Real Numbers| NCERT Solutions| Class 10

NCERT Solutions of Chapter: Real Numbers. NCERT Solutions along with worksheets and notes for Class 9.

 

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

Ans-

(i) 135 and 225

225 > 135 [225 is dividend and 135 is divisor]

Therefore,

[Divide 225 by 135, get 1 as a quotient and 90 as remainder]

225 = 135 × 1 + 90

[Divide 135 by 90, get 1 as a quotient and 45 as remainder]

135 = 90 × 1 + 45

[Divide 90 by 45, get 2 as a quotient and 0 as remainder]

90 = 45 × 2 + 0

HCF (225, 135) = 45

(ii) 196 and 38220

[Divide 38220 by 196, get 195 as a quotient and 0 as remainder]

38220 = 196 × 195 + 0

HCF (38220, 196) = 196

(iii) 867 and 255

[Divide 8667 by 255, get 3 as a quotient and 102 as remainder]

867 = 255 × 3 + 102

[Divide 225 by 102, get 2 as a quotient and 51 as remainder]

255 = 102 × 2 + 51

[Divide 102 by 51, get 2 as a quotient and 0 as remainder]

102 = 51 × 2 + 0

HCF(867, 225) = 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.

Ans-

Let a is positive odd integer

Euclid’s division algorithm

a = bq + r, 0 ≤ r < b.

Here, q and r are quotient and remainder.

a = dividend and

b = divisor

Apply Euclid’s division algorithm with a and b = 6

Since 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.

Therefore, possible form of Euclid’s division lemma with remainder 0 and b= 6 is

a = 6q + 0

Here, q is quotient and 6q is divisible by 2.

Hence, a is even positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 1 and b = 6 is

a = 6q + 1

Here, q is quotient and 6q + 1 is not divisible by 2.

Hence, a is odd positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 2 and b =6  is

a = 6q + 2

Here, q is quotient and 6q + 2 is divisible by 2.

Hence, a is even positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 3 and b = 6  is

a = 6q + 3

Here, q is quotient and 6q + 3 is not divisible by 2.

Hence, a is odd positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 4 and b = 6  is

a = 6q + 4

Here, q is quotient and 6q + 4 is divisible by 2.

Hence, a is even positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 5 and b = 6  is

a = 6q + 5

Here, q is quotient and 6q + 5 is not divisible by 2.

Hence, a is odd positive integer.

But, a can not be even positive integer. (Given condition)

Therefore, a can not be 6q and 6q + 2 (even integers)

Therefore, any odd integer is of the form 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 column in which they can march?

Ans-

Use Euclid’s algorithm to find HCF of 616 and 32

616 = 32 × 19 + 8

32 = 8 × 4 + 0

Therefore, HCF (616, 32) = 8

Maximum number of column in which members can march = HCF (616, 32)

=8

Hence, the maximum number of column in which members can march is 8.

 

(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 + 1.]

Ans-

Let a is positive integer

Euclid’s division algorithm

a = bq + r, 0 ≤ r < b.

Here, q and r are quotient and remainder.

a = dividend and

b = divisor

Apply Euclid’s division algorithm with a and b = 3

Since 0 ≤ r < 3, the possible remainders are 0, 1 and 2.

Therefore, possible form of Euclid’s division lemma with remainder 0 and b= 3 is

a = 3q + 0

a2 = 9q2 ( Squaring both sides)

a2 = 3(3q2)

= 3k1

Therefore, possible form of Euclid’s division lemma with remainder 1 and b= 3 is

a = 3q + 1

a2 = (3q + 1)2 ( Squaring both sides)

a2 = 9q2 + 6q + 1

= 3(3q2 + 2q) + 1

= 3k2 + 1

Therefore, possible form of Euclid’s division lemma with remainder 2 and b= 3 is

a = 3q + 2

a2 = (3q + 2)2 ( Squaring both sides)

a2 = 9q2 + 12q + 4

= 9q2 + 12q + 3 + 1

= 3(3q2 + 4q + 1) +1

=3k3 + 1

Where k1, k2 and k3 are some positive integers.

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.

Ans-

Let a is positive integer

Euclid’s division algorithm

a = bq + r, 0 ≤ r < b.

Here, q and r are quotient and remainder.

a = dividend and

b = divisor

Apply Euclid’s division algorithm with a and b = 3

Since 0 ≤ r < 3, the possible remainders are 0, 1 and 2.

Therefore, possible form of Euclid’s division lemma with remainder 0 and b= 3 is

a = 3q + 0

a3 = 27q3 (Cubing both sides)

a3 = 9(3q3)

= 9k1

Therefore, possible form of Euclid’s division lemma with remainder 1 and b= 3 is

a = 3q + 1

a3 = (3q + 1)3 (Cubing both sides)

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

=9(3q3 + 3q2 + q) + 1

= 9k2 + 1

Where, k2 is positive integer.

Therefore, possible form of Euclid’s division lemma with remainder 2 and b= 3 is

a = 3q + 2

a3 = (3q + 2)3 (Cubing both sides)

a3 = 27q3 + 54q2 + 36q +8

=9(3q3 + 6q2 + 4q) + 8

= 9k3 + 1

Where, k3 is positive integer.

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

 

Exercise 1.2

(1) Express each number as a product of its prime factors:

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

Ans-

(i) 140

Prime factorisation of 140 = 2 × 2 × 7 × 5

= 22 × 5 × 7

(ii) 156

Prime factorisation of 156 = 2 × 2 × 13 × 3

= 22 × 3 × 13

(iii) 3825

Prime factorisation of 3825 = 5 × 5 × 3 × 3 × 17

= 32 × 52 × 17

(iv) 5005

Prime factorisation of 5005 = 5 × 7 × 11 × 13

(v) 7429

Prime factorisation of 7429 =17 × 19 × 23

 

(2) Find the LCM and HCF of the following pairs of integers and verify thast LCM × HCF = Product of the two numbers.

(i) 26 and 91      (ii) 510 and 92       (iii) 336 and 54

Ans-

(i) 26 and 91

Prime factorisation of 26 = 2 × 13

Prime factorisation of 91 = 7 × 13

HCF (26, 91) = Product of the smallest power of each common prime factor in the numbers.

= 13

LCM (26, 91) = Product of the greatest power of each common prime factor in the numbers.

= 2 × 7 × 13

= 182

Verification

LCM × HCF = Product of the two numbers.

182 × 13 = 26 × 91

2366 = 2366

LHS = RHS

Hence, verified.

(ii) 510 and 92

Prime factorisation of 510 = 2 × 3 × 5 × 17

Prime factorisation of 92 = 2 × 2 × 23

HCF (510, 92) = Product of the smallest power of each common prime factor in the numbers.

= 2

LCM (510, 92) = Product of the greatest power of each common prime factor in the numbers.

= 2  2  3  5  17  23

= 23460

Verification

LCM × HCF = Product of the two numbers.

23460 × 2 = 510 × 92

46920 = 46920

LHS = RHS

Hence, verified.

(iii) 336 and 54

Prime factorisation of 336 = 2 × 2 × 2 × 2 × 3 × 7

Prime factorisation of 54 = 2 × 3 × 3 × 3

HCF (336, 54) = Product of the smallest power of each common prime factor in the numbers.

= 2 × 3 = 6

LCM (336, 54) = Product of the greatest power of each common prime factor in the numbers.

= 2 × 2 × 2 × 2 × 3 × 3

= 23460

Verification

LCM × HCF = Product of the two numbers.

23460 × 2 = 510 × 92

46920 = 46920

LHS = RHS

Hence, verified.

 

(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

Ans-

(i) 12, 15 and 21

Prime factorisation of 12 = 2 × 2 × 3

Prime factorisation of 15 = 3 × 5

Prime factorisation of 21 = 3 × 7

HCF (12, 15, 21) = Product of the smallest power of each common prime factor in the numbers.

= 3

LCM (12, 15, 21) = Product of the greatest power of each common prime factor in the numbers.

= 2 × 2 × 3 × 5 × 7

= 420

Hence, LCM (12, 15, 21) = 420 and HCF (12, 15, 21) = 3

(ii) 17, 23 and 29

Prime factorisation of 17 = 1 × 17

Prime factorisation of 23 = 1 × 23

Prime factorisation of 29 = 1 × 29

HCF (17, 23, 29) = Product of the smallest power of each common prime factor in the numbers.

= 1

LCM (17, 23, 29) = Product of the greatest power of each common prime factor in the numbers.

= 17 × 23 × 29

= 11339

Hence, LCM (17, 23, 29) = 11339 and HCF (17, 23, 29) = 1

(iii) 8, 9 and 25

Prime factorisation of 8 =1 × 2 × 2 × 2

Prime factorisation of 9 =1× 3 × 3

Prime factorisation of 25 =1 × 5 × 5

HCF (8, 9, 25) = Product of the smallest power of each common prime factor in the numbers.

= 1

LCM (8, 9, 25) = Product of the greatest power of each common prime factor in the numbers.

= 2 × 2 × 2 × 3 × 3 × 5 × 5

= 1800

Hence, LCM (8, 9, 25) = 1800 and HCF (8, 9, 25) = 1

 

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

Ans- If a number will end up with digit 0,

Therefore, that digit will be divisible by 10 or its prime factors, ie, 5 and 2.

But,

Prime factors of 6n = (3  2)n

Therefore, 6n can be divisible by only 3 and 2.

But, according to Uniqueness of the Fundamental Theorem of Arithmetic, there could not be any other prime factor in the prime factorisation of 6n .

Hence, there is no natural number n for which 6n ends with digit 0.

 

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

Ans-

7 × 11 × 13 + 13

= 13(77 + 1)

= 13 × 78

= 13 × 13 × 6

7 × 11 × 13 + 13 is a composite number because it can be written as product of 13 × 13 × 6. Composite numbers are those numbers which has two or more than two factors other than 1.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5(1009)

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number because it can be written as product of 5  1009. Composite numbers are those numbers which has two or more than two factors other than 1.

 

(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?

Ans- Time taken by Ravi = 12 min

Time taken by Sonia = 18 min

To calculate time when Ravi will meet Sonia we will take LCM of time taken by both to complete a round.

Prime factorisation of 12 = 2 × 2 × 3

Prime factorisation of 18 = 2 × 3 × 3

LCM(12, 18) = 2 × 2 × 3 × 3 = 36 min

Hence, after 36 minutes they will meet again at the starting point.

 

Exercise 1.3

(1) Prove that √5 is irrational.

Ans-

Therefore, 5 divides b2.

So, 5 divides b. (If p divides a2, then p divides a)

Therefore, a and b have at least 5 as a common factor.

But, a and b have no common factor other than 1 because both are coprime numbers. Therefore, it contradicts.

Hence, √5 is irrational.

 

(2) Prove that 3 + 2√5  is irrational.

Ans-

Therefore, 245 divides c2.

So, 245 divides c. (If p divides a2, then p divides a)

Therefore, a and b have at least 245 as a common factor.

But, a and b have no common factor other than 1 because both are coprime numbers. Therefore, it contradicts.

Hence, 7√5 is irrational.

Therefore, √2  is rational.

But, this contradicts the fact that √2 is irrational.

Therefore, our assumption that 6 + √2 is rational is incorrect.

So, we conclude that 6 + √2 is irrational.

 

Exercise 1.4

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

Ans-

(i)

(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 factors of q?

Ans-

(i) 43.123456789

It is a terminating decimal expansion. Therefore, it is a rational number.

Hence, prime factor of q will either 2 or 5 or both.

(ii) 0.120120012000120000…

It is non- terminating non- recurring decimal expansion. Therefore, it is not rational number,ie, irrational number.

It is a terminating decimal expansion. Therefore, it is a rational number.

Hence, prime factor of q will not be of the form 2n 5m.

 

Helping Topics

Real Numbers

Worksheet Class 10