NCERT Solutions for Chapter 9 Rational Numbers Class 7 Maths
Book Solutions1
Answer
Let us write β1 and 0 as rational numbers with denominator 6.
β β1=β6/6 and 0 = 0/6
β΄ β6/6<β5/6<β4/6<β3/6<β2/6<β1/6<0
β β1<β5/6<β2/3<β1/2<β1/3<β1/6<0
Therefore, five rational numbers between β1 and 0 would be
β56,β23,β12,β13,β16
(ii) β2 and β1
Let us write β2 and β1 as rational numbers with denominator 6.
β β2=β12/6 and β1=β6/6
β΄ β12/6<β11/6<β10/6<β9/6<β8/6<β7/6<β6/6
Β β2<β11/6<β5/3<β3/2<β4/3<β7/6<β1
Therefore, five rational numbers between β2 and β1 would be
β11/6,β5/3,β3/2,β4/3,β7/6
(iii) β4/5and β2/3
Let us write β4/5 and β2/3 as rational numbers with the same denominators.
β β4/5=β36/45 and β2/3=β30/45
β΄ β36/45<β35/45<β34/45<β33/45<β32/45<β31/45<β30/4
β β4/5<β7/9<β34/45<β11/15<β32/45<β31/45<β2/3
Therefore, five rational numbers between β45β45 and β23β23 would be
β7/9,β34/45,β11/15,β32/45,β31/45,β2/3
(iv) β1/2 and 2/3
Let us write β1/2and 2/3 as rational numbers with the same denominators.
β β1/2=β3/6 and 2/3=4/6
β΄ β3/6<β2/6<β1/6<0<1/6<2/6<3/6<4/6
β β1/2<β1/3<β1/6<0<1/6<1/3<1/2<2/3
Therefore, five rational numbers between β1/2 and 2/3 would be β1/3,β1/6,0,1/6,1/3
2
Answer
β (β3Γ1)/(5Γ1),(β3Γ2)/(5Γ2),(β3Γ3)/(5Γ3),(β3Γ4)/(5Γ4),........
Therefore, the next four rational numbers of this pattern would be
(β3Γ5)/(5Γ5),(β3Γ6)/(5Γ6),(β3Γ7)/(5Γ7),(β3Γ8)/(5Γ8) =
β15/25,β18/30,β21/35,β24/40
(ii) β1/4,β2/8,β3/12,..........
β (β1Γ1)/(4Γ1),(β1Γ2)/(4Γ2),(β1Γ3)/(4Γ3),..........
Therefore, the next four rational numbers of this pattern would be
(β1Γ4)/(4Γ4),(β1Γ5)/(4Γ5),(β1Γ6)/(4Γ6),(β1Γ7)/(4Γ7) =
β4/16,β5/20,β6/24,β7/28
(iii) β1/6,2/β12,3/β18,4/β24,.........
β (β1Γ1)/(6Γ1),(1Γ2)/(β6Γ2),(1Γ3)/(β6Γ3),(1Γ4)/(β6Γ4),.........
Therefore, the next four rational numbers of this pattern would be
(1Γ5)/(β6Γ5),(1Γ6)/(β6Γ6),(1Γ7)/(β6Γ7),(1Γ8)/(β6Γ8) =
5/β30,6/β36,7/β42,8/β48
(iv) β2/3,2/β3,4/β6,6/β9,........
β(β2Γ1)/(3Γ1),(2Γ1)/(β3Γ1),(2Γ2)/(β3Γ2),(2Γ3)/(β3Γ3),.........
Therefore, the next four rational numbers of this pattern would be
(2Γ4)/(β3Γ4),(2Γ5)/(β3Γ5),(2Γ6)/(β3Γ6),(2Γ7)/(β3Γ7) =
8/β12,10/β15,12/β18,14/β21
3
Answer
(β2Γ2)/(7Γ2)=β4/14,(β2Γ3)/(7Γ3)
=β6/21,(β2Γ4)/(7Γ4)
=β8/28,(β2Γ5)/(7Γ5)
=β10/35
Therefore, four equivalent rational numbers are β4/14,β6/21,β8/28,β10/35
(ii) 5/β3
(5Γ2)/(β3Γ2)
=10/β6,(5Γ3)/(β3Γ3)
=15/β9, (5Γ4)/(β3Γ4)
=20/β12,(5Γ5)/(β3Γ5)
=25/β15
Therefore, four equivalent rational numbers are 10/β6,15/β9,20/β12,25/β15
(iii) 4/9
(4Γ2)/(9Γ2)
=8/18,(4Γ3)/(9Γ3)
=12/27,(4Γ4)/(9Γ4)
16/36,(4Γ5)/(9Γ5)
=20/45
Therefore, four equivalent rational numbers are 8/18,12/27,16/36,20/45
4
Answer
Β

(ii) β5/8
Β

(iii) β7/4
Β

(iv) 7/8
Β

5

Answer
Therefore, A = 6/3, P = 7/3,Q = 8/3 and B = 9/3
Similarly T = β3/3, R = β4/3, S = β5/3 and U = β6/3
Thus, the rational numbers represented P, Q, R and S are 7/3,8/3,β4/3 and β5/3 respectively.
6
Answer
β β7/21= β1/3 and 3/9 = 1/3 [Converting into lowest term]
β΅ β1/3β 1/3
β΄ β7/21β 3/9
(ii) β16/20 and 20/β25
β β16/20 = β4/5 and 20/β25 = 4/β5 [Converting into lowest term]
β΅ β4/5 = β4/5
β΄ β16/20 = 20/β25
(iii) β2/β3 and 2/3
β β2/β3 = 2/3 and 2/3 = 2/3 [Converting into lowest term]
β΅ 2/3 = 2/3
β΄ β2/β3 = 2/3
(iv) β3/5 and β12/20
β β3/5 = β3/5 and β12/20 = β3/5 [Converting into lowest term]
β΅ β3/5 = β3/5
β΄ β3/5 = β12/20
(v) 8/β5 and β24/15
β 8/β5 = β8/5 and β24/15 = β8/5 [Converting into lowest term]
β΅ β8/5 = β8/5
β΄ 8/β5 = β24/15
(vi) 1/3 and β1/9
β 1/3 = 1/3 and β1/9 = β1/9 [Converting into lowest term]
β΅ 1/3β β1/9
β΄ 1/3β β1/9
(vii) β5/β9 and 5/β9
β β5/β9 = 5/9 and 5/β9 = 5/9 [Converting into lowest term]
β΅ 5/9β 5/9
β΄ β5/β9 β 5/β9
7
Answer
(ii)25/45 = (25Γ·5)/(45Γ·5) = 5/9
[H.C.F. of 25 and 45 is 5]
(iii)β44/72= (β44Γ·4)/(72Γ·4) = β11/18Β [H.C.F. of 44 and 72 is 4]
(iv) β8/10= (β8Γ·2)/(10Γ·2) = β4/5 [H.C.F. of 8 and 10 is 2]
8
Answer
Since, the positive number if greater than negative number.
(ii) (β4Γ7)/(5Γ7)___ (β5Γ5)/(7Γ5)
β β28/35 < β25/35
β β4/5 < β5/7
(iii) (β7Γ2)/(8Γ2)___ {14Γ(β1)}/{β16Γ(β1)}
β β14/16 = β14/16
ββ7/8 = 14/β16
(iv) (β8Γ4)/(5Γ4)___ (β7Γ5)/(4Γ5)
β β32/20 > β35/20
β β8/5 > β7/4
(v) (1/β3) ___ (β1/4)
β1/β3 < β1/4
(vi) (5/β11)___ (β5/11)
β5/β11 = β5/11
(vii) 0 > β7/6
Since, 0 is greater than every negative number.
9
Answer
Since 4/6 < 15/6
Therefore 2/3 < 5/2
(ii) (β5Γ1)/(6Γ1)=β5/6 and (β4Γ2)/(3Γ2)=β8/6
Since β5/6 > β8/6 Therefore β5/6 > β4/3
(iii) (β3Γ3)/(4Γ3)=β9/12 and {2Γ(β4)}/{β3Γ(β4)}=β8/122
Since β9/12 < β8/12
Therefore β3/4 < 2/β3
(iv) β1/4 < 1/4 Since positive number is always greater than negative number.
(v) β3.2/7=β2.3/7=(β23Γ5)/(7Γ5)=β115/35 and β3.4/5=β19/5=(β19Γ7)/(5Γ7)=β133/35
Since β115/35 > β133/35
Thereforeβ3.2/7 > β3.4/5
10
Answer
(ii) 1/3,β2/9,β4/3β3/9,β2/9,β12/9 [Converting into same denominator]
Now β12/9<β2/9<3/9β β4/3<β2/9<1/3
(iii) β3/7,β3/2,β3/4
β β3/2<β3/4<β3/7
1
Answer
= (5β11)/4
= β6/4=β3/2
(ii) 5/3+3/5
= (5Γ5)/(3Γ5)+(3Γ3)/(5Γ3)
= 25/15+9/15 [L.C.M. of 3 and 5 is 15]
= (25+9)/15=34/15=2.4/15
(iii) β9/10+22/15
= (β9Γ3)/(10Γ3)+(22Γ2)/(15Γ2)
= β27/30+44/30 [L.C.M. of 10 and 15 is 30]
= (β27+44)/30=17/30
(iv) β3/β11+5/9
= (β3Γ9)/(β11Γ9)+(5Γ11)/(9Γ11)
= 27/99+55/99 [L.C.M. of 11 and 9 is 99]
= (27+55)/99=82/99
(v) β8/19+(β2)/57
= (β8Γ3)/(19Γ3)+{(β2)Γ15}/{7Γ1}
= β24/57+(β2)/57 [L.C.M. of 19 and 57 is 57]
= (β24β2)/57 = β26/57
2
Answer
(i) 7/24β17/36
= (7Γ3)/(24Γ3)β(17Γ2)/(36Γ2)
= 21/72β34/72 [L.C.M. of 24 and 36 is 72]
= (21β34)/72Β
= β13/72
Β
(ii) 5/63β(β6/21)= (5Γ1)/(63Γ1)β{(β6Γ3)/(21Γ3)}Β
= 5/63β(β18)/63 [L.C.M. of 63 and 21 is 63]
= {5β(β18)}/63Β
= 23/63
Β
(iii) β6/13β(β7/15)
= (β6Γ15)/(13Γ15)β{(β7Γ13)/(15Γ13)}
= (β90/195)β(β91/195) [L.C.M. of 13 and 15 is 195]
= {β90β(β91)}/195Β
= (β90+91)/195
=1/195
Β
(iv) β3/8β7/11Β
= (β3Γ11)/(8Γ11)β(7Γ8)/(11Γ8)Β
= β33/88β56/88
[L.C.M. of 8 and 11 is 88]
= (β33β56)/88Β
= β89/88
=β1.1/88
Β
(v) β2.1/9β6Β
= (β19/9)(β6/1)Β
= (β19Γ1)/(9Γ1)β(6Γ9)/(1Γ9) [L.C.M. of 9 and 1 is 9]
= (β199β54)/9Β
= (β19β54)/9Β
= β73/9=β8.1/9
3
Answer
= {9Γ(β7)}/{2Γ4}
= β63/8
=β7.7/8
(ii)3/10Γ(β9)
={3Γ(β9)}/10
=β27/10
β2.7/10
(iii) β6/5Γ9/11
={(β6)Γ9}/{5Γ11}
=β54/55
(iv) 3/7Γ(β2/5)
={3Γ(β2)}/{7Γ5}
=β6/35
(v) 3/11Γ2/5
=(3Γ2)/(11Γ5)
=6/55
(vi) 3/β5Γ(β5/3)
={13β5}Γ{(β5/3)}
=1
4
Answer
= (β4)Γ3/2
=(β2)Γ3
=β6
(ii) β3/5Γ·2
= β3/5Γ1/2
={(β3)Γ1}/{5Γ2}
=β3/10
(iii) β4/5Γ·(β3)
= (β4)/5Γ1/(β3)
={(β4)Γ1}/{5Γ(β3)}
=4/15
(iv) β1/8Γ·3/4
= β1/8Γ4/3
= ((β1)Γ1)/(2Γ3)
=β1/6
(v) β2/13Γ·1/7
= β2/13Γ7/1
=((β2)Γ7)/(13Γ1)
=β14/13
=β1.1/13
(vi) β7/12Γ·(β2/13)
= β7/12Γ13/(β2)
=((β7)Γ13)/(12Γ(β2))
=β91/24
=3.19/24
(vii) 3/13Γ·(β4/65)
= 3/13Γ65/(β4)
=(3Γ(β5))/(1Γ4)
=β15/4
=β3.3/4