# NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers

**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**

**Sol. (i)** Since 225 > 135, applying division lemma to 225 and 135

$$135)225(1\\\frac{135}{\qquad90)135(1}\\\qquad\qquad\qquad45)90(2\\\qquad\qquad\qquad\space\frac{90}{×}$$

The remainder has now become zero.

Least divisor at 45.

The HCF of 135 and 225 is 45.**Second Method**

Since

225 > 135 we apply the division lemma to 225 and 135, to get

225 = 135 × 1 + 90

Remainder

90 ≠ 0, Again apply division lemma 135 and 90.

135 = 90 × 1 + 45

Remainder

45 ≠ 0, Again apply division lemma 90 and 45

90 = 45 × 2 + 0

Remainder = 0, so our procedure stops.

Then HCF of 225 and 135 = 45.

**(ii)** Since 38220 > 196, we apply the division lemma to 38220 and 196.

38220 = 196 × 195 + 0

Remainder = 0, so our procedure stops

Then HCF of 38220 and 196 = 195.

**(iii)** Since 867 > 255, we apply the division lemma to 867 and 255.

867 = 255 × 3 + 102

Then

255 = 102 × 2 + 51

102 = 51 × 2 + 0

Remainder = 0, so our procedure stops then HCF of 867 and 255 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.**

**Sol.** By Euclid's division lemma.

a = bq + r, 0 ≤ r < b

According to question, b = 6

a = 6q + r ...(i)

0 ≤ r < 6

Then r = 0, 1, 2, 3, 4, 5.

Put the value r in equation (i)

a = 6q

a = 6q + 1

a = 6q + 2

a = 6q + 3

a = 6q + 4

a = 6q + 5

6q, 6q + 2, 6q + 4 are the even integer.

Remaining integer 6q + 1, 6q + 3 and 6q + 5 are odd integer.

**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?**

**Sol.** Maximum number of columns

= HCF of 616 and 32

Let us use Euclid's algorithm to find their HCF.

616 = 32 × 9 + 8

32 = 8 × 4 + 0

So, the HCF of 616 and 32 is 8.

Maximum number of columns = 8 columns.

**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 than 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].**

**Sol.** Positive integer a = 3q + r, 0 ≤ r < 3

Then r = 0, 1, 2

Positive integer are 3q, 3q + 1 and 3q + 2.

Take

a = 3q

a^{2} = 9q^{2}

⇒ a^{2} = 3. 3q^{2}

Let be 3q^{2} = m

a^{2} = 3m

Again take

a = 3q + 1

a^{2} = (3q + 1)^{2}

Squaring both side

⇒ a^{2} = 9q^{2} + 1 + 6q

⇒ a^{2} = 9q^{2} + 6q + 1

= 3(3q^{2} + 2q) + 1

Let 3q^{2} + 2q = m

= 3m + 1

Again take

a = (3q + 2)

a^{2} = (3q + 2)^{2}

Squaring both side

⇒ a^{2} = 9q^{2} + 4 + 12q

⇒ a^{2} = 9q^{2} + 12q + 3 + 1

⇒ a^{2} = 3(3q^{2} + 4q + 1) + 1

Let (3q^{2} + 4q + 1) = m

a^{2} = 3m + 1

So, it is clear 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.**

**Sol.** Use Euclid's division lemma.

We know that positive integers are 3q, 3q + 1 and 3q + 2.

Take a = 3q

a^{3} = (3q)^{3}

⇒ a^{3} = 27q^{3}

⇒ a^{3} = 9 × 3q^{3}

Let 3q^{3} = m

a^{3} = 9 m

Again take

a = 3q + 1

a^{3} = (3q + 1)^{3}

[(a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}]

⇒ a^{3} = 27q^{3} + 1 + 27q^{2} + 9q

⇒ a^{3} = 27q^{3} + 27q^{2} + 9q + 1

⇒ a^{3} = 9(3q^{3} + 3q^{2} + q) + 1

Let

m = 3q^{3} + 3q^{2} + q

a^{3} = 9m + 1

Let

m = 3q^{3} + 6q^{2} + 4q

a^{3} = 9m + 8

So,

it is clear that the 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**

**Sol. **(i) 140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7 × 1

(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**

**Sol.** (i) We have

26 = 2 × 13

91 = 7 × 13

LCM = Greatest powers of the prime factors

LCM = 2 × 7 × 13

= 182

HCF = Smallest powers of the common factors

HCF = 13

Product of two numbers = 26 × 91

= 2366 ...(i)

LCM × HCF

= 182 × 13

= 2366 ...(ii)

By equations (i) and (ii)

Product of the two numbers = LCM × HCF

**(ii)** We have

510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

= 2^{2} × 23

LCM = Greatest powers of the prime factors

LCM = 2^{2} × 3 × 5 × 17 × 23

= 1020 × 23

= 23460

HCF = Smallest powers of the common factors

= 2

Product of the two numbers = 510 × 92

= 46,290 ...(i)

LCM × HCF

= 23,460 × 2

= 46,290 ...(ii)

By equations (i) and (ii)

Product of the two numbers = LCM × HCF

**(iii)** We have,

336 = 2 × 2 × 2 × 3 × 7

= 2^{4} × 3 × 7

54 = 2 × 3 × 3 × 3

= 2 × 3^{3}

LCM = Greatest powers of the prime factors

= 2^{4} × 3^{3} × 7

= 16 × 27 × 7

= 3024

HCF = Smallest powers of the common factors

= 2 × 3

= 6

Product of the two numbers = 336 × 54

= 18144 ...(i)

LCM × HCF

= 3024 × 6

= 18144 ...(ii)

By equations (i) and (ii)

Product of the two numbers = LCM × HCF

**3. Find the LCM and HCF of the following integers by applying the prime factorisation method.(i) 12, 15 and 21(ii) 17, 23 and 29(iii) 8, 9 and 2**

**Sol.** **(i)** We have

12 = 2 × 2 × 3

15 = 3 × 5

21 = 3 × 7

LCM = 2^{2} × 3 × 5 × 7

LCM

= 4 × 3 × 5 × 7

= 12 × 5 × 7

= 60 × 7

= 420

HCF = 3

So,

LCM (12, 15, 21) = 420

HCF (12, 15, 21) = 3

**(ii)** We have

17 = 17 × 1

23 = 23 × 1

29 = 29 × 1

LCM

= 17 × 23 × 29

= 391 × 29

= 11339

HCF = 1

LCM (17, 23, 29) = 11,339

HCF (17, 23, 29) = 1

(iii) We have,

8 = 2 × 2 × 2 × 1 = 2^{3}

9 = 3 × 3 × 1 = 3^{2}

25 = 5 × 5 × 1 = 5^{2}

LCM = 2^{3} × 3^{2} × 5^{2}

= 8 × 9 × 25

= 200 × 9

= 1800

HCF = 1

So, LCM ( 8, 9, 25) = 1800

HCF (8, 9, 25) = 1

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

**Sol.** We know that

LCM of two numbers × HCF of two numbers

= Produced of Numbers

LCM × 9 = 306 × 657

LCM =(306×657)/9

LCM = 34 × 657

LCM = 22338

LCM (306, 657) = 22338

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

**Sol.** If the numbers 6^{n}, for any n, were to end with the digit 5, then it would be divisible by 5.

The prime factorisation of 6^{n} = (2 × 3)^{n}

Therefore, the prime factorisation of 6^{n} does not contain the prime number 5.

Hence, it is clear that for any Natural number n, 6^{n} is not divisible by 5 and thus it proves that 6^{n} cannot end with digit 0 for any natural number n.

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

**Sol.** By the definition of composite number, we know if a number is composite means it have two or more than two factors.

Take Number

7 × 11 × 13 + 13

⇒ 13 (7 × 11 + 1)

⇒ 13 × 78

Hence, Number is composite.

Again take,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 (5 is common)

⇒ 5(7 × 6 × 4 × 3 × 2 × 1 + 1)

⇒ 5(1008 + 1)

⇒ 5 × 1009

Hence, the number is composite.

**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?**

**Sol.** Number of minutes will they meet again at the starting point = LCM of 18 and 12

18 = 2 × 3 × 3 = 2 × 32

12 = 2 × 2 × 3 = 22 × 3

LCM = 22 × 32

= 4 × 9

= 36 minutes.

**Exercise 1.3**

**1. Prove that √ 5 is irratonal.**

**Sol. **Let us assume, to the contrary, that √ 5 is rational.

That is, we can find integers a and b (b ≠ 0).

Such that √ 5=a/b

Suppose a and b have a comon factor other than 1, then we can divide by the common factor and assume that a and b are coprime.

So, √ 5b=a

Squaring both sides, then

5b^{2} = a^{2} ...(i)

Therefore, a^{2} is divisible by 5, it follows that a is also divisible by 5.

So, a = 5c for some integer c

Put the value a in equation (i)

5b^{2} = 25c^{2}

⇒ b^{2} = 5c^{2}

This means that b^{2} is divisible by 5 and so b is also divisible by 5.

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

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √5 is rational.

So, we conclude that √5 is irrational.

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

**Sol.** Let us assume, to the contrary, that 3+2 **√**5+ is rational.

That is, we can find coprime a and b (b ≠ 0) such that

$$3+2\sqrt{5}=\frac{a}{b}\\2\sqrt{5}=\frac{a}{b}-3\\2\sqrt{5}=\frac{a-3b}{b}\\⇒\sqrt{5}=\frac{a-3b}{2b}\\\text{Since, a and b are integers, we get}\space\frac{a-3b}{2b} \space is\space rational\space and\space so\space\sqrt{5}\space is\space rational.\text{But this contradicts the fact that}\sqrt{5}\space\text{is irrational. So, we conclude that}\space3+2\sqrt{5}\text{is irrational.}$$

**3. Prove that the following are irrationals :**

$$(i)\space\frac{1}{\sqrt{2}}\\(ii)\space7\sqrt{5}\\\space(iii)\space6+\sqrt{2}$$

**Sol.** **(i)** Let us assume, to the contrary that 1/√2 is rational.

That is, we can find, integers a and b (b ≠ 0) Such that

(1/√2)=a/b

b = √2a

(a and b are coprime)

Squaring both sides

b^{2} = 2a^{2} ...(i)

Therefore, b^{2} is divisible by 2, then we know that b is divisible by 2.

Then b = 2c for some integer c

Put the value b in equation (i)

(2c)^{2} = 2a^{2}

⇒ 4c^{2} = 2a^{2}

⇒ a^{2} = 2c^{2}

This means that a^{2} is divisible by 2 and so a is also divisible by 2.

Therefore, a and b have at least 2 is a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that 1/√2 is rational.

So, we conclude that 1/√2 is irrational.

**(ii)** Let us assume, to the contrary, that 7√5 is rational.

That is, we can find coprime a and b (b ≠ 0) Such that

7√5=a/b

Then √5=a/7b

Since, a and b are integers, a/7b is rational, and so √5 is rational.

But this contradicts the fact that √5 is irrational.

So, we conclude that 7√5 is irrational.

**(iii)** Let us assume, to the contrary, that 6+√2 is rational.

That is, we can find coprime a and b (b ≠ 0) Such that

$$\space6+\sqrt{2}=\frac{a}{b}\\\sqrt{2}=\frac{a}{b}-6\\\sqrt{2}=\frac{a-6b}{b}\\\text{Since, a and b are integers, we get}\space\frac{a-6b}{b}\space\text{is rational and so}\sqrt{2}\space is rational.\\\text{But this contradicts the fact that}\space \sqrt{2}\space\text{is irrational.}\text{This contradiction has arisen because our incorrect assumption that}\space6+\sqrt{2}\space\text{is rational.}\\\space\text{So, we conclude that}\space6+\sqrt{2}\space\text{is irrational.}$$

**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 :**

$$(i)\frac{13}{3125}\\(ii)\frac{17}{8}\\(iii)\frac{64}{455}\\(iv)\frac{15}{1600}\\(v)\frac{29}{343}\\(vi)\frac{23}{2^35^2}\\(vii)\frac{129}{2^25^77^5}\\(viii)\frac{6}{15}\\(ix)\frac{35}{50}\\(x)\frac{77}{210}$$

**Sol. **$$(i)\frac{13}{3125}=\frac{13}{5×5×5×5×5}\\=\frac{13}{5^5}=\frac{13×2^5}{5^5×2^5}\\\text{Denominator is clearly form of}\space2^n×5^m\text{So, it is terminating.}$$

$$(ii)\frac{17}{8}=\frac{17}{2^3}=\frac{17×5^3}{2^3×5^3}\\\text{Denominator is clearly form of}\space2^n×5^m\space\text{So, it is terminating.}$$

$$(iii)\frac{64}{455}=\frac{64}{5×7×13}\\\text{Denominator is clearly not of the form}\space2^n×5^m\space\text{So, it is non-terminating repeating.}$$

$$(iv)\frac{15}{1600}=\frac{15}{2^6×5^2}\\\text{Denominator is clearly form of}\space2^n×5^m\space\text{So, it is terminating.}$$

$$(v)\frac{29}{343}=\frac{29}{7×7×7}=\frac{29}{7^3}\\\text{Denominator is clearly not of the form 2n}\space2^n×5^m\space\text{So, it is non-terminating repeating.}$$

$$(vi)\frac{23}{2^35^2}=\frac{23}{2^3×5^2}\\\text{Denominator is clearly form of}\space2^n×5^m\space\text{So, it is terminating.}$$

$$(vii)\frac{129}{2^25^77^5}=\frac{129}{2^2×5^7×7^5}\\\text{Denominator is clearly not of the form}\space2^n×5^m\space\text{So, it is non-terminating repeating.}$$

$$(viii)\frac{6}{15}=\frac{6}{3×5}\\=\frac{2×2}{5×2}\\=\frac{2}{5}\\\text{Denominator is clearly form of}\space2^n×5^m\space\text{So, it is terminating.}$$

$$(ix)\frac{35}{50}=\frac{7×5}{2×2^5}\\=\frac{7}{2×5}\\\text{Denominator is clearly form of}\space2^n×5^m\space\text{So, it is terminating.}$$

$$(x)\frac{77}{210}=\frac{77}{2×3×5×7}\\\text{Denominator is clearly not of the form}\space2^n×5^m\space\text{So, it is non-terminating repeating.}$$

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

**Sol.** $$(i)\space\frac{13}{3125}=\frac{13}{5×5×5×5×5}\\=\frac{13×2^5}{5^5×2^5}\\=\frac{13×32}{10^5}\\=\frac{416}{100000}\\=0.00416$$

$$(ii)\space\frac{17}{8}=\frac{17}{2^3}\\=\frac{17×5^3}{2^3×5^3}\\=\frac{17×125}{1000}\\=\frac{2125}{1000}\\=2.125$$

(iii) Non-terminating

$$(iv)\space\frac{15}{1600}=\frac{15}{2^6×5^2}\\=\frac{15×5^4}{2^6×5^2×5^4}\\=\frac{15×625}{2^6×5^6}\\=\frac{9375}{10^6}\\=0.09375$$

(v) Non-terminating

$$(vi)\space\frac{23}{2^3×5^2}=\frac{23×5}{2^3×5^2×5}\\=\frac{115}{10^3}\\=\frac{115}{1000}\\=0.115$$

(vii) Non-terminating

$$(viii)\space\frac{6}{15}=\frac{3×2}{3×5}\\=\frac{2×2}{5×2}\\=\frac{4}{10}\\=0.4$$

$$(ix)\space\frac{35}{50}=\frac{7×5}{2×5^2}\\=\frac{7}{2×5}\\=\frac{7}{10}\\=0.7$$

(x) 77/210 has a non-terminating decimal expansion.

**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 formp/q, what can you say about the prime factor of q?**

$$(i)\space43.123456789\\(ii)\space 0.120120012000120000....\\(iii)\space\overline{\text{43123456789}}$$

**Sol. (i)** Rational Prime factor of q is either 2 or 5 or both only.**(ii)** Irrational**(iii)** Rational, prime factors of q will also have a factor other than 2 or 5.