NCERT Grade 10-Real Numbers-Answers

NCERT Solutions for Class 10 Maths

Find 100% accurate solutions for NCERT Class X Math. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

Sign Up to Download as PDF

1.   (i) 135 and 225

We have 225 > 135,

So, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Here remainder 90 ≠ 0, we apply the division lemma again to 135 and 90 to obtain

135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45 ≠ 0, and apply the division lemma to obtain

90 = 2 × 45 + 0

Since that time the remainder is zero, the process get stops.

The divisor at this stage is 45

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

We have 38220 > 196,

So, we apply the division lemma to 38220 and 196 to obtain 38220 = 196 × 195 + 0

Since we get the remainder is zero, the process stops. The divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

We have 867 > 255,

So, we apply the division lemma to 867 and 255 to obtain

867 = 255 × 3 + 102

Here remainder 102 0, we apply the division lemma again to 255 and 102 to obtain

255 = 102 × 2 + 51

Here remainder 51 0, we apply the division lemma again to 102 and 51 to obtain

102 = 51 × 2 + 0

Since we get the remainder is zero, the process stops.

The divisor at this stage is 51,

Therefore, HCF of 867 and 255 is 51.

2.    Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,

a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, where k1 is a positive integer

6q + 3 = (6q + 2) + 1 =2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5

3.    We have to find the HCF (616, 32) to find the maximum number of columns in which they can march.

To find the HCF, we can use Euclid’s algorithm.

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

4.    Let a be any positive integer and b = 3.

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 Or,

a2= (3q)2 or (3q +1)2 or(3q + 2)2

a2 = (9q)or 9q2 + 6q +1 or 9q2 +12q+4

     = 3 × (3q2) or 3(3q2 +2q)+1 or 3(3q2 +4q +1)+1

     = 3k1 or 3k 2 +1 or 3k 3 +1

Where k1, k2, and k3 are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

5.    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. We have three cases.

Case 1: When a = 3q,

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

Where m is an integer such that m =3q3

Case 2: When a = 3q + 1,

a3 = (3q +1)3

a3 = 27q3+ 27q2+ 9q + 1

a= 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.

6.    (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

7.     (i) 26 and 91

26 = 2 × 13

91 = 7 × 13

HCF = 13

LCM = 2 × 7 × 13 = 182

Product of two numbers 26 and 91 = 26 × 91 = 2366 HCF × LCM = 13 × 182 = 2366

Hence, product of two numbers = HCF × LCM

(ii) 510 and 92

510= 2 × 3 × 5 × 17

92 = 2 × 2 × 23

HCF = 2

LCM = 2×2×3×5×17×23 = 23460

Product of two numbers 510 and 92 = 510 × 92 = 46920

HCF × LCM = 2 × 23460 = 46920

Hence, product of two numbers = HCF × LCM

(iii)  336 and 54

336 = 2×2×2×2×3× 7 = 24×3×7

54=2×3×3×3 = 2×33

HCF = 2 × 3 = 6

LCM = 24 × 33 × 7 = 3024

Product of two numbers 336 and 54 = 336 × 54 = 18144

HCF × LCM = 6 × 3024 = 18144

Hence, product of two numbers = HCF × LCM

8.    (i) 12, 15 and 21

12 = 22 × 3

15 = 3 × 5

21 = 3 × 7

HCF = 3

LCM = 22 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

17 =1 × 17

23 = 1 × 23

29 = 1 × 29

HCF = 1

LCM = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

8 = 2 × 2 × 2 = 23

9= 3 × 3 = 32

25 =5 x 5 =52

HCF = 1

LCM = 23 × 32 × 52= 1800

9.    HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers

∴ LCM × HCF = 306 × 657

LCM = 22338

10.   If any number ends with the digit 0, it should be divisible by 10. In other words, it will also be divisible by 2 and 5 as 10 = 2 × 5 Prime factorisation of 6n = (2 × 3)n

It can be observed that 5 is not in the prime factorisation of 6n. Hence, for any value of n, 6n will not be divisible by 5.

Therefore, 6n cannot end with the digit 0 for any natural number n.

11.   Numbers are of two types – prime and composite.

Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.

It can be observed that

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)

    = 13 × (77 + 1) = 13 × 78 = 13 × 13 × 6

The given expression has 6 and 13 as its factors.

Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

=5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1) = 5 × 1009

1009 cannot be factorized further

Therefore, the given expression has 5 and 1009 as its factors.

Hence, it is a composite number.

12.   It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

18 = 2 × 3 × 3 And, 12 = 2 × 2 × 3

LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

13.   Let us prove √5 irrational by contradiction.

Let us suppose that √5 is rational. It means that we have co-prime integers a and b (b ≠ 0) such that

⇒ b√5 = a

Squaring both sides, we get

⇒ 5b= a2                                                                     (1)

It means that 5 is factor of a2

Hence, 5 is also factor of a by Theorem.          (2)

If, 5 is factor of a, it means that we can write a = 5c for some integer c.

Substituting value of a in (1),

5b2 = 25c2        ⇒         b2 = 5c2

It means that 5 is factor of b2.

Hence, 5 is also factor of b by Theorem.            (3)

From (2) and (3), we can say that 5 is factor of both a and b. But, a and b are co-prime.

Therefore, our assumption was wrong. √5 cannot be rational. Hence, it is irrational.

14.   We will prove this by contradiction.

Let us suppose that (3+2√5 ) is rational.

It means that we have co-prime integers a and b (b ≠ 0) such that

a and b are integers.

It means L.H.S of (1) is rational but we know that A/3 is irrational. It is not possible. Therefore, our supposition is wrong. (3+2√5) cannot be rational.

Hence, (3+2 √5) is irrational.

15.   (i) We can prove  rational by contradiction.

Let us suppose that  is rotational.

It means we have some co-prime integers a and b (b ≠ 0) such that

R.H.S of (1) is rational but we know that √2 is irrational. It is not possible which means our supposition is wrong.

Therefore,  cannot be rotational.

Hence, it is irrational.

(ii) We can prove 7√5 irrational by contradiction.

Let us suppose that 7√5 is rational.

It means we have some co-prime integers a and b (b ≠ 0) such that

R.H.S of (1) is rational but we know that √5 is irrational. It is not possible which means our supposition is wrong. Therefore, 7√5 cannot be rational.

Hence, it is irrational.

(iii)  We will prove 6+ √2 irrational by contradiction.

Let us suppose that (6 + √2) is rational.

It means that we have co-prime integers a and b (b ≠ 0) such that

a and b are integers.

It means L.H.S of (1) is rational but we know that √2 is irrational. It is not possible.

Therefore, our supposition is wrong. (6 + √2) cannot be rational.

Hence, (6 + √2) is irrational.

16.   According to Theorem, any given rational number of the form  where p and q are q co-prime, has a terminating decimal expansion if q is of the form 2n × 5m, where m and n are non-negative integers.

(i) 

q = 3125 =5 × 5 × 5 × 5 × 5 =S5

Here, denominator is of the form 2n × 5m, where m = 5 and n = 0.

It means rational number  has a terminating decimal expansion.

(ii)  

q = 8 = 2 × 2 × 2 = 23

Here, denominator is of the form 2n × 5m, where m = 0 and n = 3.

It means rational number  has a terminating decimal expansion.

(iii) 

q = 455 = 5 × 91

Here, denominator is not of the form 2n × 5m, where m and n are non-negative integers.

It means rational number   has a non-terminating repeating decimal expansion.

(iv) 

 q = 320 =2 × 2 × 2 × 2 × 2 × 2 × 5= 2× 5

Here, denominator is of the form 2n × 5m, where m = 1 and n = 6.

It means rational number  has a terminating decimal expansion.

(v) 

q = 343 = 7 × 7 × 7

Here, denominator is not of the form 2n × 5m, where m and n are non-negative integers.

It means rational number  has non-terminating repeating decimal expansion.

(vi) 

q = 23 × 52

Here, denominator is of the form 2n × 5m, where m = 2 and n = 3 are non-negative integers.

It means rational number  has terminating decimal expansion.

(vii) 

q = 22 × 57× 75

Here, denominator is of the form 2n × 5m, where m = 2 and n = 3 are non-negative integers.

It means rational number has non-terminating repeating decimal expansion.

(viii) 

q = 5 = 51

Here, denominator is of the form 2n × 5m, where m = 1 and n = 0.

It means rational number  has terminating decimal expansion.

(ix) 

q = 10 = 2 × 5 = 21 × 51

Here, denominator is of the form 2n × 5m, where m = 1 and n = 1.

It means rational number  has terminating decimal expansion.

(x) 

q = 30 = 5 × 3 × 2

Here, denominator is not of the form 2n × 5m, where m and n are non-negative integers.

It means rational number has non-terminating repeating decimal expansion.

17.

18.   (i) 43.123456789

It is rational because decimal expansion is terminating. Therefore, it can be expressed in  form where factors of q are of the form 2n × 5m where n and m are non-negative integers.

(ii) 0.1201120012000120000…

It is irrational because decimal expansion is neither terminating nor non-terminating repeating.

(iii) 

It is rational because decimal expansion is non-terminating repeating.

Therefore, it can be expressed in  form where factors of q are not of the q form 2n × 5m where n and m are non-negative integers.

 

 

MySchoolPage connects you with exceptional, certified math tutors who help you stay focused, understand concepts better and score well in exams!




Mathematics - Videos


Physics - Videos


Biology - Videos


Chemistry - Videos

Added to Cart