Rational Numbers

NCERT Solutions for Chapter 9 Rational Numbers Class 7 Maths

Book Solutions

1

List five rational numbers between:
(i) βˆ’1 and 0
(ii) βˆ’2 and βˆ’1
(iii) βˆ’4/5 and βˆ’2/3
(iv) βˆ’1/2 and 2/3

Answer

(i) βˆ’1 and 0
Let us write βˆ’1 and 0 as rational numbers with denominator 6.
β‡’ βˆ’1=βˆ’6/6 and 0 = 0/6
∴ βˆ’6/6<βˆ’5/6<βˆ’4/6<βˆ’3/6<βˆ’2/6<βˆ’1/6<0
β‡’ βˆ’1<βˆ’5/6<βˆ’2/3<βˆ’1/2<βˆ’1/3<βˆ’1/6<0
Therefore, five rational numbers between βˆ’1 and 0 would be
βˆ’56,βˆ’23,βˆ’12,βˆ’13,βˆ’16

(ii) βˆ’2 and βˆ’1
Let us write βˆ’2 and βˆ’1 as rational numbers with denominator 6.
β‡’ βˆ’2=βˆ’12/6 and βˆ’1=βˆ’6/6
∴ βˆ’12/6<βˆ’11/6<βˆ’10/6<βˆ’9/6<βˆ’8/6<βˆ’7/6<βˆ’6/6
Β βˆ’2<βˆ’11/6<βˆ’5/3<βˆ’3/2<βˆ’4/3<βˆ’7/6<βˆ’1
Therefore, five rational numbers between βˆ’2 and βˆ’1 would be
βˆ’11/6,βˆ’5/3,βˆ’3/2,βˆ’4/3,βˆ’7/6

(iii) βˆ’4/5and βˆ’2/3
Let us write βˆ’4/5 and βˆ’2/3 as rational numbers with the same denominators.
β‡’ βˆ’4/5=βˆ’36/45 and βˆ’2/3=βˆ’30/45
∴ βˆ’36/45<βˆ’35/45<βˆ’34/45<βˆ’33/45<βˆ’32/45<βˆ’31/45<βˆ’30/4
β‡’ βˆ’4/5<βˆ’7/9<βˆ’34/45<βˆ’11/15<βˆ’32/45<βˆ’31/45<βˆ’2/3
Therefore, five rational numbers between βˆ’45βˆ’45 and βˆ’23βˆ’23 would be
βˆ’7/9,βˆ’34/45,βˆ’11/15,βˆ’32/45,βˆ’31/45,βˆ’2/3

(iv) βˆ’1/2 and 2/3
Let us write βˆ’1/2and 2/3 as rational numbers with the same denominators.
β‡’ βˆ’1/2=βˆ’3/6 and 2/3=4/6
∴ βˆ’3/6<βˆ’2/6<βˆ’1/6<0<1/6<2/6<3/6<4/6
β‡’ βˆ’1/2<βˆ’1/3<βˆ’1/6<0<1/6<1/3<1/2<2/3
Therefore, five rational numbers between βˆ’1/2 and 2/3 would be βˆ’1/3,βˆ’1/6,0,1/6,1/3
Exercise 9.1 Page Number 182

2

Write four more rational numbers in each of the following patterns:
(i) βˆ’3/5,βˆ’6/10,βˆ’9/15,βˆ’12/20,.........
(ii) βˆ’1/4,βˆ’2/8,βˆ’3/12,.........
(iii) βˆ’1/6,2/βˆ’12,3/βˆ’18,4/βˆ’24,.........
(iv) βˆ’2/3,2/βˆ’3,4/βˆ’6,6/βˆ’9,..........

Answer

(i) βˆ’3/5,βˆ’6/10,βˆ’9/15,βˆ’12/20,.......
β‡’ (βˆ’3Γ—1)/(5Γ—1),(βˆ’3Γ—2)/(5Γ—2),(βˆ’3Γ—3)/(5Γ—3),(βˆ’3Γ—4)/(5Γ—4),........
Therefore, the next four rational numbers of this pattern would be
(βˆ’3Γ—5)/(5Γ—5),(βˆ’3Γ—6)/(5Γ—6),(βˆ’3Γ—7)/(5Γ—7),(βˆ’3Γ—8)/(5Γ—8) =
βˆ’15/25,βˆ’18/30,βˆ’21/35,βˆ’24/40

(ii) βˆ’1/4,βˆ’2/8,βˆ’3/12,..........
β‡’ (βˆ’1Γ—1)/(4Γ—1),(βˆ’1Γ—2)/(4Γ—2),(βˆ’1Γ—3)/(4Γ—3),..........
Therefore, the next four rational numbers of this pattern would be
(βˆ’1Γ—4)/(4Γ—4),(βˆ’1Γ—5)/(4Γ—5),(βˆ’1Γ—6)/(4Γ—6),(βˆ’1Γ—7)/(4Γ—7) =
βˆ’4/16,βˆ’5/20,βˆ’6/24,βˆ’7/28

(iii) βˆ’1/6,2/βˆ’12,3/βˆ’18,4/βˆ’24,.........
β‡’ (βˆ’1Γ—1)/(6Γ—1),(1Γ—2)/(βˆ’6Γ—2),(1Γ—3)/(βˆ’6Γ—3),(1Γ—4)/(βˆ’6Γ—4),.........
Therefore, the next four rational numbers of this pattern would be
(1Γ—5)/(βˆ’6Γ—5),(1Γ—6)/(βˆ’6Γ—6),(1Γ—7)/(βˆ’6Γ—7),(1Γ—8)/(βˆ’6Γ—8) =
5/βˆ’30,6/βˆ’36,7/βˆ’42,8/βˆ’48

(iv) βˆ’2/3,2/βˆ’3,4/βˆ’6,6/βˆ’9,........
β‡’(βˆ’2Γ—1)/(3Γ—1),(2Γ—1)/(βˆ’3Γ—1),(2Γ—2)/(βˆ’3Γ—2),(2Γ—3)/(βˆ’3Γ—3),.........
Therefore, the next four rational numbers of this pattern would be
(2Γ—4)/(βˆ’3Γ—4),(2Γ—5)/(βˆ’3Γ—5),(2Γ—6)/(βˆ’3Γ—6),(2Γ—7)/(βˆ’3Γ—7) =
8/βˆ’12,10/βˆ’15,12/βˆ’18,14/βˆ’21
Exercise 9.1 Page Number 182

3

Give four rational numbers equivalent to:
(i) βˆ’2/7
(ii) 5/βˆ’3
(iii) 4/9

Answer

(i) βˆ’2/7
(βˆ’2Γ—2)/(7Γ—2)=βˆ’4/14,(βˆ’2Γ—3)/(7Γ—3)
=βˆ’6/21,(βˆ’2Γ—4)/(7Γ—4)
=βˆ’8/28,(βˆ’2Γ—5)/(7Γ—5)
=βˆ’10/35
Therefore, four equivalent rational numbers are βˆ’4/14,βˆ’6/21,βˆ’8/28,βˆ’10/35

(ii) 5/βˆ’3
(5Γ—2)/(βˆ’3Γ—2)
=10/βˆ’6,(5Γ—3)/(βˆ’3Γ—3)
=15/βˆ’9, (5Γ—4)/(βˆ’3Γ—4)
=20/βˆ’12,(5Γ—5)/(βˆ’3Γ—5)
=25/βˆ’15
Therefore, four equivalent rational numbers are 10/βˆ’6,15/βˆ’9,20/βˆ’12,25/βˆ’15

(iii) 4/9
(4Γ—2)/(9Γ—2)
=8/18,(4Γ—3)/(9Γ—3)
=12/27,(4Γ—4)/(9Γ—4)
16/36,(4Γ—5)/(9Γ—5)
=20/45
Therefore, four equivalent rational numbers are 8/18,12/27,16/36,20/45
Exercise 9.1 Page Number 183

4

Draw the number line and represent the following rational numbers on it:
(i) 3/4
(ii) βˆ’5/8
(iii) βˆ’7/4
(iv) 7/8

Answer

(i) 3/4
Β 

(ii) βˆ’5/8
Β 

(iii) βˆ’7/4
Β 

(iv) 7/8
Β 
Exercise 9.1 Page Number 183

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.
Β 

Answer

Each part which is between the two numbers is divided into 3 parts.
Therefore, A = 6/3, P = 7/3,Q = 8/3 and B = 9/3
Similarly T = βˆ’3/3, R = βˆ’4/3, S = βˆ’5/3 and U = βˆ’6/3
Thus, the rational numbers represented P, Q, R and S are 7/3,8/3,βˆ’4/3 and βˆ’5/3 respectively.
Exercise 9.1 Page Number 183

6

Which of the following pairs represent the same rational numbers:
(i) βˆ’7/21 and 3/9
(ii) βˆ’16/20 and 20/βˆ’25
(iii) βˆ’2/βˆ’3and 2/3
(iv) βˆ’3/5 and βˆ’12/20
(v) 8/βˆ’5 and βˆ’24/15
(vi) 1/3 and βˆ’1/9
(vii) βˆ’5/βˆ’9 and 5/βˆ’9

Answer

(i) βˆ’7/21 and 3/9
β‡’ βˆ’7/21= βˆ’1/3 and 3/9 = 1/3 [Converting into lowest term]
∡ βˆ’1/3β‰ 1/3
∴ βˆ’7/21β‰ 3/9

(ii) βˆ’16/20 and 20/βˆ’25
β‡’ βˆ’16/20 = βˆ’4/5 and 20/βˆ’25 = 4/βˆ’5 [Converting into lowest term]
∡ βˆ’4/5 = βˆ’4/5
∴ βˆ’16/20 = 20/βˆ’25

(iii) βˆ’2/βˆ’3 and 2/3
β‡’ βˆ’2/βˆ’3 = 2/3 and 2/3 = 2/3 [Converting into lowest term]
∡ 2/3 = 2/3
∴ βˆ’2/βˆ’3 = 2/3

(iv) βˆ’3/5 and βˆ’12/20
β‡’ βˆ’3/5 = βˆ’3/5 and βˆ’12/20 = βˆ’3/5 [Converting into lowest term]
∡ βˆ’3/5 = βˆ’3/5
∴ βˆ’3/5 = βˆ’12/20

(v) 8/βˆ’5 and βˆ’24/15
β‡’ 8/βˆ’5 = βˆ’8/5 and βˆ’24/15 = βˆ’8/5 [Converting into lowest term]
∡ βˆ’8/5 = βˆ’8/5
∴ 8/βˆ’5 = βˆ’24/15

(vi) 1/3 and βˆ’1/9
β‡’ 1/3 = 1/3 and βˆ’1/9 = βˆ’1/9 [Converting into lowest term]
∡ 1/3β‰ βˆ’1/9
∴ 1/3β‰ βˆ’1/9

(vii) βˆ’5/βˆ’9 and 5/βˆ’9
β‡’ βˆ’5/βˆ’9 = 5/9 and 5/βˆ’9 = 5/9 [Converting into lowest term]
∡ 5/9β‰  5/9
∴ βˆ’5/βˆ’9 β‰  5/βˆ’9
Exercise 9.1 Page Number 183

7

Rewrite the following rational numbers in the simplest form:
(i) βˆ’8/6
(ii) 25/45
(iii) βˆ’44/72
(iv) βˆ’8/10

Answer

(i) βˆ’8/6 = (βˆ’8Γ·2)/(6Γ·2) = βˆ’43 [H.C.F. of 8 and 6 is 2]

(ii)25/45 = (25Γ·5)/(45Γ·5) = 5/9
[H.C.F. of 25 and 45 is 5]

(iii)βˆ’44/72= (βˆ’44Γ·4)/(72Γ·4) = βˆ’11/18Β  [H.C.F. of 44 and 72 is 4]

(iv) βˆ’8/10= (βˆ’8Γ·2)/(10Γ·2) = βˆ’4/5 [H.C.F. of 8 and 10 is 2]
Exercise 9.1 Page Number 183

8

Fill in the boxes with the correct symbol out of <, > and =:
(i) βˆ’5/7 ___ 2/3
(ii) βˆ’4/5 ___ βˆ’5/7
(iii) βˆ’7/8___ 14/βˆ’16
(iv) βˆ’8/5 ___ βˆ’7/4
(v) 1/βˆ’3 ___ βˆ’1/4
(vi) 5/βˆ’11 ___ βˆ’5/11
(vii) 0 ___βˆ’7/6

Answer

(i) βˆ’5/7 < 2/3
Since, the positive number if greater than negative number.

(ii) (βˆ’4Γ—7)/(5Γ—7)___ (βˆ’5Γ—5)/(7Γ—5)
β‡’ βˆ’28/35 < βˆ’25/35
β‡’ βˆ’4/5 < βˆ’5/7

(iii) (βˆ’7Γ—2)/(8Γ—2)___ {14Γ—(βˆ’1)}/{βˆ’16Γ—(βˆ’1)}
β‡’ βˆ’14/16 = βˆ’14/16
β‡’βˆ’7/8 = 14/βˆ’16

(iv) (βˆ’8Γ—4)/(5Γ—4)___ (βˆ’7Γ—5)/(4Γ—5)
β‡’ βˆ’32/20 > βˆ’35/20
β‡’ βˆ’8/5 > βˆ’7/4

(v) (1/βˆ’3) ___ (βˆ’1/4)
β‡’1/βˆ’3 < βˆ’1/4

(vi) (5/βˆ’11)___ (βˆ’5/11)
β‡’5/βˆ’11 = βˆ’5/11

(vii) 0 > βˆ’7/6
Since, 0 is greater than every negative number.
Exercise 9.1 Page Number 183

9

Which is greater in each of the following:
(i) 2/3,5/2
(ii) βˆ’5/6,βˆ’4/3
(iii) βˆ’3/4,2/βˆ’3
(iv) βˆ’1/4,1/4
(v) βˆ’3.2/7,βˆ’3.4/5

Answer

(i) (2Γ—2)/(3Γ—2)=4/6 and (5Γ—3)/(2Γ—3)=15/6
Since 4/6 < 15/6
Therefore 2/3 < 5/2

(ii) (βˆ’5Γ—1)/(6Γ—1)=βˆ’5/6 and (βˆ’4Γ—2)/(3Γ—2)=βˆ’8/6
Since βˆ’5/6 > βˆ’8/6 Therefore βˆ’5/6 > βˆ’4/3

(iii) (βˆ’3Γ—3)/(4Γ—3)=βˆ’9/12 and {2Γ—(βˆ’4)}/{βˆ’3Γ—(βˆ’4)}=βˆ’8/122
Since βˆ’9/12 < βˆ’8/12
Therefore βˆ’3/4 < 2/βˆ’3

(iv) βˆ’1/4 < 1/4 Since positive number is always greater than negative number.

(v) βˆ’3.2/7=βˆ’2.3/7=(βˆ’23Γ—5)/(7Γ—5)=βˆ’115/35 and βˆ’3.4/5=βˆ’19/5=(βˆ’19Γ—7)/(5Γ—7)=βˆ’133/35
Since βˆ’115/35 > βˆ’133/35
Thereforeβˆ’3.2/7 > βˆ’3.4/5
Exercise 9.1 Page Number 184

10

Write the following rational numbers in ascending order:
(i) βˆ’3/5,βˆ’2/5,βˆ’1/5
(ii) 1/3,βˆ’2/9,βˆ’4/3
(iii) βˆ’3/7,βˆ’3/2,βˆ’3/4

Answer

(i) βˆ’3/5,βˆ’2/5,βˆ’1/5β‡’ βˆ’3/5<βˆ’2/5<βˆ’1/5

(ii) 1/3,βˆ’2/9,βˆ’4/3β‡’3/9,βˆ’2/9,βˆ’12/9 [Converting into same denominator]
Now βˆ’12/9<βˆ’2/9<3/9β‡’ βˆ’4/3<βˆ’2/9<1/3

(iii) βˆ’3/7,βˆ’3/2,βˆ’3/4
β‡’ βˆ’3/2<βˆ’3/4<βˆ’3/7
Exercise 9.1 Page Number 184

1

Find the sum:
(i) 5/4+(βˆ’11/4)
(ii) 5/3+3/5
(iii) βˆ’9/10+22/15
(iv) βˆ’3/βˆ’11+5/9
(v) βˆ’8/19+(βˆ’2)/57
(vi) βˆ’2/3+0
(vii) βˆ’2.1/3+4.3/5

Answer

(i) 5/4+(βˆ’11/4)
= (5βˆ’11)/4
= βˆ’6/4=βˆ’3/2

(ii) 5/3+3/5
= (5Γ—5)/(3Γ—5)+(3Γ—3)/(5Γ—3)
= 25/15+9/15 [L.C.M. of 3 and 5 is 15]
= (25+9)/15=34/15=2.4/15

(iii) βˆ’9/10+22/15
= (βˆ’9Γ—3)/(10Γ—3)+(22Γ—2)/(15Γ—2)
= βˆ’27/30+44/30 [L.C.M. of 10 and 15 is 30]
= (βˆ’27+44)/30=17/30

(iv) βˆ’3/βˆ’11+5/9
= (βˆ’3Γ—9)/(βˆ’11Γ—9)+(5Γ—11)/(9Γ—11)
= 27/99+55/99 [L.C.M. of 11 and 9 is 99]
= (27+55)/99=82/99

(v) βˆ’8/19+(βˆ’2)/57
= (βˆ’8Γ—3)/(19Γ—3)+{(βˆ’2)Γ—15}/{7Γ—1}
= βˆ’24/57+(βˆ’2)/57 [L.C.M. of 19 and 57 is 57]
= (βˆ’24βˆ’2)/57 = βˆ’26/57

(vi) βˆ’2/3+0=βˆ’2/3

(vii) βˆ’2.1/3+4.3/5 = (βˆ’7/3+23/5) = (βˆ’7Γ—5)/(3Γ—5)+(23Γ—3)/(5Γ—3) = βˆ’35/15+69/15 [L.C.M. of 3 and 5 is 15]
= (βˆ’35+69)/15 = 34/15=2.4/15
Exercise 9.2 Page Number 190

2

Find:
(i) 7/24βˆ’17/36
(ii) 5/63βˆ’(βˆ’6/21)
(iii) βˆ’6/13βˆ’(βˆ’7/15)
(iv) βˆ’3/8βˆ’7/11
(v) βˆ’2.1/9βˆ’6

Answer

(i) 7/24βˆ’17/36

= (7Γ—3)/(24Γ—3)βˆ’(17Γ—2)/(36Γ—2)

= 21/72βˆ’34/72 [L.C.M. of 24 and 36 is 72]

= (21βˆ’34)/72Β 

= βˆ’13/72

Β 

(ii) 5/63βˆ’(βˆ’6/21)= (5Γ—1)/(63Γ—1)βˆ’{(βˆ’6Γ—3)/(21Γ—3)}Β 

= 5/63βˆ’(βˆ’18)/63 [L.C.M. of 63 and 21 is 63]

= {5βˆ’(βˆ’18)}/63Β 

= 23/63

Β 

(iii) βˆ’6/13βˆ’(βˆ’7/15)

= (βˆ’6Γ—15)/(13Γ—15)βˆ’{(βˆ’7Γ—13)/(15Γ—13)}

= (βˆ’90/195)βˆ’(βˆ’91/195) [L.C.M. of 13 and 15 is 195]

= {βˆ’90βˆ’(βˆ’91)}/195Β 

= (βˆ’90+91)/195

=1/195

Β 

(iv) βˆ’3/8βˆ’7/11Β 

= (βˆ’3Γ—11)/(8Γ—11)βˆ’(7Γ—8)/(11Γ—8)Β 

= βˆ’33/88βˆ’56/88

[L.C.M. of 8 and 11 is 88]

= (βˆ’33βˆ’56)/88Β 

= βˆ’89/88

=βˆ’1.1/88

Β 

(v) βˆ’2.1/9βˆ’6Β 

= (βˆ’19/9)(βˆ’6/1)Β 

= (βˆ’19Γ—1)/(9Γ—1)βˆ’(6Γ—9)/(1Γ—9) [L.C.M. of 9 and 1 is 9]

= (βˆ’199βˆ’54)/9Β 

= (βˆ’19βˆ’54)/9Β 

= βˆ’73/9=βˆ’8.1/9

Exercise 9.2 Page Number 190

3

Find the product:
(i) 9/2Γ—(βˆ’7/4)
(ii) 3/10Γ—(βˆ’9)
(iii) βˆ’6/5Γ—9/11
(iv) 3/7Γ—(βˆ’2/5)
(v) 3/11Γ—2/5
(vi) 3/βˆ’5Γ—5/3

Answer

(i) 9/2Γ—(βˆ’7/4)
= {9Γ—(βˆ’7)}/{2Γ—4}
= βˆ’63/8
=βˆ’7.7/8

(ii)3/10Γ—(βˆ’9)
={3Γ—(βˆ’9)}/10
=βˆ’27/10
βˆ’2.7/10

(iii) βˆ’6/5Γ—9/11
={(βˆ’6)Γ—9}/{5Γ—11}
=βˆ’54/55

(iv) 3/7Γ—(βˆ’2/5)
={3Γ—(βˆ’2)}/{7Γ—5}
=βˆ’6/35

(v) 3/11Γ—2/5
=(3Γ—2)/(11Γ—5)
=6/55

(vi) 3/βˆ’5Γ—(βˆ’5/3)
={13βˆ’5}Γ—{(βˆ’5/3)}
=1
Exercise 9.2 Page Number 190

4

Find the value of:
(i) (βˆ’4)Γ·2/3
(ii) βˆ’3/5Γ·2
(iii) βˆ’4/5Γ·(βˆ’3)
(iv) βˆ’1/8Γ·3/4
(v) βˆ’2/13Γ·1/7
(vi) βˆ’7/12Γ·(2/13)
(vii) 3/13Γ·(βˆ’4/65)

Answer

(i) (βˆ’4)Γ·2/3
= (βˆ’4)Γ—3/2
=(βˆ’2)Γ—3
=βˆ’6

(ii) βˆ’3/5Γ·2
= βˆ’3/5Γ—1/2
={(βˆ’3)Γ—1}/{5Γ—2}
=βˆ’3/10

(iii) βˆ’4/5Γ·(βˆ’3)
= (βˆ’4)/5Γ—1/(βˆ’3)
={(βˆ’4)Γ—1}/{5Γ—(βˆ’3)}
=4/15

(iv) βˆ’1/8Γ·3/4
= βˆ’1/8Γ—4/3
= ((βˆ’1)Γ—1)/(2Γ—3)
=βˆ’1/6

(v) βˆ’2/13Γ·1/7
= βˆ’2/13Γ—7/1
=((βˆ’2)Γ—7)/(13Γ—1)
=βˆ’14/13
=βˆ’1.1/13

(vi) βˆ’7/12Γ·(βˆ’2/13)
= βˆ’7/12Γ—13/(βˆ’2)
=((βˆ’7)Γ—13)/(12Γ—(βˆ’2))
=βˆ’91/24
=3.19/24

(vii) 3/13Γ·(βˆ’4/65)
= 3/13Γ—65/(βˆ’4)
=(3Γ—(βˆ’5))/(1Γ—4)
=βˆ’15/4
=βˆ’3.3/4
Exercise 9.2 Page Number 190