A number which is a continuous quantity that represents a distance along a line and can be used for a variety of different calculations such as velocity, mass and energy. Real numbers can be of either positive numbers, negative numbers, decimals, fractions or even pi. There are quite a few applications of Real numbers such as Physics, Computation, Logic, Computation and Set Theory.

Real numbers include the following types of numbers:

- Whole Numbers
- Irrational Numbers
- Rational Numbers

Check out the RS Aggarwal Class 10 Solutions Chapter 1 Real Numbers are available below:

Q.1: What do you mean by Euclidâ€™s division lemma?

Sol:

Euclid’s division algorithm states that for any two positive integers a and b, there exist unique integers Q. and r, such that a = bQ + r, where Q â‰ 0 and r < b.

Q.2: A number when divided by 61 gives 27 as Quotient and 32 as remainder. Find the number.

Sol:

We know, Dividend = Divisor x Quotient + Remainder

Given: Divisor = 61, Quotient = 27, Remainder = 32

Let the Dividend be x.

Therefore, x = 61 x 27 + 32 = 1679

Hence, the required number is 1679.

Q.3: By what number should 1365 be divided to get 31 as Quotient and 32 as remainder?

Sol:

Given: Dividend = 1365, Quotient = 31, Remainder = 32

Let the divisor be x.

Dividend = Divisor x Quotient + Remainder

1365 = 31x + 32

\(\Rightarrow\) 1365 – 32 = 31x

\(\Rightarrow\) 1333 = 31x

\(\Rightarrow\) x = \(\frac{ 1333 }{ 31 }\) = 43

Hence, 1365 should be divided by 43 to get 31 as Quotient and 32 as remainder.

Q.4: Using Euclidâ€™s division algorithm, find the HCF of:

(i) 405 and 2520 (ii) 504 and 1188 (iii) 960 and 1575

Sol:

(i)

On applying Euclid’s algorithm, i.e. dividing 2520 by 405, we get:

Quotient = 6, Remainder = 90

Therefore, 2520 = 405 x 6 + 90

Again on applying Euclid’s algorithm, i.e. dividing 405 by 90, we get:

Quotient = 4, Remainder = 45

Therefore, 405 = 90 x 4 + 45

Again on applying Euclid’s algorithm, i.e. dividing 90 by 45, we get:

Therefore, 90 = 45 x 2 + 0

Hence, the HCF of 2520 and 405 is 45.

(ii)

On applying Euclid’s algorithm, i.e. dividing 1188 by 504, we get:

Quotient = 2, Remainder = 180

Therefore, 1188 = 504.2 +180

Again on applying Euclid’s algorithm, i.e. dividing 504 by 180, we get:

Quotient = 2, Remainder = 144

Therefore, 504 = 180x 2 + 144

Again on applying Euclid’s algorithm, i.e. dividing 180 by 144, we get:

Quotient = 1, Remainder = 36

Therefore, 180 = 144 x 1 + 36

Again on applying Euclid’s algorithm, i.e. dividing 144 by 36, we get:

Therefore, 144 = 36 x 4 + 0

Hence, the HCF of 1188 and 504 is 36.

(iii)

On applying Euclid’s algorithm, i.e. dividing 1575 by 960, we get:

Quotient = 1, Remainder = 615

Therefore, 1575 = 960 x 1 + 615

Again on applying Euclid’s algorithm, i.e. dividing 960 by 615, we get:

Quotient = 1, Remainder = 345

Therefore, 960 = 615 x 1 + 345

Again on applying Euclid’s algorithm, i.e. dividing 615 by 345, we get:

Quotient = 1, Remainder = 270

Therefore, 615 = 345 x 1 + 270

Again on applying Euclid’s algorithm, i.e. dividing 345 by 270, we get:

Quotient = 1, Remainder = 75

Therefore, 345 = 270 x 1 + 75

Again on applying Euclid’s algorithm, i.e. dividing 270 by 75, we get:

Quotient = 3, Remainder = 45

Therefore, 270 = 75 x 3 + 45

Again on applying Euclid’s algorithm, i.e. dividing 75 by 45, we get:

Quotient = 1, Remainder = 30

Therefore, 75 = 45 x 1 + 30

Again on applying Euclid’s algorithm, i.e. dividing 45 by 30, we get:

Quotient = 1, Remainder = 15

Therefore, 45 = 30 x 1 + 15

Again on applying Euclid’s algorithm, i.e. dividing 30 by 15, we get:

Quotient = 2, Remainder = 0

Therefore, 30 = 15 x 2 + 0

Hence, the H.C.F of 960 and 1575 is 15.

Q.5: Show that every positive integer is either even or odd.

Sol:

Let us assume that there exists a smallest positive integer that is neither odd nor even, say n. Since n is the least positive integer which is neither even nor odd, n – 1 must be either odd or even.

Case 1: If n – 1 is even, n – 1 = 2k for some k.

But this implies n = 2k + 1

This implies n is odd.

Case 2: If n – 1 is odd, n – 1 = 2k + 1 for some k.

But this implies n = 2k + 2 = 2(k + 1)

This implies n is even.

In both ways, we have a contradiction.

Thus, every positive integer is either even or odd.

Q.6: Show that any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

Sol:

Let n be any arbitrary positive odd integer.

On dividing n by 6, let m be the Quotient and r be the remainder. So, by Euclid’s division lemma, we have

n = 6m + r, where m â‰ 0 and Â r < 6.

As m â‰ 0 and Â r < 6 and r is an integer, r can take values 0, 1, 2, 3, 4, 5.

\(\Rightarrow\)n= 6m or n = 6m + 1 or n = 6m + 2 or n = 6m + 3 or n = 6m + 4 or n = 6m + 5

\(\Rightarrow\)n = 6m + 1 or n = 6m + 3 or n = 6m + 5

Thus, any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.

Q.7: Show that any positive odd integer is of the form ( 4m + 1 ) or ( 4m + 3),, where m is some integer.

Sol:

Let n be any arbitrary positive odd integer.

On dividing n by 4, let m be the Quotient and r be the remainder. So, by Euclid’s division lemma, we have

n = 4m + r, where m â‰ 0 and Â r < 4.

As m â‰ 0 and Â r < 4 and r is an integer, r can take values 0, 1, 2, 3.

\(\Rightarrow\) n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3

\(\Rightarrow\)n = 4m + 1 or n = 4m + 3

Thus, any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.

Q.8: Using prime factorization, find the HCF and LCF of:

(i) 36, 84 (ii) 23, 31 (iii) 96, 404 (iv) 144, 198 (v) 396, 1080

(vi) 1152, 1664

In each case, verify that:

HCF x LCM = product of given numbers.

Sol:

(i) 36, 84

Prime factorisation:

36 = 2^{2} x 3^{2}

84 = 2^{2} x 3 x 7

HCF = product of smallest power of each common prime factor in the numbers = 2^{2} x 3 = 12

LCM = product of greatest power of each prime factor involved in the numbers = 2^{2} x 3^{2} x 7 = 252

(ii) 23, 31

Prime factorization:

23 = 23

31 = 31

HCF = product of smallest power of each common prime factor in the numbers = 1

LCM = product of greatest power of each prime factor involved in the numbers = 23 x 31 = 713

(iii) 96, 404

Prime factorization:

96 = 2^{5} x 3

404 = 2^{2} x 101

HCF = product of smallest power of each common prime factor in the numbers = 2^{2} = 4

LCM = product of greatest power of each prime factor involved in the numbers = 2^{5} x 3 x 101 = 9696

(iv) 144, 198

Prime factorization:

144 = 2^{4} x 3^{2}

198 = 2 x 3^{2} x 11

HCF = product of smallest power of each common prime factor in the numbers = 2 x 3^{2} = 18

LCM = product of greatest power of each prime factor involved in the numbers = 2^{4} x 3^{2} x 11 = 1584

(v) 396, 1080

Prime factorization:

396 = 2^{2} x 3^{2} x 11

1080 = 2^{3} x 3^{3} x 5

HCF = product of smallest power of each common prime factor in the numbers = 2^{2} x 3^{2} = 36

LCM = product of greatest power of each prime factor involved in the numbers = 2^{3} x 3^{3} x 5 x 11 = 11880

(vi) 1152, 1664

Prime factorization:

1152 = 2^{7} x 3^{2}

1664 = 2^{7} x 13

HCF = product of smallest power of each common prime factor involved in the numbers = 2^{7} = 128

LCM = product of greatest power of each prime factor involved in the numbers = 2^{7} x 3^{2} x 13 = 14976

Q.9: Using prime factorization, find the HCF and LCF of:

(i) 8, 9, 25 (ii) 12, 15, 21 (iii) 17, 23, 29 (iv) 24, 36, 40

(iv) 24, 36, 40 (v) 30, 72, 432 (vi) 21, 28, 36, 45

Sol:

(i) 8 = 2 x 2 x 2 = 2^{3}

9 = 3 x 3 = 3^{2}

25 = 5 x 5 = 5^{2}

HCF = Product of smallest power of each common prime factor in the numbers = 1

LCM = Product of the greatest power of each prime factor involved in the numbers = 2^{3} x 3^{2} x 5^{2} = 1800

(ii) 12 = 2 x 2 x 3 = 2^{2} x 3

15 = 3 x 5

21 = 3 x 7

HCF = Product of smallest power of each common prime factor in the numbers = 3

LCM = Product of the greatest power of each prime factor involved in the numbers = 2^{2} x 3 x 5 x 7 = 420

(iii) 17 = 17

23 = 23

29 = 29

HCF = Product of smallest power of each common prime factor in the numbers = 1

LCM = Product of the greatest power of each prime factor involved in the numbers = 17 x 23 x 29 = 11339

(iv) 24 = 2 x 2 x 2 x 3 = 2^{3} x 3

36 = 2 x 2 x 3 x 3 = 2^{2} x 3^{2}

40= 2 x 2 x 2 x 5 = 2^{3} x 5

Therefore, HCF = Product of smallest power of each common prime factor in the numbers = 2^{2} = 4

Therefore, LCM = Product of the greatest power of each prime factor involved in the numbers = 2^{3} x 3^{2} x 5 = 360

(v) 30 = 2 x 3 x 5

72=2x2x2x3x3=2^{3}x3^{2}

432.2x2x2x2x3x3x3=2^{4} x 3^{3}

Therefore, HCF = Product of smallest power of each common prime factor in the numbers = 2 x 3 = 6

Therefore, LCM = Product of the greatest power of each prime factor involved in the numbers = 2^{4} x 3^{3} x 5 = 2160

(vi) 21 = 3 x 7

28 = 28 = 2 x 2 x 7 = 2^{2} x 7

36 = 2 x 2 x 3 x 3 = 2^{2} x 3^{2}

45 = 5 x 3 x 3 = 5 x 3^{2}

Therefore, HCF = Product of smallest power of each common prime factor in the numbers = 1

Therefore, LCM = Product of the greatest power of each prime factor involved in the numbers = 2^{2Â }x 3^{2} x 5 x 7 = 1260

Q.10: The HCF of two numbers is 23 and their LCM is 1449. If one of the numbers is 161, find the other.

Sol:

Let, the two numbers be a and b.

Let the value of a be 161.

Given: HCF = 23 and LCM = 1449

we know, a x b = HCF x LCM

\(\Rightarrow\) 161 x b = 23 x 1449

\(\Rightarrow\) Therefore, b = \(\frac{23X1449}{161}\) = \(\frac{33327}{161}\)== 207

Hence, the other number b is 207.

Access CBSE Sample Paper for class 10Â MathsÂ here.

Access NCERT Book for class 10Â MathsÂ here.