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NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

 

Here, NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers is provided which will be very helpfull for students. These NCERT Solutions are up to dated according to CBSE latest syllabus 2020-21. As the NCERT Solutions are prepared by StudyMates91, we have explained every steps so you can easily understand the concepts behind every Questions without any difficulty. These Real Numbers Class 10 solutions can be really very beneficial for the preparation of Board exams.

We have solved every questions stepwise so that you don't have to face any difficulty in understanding the solutions. Class 10 Maths Real Number NCERT Solutions will be also helpful in higher classes because variety of questions related to these topics can be asked so you must study and understand them properly.

Page No: 7

Exercise 1.1

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

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Answer

(i) Here 225 is greater than 135 and we always divide greater number with smaller one.
So, By Dividing 225 by 135 we get 1 as quotient and 90 as a remainder so that
225= 135 × 1 + 90

Dividing 135 by 90 we get 1 as quotient and 45 as a remainder so that
135= 90 × 1 + 45

Now, by Dividing 90 by 45 we get 2 as quotient and no remainder so we can write it as
90 = 2 × 45+ 0

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

(ii) 38220 is greater than 196 and we always divide greater number with smaller one.

So, by Dividing 38220 by 196 we get 195 as quotient and no remainder so we can write it as
38220 = 196 × 195 + 0

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

(iii) 867 is greater than 255 and we always divide greater number with smaller one.

Therefore, by Dividing 867 by 255 then we get 3 as quotient and remainder is 102 so we can write it as
867 = 255 × 3 + 102

Now, Dividing 255 by 102 then we get 2 as quotient and the remainder is 51 so we can write it as
255 = 102 × 2 + 51

Again, By Dividing 102 by 51 we get quotient 2 and no remainder so we can write it as
102 = 51 × 2 + 0

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

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer

Let  a be any positive integer  and b = 6.

Then using Euclid’s algorithm we get a = 6q + r  where r is remainder and q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6 

So total possible forms of a will be 6q + 0 or 6q , 6q + 1 , 6q + 2,6q + 3, 6q + 4, 6q + 5

Case I : When a = 6q

6 is divisible by 2 so it is a even number 

Case 2 : When a = 6q + 1 

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

Case 3 : When a = 6q + 2 

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

 

Case 4 : When a = 6q  +3 

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

Case 5 : When a = 6q + 4 

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

Case 6 : When a = 6q + 5 

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

So, Any positive odd numbers are in the 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?

Answer

The HCF of 616 and 32 will gives the maximum number of columns in which they can march.
We can use Euclid's division lemma 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.]

Answer

Let a be any positive integer is of the form bq + r where b = 3 and q ≥ 0.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
So, we have the following three cases.

Case 1 :  When a = 3q
a² = (3q)²

a² = (9q)²= 3m where m = 3(3q²)    3k₂ + 1 or 3k₃ + 1

Case 1 :  When a = 3q + 1

a² = (3q + 1)² 

a² = 9q² + 6q + 1 = 3(3q² + 2q)  + 1

a² = 3m +1 where m = 3(3q² + 2q)

 

Case 1 :  When a = 3q + 2

a² = (3q + 2)² 

a² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 2q + 1) +1

a² = 3m +1 where m = 3(3q² + 2q + 1)

Hence, we can say that the square of any positive integer is either of the form 3m or 3m + 1.

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.

Answer

Let a be any positive integer is of the form bq + r where b = 3 and q ≥ 0.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
So, we have the following three cases.

 

Case 1: When a = 3q,

a³ = (3q)³ = 27q³ = 9(3q)³ = 9m, Where  m = 3q³

Case 2: When a = 3q + 1,
a³ = (3q +1)³
a³= 27q³ + 27q² + 9q + 1
a³ = 9(3q³ + 3q² + q) + 1
a³ = 9m + 1, Where 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 = (3q³ + 6q² + 4q)

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

Page No: 11

Exercise 1.2

1. Express each number as product of its prime factors:

(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Answer

(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

Answer

(i) Expressing 26 and 91 as product of its prime factors, we get,

26 = 2 × 13
91 =7 × 13
So, HCF = 13
LCM =2 × 7 × 13 =182
Product of two numbers 26 × 91 = 2366
Product of HCF and LCM 13 × 182 = 2366
Hence, product of two numbers = product of their HCF and LCM

(ii) Prime Factorization of

 510 = 2 × 3 × 5 × 17
92 =2 × 2 × 23
HCF = 2
LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of two numbers 510 × 92 = 46920
Product of HCF and LCM 2 × 23460 = 46920
Hence, product of two numbers = product of their HCF and LCM.

(iii) Prime Factorization of

 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 2 × 3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024
Product of two numbers 336 × 54 =18144
Product of HCF and LCM 6 × 3024 = 18144
Hence, product of two numbers = product of their HCF and LCM.

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

Answer

(i) Prime Factorization of

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

(ii) Prime Factorization of

17 = 1 × 17
23 = 1 × 23
and 29 = 1 × 29
HCF = 1
LCM = 1 × 17 × 19 × 23 = 11339

(iii) Prime Factorization of

8 =1 × 2 × 2 × 2
9 =1 × 3 × 3
and 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).

Answer

We  know that the Product of LCM and HCF = product of two number.

Therefore,
LCM × 9 = 306 × 657    [Divide both side by 9]
LCM = (306 × 657) / 9 = 22338
Thus, the LCM of 306 and 657 is 22338.

5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

Answer
We know that If any number has its prime factor in the form of 2ⁿ × 5^m then the number with any exponent ends with digit zero.

So the value of 6ⁿ should be divisible by 2 and 5 both, but 6ⁿ is divisible by only 2  not with 5. So it can’t end with digit 0.

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

Answer
We have,
From the definition of composite number, we know, a composite number has factors other than 1 and itself. Therefore, from the given expression;

7 × 11 × 13 + 13

Taking 13 as a common factor, we get,

= 13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

Hence, 7 × 11 × 13 + 13 is a composite number.

Now, from the second expression we have,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 as a common factor, 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

Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is also 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?

Answer

Both Sonia and Ravi move in the same direction and at the same time, therefore to find the time when they will meet again at the starting point we have to find the LCM of 12 and 18.

Therefore, Expressing 12 and 18 as product of its prime factors, we get,

18 = 2 × 3 × 3
12 = 2 × 2 × 3

LCM(18,12) = 2×3×3×2×1=36

Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

Exercise 1.3

1. Prove that √5 is irrational.

Answer
If possible, Let us assume, that √5 is a rational number.

So, we can express it in the form of p/q, where p and q are co-primes and q is not equal to zero.

We can write √5 = p/q

Multiply by q both side we get
q√5 = p

Squaring both the sides, we get,

(q√5)² = p²

⇒ 5q² = p        …………….. (1)

Thus, p² is divisible by 5, so p is also divisible by 5.

Now, Let us say p = 5k, for some positive integer k and substituting the value of p in equation (1) we get,

5q² = (5k)²

⇒5q² = 25k²

Dividing both side by 5 we get

⇒q² = 5k²

Here q² is divisible by 5 it means q  is also divisible by 5.

Therefore, p and q both are divisible by 5 But, p and q co-primes .Since, our assumption is wrong.

Hence, √5 is an irrational number.

2. Prove that 3 + 2√5 is irrational.

Answer
If possible, Let us assume, that 3 + 2√5 is rational.

So, we can express it in the form of p/q, where p and q are co-primes and q is not equal to zero.

So we can write this number as
3 + 2√5 = p/q     [Subtract 3 from both sides]
2√5 = p/q – 3
2√5 = (p-3q)/q
Now, dividing both sides by 2, we get
√5 = (p-3q)/2q
Here p and q are integer so (p-3q)/2q is a rational number therefore √5 should be a rational number But √5 is a irrational number So this contradicts the fact that 3 + 2√5 is a irrational number.
Hence, 3 + 2√5 is an irrational number.

3. Prove that the following are irrationals:
(i) 1/√2 

(ii) 7√5 

(iii) 6 + √2

Answer

(i) If possible, Let us assume, that  1/√2 is a rational number.

So, we can express it in the form of p/q, where p and q are co-primes and q is not equal to zero
So, we can write this number as

1/√2 = p/q
Multiplying both sides by √2, we get
1 = (p√2)/q
Now, Multiplying both sides by q we get,
q = p√2
Dividing both sides by p we get

q/p = √2

Here p and q are integer so q/p is a rational number therefore √2 should be a rational number But √2 is an irrational number So this contradicts the fact that 1/√2 is a irrational number.
Hence, 1/√2 is an irrational number.

(ii) Let us assume 7√5 is a rational number.

Then we can find co-prime p and q (q ≠ 0) such that 7√5 = p/q

Dividing both sides by 7 we get,

√5 = p/7q

Since, p and q are integers, So p/7q  is a rational number thus, √5 is also a rational number, which contradicts the fact that √5 is an irrational.

Hence, 7√5 is an irrational.

(iii) Let us assume that 6 +√2 is a rational number.

Then we can find co-primes p and q (q ≠ 0) such that 6 +√2 = p/q⋅

Subtracting 6 from both sides, we get,

√2 = (p/q) – 6

= √2 = (p-6q/q)

Since, p and q are integers, thus (p-6q/q) is a rational number and therefore, √2 is also rational. This contradicts the fact that √2 is an irrational number.

Hence, that 6 +√2 is an irrational.

 

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) 17/8 

(iii) 64/455 

(iv) 15/1600 

(v) 29/343 

(vi) 23/2³ × 5² 

(vii) 129/2² × 5⁷ × 7⁵ 

(viii) 6/15 

(ix) 35/50 

(x) 77/210

Answer

(i) 13/3125
Factorization of the denominator 3125 is
3125 =5 × 5 × 5 × 5 × 5 = 5⁵

So, the denominator 3125 is in the form of 5^m Thus, it has terminating decimal expansion.
(ii) 17/8
Factorization of the denominator 8 is
8 =2 × 2 × 2 = 2³
So, the denominator 8 is in the form of 2ⁿ Thus, it has terminating decimal expansion.

(iii) 64/455
Factorizing the denominator 455 we get,
455 =5 × 7 × 13
The prime factors of the denominator 455 is not in the form of 2^m × 5ⁿ . Therefore, it has non-terminating decimal expansion.

(iv) 15/1600
Factorizing the denominator 1600 we get,
1600 =2 × 2 × 2 ×2 × 2 × 2 × 5 × 5 = 2⁶ × 5²
The prime factors of the denominator 1600 is in the form of 2^m × 5ⁿ . Therefore, it has terminating decimal expansion.

(v) 29/343
Factorizing the denominator 343 we get,
343 = 7 × 7 × 7 = 7³
The prime factors of the denominator 343 is not in the form of 2^m × 5ⁿ . Therefore, it has non-terminating decimal expansion.

(vi) 23/(2³ × 5²)
The denominator is in the form of 2^m × 5ⁿ . Therefore, it has terminating decimal expansion.

(vii) 129/(2² × 5⁷ × 7⁵ )
The denominator  is not in the form of 2^m × 5ⁿ  only it has 7^a also. Therefore, it has non-terminating decimal expansion.

(viii) 6/15
By Simplifying nominator and denominator both we get 2/5
and the Denominator is in form of 5^m, therefore it is has terminating decimal expansion.

(ix) 35/50

By Simplifying nominator and denominator both we get 7/10
Factorizing the denominator 10 we get
10=2 × 5
So denominator is in form of 2^m × 5ⁿ therefore it is has terminating decimal expansion.

(x) 77/210
By Simplifying nominator and denominator both we get 11/30
Factorizing the denominator 30 we get
30=2 × 3 × 5
Denominator has 3 also in its factor So, denominator is not in  the form of 2^m × 5ⁿ

Hence it is has non-terminating decimal expansion.

Page No: 18

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

Answer

(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

Answer

(i) Since the given number has a terminating decimal expansion, therefore it is a rational number of the form p/q, and the denominator q is in the form of  2^m × 5ⁿ.

 

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

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

FAQ on Chapter 1 Real Numbers

Q. How many exercises in Chapter 1 Real Numbers?

Answer:

There are total four exercises in this chapter. In the first exercise, there are only four questions and most of them are based on the topic Euclid's division lemma. The second exercise of this chapter is consists of HCF and LCM related questions. The third exercise has only three questions in which you have to prove any number rational or irrational. The fourth and last exercise has three questions in which you have to expand fractions into decimals and write decimals in the form of fraction.

Q. What are the main topics covered in the NCERT Solutions for Class 10 Maths Chapter 1?

Answer:

The main topics that are covered in the NCERT Solutions for Class 10 Maths Real Numbers are : 

1. Euclid's division algorithm,

2. The Fundamental Theorem of Arithmetic, 

3. Prove of Irrational numbers 

4. Decimal expansions.

Q. What is real no. class 10?

Answer:

Simply Real numbers are the combined set of rational and irrational numbers, in the number system

Q. What is an odd number?

Answer:

The whole numbers that leaves 1 as a remainder when divided by 2 are called as Odd Numbers