Hence it is terminating.

(vii) 129/(22 × 57 × 75 )
Denominator has 7 in denominator so denominator is not in form of 2m × 5n
Hence it is none terminating.

(viii) 6/15
divide nominator and denominator both by 3 we get 2/5
Denominator is in form of 5m so it is terminating.

(ix) 35/50 divide denominator and nominator both by 5 we get 7/10
Factorize the denominator we get
10=2 × 5
So denominator is in form of 2m × 5n so it is terminating.

(x) 77/210
simplify it by dividing nominator and denominator both by 7 we get 11/30
Factorize the denominator we get
30=2 × 3 × 5
Denominator has 3 also in denominator so denominator is not in form of 2m × 5n
Hence it is none terminating.

Page No: 18

2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Solution

(i) 13/3125 = 13/55 = 13×25/55×25 = 416/105 = 0.00416 (ii) 17/8 = 17/23 = 17×53/23×53 = 17×53/103 = 2125/103 = 2.125

(iv) 15/1600 = 15/24×102 = 15×54/24×54×102 = 9375/106 = 0.009375

(vi) 23/2352 = 23×53×22/23 52×53×22 = 11500/105 = 0.115

(viii) 6/15 = 2/5 = 2×2/5×2 = 4/10 = 0.4

(ix) 35/50 = 7/10 = 0.7.

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 you say about the prime factors of q?

(i) 43.123456789
(ii) 0.120120012000120000...
(iii) 43.123456789

Solution

(i) Let x = 43.123456789
x has a terminating decimal expansion, it is a rational number.
Now remove the decimal from the number
∴  x = 43123456789/1000000000
⇒ x = 43123456789/109
From (2) x is a rational number and of the form p/q where p = 43123456789 and 1 = 109
Now, Prime factors of q = 109 = (2 × 5)9
⇒ Prime factors of q are 29 × 59

(ii) The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational number.

(iii) Since the decimal expansion is non-terminating recurring, the given number is a rational number of the form p/q, and q is not of the form 2m × 5n.

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