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)

 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

(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



(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]

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 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 3

    

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

(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 6

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]

(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

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

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

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

1 3



8 4

7  2 

13

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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Tuesday, December 19, 2023