Class 10 - Maths - Real Numbers

Exercise 1.1

Question 1:

Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225                   (ii) 196 and 38220                 (iii) 867 and 255

(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, then he 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, then he process stops.

Since the divisor at this stage is 196, therefore the 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, then he process stops.

Since the divisor at this stage is 51, therefore the HCF of 867 and 255 is 51

Question 2:

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Let a be any positive integer and b = 6

Then by Euclid’s lemma, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because

0 ≤ r < 6

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Now, 6q + 1 = 2 * 3q + 1 = 2k1 + 1, where k1 is a positive integer.

6q + 3 = (6q + 2) + 1 = 2 * (3q + 1) + 1 = 2k2 + 1, where k2 is a positive integer.

6q + 5 = (6q + 4) + 1 = 2 * (3q + 2) + 1 = 2k3 + 1, where k3 is a positive integer.

Clearly, 6q + 1 or 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1 or 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.

Question 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?

HCF(161, 32) will give the maximum number of columns in which they can march.

We can use Euclid’s lemma to find the HCF.

616 = 32 * 19 + 8

32 = 8 * 4 + 0

The HCF(616, 32) is 8.

Therefore, they can march in 8 columns each.

Question 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 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.]

Let a be any positive integer and b = 3

Then by Euclid’s lemma, a = 3q + r for some integer q ≥ 0 and r = 0, 1, 2because 0 ≤ r < 3

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

Now, a2 = (3q)2 or (3q + 1)2 or (3q + 2)2

=> a2 = 3(3q2) or 9q2 + 6q + 1 or 9q2 + 12q + 4

=> a2 = 3(3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1

=> a2 = 3k1 or 3k2 + 1 or 3k2 + 1  where k1, k2 and k3 are some positive integers.

Hence, the square of any positive integer is either in the form of 3m or 3m + 1.

Question 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.

Let a be any positive integer and b = 3

Now, using Euclid division lemma, a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

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

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

Case 1: When a = 3q

a3 = (3q) 3 = 27q3 = 9(3q3) = 9m

Where m is an integer such that m = 3q3

Case 2: when a = 3q + 1

=> a3 = (3q + 1) 3

=> a3 = 27q3 + 27q2 + 9q + 1

=> a3 = 9(3q3 + 3q2 + q) + 1

=> a3 = 9m + 1

Where m is an integer such that m = 3q3 + 3q2 + q

Case 3: when a = 3q + 2

=> a3 = (3q + 2) 3

=> a3 = 27q3 + 54q2 + 36q + 8

=> a3 = 9(3q3 + 6q2 + 4q) + 8

=> a3 = 9m + 8

Where m is an integer such that m = 3q3 + 6q2 + 4q

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

Exercise 1.2

Question 1:

Express each number as a product of its prime factors:

(i) 140                  (ii) 156                 (iii) 3825                (iv) 5005                    (v) 7429

(i) 140 = 2 * 2 * 5 * 7 = 22 * 5 * 7

(ii) 156 = 2 * 5 * 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

Question 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

(i) 26 and 91

26 = 2 * 13

91 = 7 * 13

HCF(26, 91) = 13

LCM(26, 91) = 2 * 7 * 13 = 812

Product of two numbers = 26 * 91 = 2366

LCM(26, 91) * HCF(26, 91) = 13 * 182 = 2366

Hence, LCM * HCF = product of the two numbers

(ii) 510 and 92

510 = 2 * 3 * 5 * 18

92 = 2 * 2 * 23

HCF(510, 92) = 2

LCM(510, 92) = 2 * 2 * 3 * 5 * 17 * 23 = 23460

Product of two numbers = 510 * 92 = 46920

LCM(510, 92) * HCF(510, 92) = 2 * 23460 = 46920

Hence, LCM * HCF = product of the two numbers

(iii) 336 and 54

336 = 2 * 2 * 2 * 2 * 3 * 7

54 = 2 * 3 * 3 * 3

HCF(336, 54) = 2 * 3 = 6

LCM(336, 54) = 2 * 2 * 2 * 2 * 3 * 3 * 3 * 7 = 3024

Product of two numbers = 336 * 54 = 18144

LCM(336, 54) * HCF(336, 54) = 6 * 3024 = 18144

Hence, LCM * HCF = product of the two numbers

Question 3:

Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21                  (ii) 17, 23 and 29                    (iii) 8, 9 and 25

(i) 12, 15 and 21

12 = 2 * 2 * 3

15 = 3 * 5

21 = 3 * 7

HCF(12, 15, 21) = 3

LCM(12, 15, 21) = 2 * 2 * 3 * 5 * 7 = 420

(ii) 17, 23 and 29

17 = 1 * 17

23 = 1 * 23

29 = 1 * 29

HCF(17, 23, 29) = 1

LCM(17, 23, 29) = 17 * 23 * 29 = 11339

(iii) 8, 9 and 25

8 = 2 * 2 * 2

9 = 3 * 3

25 = 5 * 5

HCF(8, 9, 25) = 1

LCM(8, 9, 25) = 2 * 2 * 2 * 3 * 3 * 5 * 5 = 8 * 9 * 25 = 1800

Question 4:

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

We know that

LCM * HCF = product of the two numbers

=> LCM (306, 657) * HCF (306, 657) = 306 * 657

=> LCM (306, 657) * 9 = 306 * 657

=> LCM (306, 657) = (306 * 657)/9       => LCM (306, 657) = 306 * 73

=> LCM (306, 657) = 22338

Question 5:

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

If any number ends with the digit o, it should be divisible by 10.

In other words, it will also be divisible by 2 and 5 as 2 * 5 = 10

Now, prime factor of 6n = (2 * 3)n

It can be observed that 5 is not in the prime factorization of 6n

Hence, for any value of n, 6n will not be divisible by 5.

Therefore, 6n can not end with the digit 0 for any natural number n.

Question 6:

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

Numbers are of two types-prime and composite. Prime numbers can be divided by 1 and one

itself whereas composite numbers have factors other than 1 and itself.

It can be observed that

7 * 11 * 13 + 13 = 13 * (7 * 11 + 1)

= 13 * (77 + 1)   = 13 * 78     = 13 * 13 * 6

The given expression has 6 and 13 as its factor. Therefore, it is a composite number.

Now, 7 * 6 * 5 * 4 * 3 * 2 * 1 + 5 = 5 * (7 * 6 * 4 * 3 * 2 * 1 + 1)

= 5 * (1008 + 1)

= 5 * 1009

1009 cannot be factorized further. Therefore, the given expression has 5 and 1009 as its

factors. Hence, it is a composite number.

Question 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?

Required number of minutes is the LCM of 18 and 12

We have,

18 = 2 * 3 * 3

12 = 2 * 2 * 3

LCM(12, 18) = 2 * 2 * 3 * 3 = 36

Hence, Ravi and Sonia will meet again at the starting point after 36 minutes.

Exercise 1.3

Question 1:

Prove that √5 is irrational.

Let us suppose that √5 is a rational number. So, it can be represented as p/q form

=> √5 = p/q

Take square on both side, we get

=> 5 = p2/q2

=> 5q2 = p2 ..............1

So p2 is divisible by 5.

=> p is divisible by 5.

Let p = 5x          (x is a positive integer)

Now p2 = 25c2

From equation 1, we get

5q2 = 25c2

=> q2 = 5c2

So, q is divisible by 5

Thus p and q has a common factor 5. It is contradiction of our assumption.

So, √5 is an irrational number.

Question 2:

Prove that 3 + 2√5 is irrational.

Let take that 3 + 2√5 is a rational number.

So, we can write this number as

3 + 2√5 = a/b

Here a and b are two co prime number and b is not equal to 0

Subtract 3 both sides we get

2√5 = a/b – 3

2√5 = (a - 3b)/b

Now divide by 2 we get

√5 = (a - 3b)/2b

Here, a and b are integer. So (a - 3b)/2b is a rational number.

It means √5 should be a rational number.

But √5 is a irrational number. So, it contradicts the fact.

Hence, 3 + 2√5 is a irrational number.

Question 3:

Prove that the following are irrationals:

(i) 1/√2                                    (ii) 7√5                                    (iii) 6 + √2

(i) Let take that 1/√2 is a rational number.

So we can write this number as

1/√2 = a/b

Here a and b are two co prime number and b is not equal to 0

Multiply by √2 both sides we get

1 = (a√2)/b

Now multiply by b

b = a√2

divide by a, we get

b/a = √2

Here, a and b are integer, so b/a is a rational number.  It means √2 should be a rational

number. But √2 is an irrational number. Hence, it is contradict the fact.

Hence, 1/√2 is an irrational number

(ii) Let take that 7√5 is a rational number.

So, we can write this number as

7√5 = a/b

Here a and b are two co prime number and b is not equal to 0

Divide by 7 we get,

√5 =a/7b

Here, a and b are integer, so a/7b is a rational number.

It means √5 should be a rational number but √5 is an irrational number.

So, it is contradict the fact.

Hence, 7√5 is an irrational number.

(iii) Let take that 6 + √2 is a rational number.

So, we can write this number as

6 + √2 = a/b

Here a and b are two co prime number and b is not equal to 0

Subtract 6 both side, we get

√2 = a/b – 6

√2 = (a - 6b)/b

Here ,a and b are integer, so (a - 6b)/b is a rational number.

It means √2 should be a rational number but √2 is an irrational number.

So, it is contradict the fact.

Hence, 6 + √2 is an irrational number.

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