# R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 1 - Real Numbers

## Chapter 1 - Real Numbers FA

If possible (4 + 3√5) is rational.

Now, (4 + 3√5) - 4 = 3√5 is rational (∵ difference of two rationals is rational)

Also, × 3√5 = √ is rational. (∵ Product of two rationals is rational)

Therefore, √5 is rational.

This contradicts the fact √5 is irrational.

Since the contradiction arises by assuming (4 + 3√5) rational.

Hence, (4 + 3√5) is irrational.

## Chapter 1 - Real Numbers Ex. 1A

For any two given positive integers a and b there exist unique whole numbers q and r such that

Here, we call 'a' as dividend, b as divisor, q as quotient and r as remainder.

Dividend = (divisor quotient) + remainder

By Euclid's Division algorithm we have:

Dividend = (divisor × quotient) + remainder

= (61 27) + 32 = 1647 + 32 = 1679

By Euclid's Division Algorithm, we have:

Dividend = (divisor quotient) + remainder

Here, 612 < 1314

Applying Euclid's division algorithm, we get

1314 = 612 × 2 + 90 ….. r ≠ 0

612 = 90 × 6 + 72 ….. r ≠ 0

90 = 72 × 1 + 18 ….. r ≠ 0

72 = **18** × 4 +
0

Since remainder is zero.

Hence, HCF of 612 and 1314 is 18.

Here, 1260 < 7344

Applying Euclid's division algorithm, we get

7344 = 1260 × 5 + 1044 ….. r ≠ 0

1260 = 1044 × 1 + 216 ….. r ≠ 0

1044 = 216 × 4 + 180 ….. r ≠ 0

216 = 180 × 1 + 36 ….. r ≠ 0

180
= **36** × 5 + 0

Since remainder is zero.

Hence, HCF of 1260 and 7344 is 36.

Here, 4052 < 12576

Applying Euclid's division algorithm, we get

12576 = 4052 × 3 + 420 ….. r ≠ 0

4052 = 420 × 9 + 272 ….. r ≠ 0

420 = 272 × 1 + 148 ….. r ≠ 0

272 = 148 × 1 + 148 ….. r ≠ 0

148 = 124 × 1 + 24 ….. r ≠ 0

124 = 24 × 5 + 4 ….. r ≠ 0

24 = **4** × 6 +
0

Since remainder is zero.

Hence, HCF of 4052 and 12576 is 4.

Here, 650 < 1170

Applying Euclid's division algorithm, we get

1170 = 650 × 1 + 520 ….. r ≠ 0

650 = 520 × 1 + 130 ….. r ≠ 0

520 = **130** × 4 +
0

Since remainder is zero.

Therefore, HCF of 650 and 1170 is 130.

Hence, the largest number which divides 650 and 1170 is 130.

We know that,

Smallest prime number = 2

Smallest composite number = 4

HCF of 2 and 4 is 2.

Hence, the HCF of the smallest prime number and the smallest composite number is 2.

Any positive integer is of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 for some positive integer n.

When n = 6m,

n^{3} - n

= (6m)^{3 }- 6m

= 216 m^{3 }- 6m

= 6m(36m^{2 }- 1)

= 6q, where q = m(36m^{2}
-1)

_{}n^{3} - n is divisible
by 6

When n = 6m + 1,

n^{3} - n

= n(n^{2 }- 1)

= n (n - 1) (n + 1)

= (6m + 1) (6m) (6m + 2)

= 6m(6m + 1) (6m + 2)

= 6q, where q = m(6m + 1) (6m + 2)

∴n^{3}
- n is divisible by 6

When n = 6m + 2,

n^{3}- n

= n (n - 1) (n + 1)

= (6m + 2) (6m + 1) (6m + 3)

= (6m + 1) (36 m^{2} +
30m + 6)

= 6m (36 m^{2} + 30m +
6) + 1 (36m^{2} + 30m + 6)

= 6[m (36m^{2 }+ 30m +
6)] + 6 (6m^{2} + 5m + 1)

= 6p + 6q,

Where p = m (36m^{2} +
30m + 6)

q = 6m^{2} + 5m + 1

∴ n^{3
}- n is divisible by 6

When n = 6m + 3

n^{3} - n

= (6m + 3)^{3} - (6m +
3)

= (6m + 3) [(6m + 3)^{2}
- 1]

= 6m [6m + 3)^{2} - 1]
+ 3 [(6m + 3)^{2} - 1]

= 6 [m [(6m + 3)^{2 }-
1] + 3 [36m^{2 }+ 36m + 8]

= 6 [m [(6m + 3)^{2} - 1]
+ 6 [18m^{2} + 18m + 4]

= 6p + 3q,

Where p = m[(6m + 3)^{2}
- 1]

q = 18m^{2} + 18m + 4

∴ n^{3}
- n is divisible by 6.

Let the two odd positive no. be x = 2k + 1 and y = 2p + 1

Hence, x^{2}
+ y^{2} = (2k + 1)^{2} +(2p + 1)^{2}

= 4k^{2 }+ 4k +
1 + 4p^{2 }+ 4p + 1

= 4k^{2 }+ 4p^{2
}+ 4k + 4p + 2

= 2(2k^{2 }+ 2p^{2}
+ 2k + 2p + 1)

Therefore, x^{2}
+ y^{2} = 2m, where m = 2k^{2 }+ 2p^{2 }+ 2k + 2p +
1.

So, x^{2 }+
y^{2 }is an even number.

Also, it does not have any multiple of 4 as a factor, hence

x^{2} + y^{2 }is even but not divisible by 4.

Her, 1190 < 1445

1445 = 1190 x 1 + 255 ….. r ≠ 0

1190 = 255 x 4 +170 ….. r ≠ 0

255 = 170 x 1 + 85 ….. r ≠ 0

170 = **85** x 2 + 0

Since remainder is zero

Therefore, HCF of 1190 and 1445 is 85.

85 = 255 - 170 x 1

= (1445 - 1190 x 1) - (1190 - 255 x 4)

= (1445 - 1190) - [1190 - (1445 - 1190) x 4]

= (1445 - 1190) - (1190 - 1445 x 4 - 1190 x 4)

= 1445 - 1190 - (1190 x 5 - 1445 x 4)

= 1445 - 1190 - 1190 x 5 + 1445 x 4

= 1445 x 5 - 1190 x 6

= 1190 x (- 6) + 1445 x 5

= 1190m + 1445n, where m= -6 and n = 5

Here, 441< 567

Applying Euclid's division algorithm, we get

567 = 441 × 1 + 126 ….. r ≠ 0

441 = 126 × 3 + 63 ….. r ≠ 0

126
= **63** × 2 + 0

Since remainder is zero.

Therefore, HCF of 441 and 567 is 63. ….(i)

Here, 63 < 693

Applying Euclid's division algorithm, we get

693
= **63** × 11 + 0

Since remainder is zero.

Therefore, HCF of 63 and 693 is 63. ….(ii)

From (i) and (ii),

HCF of 441, 567 and 693 is 63.

1251 - 1 = 1250

9377 - 2 = 9375

15628 - 3 = 15625

Now, we find the HCF of 1250, 9375 and 15625.

Here, 1250 < 9375

Applying Euclid's division algorithm, we get

9375 = 1250 × 7 + 625 ….. r ≠ 0

1250
= **625** × 2 + 0

Since remainder is zero.

Therefore, HCF of 1250 and 9275 is 625.….(i)

Here, 625 < 15625

Applying Euclid's division algorithm, we get

15625
= **625** × 25 + 0

Since remainder is zero.

Therefore, HCF of 625 and 15625 is 625. ….(ii)

From (i) and (ii),

HCF of 1250, 9375 and 15625 is 625.

Hence, the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively is 625.

## Chapter 1 - Real Numbers Ex. 1B

429 = 3 × 11 × 13

5005 = 5 × 7 × 11 × 13

2431 = 11 × 13 × 17

21 = 3 × 7

28 = 2
× 2 × 7 = 2^{2} × 7

36 = 2
× 2 × 3 × 3 = 2^{2} × 3^{2}

HCF(21, 28, 36) = 1

LCM(21, 28, 36) = 2^{2}
× 3^{2} × 7 = 4 × 9 × 7 = 252

Prime factorization of 404 and 96 is given by

404 = 2 × 2 × 101 = 2^{2
}× 101

96 = 2 × 2 × 2 × 2 × 2 × 3
= 2^{5} × 3

HCF(404, 96) = 2 × 2 = 4

LCM(404, 96) = 2^{5}
× 3 × 101 = 9696

We have,

HCF × LCM = Product of the two numbers

⇒ 4 × 9696 = 404 × 96

⇒ 38784 = 38784

⇒ LHS = RHS

Hence, it is verified.

a = x^{3}y^{2}
= x × x × x × y × y

b = xy^{3}
= x × y × y × y

HCF(a, b) = x × y ×
y = xy^{2}

LCM(a, b) = x^{3}y^{3}

We know that,

Product of the two numbers = HCF × LCM

⇒ ab = 5 × 200

⇒ ab = 1000

Let a and b be two numbers.

According to the condition,

LCM(a, b) = 9 × HCF(a, b) …. (i)

LCM(a, b) + HCF(a, b) = 500 …. (ii)

From (i) and (ii), we get

9 × HCF(a, b) + HCF(a, b) = 500

⇒ 10 HCF(a, b) = 500

⇒ HCF(a, b) = 50

We know that,

LCM = Product of highest power of each factor involved in the numbers.

HCF = Product of smallest power of each common factor

⇒ LCM is always a multiple of HCF.

⇒ LCM = k × HCF

Here, LCM = 175 and HCF = 15

But in this case, LCM ≠ k × HCF.

Therefore, two numbers can't have LCM as 175 and HCF as 15.

The prime factorization of 42, 49 and 63 are

42 = 2 × 3 × 7

49 = 7 × 7

63 = 3 × 3 × 7

⇒ HCF(42, 49, 63) = 7

Hence, greatest possible length of each plank is 7m.

We have total length of timber = 42 + 49 + 63 = 154

Number of planks = 154/7 = 22

Therefore, 22 planks were formed.

Subtracting 5 and 7 from 320 and 457 respectively:

320 - 5 = 315,

457 - 7 = 450

Let us now find the HCF of 315 and 405 through prime factorization:

The required number is 45.

By going upward

5 11= 55

55 3= 165

1652 = 330

330 2 = 660

Let us find the HCF of 336, 240 and 96 through prime factorization:

Each stack of book will contain 48 books

Number of stacks of the same height

7 m = 700cm, 3m 85cm = 385 cm

12 m 95 cm = 1295 cm

Let us find the prime factorization of 700, 385 and 1295:

Greatest possible length = 35cm

Let us find the prime factorization of 1001 and 910:

1001 = 11 7 13

910 = 2 5 7 13

H.C.F. of 1001 and 910 is 7 13 = 91

Maximum number of students = 91

Let us find the LCM of 64, 80 and 96 through prime factorization:

L.C.M of 64, 80 and 96

=

Therefore, the least length of the cloth that can be measured an exact number of times by the rods of 64cm, 80cm and 96cm = 9.6m

Interval of beeping together = LCM (60 seconds, 62 seconds)

The prime factorization of 60 and 62:

60 = 30 2, 62 = 31 2

L.C.M of 60 and 62 is 30 31 2 = 1860 sec = 31min

electronic device will beep after every 31minutes

After 10 a.m., it will beep at 10 hrs 31 minutes

## Chapter 1 - Real Numbers Ex. 1D

√2 = 1.414….and √3 = 1.732…

We know that, 1.414 < 1.5 < 1.732

Therefore a rational number between √2 and √3 is 1.5.

**Proof:**

Let us assume that √6 is a rational number.

, where a and b are integers having no common factor other than 1, and b ≠ 0.

Now, ….( on squaring both sides)

⇒
6b^{2} = a^{2} ……(1)

⇒ 6 divides a^{2}
….. (∵ 6 divides 6b^{2})

⇒ 6 divides a

Let a = 6c for some integer c

Putting a = 6c in (1), we get

6b^{2} =
36c^{2}

⇒ b^{2} =
6c^{2}

⇒ 6 divides b^{2}
….. (∵ 6 divides 6c^{2})

⇒ 6 divides b …..
(∵ 6 divides b^{2} = 6 divides b)

Thus, 6 is a common factor of a and b

But, this contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming that √6 is rational.

Hence, √6 is irrational.

If possible (2 + √3) is rational.

⇒ (2 + √3) - 2 is rational…. (∵ difference of two rationals is rational)

Therefore, √3 is rational.

This contradicts the fact √3 is irrational.

Since the contradiction arises by assuming (2 + √3) rational.

Hence, (2 + √3) is irrational.

If possible (4 - √3) is rational.

⇒ 4 -(4 - √3) is rational….(∵ difference of two rationals is rational)

Therefore, √3 is rational.

This contradicts the fact √3 is irrational.

Since the contradiction arises by assuming (4 - √3) rational.

Hence, (4 - √3) is irrational.

If possible (3 + 5√2) is rational.

Now, (3 + 5√2) - 3 = 5√2 is rational (∵ difference of two rationals is rational)

Also, × 5√2 = √2 is rational (∵ Product of two rationals is rational)

Therefore, √2 is rational.

This contradicts the fact √2 is irrational.

Since the contradiction arises by assuming (3 + 5√2) rational.

Hence, (3 + 5√2) is irrational.

If possible (2 + 3√5) is rational.

Now, (2 + 3√5) - 2 = 3√5 is rational (∵ difference of two rationals is rational)

Also, × 3√5 = √5) is rational (∵ Product of two rationals is rational)

Therefore, √5 is rational.

This contradicts the fact √5 is irrational.

Since the contradiction arises by assuming (2 + 3√5) rational.

Hence, (2 + 3√5) is irrational.

If possible is rational.

Now, (∵ difference of two rationals is rational)

Also, (∵ Product of two rationals is rational)

Therefore, √2 is rational.

This contradicts the fact √2 is irrational.

Since the contradiction arises by assuming rational.

Hence, is irrational.

Let us assume that from 1 is a rational number.

, where a and b are non - zero integers having no common factor other than 1, and b ≠ 0.

Now,

But, 5a and 3b are non - zero integers.

is rational.

Thus, from (2), it follows that √5 is rational.

This contradicts the fact that √5 is irrational.

The contradiction arises by assuming that is rational.

Hence, is irrational.

**Proof:**

Let us assume that (√3 + √5) is a rational number.

⇒ (√3 + √5) = a, where a is rational.

⇒ 3 = (a - √5)

On squaring both sides, we get

⇒ 3 = (a - √5)^{2}
= a^{2} + 5 - 2a√5

⇒a^{2} + 2
- 2a√5 = 0

But, is a rational number.

Thus, from (1), √5 is rational.

But, this contradicts the fact that √5 is irrational.

The contradiction arises by assuming that (√3 + √5) √6 is rational.

Hence, (√3 + √5) is irrational.

(i) The sum of two rationals is always rational - True

(ii) The product of two rationals is always rational - True

(iii) The sum of two irrationals is an irrational - False

(iv) The product of two irrationals is an irrational - False

(v) The sum of a rational and an irrational is irrational - True

(vi) The product of a rational and an irrational is irrational - True

## Chapter 1 - Real Numbers Ex. 1E

A whole number that can be divided evenly by numbers other than 1 or itself.

## Chapter 1 - Real Numbers Ex. 1C

## Chapter 1 - Real Numbers MCQ

### Other Chapters for CBSE Class 10 Maths

Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 15- Perimeter and Areas of Plane Figures Chapter 16- Areas of Circle, Sector and Segment Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- Probability### R S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects

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