Class – 7 CH-9 RATIONAL NUMBERS MATHS NCERT SOLUTIONS

 

Class – 7 CH-9 RATIONAL NUMBERS 

MATHS NCERT SOLUTIONS

 Exercise 9.1 

     
 
Question 1:   
(i) 1 and 0 (ii) 2 and 1  
(iii) 4 2
and     (iv) 1 2
and  
List five rational numbers between: 
5 3 2 3
  SOLUTION 1:  
(i) 1 and 0 
Let us write 1 and 0 as rational numbers with denominator 6. 
   1  6 and 0 =  0  
6 6
6 5 4 3 2 1
   0  
6 6 6 6 6 6
 1 5  2  1 1 1 0 
6 3 2 3 6
 
Therefore, five rational numbers between 1 and 0 would be 
   5 2 1 1 1
, , , ,  
6 3 2 3 6
 
(ii) 2 and 1  
Let us write 2 and 1 as rational numbers with denominator 6. 
   2  12 and  1  6  
6 6
12 11 10    9 8 7 6
          6 6 6 6 6 6 6
 2 11 5  3  4  7 1 
6 3 2 3 6
 
Therefore, five rational numbers between 2 and 1 would be 
   11 5 3 4 7
, , , ,  
6 3 2 3 6
 
 
 

   
 
(iii) 
 
(iv) 
 
 
 
   
4 2
and   
5 3
4 2
Let us write and as rational numbers with the same denominators. 
5 3
4 36 2 30   and   
5 45 3 45
36 35 34 33 32 31 30
    
45 45 45 45 45 45 45
4 7 34 11 32 31 2
 
5 9 45 15 45 45 3
 
4 2
Therefore, five rational numbers between and would be 
5 3
 7 34 11 32 31 2
, , , , ,  
9 45 15 45 45 3
 
1 2
and   
2 3
1 2
Let us write and as rational numbers with the same denominators. 
2 3
1 3 2 4
  and   
2 6 3 6
3 2 1 1 2 3 4         0    
6 6 6 6 6 6 6
  1 1 1 1 1 1 2        0   
2 3 6 6 3 2 3
 
1 2
Therefore, five rational numbers between and would be 
2 3
 1 1 1 1
, ,0, ,
3 6 6 3
 
 
Question 2:   
Write four more rational numbers in each of the following patterns: 
  3 6 9 12
(i) , , , ,.........   
1 2 3
(ii) , , ,..........  
1 2 3 4
(iii) , , , ,.........   
6   12 18 24
2 2 4 6
(iv) , , , ,.......... 
3   3 6 9
  SOLUTION 2:  
   3 6 9 12
(i) , ,  , ,.........  
3 1 3 2 3 3 3 4
, , , ,......... 
Therefore, the next four rational numbers of this pattern would be 
3 5,  3 6,  3 7,  3 8 = 15, 18, 21, 24 
5 5 5 6 5 7 5 8 25 30 35 40
 
 1 2 3
(ii) , , ,.......... 
1 1 1 2 1 3
, , ,.......... 
  Therefore, the next four rational numbers of this pattern would be 
1 4,  1 5,  1 6,  1 7 =    4, 5, 6, 7 
4 4 4 5 4 6 4 7 16 20 24 28
 
1 2 3 4
(iii) , , , ,.........  
6 12 18 24
1 1 1 2 1 3 1 4
, , , ,......... 
6 1  6 2    6 3 6 4
Therefore, the next four rational numbers of this pattern would be 
1 5 1 6 1 7 1 8 5 6 7 8
  , , , = , , ,  
6 5  6 6  6 7  6 8 30 36 42 48
 
 
2 2 4 6
(iv) , , , ,..........  
3   3 6 9
2 1 2 1 2 2 2 3
, , , ,.......... 
3 1    3 1 3 2  3 3
Therefore, the next four rational numbers of this pattern would be 
2 4 2 5 2 6 2 7 8 10 12 14
, , , = , , ,  
3 4  3 5  3 6  3 7 12 15 18 21
 
 
Question 3:   
Give four rational numbers equivalent to: 
2 5 4
(i)     (ii)     (iii)   
7 3 9
  SOLUTION 3:  
2
(i)    
7
2 2 4  2 3 6  2 4 8  2 5 10
 ,  ,  ,  
7 2 14 7 3 21 7 4 28 7 5 35
  4 6 8 10
Therefore, four equivalent rational numbers are , , ,
14 21 28 35
5
(ii)    
3
5 2 10 5 3 15 5 4 20 5 5 25
, , ,  
3 2 6  3 3 9  3 4 12  3 5 15
10 15 20 25
Therefore, four equivalent rational numbers are , , ,
 6 9 12 15
(iii) 
 
8 12 16 20
Therefore, four equivalent rational numbers are , , ,
18 27 36 45
 
 
 
Question 4:   
Draw the number line and represent the following rational numbers on it: 
3 5 7 7
(i)     (ii)     (iii)     (iv)   
4 8 4 8
  SOLUTION 4:  
(i)    
 
(iv)    
   
Question 5:   
The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S. 
  
Each part which is between the two numbers is divided into 3 parts. 
Therefore, A =  , P = , Q =   and B =    
Similarly T =  3, R = 4, S = 5 and U =  6  
3 3 3 3
Thus, the rational numbers represented P, Q, R and S are 7 8, , 4 and  5 
3 3 3 3 respectively. 
 
 
Question 6:   
Which of the following pairs represent the same rational numbers: 
7 3
(i) and    
21 9
16 20
(ii) and    
20 25
2 2
(iii) and   
3 3
3 12
(iv) and      
5 20
8 24
(v) and      
5 15
1 1
(vi)   and   
3
(vii) and  
9 9
 
 

   
  SOLUTION 6
(i) 
  
   
  
 
(ii) 
  
   
  
 
(iii) 
 
  
  
 
(iv) 
  
  
 
   
:  
7 3
and   
21 9
7 1 3 1
= and =    
21 3 9 3
1 1
 
3 3
7 3
 
21 9
16 20
and   
20 25
16 4 20 4 4
  =   and =    
20 5 25 5 5
4 4
=   
5 5
16 20
 = 20 25
2 2
and   
3 3
2 2 2 2
=   and =    
3 3 3 3
  =   
2 2
=  
3 3
3 12
and   
5 20
3 3 12 3  =   and   =  
5 5 20 5
3 3
=  
5 5
3 12
  =  
5 20
  [Converting into lowest term] 
[Converting into lowest term] 
[Converting into lowest term] 
[Converting into lowest term] 

   
8 24
(v) and   
5 15
8 8 24 8
=   and   = [Converting into lowest term] 
5 5 15 5
8 8
  = 5 5
8 24
  =  
5 15
 
1 1
(vi) and   
3 9
1 1 1 1
=   and = [Converting into lowest term] 
3 3 9 9
1 1   
3 9
1 1    
3 9
 
5 5
(vii) and   
9 9
5 5 5 5
=   and = [Converting into lowest term] 
9 9 9 9
5  
9 9
5 5    
9 9
 
 
Question 7:   
Rewrite the following rational numbers in the simplest form: 
8 25 44 8
(i)     (ii)     (iii)     (iv)   
6 45 72 10
  SOLUTION 7:  
8  8 2 4
(i) = =   [H.C.F. of 8 and 6 is 2] 
6 62 3
 
 
(ii)    =     [H.C.F. of 25 and 45 is 5] 
44  44 4 11
72 724 18
(iv) 8  8 2 4
= =   [H.C.F. of 8 and 10 is 2] 
(iii)   = =   [H.C.F. of 44 and 72 is 4] 
10 102 5
 
 
Question 8:   
Fill in the boxes with the correct symbol out of <, > and =: 
    
    
    
    
52 45 714 87
(i)     (ii)   (iii)   (iv)   
73 57 816 54
    
    
    
11 557
(v)   (vi)   (vii) 0   
34 11116
  SOLUTION 8:  
 
5 2
(i) Since, the positive number if greater than negative number. 
7
   
  
5 7 7 5 35 35 5 7
7 2 14 1 14 14 7 14
(ii) 4 75 5    28  < 25  4  <  5 
   
  
82   16  1 16 16 8 16
8 4 7 5 32 35 8 7
(iii)     =      =  
   
  
(iv)     >     >   
5 44 5 20 20 5 4
    
11 1 1
(v)   <   
34 3 4
    
55 5 5
11 11
7 11 11
(vi)   =   
 
(vii) 0  >  Since, 0 is greater than every negative number. 6
 
 
 
Question 9:   
Which is greater in each of the following: 
2 5  5 4 3 2 1 1
(i) ,     (ii) ,   (iii) ,   (iv) ,   
3 2 6 3 4 3 4 4
2 4
(v) 3 , 3  
7 5
  SOLUTION 9:  
(i)   and    
4 15 2 5
Since <   Therefore   
6 6 3 2
  5 1 5  4 2 8
(ii)  and   6 1 6 3 2 6
Since 5 > 8  Therefore 5 > 4  
6 6 6 3
 3 3 9 2  4 8
(iii)  and   
4 3 12   3  4 12
Since 9 < 8  Therefore 3  2  
12 12 4 3
1 1
(iv) < Since positive number is always greater than negative 
4 4 number. 
(v) 32  23   23 5  115 and 34  19   19 7  133  
7 7 7 5 35 5 5 5 7 35
Since 115 > 133  Therefore 3 2 > 3 4  
35 35 7 5
 
 
Question 10:   
Write the following rational numbers in ascending order: 
(i)   3, 2, 1  
1 2 4
(ii)  
(iii)   3, 3, 3 
7 2 4
 
  SOLUTION 10:  
  3 2 1
(i) , ,   
5 5 5
3 2 1
   
5 5 5
1  2 4 3  2 12
(ii) , ,    , ,   [Converting into same denominator] 
3 9 3 9 9 9
12 2 3 4 2 1    Now    
9 9 9 3 9 3
  3 3 3
(iii) , ,    
7 2 4
3 3 3
   
2 4 7 

 Exercise 9.2  

 
Question 1:   
Find the sum: 
5 11 5 3
(i) 4 4   (ii) 3  5  
9 22 3 5
(iii)   (iv)   
10 15 11 9
8 2 2
(v)   (vi) 0  
19 57 3
1 3
(vii) 2 4  
3 5
  SOLUTION 1:  
5 11 5 11 6 3
(i)  4  = 4 = 4  2  4 
5 3 25 9
(ii)  =     [L.C.M. of 3 and 5 is 15] 
3 5 15 15
259 34 4
  =  2   
15 15 15
9 22  9 3 22 2 27 44
(iii) = = [L.C.M. of 10 and 15 is 30] 
10 15 10 3 15 2 30 30
 27 44 17
     =   
30 30
3 5  3 9 5 11 27 55
(iv)  = = [L.C.M. of 11 and 9 is 99] 
11 9  11 9 9 11 99 99
 
2755 82
     =   
99 99
8 2  8 3  2 1 24 2
(v) = = [L.C.M. of 19 and 57 is 57] 
19 57 19 3 57 1 57 57
 24 2 26
         =   =   
57 57
2 2
(vi)  0   
3 3
1
 
(vii) 21 43 = 7  23 =  7 5  23 3 = 35  69 [L.C.M. of 3 and 5 is 15] 
3 5 3 5 3 5 5 3 15 15
34 4
       =  =  2 
15 15
 
 
Question 2:   
(i) 7 17    
24 36 (ii) 5 6
 21  
63 
(iii) 6 7
15  
13  (iv) 3 7    
8 11
Find: 
(v)  2
  SOLUTION 2:  
7 1721 34
(i) =  =   [L.C.M. of 24 and 36 is 72] 
24 3672 72
21 34 13
=   =   
72 72
5 6 5 1  6 3 5 18
(ii) 63 21 = 63 1  21 3  = 63  63  [L.C.M. of 63 and 21 is 63] 
5  18 5 18 23
  =   =   
63 63 63
6 7  6 15  7 13 90 91
(iii) 13 15 = 13 15 15 13  = 195 195 [L.C.M. of 13 and 15 is 195] 
  90  91  90 91 1
195 195 195
3 7  3 11 7 8 33 56 (iv) = =   
8 11 8 11 11 8 88 88
 33 56 89 1
  = = 1   
88 88 88 [L.C.M. of 8 and 11 is 88] 
1
(v) 2 6 = 19  6 =  19 1 6 9  
9 9 1 9 1 1 9
19 54  19 54 73 1 [L.C.M. of 9 and 1 is 9] 
  =   =   
  = = = 8  
9 9 9 9 9
2

   
Question 3:   
Find the product: 
9 7
(i)  4    
2 
6 9
(iii)    
5 11
3 2
(v)    
11 5
  SOLUTION 3:  
9 7 9  7 63 7
(i) 2 4  = 2 4 = 8 7 8  
3 3  9
(ii)    9 272 7  
10 10 10 10
6 9  6 9 54
(iii)     
5 11 5 11 55
3 2 3  2 6 (iv)  5  7 5  35  
7
3 2 3 2 6 (v)     
11 5 11 5 55
3 5 3  5
(vi)   5 3 1 
5  3   
 
 
Question 4:   
Find the value of: 
(i)  4     
4
(iii)   3   
5
2 1
(v)    
13 7
3 4
(vii)   
13 65
 
 
  (ii) 
(iv) 
(vi) 
(ii) 
(iv) 
(vi)   9  
3 2
 5   
7 
3 5
 5 3
3
 2  
5
1 3
  
8 4
7  2 
13  
12

   
  SOLUTION 4:  
(i)  4   =      4   2 3 6  
3 3 1  3 1
(ii) 2 =    3  
5 5 2 5 2 10
4 4 1  4 1 4 (iii)   3 =   5 5 3 5   3 15
1 3 1 4  1 1 1 (iv)  =  =   
8 4 8 3 2 3 6
2 1 2 7  2 7 14 1
(v)  =   1  
13 7 13 1 13 1 13 13
7 2 7 13  7 13 91 19
(vi) 13 = 12 2 = 12  2  24 324  
12 
3 4 3 65 3  5 15 3
(vii) 65 = 134  1 4  4 34  13 
 
 
 

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