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

Real Numbers Exercise Ex. 1A

Solution 1

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

Solution 2

By Euclid's Division algorithm we have:

Dividend = (divisor × quotient) + remainder

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

Solution 3

By Euclid's Division Algorithm, we have:

Dividend = (divisor quotient) + remainder

Solution 4(i)

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.

Solution 4(ii)

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.

Solution 4(iii)

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.

Solution 5

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.

Solution 6

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.

Solution 7

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,

n3 - n

= (6m)3 - 6m

= 216 m3 - 6m

= 6m(36m2 - 1)

= 6q, where q = m(36m2 -1)

 n3 - n is divisible by 6

When n = 6m + 1,

n3 - n

= n(n2 - 1)

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

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

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

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

n3 - n is divisible by 6

When n = 6m + 2,

n3- n

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

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

= (6m + 1) (36 m2 + 30m + 6)

= 6m (36 m2 + 30m + 6) + 1 (36m2 + 30m + 6)

= 6[m (36m2 + 30m + 6)] + 6 (6m2 + 5m + 1)

= 6p + 6q,

Where p = m (36m2 + 30m + 6)

q = 6m2 + 5m + 1

n3 - n is divisible by 6

When n = 6m + 3

n3 - 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 [36m2 + 36m + 8]

= 6 [m [(6m + 3)2 - 1] + 6 [18m2 + 18m + 4]

= 6p + 3q,

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

q = 18m2 + 18m + 4

n3 - n is divisible by 6. 

Solution 8

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

Hence, x2 + y2 = (2k + 1)2 +(2p + 1)2

 = 4k2 + 4k + 1 + 4p2 + 4p + 1

 = 4k2 + 4p2 + 4k + 4p + 2

 = 2(2k2 + 2p2 + 2k + 2p + 1)

Therefore, x2 + y2 = 2m, where m = 2k2 + 2p2 + 2k + 2p + 1.

So, x2 + y2 is an even number.

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

x2 + y2 is even but not divisible by 4. 

Solution 9

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

Solution 10

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.

Solution 11

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.

Real Numbers Exercise Ex. 1B

Solution 1

429 = 3 × 11 × 13

Solution 2

5005 = 5 × 7 × 11 × 13

Solution 3

2431 = 11 × 13 × 17

Solution 4(i)

Solution 4(ii)

Solution 4(iii)

96 equals 2 cross times 2 cross times 2 cross times 2 cross times 2 cross times 3
404 equals 2 cross times 2 cross times 101
Therefore comma
HCF left parenthesis 96 comma 404 right parenthesis equals 2 cross times 2 equals 4
LCM left parenthesis 96 comma 404 right parenthesis equals 2 cross times 2 cross times 2 cross times 2 cross times 2 cross times 3 cross times 101 equals 9696
Now comma space HCF cross times LCM equals 4 cross times 9696 equals 38784
And comma space 96 cross times 404 equals 38784
therefore HCF cross times LCM equals Product space of space 96 space and space 404

 

Hence, verified.

Solution 4(iv)

Solution 4(v)

Solution 4(vi)

Solution 5(i)

Solution 5(ii)

Solution 5(iii)

Solution 5(iv)

Solution 5(v)

Solution 5(vi)

21 = 3 × 7

28 = 2 × 2 × 7 = 22 × 7

36 = 2 × 2 × 3 × 3 = 22 × 32

HCF(21, 28, 36) = 1

LCM(21, 28, 36) = 22 × 32 × 7 = 4 × 9 × 7 = 252

Solution 6

Prime factorization of 404 and 96 is given by

404 = 2 × 2 × 101 = 22 × 101

96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3

HCF(404, 96) = 2 × 2 = 4

LCM(404, 96) = 25 × 3 × 101 = 9696

We have,

HCF × LCM = Product of the two numbers

4 × 9696 = 404 × 96

38784 = 38784

LHS = RHS

Hence, it is verified.

Solution 7

a = x3y2 = x × x × x × y × y

b = xy3 = x × y × y × y

HCF(a, b) = x × y × y = xy2

LCM(a, b) = x3y3

Solution 8

Solution 9

Solution 10

We know that,

Product of the two numbers = HCF × LCM

ab = 5 × 200

ab = 1000

Solution 11

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

Solution 12

Solution 13

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.

Solution 14(i)

Solution 14(ii)

Solution 14(iii)

Solution 14(iv)

Solution 15

Clearly, the required number divides open parentheses 438 minus 6 close parentheses equals 432 and open parentheses 606 minus 6 close parentheses equals 600 exactly.

Therefore, required number = HCF(432, 600).

Now,

 

432 equals 2 cross times 2 cross times 2 cross times 2 cross times 3 cross times 3 cross times 3 equals 2 to the power of 4 cross times 3 cubed
600 equals 2 cross times 2 cross times 2 cross times 3 cross times 5 cross times 5 equals 2 cubed cross times 3 cross times 5 squared
HCF left parenthesis 432 comma 600 right parenthesis equals 2 cubed cross times 3 equals 24

 

Hence, 24 is the required number.

Solution 16

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.

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Largest four-digit number = 9999

Now,

4 = 2 × 2

7 = 7 × 1

13 = 13 × 1

Therefore, LCM (4, 7, 13) = 2 × 2 × 7 × 13 = 364

On dividing 9999 by 364, the remainder = 171

That means, 9999 - 171 = 9828 is exactly divided by 4, 7 and 13.

And, largest four-digit number leaving remainder 3 when divided by 4, 7 and 13 = 9828 + 3 = 9831

Solution 22

5 = 5 × 1

6 = 2 × 3

4 = 2 × 2

3 = 3 × 1

Therefore, LCM(5, 6, 4, 3) = 5 × 2 × 2 × 3 = 60

On dividing 2497 by 60, the remainder = 37

Hence, number to be added = 60 - 37 = 23

Solution 23

Difference between 43 and 91 = 91 - 43 = 48

Difference between 91 and 183 = 183 - 91 = 92

Difference between 183 and 43 = 183 - 43 = 140

 

Now,

48 = 2 × 2 × 2 × 2 × 3

92 = 2 × 2 × 23

140 = 2 × 2 × 5 × 7

Now, HCF (48, 92, 140) = 2 × 2 = 4

Hence, 4 is the greatest number that will divide 43, 91 and 183 leaving the same remainder.

Solution 24

Now,

20 - 14 = 6

25 - 19 = 6

35 - 29 = 6

40 - 34 = 6

 

And,

20 = 2 × 2 × 5 = 22 × 5

25 = 5 × 5 = 52

35 = 5 × 7

40 = 2 × 2 × 2 × 5 = 23 × 5

Therefore, LCM (20, 25, 35, 40) = 23 × 52 × 7 = 1400

 

Hence, required number = 1400 - 6 = 1394

Solution 25

Solution 26

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

Solution 27

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.

Solution 28

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

Solution 29

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

Solution 30

 

 

Solution 31

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

Solution 32

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

Solution 33

 

Therefore, all three lights will again change simultaneously after 7 min 12 sec, that is at 8:7:12 hours.

Solution 34

Solution 35

Now, a number ending with 0 should have 5 as factor.

But, 5 is not a factor of 6n.

Hence, 6n can never end with 0 for any natural number n.

Real Numbers Exercise Ex. 1C

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 2(v)

Solution 2(vi)

Solution 2(vii)

Solution 2(viii)

Solution 3


Real Numbers Exercise Ex. 1D

Solution 1

Rational Numbers:

The numbers of the form p over q, where p and q are integers and q not equal to 0, are called rational numbers.

Irrational numbers:

The numbers that are expressible as non-terminating and non-repeating decimals when expressed in decimal form are known as irrational numbers.

Real numbers:

Rational numbers and irrational numbers together form a set of real numbers.

Solution 2

Solution 3

√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.

Solution 4

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)

6b2 = a2 ……(1)

⇒ 6 divides a2 ….. (∵ 6 divides 6b2)

⇒ 6 divides a

Let a = 6c for some integer c

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

6b2 = 36c2

⇒ b2 = 6c2

⇒ 6 divides b2 ….. (∵ 6 divides 6c2)

⇒ 6 divides b ….. (∵ 6 divides b2 = 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.

Solution 5

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.

Solution 6

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.

Solution 7

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.

Solution 8

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.

Solution 9

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.

Solution 10

Solution 11

Solution 12

 

 

Solution 13

Solution 14

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.

Solution 15

Let   be rational.

Then, there exist co-primes   and   such that

Since   and   are integers, so   is rational.

But,  is irrational.

Thus, we arrive at a contradiction.

Since the contradiction arises by assuming that   is rational, hence,   is irrational.

Solution 16

 

Solution 17

(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 always an irrational - False

(iv) The product of two irrationals is always 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

Real Numbers Exercise Ex. 1E

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

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

Solution 8

Solution 9

Solution 10

 

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Real Numbers Exercise MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Correct option: (d)

LCM space of space open parentheses 2 cubed cross times 3 cross times 5 close parentheses space and space open parentheses 2 to the power of 4 cross times 5 cross times 7 close parentheses equals 2 to the power of 4 cross times 3 cross times 5 cross times 7 equals 1680


Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Real Numbers Exercise Test Yourself

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

A number is a terminating deicmal, if its denominator is of the form 2 to the power of m cross times 5 to the power of n, where m and n are non-negative integers.

Now,


17 over 30 equals fraction numerator 17 over denominator 2 cross times 3 cross times 5 end fraction


Since the denominator of 17 over 30 is not of the form 2 to the power of m cross times 5 to the power of n as it has an additional factor 3 also.


Therefore, 17 over 30 is not a terminating decimal.

Solution 8

Solution 9

Solution 10

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.

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

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