Chapter 1 : Real Numbers

Exercise 1.1  Complete solution  Exercise 1.2 Complete solution  Exercise 1.3 Complete solution  Exercise 1.4 Complete solution

Algebraic Identities  (i)    (ii)  (iii)  (iv)                      (v)

Important notes  (i)  Euclid’s Division Lemma : Given positive integers  and , there exist unique integers  and  satisfying , (ii) Fundamental Theorem of Arithmetic : Every composite number can be expressed ( factorised) as a  product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.  (iii) The prime factorisation of a natural number is unique, except for the order of its factors.  (iv) HCF = Product of the smallest power of each common prime factor in the numbers.  (v) LCM = Product of the greatest power of each prime factor, involved in the numbers .  (vi) If two positive integers  and ,then . i.e ., HCF of two numbers × LCM of two numbers = One number × Other number (vii) The product of three numbers is not equal to the product of their HCF and LCM. (viii) A number  is called rational if it can be written in the form ,  where p and q are integers . Example : 0 ,  – 5 , , …… etc . (ix) A number  is called irrational if it cannot be written in the form ,   where p and q are integers  . Example :   ……… etc. (x) Let  be a prime number. If  divides , then  divides , where  is a positive integer. (xi)  The sum or difference of a rational and an irrational number is irrational . (xii) The product and quotient of a non-zero rational and irrational number is irrational. (xiii) Let  be a rational number whose decimal expansion terminates. Then we can express  in the form , where  and  are coprime, and the prime factorisation of  is of the form  , where  are non-negative integers. (xiv). Let be a rational number, such that the prime factorisation of  is of the form , where  are non-negative integers. Then  has a decimal expansion which terminates. (xv). Let   be a rational number, such that the prime factorisation of  is not of the form  , where  are non-negative integers. Then  has a decimal expansion which is non-terminating repeating (recurring).

Class 10 Maths Chapter 1 : Real Numbers Exercise 1.1

1. Use Euclid’s division algorithm to find the HCF of :  (i) 135 and 225      (ii) 196 and 38220     (iii) 867 and 225   (iv) 272 and 1032   (v) 405 and 2520    (vi) 155 and 1385    (vii) 384 and 1296   (viii) 1848 and 3058

Solution : (i) Since   225 > 135

Using Euclid’s division algorithm , we have

 

  Therefore, the HCF of 135 and 225 is 45 .

 (ii) Since 38220 > 196

 Using Euclid’s division algorithm , we have

 

Therefore, the HCF of 196 and 38220 is 195 .

(iii)  Since  867 > 225

  Using Euclid’s division algorithm , we have

   

  

                               

  Therefore , HCF(867 , 255) is 51 .      

(iv) Since,  1032 > 272  

Using Euclid’s division algorithm , we have

1032 = 272 3 + 216

272 = 216 1 + 56

216 = 56 3 + 48

56 = 48 1 + 8

48 = 8 6 + 0

Therefore , HCF(272 , 1032) is 8 .      

(v) Since , 2520 > 405 

Using Euclid’s division algorithm , we have

2520 = 405 6 + 90

405 = 90 4 + 45

90 = 45 2 + 0

Therefore, HFC (405 , 2520) is 45 .

(vi) Since,  155 < 1385 

Using Euclid’s division algorithm , we have

1385 = 155 8 + 145

155 = 145 1 + 10

145 = 10 14 + 5

10 = 5 2 + 0

Therefore, HCF(155,1385) is 5 .

(vii)  Since , 384 < 1296 

Using Euclid’s division algorithm , we have

1296 = 384 3 + 144

384 = 144 2 + 96

144 = 96 1 + 48

96 = 48 2 + 0

Therefore, HCF(384,1296) is 48 .

(viii)  Since , 1848 < 3058

Using Euclid’s division algorithm , we have

3058 = 1848 1 + 1210

1848 = 1210 1 + 638

1210 = 638 1 + 572

638 = 572 1 + 66

572 = 66 8 + 44

66 = 44 1 + 22

44 = 22 2 + 0

Therefore, HCF(1848 ,3058) is 22 .

2. Show that any positive odd integers is of the form   or   , where is some integer.

Solution :

let ,  be any positive odd integers and  .

 Using  Euclid’s algorithm , we have

 ,  

 ,  

So,  0 , 1 , 2 , 3 , 4 or 5 .  

If  then  is an even numbers .

If  then  is an odd numbers .

If  then  is an even numbers .

If  then  is an odd numbers .

If then  is an even numbers .

If  then   is an odd numbers .

Therefore, any positive odd integer is of the form  ,   or  .

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 ?

Solution :  We find the maximum number of column of (616 , 32) .

Using Euclid’s  division algorithm  , we have

Therefore , the HCF (616 , 32) is 8 .

Thus , the maximum number of column is  8 .

4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form  orfor some integer . [ Let   be any positive integer then it is of the form  or . Now square each of these and show that they can be rewritten in the form or  .]

Solution :

let,  be any positive integers and  .

We apply the Euclid’s division algorithm ,

   ,    

 , 

So ,  ,  or   .

If    then    

 , where 

If    then  

 , where 

If    then

,where  

Thus, the square of any positive integer is either of the form or  for some integer .

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form  or  .

Solution :

let,  be any positive integer and.

Using  Euclid’s algorithm , we have

 ,   

 , 

Therefore ,  0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 or 8 .

If  then  

 ,where  for some integer .

If then  

 ,where  for some integer q .

If then

   ,where  for some integer  q.

Therefore ,the cube of any positive integer is of the form  , or  , for some integer  .

6. Himadri  has a collection of 625 Indian postal stumps and 325 International postal stumps . She wants to display them in identical groups of Indian and International stumps with no stump left out . What is the greatest number of groups Himadri can display the stumps ?

Solution : The HCF of 625 and 325 can be found using the Euclid’s algorithm, We have

625 = 325 1 + 300

325 = 300 1 + 25

300 = 25 12 + 0 

The HCF of 625 and 325 is 25.

So, the greatest number of identical groups Himadri can display the stamps is 25 groups.

7. Two ropes are of length 64 cm and 80 cm . Both are to be cut into pieces of equal length . What should be the maximum length of the pieces ?  

Solution : We need to find the HFC of 64 and 80 .

Using  Euclid’s algorithm , we have

80 = 64 1 + 16

64 = 16 4 + 0

The HCF of 64 and 80 is 16.

So, the maximum length of the pieces into which both ropes can be cut while maintaining equal length is 16 cm

 Class 10 Maths Chapter 1 : Real Numbers Exercise 1.2   

1. Express each number as a product of its prime factors :   (i) 140       (ii) 156        (iii) 3825      (iv) 5005        (v) 7429

Solution : (i) We have,

                                         

   (ii) We have,

                               

 (iii) We have,

                                

 (iv)  We have ,  

 (v)  We have ,

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

Solution :    (i)   26 and 91 

 We have ,  and     

   HCF (26 , 91)  13   

and LCM (26 , 91)

 Now ,

 

 

     Verified .

  (ii)    510 and 92 

 We have ,  

 and       

 HCF(510 , 92)    

 and  LCM(510 , 92)  

  Now , 

  

  

      Verified .

(iii)   336 and 54

We have,

                       

 and     

 HCF(336 , 54)    

 and LCM(336 , 54)

Now , 

 

 

     Verified . 

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

Solution :    (i) 12 , 15  and 21     

  We have ,

 

   HCF (12 , 15 , 21)  3   

and LCM (12 , 15 , 21)  

(ii)   17 , 23  and 29   

  We have , 

  

  

    HCF (17 , 23 , 29)   

 and LCM (17 , 23 , 29)  

(iii)   8 , 9  and 25

 We have ,

 

  

    HCF(8 , 9 , 25)   

  and  LCM (8 , 9 , 25)  

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

Solution :  We have ,

 

                               

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

Solution :  We have , 

 

The prime factors of  does not contain  in factor , where  are positive integers .Therefore,  does not end with the digit 0 .

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

Solution:  We have ,

 is a composite number.

and 

 

 

  is 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 ?

Solution : The required  number of minutes  is  LCM (18 , 12) . We find the LCM by prime factorization method ,

  

and  

Therefore, the LCM (18 , 12)  

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

8. (i) The soldiers in a regiment can be stood in some raws consisting of 15 , 20 or 25 numbers of soldiers . Find the least number of soldiers in the regiment .

Solution :  We need to determine the LCM of the numbers 15, 20, and 25 .

( Because the regiment can arrange soldiers in rows consisting of 15 , 20 and 25 .)

We have ,   ,  ,

LCM

So, the least number of soldiers in the regiment is 300.

(ii) A bell rings at every 18 seconds , another bell rings at every 60 seconds . If these two bells ring simultenously  an instant , then find after how many seconds will the bells ring simultenously again .

Solution : We need to determine the LCM of the time intervals between the bell rings.

The first bell rings every 18 seconds, and the second bell rings every 60 seconds.

Find the LCM of 18 and 60.

We have,

LCM

So, the bells will ring simultaneously again after 180 seconds

(iii) A radio station plays ‘Assam Sangeet ’ once every two days . Another radio station plays the same song once every three days . How many times in 30 days will both the radio stations play the same song on the same day .

Solution : To find out how many times both radio stations will play the same song on the same day in 30 days, we need to determine the LCM of their respective schedules.

One radio station plays the song once every two days, and the other plays it once every three days.

We have,

  

 

LCM

So, the LCM of the two radio stations' schedules is 6 days.

This means they will play the same song on the same day every 6 days.

In 30 days, they will play the same song on the same day times.

Class 10 Maths Chapter 1 : Real Numbers Exercise 1.3       

1. Prove that  is irrational .

Solution : let, us assume to the contrary that  is rational .There exists co-prime integers and  () such that

  

 Therefore ,  is divisible by 5 . So,  is also divisible by 5 .

 Let  , for some integer c  .

 From  and  , we get

 

 

So,  is divisible by 5 . So,  is also divisible by 5 . 

Therefore,  and  have at least as a common factor . But this contradicts the fact that and  are co-prime. This contradiction has arisen because of our incorrect assumption that   is rational . So, we conclude that  is irrational .

2. Prove that  is irrational .

Solution :  let us assume , to the contrary  that   is rational .

We can find co-prime  and  ( ) such that

Since, 2 ,  and  are integers , is rational and so, is rational . But this contradicts the fact that   is irrational . So , we conclude that is irrational .

3. Prove that the following are irrationals :  (i)         (ii)       (iii)   

Solution :  (i)  Let us assume , to the contrary , that   is rational . We can find co-prime and  ( ) such that

 

Since  2 , and  are integers, is rational and so, is rational . But this contradicts the fact that  is irrational . So , is an irrational .

(ii)  Let us assume , to the contrary , that   is rational .  We can find co-prime and  ( ) such that 

 Since  7 , and  are integers ,  is rational and so, is rational . But this contradicts the fact that  is irrational .So , is an irrational .

(iii)  Let us assume , to the contrary , that   is rational . We can find co-prime and  ( ) such that

 Since and  are integers ,  is rational and so, is rational . But this contradicts the fact that  is irrational . So , is an irrational  .

 Class 10 Maths Chapter 1 : Real Numbers 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 repeating decimal expansion : (i)     (ii)     (iii)     (iv)     (v)     (vi)      (vii)     (viii     (ix)      (x)

Solution :  (i)  We have ,

 The denominator of the fraction is of the form  , where are non negative integers . Therefore , is a terminating decimal expansion .          

 (ii)  We have ,

 The denominator of the fraction is of the form  , where  are non negative integers . Therefore ,   is a terminating decimal expansion .                      

 (iii)  We have ,  

 The denominator of the fraction is not of the form  ,  where  are non negative integers . Therefore ,  is a non-terminating repeating decimal expansion .                

 (iv) We have, 

 The denominator of the fraction is of the form  , where are non negative integers . Therefore ,  is a terminating decimal expansion .                 

 (v)  We have,       

 The denominator of the fraction is not of the form  , where  are non negative integers .Therefore , is a non-terminating repeating decimal expansion .          

  (vi)  We have , 

  The denominator of the fraction is of the form  , where are non negative integers . Therefore, is a terminating decimal expansion .            

 (vii) We have,

 The denominator of the fraction is not of the form  ,  where  are non negative integers .Therefore,   is a non-terminating repeating decimal expansion .          

 (viii)  We have, 

The denominator of the fraction is of the form , where  are non negative integers.Therefore, is a terminating decimal expansion.                  

 (ix)  We have, 

  The denominator of the fraction is of the form  ,  where  are non negative integers .Therefore,  is a terminating decimal expansion .                     

 (x) We have ,

 The denominator of the fraction is not of the form  ,where  are non negative integers .Therefore,   is a non-terminating repeating decimal expansion .          

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

Solution :  (i) We have, 

 (ii) We have, 

      

(iii) We have,

(iv) We have,

(v) We have ,

(vi) We have,

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  ?   (i) 43.123456789       (ii) 0.120120012000120000…...   (iii)  

Solution:  (i) 43.123456789 is a rational number and it is of the form   . So , the prime factors of  is of the form (It is terminating decimal expansion),where  are non-negative integers .  

  (ii) 0.120120012000120000…………. is an irrational number . (It is non-terminating and non-repeating decimal expansion) . So, the given number is not of the form .

(iii)   is a rational number and it is of the form  . So , the prime factors of  is not of the form , (It is non-terminating and repeating decimal expansion) , where  are non-negative integers .