CLASS 10 MATHS CHAPTER-1 REAL NUMBERS EXERCISE 1.1 to 1.4

EXERCISE 1.1

Solve the followings Questions.

1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255

Answer:

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

90 = 2 × 45 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain

38220 = 196 × 195 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

867 = 255 × 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain

255 = 102 × 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

102 = 51 × 2 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51, Therefore, 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.

Answer:

Let a be a given integer.

On dividing a by 6 , we get q as the quotient and r as the remainder such that,

a = 6q + r, r = 0,1,2,3,4,5

when r =0

a = 6q,even no
where r = 1

a = 6q + 1, odd no
where r = 2

a = 6q + 2, even no
where r = 3

a=6q + 3,odd no
where r = 4

a = 6q + 4, even no
where r = 5

a = 6q + 5, odd no
where r = 6

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in 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 columns in which they can march?

Answer:

For the above problem, the maximum number of coulmns would be the HCF of 616 and 32

We can find the HCF of 616 and 32 by using Euclid Division algorithm.

Therefore

616 = 19 x 32 + 8

32 = 4 x 8 + 0

8 = 8 x 1 + 0

Therefore HCF (616, 32) = HCF of (32, 8) = 8

Therefore the maximum number of columns in which they can march is 8.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of 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.

Answer:

If a and b are two positive integers, then,

Therefore, r = 0, 1, 2

Therefore, a = 3q or a = 3q + 1 or a = 3q + 2

Therefore, 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.

Answer:

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

Where m is an integer such that m = (3q)^{3}

Case 2: When a = 3q + 1,

Case 3: When a = 3q + 2,

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

EXERCISE 1.2

1. Express each number as product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer:

(i) 140 = 2 x 2 x 5 x 7 = 2^{2} x 5 x 7

(ii) 156 = 2 x 2 x 3 x 13 = 2^{2} x 3 x 13

(iii) 3825 = 3 x 3 x 5 x 17 = 3^{2} x 5^{2} x x 17

(iv) 5005 = 5 x 7 x 11 x 13

(v) 7429 = 17 x 19 x 23

2. Find the LCM and HCF of the following pairs of integers and verify that LCM xHCF = product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Answer:

(i) 26 and 91

a = 26, b = 91

∴ H.C.F = 13

L.C.M = 2 × 7 × 13

= 14 × 13 = 182

∴ H.C.F × L.C.M = a × b

13 × 182 = 26 × 96

2366 = 2366

(ii) 510 and 92

a = 510, b = 92

(iii) 336 and 54

a = 336, b = 54Read more on Sarthaks.com – https://www.sarthaks.com/661852/find-the-lcm-and-the-following-pairs-integers-and-verify-that-lcm-hcf-product-the-two-numbers

336= 2 x 2 x 2 x 2 x 3 x 7 = 2^{4} x 3 x 7

54 = 2 x 3 x 3 x 3 = 2 x 3^{3}

HCF = 2 x 3 = 6

LCM = 2^{4} x 3^{3} x 7= 3024

Product of two numbers 336 and 54 = 336 x 54= 18144

3024 x 6= 18144

Hence, product of two numbers = 18144

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 25

Answer:

(i) 12, 15 and 21

12 = 2 × 2 × 3 = 2^{2}× 3

15 = 3 × 5

and 21 = 3 × 7

For HCF, we find minimum power of prime factor

H.C.F. = (3)^{1}= 3

For LCM, taking maximum power of prime factors

L.C.M. = 2^{2} × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420

So,H.C.F. = (3)^{1}= 3

and L.C.M. = 420

(ii) 17, 23 and 29

17 = 1 × 17

23 = 1 x 23

29 = 1 x 29

For HCF, common factor is 1

HCF = 1

For LCM taking maximum power of prime factor.

L.C.M. = 1 × 17 × 23 × 29 = 11339

So H.C.F. = 1

L.C.M. = 11339

(iii) 8, 9 and 25

8 = 2 × 2 × 2 × 1 = 2^{3}× 1

9 = 3 × 3 = 3^{2}

and 25 = 5 × 5 = 5^{2}

For HCF common factor is 1

H.C.F. = 1

For LCM, taking maximum power of prime factors

L.C.M. = 2^{3}× 3^{2} × 5^{2}

= 8 × 9 × 25 = 1800

So H.C.F. = 1

L.C.M. = 1800

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

Answer:

HCF (306, 657) = 9

We know that, LCM × HCF = Product of two numbers

L.C.M x H.C.F = first Number x Second Number

L.C.M x 9 = 306 x 657

LCM = 22338

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

Answer:

TO CHECK: Whether 6^{2}can end with the digit 0 for any natural number n.

We know that

6^{2} = (2 × 3)^{n}

6^{2} = (2)^{n} ×(3)^{n}

Therefore, prime factorization of 6^{n}does not contain 5 and 2 as a factor together.

Hence 6^{n}can never end with the digit 0 for any natural number n

6. Explain why 7 x 11 x 13 and 7 x 6 x 5 x 3 x 2 x 1 + 5 are composite numbers.

Answer:

So, the given expression has 6 and 13 as its factors. Therefore, we can conclude that it is a composite number.

Similarly,

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

= 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1) [taking 5 out- common]

= 5 x (1008 + 1)

= 5 x 1009

Since, 1009 is a prime number the given expression has 5 and 1009 as its factors other than 1 and the number itself.

Hence, it is also a composite number.

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?

Answer:

It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.

LCM of 12 and 18= 2 x 2 x 3 x 3 = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Exercise 1.3

1. Prove that √2 is irrational.

Answer:

Let us prove √2 irrational by contradiction.

Let us suppose that √2 is rational.

So it can be expressed in the form p/q where p, q are co-prime integers and q≠0

√2 = p/q

Here p and q are coprime numbers and q ≠ 0

Solving

√2 = p/q

On squaring both the side we get,

=>2 = (p/q)^{2}

=>2q^{2} = p^{2} ……..(1)

=> p^{2} = q^{2}

So 2 divides p and p is a multiple of 2.

⇒ p = 2m

⇒ p² = 4m² ………………………………..(2)

From equations (1) and (2), we get,

2q² = 4m²

⇒ q² = 2m²

⇒ q² is a multiple of 2

⇒ q is a multiple of 2

Hence, p, q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√2 is an irrational number.

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

Answer:

We will prove this by contradiction.

Let us suppose that (3+2√5) is rational.

It means that we have co-prime integers aand b(b ≠ 0) such that

So, it can be written in the form a/b

3 + 2√5 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving 3 + 2√5 = a/b we get,

=>2√5 = a/b – 3

=>2√5 = (a-3b)/b

=>√5 = (a-3b)/2b

This shows (a-3b)/2b is a rational number. But we know that √5 is an irrational number.

So, it contradicts our assumption. Our assumption of 3 + 2√5 is a rational number is incorrect.

3 + 2√5 is an irrational number

Hence proved

3. Prove that the following are irrationals.

(i)

(ii)

(iii)

Answer:

(i) We can proveirrational by contradiction.

Let us suppose thatis rational.

It means we have some co-prime integers a and b (b ≠ 0) such that

1/√2 = p/q

√2 = q/p

By Squaring on both sides

2 × p^{2}= q^{2}

2, divides q^{2}

∴ 2, divides q

∵ q is an even number.

Similarly ‘p’ is an even number.

∴ p and q are even numbers.

∴ Common factor of p and q is 2.

This contradicts the fact that p and q also irrational.

∴ √2 is an irrational number.

∴ is an irrational number.

(ii) We can proveirrational by contradiction.

Let us suppose thatis rational.

It means we have some co-prime integers a and b (b ≠ 0) such that

It means √5 which is equal also a rational number.

This contradicts to the fact that √5 is an irrational number.

This contradicts to the fact that 7√5 is rational number.

∴ 7√5 is a rational number.

(iii) We will proveirrational by contradiction.

Let us suppose that () is rational.

It means that we have co-prime integers aand b(b ≠ 0) such that

∴ √2 is also rational number.

This contradicts to the fact that √2 is an irrational number.

This contradicts to the fact that 6 + √2 is a rational number.

∴ 6 + √2 is an irrational number.

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 decimal expansion.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

Answer:

(i)

q = 3125 = 5 x 5 x 5 x 5 x 5 = 5^{5}

Here, denominator is of the form , where m = 5 and n = 0.

It means rational number has a terminating decimal expansion.

(ii)

q = 8 = 2 x 2 x 2 = 2^{3}

Here, denominator is of the form , where m = 0 and n = 3.

It means rational number has a terminating decimal expansion.

(iii)

q =

Here, denominator is not of the, where m and n are non-negative integers.

It means rational number has a non-terminating repeating decimal expansion.

(iv)

q = 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 =2^{6} x 5 x 5

Here, denominator is of the form , where m = 1 and n = 6.

It means rational number has a terminating decimal expansion.

(v)

q = 343 = 7 x 7 x 7

Here, denominator is not of the form , where m and n are non-negative integers.

It means rational number has non-terminating repeating decimal expansion.

(vi)

q = 2 x 2 x 2 x 5 x 5

Here, denominator is of the form , where m = 2 and n = 3 are non-negative integers.

It means rational number has terminating decimal expansion.

(vii)

q = 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5 x 5 x 7 x 7 x 7 x 7 x 7

Here, denominator is not of the form , where m and n are non-negative integers.

It means rational number has non-terminating repeating decimal expansion.

(viii)

Here, denominator is of the form , where m = 1 and n = 0.

It means rational number has terminating decimal expansion.

(ix)

q = 10 = 2 x 5

Here, denominator is of the form , where m = 1 and n = 1.

It means rational number has terminating decimal expansion.

(x)

q = 30 = 2 x 3 x 5

Here, denominator is not of the form , where m and n are non-negative integers.

It means rational number has non-terminating repeating decimal expansion.

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

Answer:

(iv)

x

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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 , what can you say about the prime factors of q?

(i) 43.123456789

(ii) 0.120112001200 0120000…

(iii)

Answer:

(i) 43.123456789

It is rational because decimal expansion is terminating. Therefore, it can be expressed in form where factors of q are of the form where n and m are non-negative integers

(ii) 0.120112001200 0120000…

It is irrational because decimal expansion is neither terminating nor non-terminating repeating.

(iii)

It is rational because decimal expansion is non-terminating repeating. Therefore, it can be expressed in form where factors of q are not of the form where n and m are non-negative integers.