RD Sharma Class 8 Mathematics Chapter 1 Rational numbers

 

 

Page No 1.14:

Question 1:

Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers:
(i) −115 and 47

(ii) 49 and 7−12

(iii) −35 and −2−15

(iv) 2−7 and 12−35

(v) 4 and −35

(vi) −4 and 4−7

Answer:

 Commutativity of the addition of rational numbers means that if ab and cd are two rational numbers, then ab+cd=cd+ab.(i) We have −115 and 47.∴ −115 + 47= −11×75×7+4×57×5=−7735+2035=−77+2035=−5735  47+−115=4×57×5+−11×75×7=2035+−7735=20−7735=−5735∴ −115 + 47=47+−115  Hence verified.(ii)  We have 49 and 7−12.∴ 49+−712= 4×49×4+−7×312×3=1636+−2136=16−2136=−536−712+49=−7×312×3+4×49×4=−2136+1636=−21+1636=−536∴ 49+−712=−712+49 Hence verified.(iii) We have −35 and −2−15 or −35 and 215.∴ −35+215=−915+215=−9+215=−715215+−35=215+−915=2−915=−715∴ −35+−2−15=−2−15+−35Hence verified.(iv) We have 2−7 and 12−35.∴−27+−1235=−2×57×5+−1235=−10−1235=−223512−35+2−7=−1235+−2×57×5=−12−1035=−2235∴ 2−7+12−35=12−35+2−7Hence verified.(v) We have 4 and −35. ∴ 4+−35=4×51×5+−35=20−35=175  −35+4=−35+4×51×5=−3+205=175 ∴ 4+−35=−35+4 Hence verified.(vi) We have −4 and 4−7.∴ −41+−47=−4×71×7+−47=−28−47=−327 −47+−41=−47+−4×71×7=−4−287=−327∴−4+4−7=4−7−4 Hence verified.

Page No 1.14:

Question 2:

Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:
(i) x=12, y=23, z=−15

(ii) x=−25, y=43, z=−710
(iii) x=−711, y=2−5, z=−322
(iv) x=−2, y=35, z=−43

Answer:

We have to verify that (x+y)+z = x+(y+z).(i) x=12, y=23, z=−15(x+y)+z =(12+23)+−15=(36+46)+−15=76+−15=3530+−630=35−630=2930 x+(y+z)=12+(23+−15)=12+(1015+−315)=12+715=1530+1430=15+1430=2930∴ ( 12+23)+−15=12+(23+−15) Hence verified.(ii) x=−25, y=43, z=−710(x+y)+z= (−25+43)+−710=(−615+2015)+−710=1415+−710=2830+−2130=28−2130=730 x+(y+z)=−25+(43+−710)=−25+(4030+−2130)=−25+1930=−1230+1930=−12+1930=730∴−25+43)+−710=−25+(43+−710) Hence verified.(iii) x=−711, y=2−5, z=−322(x+y)+z =(−711+2−5)+−322=(−3555+−2255)+−322=−5755+−322=−114110+−15110=−114−15110=−129110 x+(y+z)=−711+(2−5+−322)=−711+(−44110+−15110)=−711+−59110=−70110+−59110=−70−59110=−129110∴ (−711+2−5)+−322=−711+(2−5+−322)Hence verified.(iv) x=−2, y=35, z=−43so, (x+y)+z =(−2+35)+−43=(−105+35)+−43=−75+−43=−2115+−2015=−21−2015=−4115 x+(y+z)=−2+(35+−43)=−21+(915+−2015)=−21+−1115=−3015+−1115=−30−1115=−4115∴ (−2+35)+−43=−2+(35+−43)Hence verified.

Page No 1.14:

Question 3:

Write the additive inverse of each of the following rational numbers:
(i) −217

(ii) 3−11
(iii) −175
(iv) −11−25

Answer:

(i) Additive inverse is the negative of the given number.So, additive inverse of −217 is 217.(ii) Additive inverse is the negative of the given number.So, additive inverse of 3−11 is 311.(iii) Additive inverse is the negative of the given number.So, additive inverse of −175 is 175.(iv) Additive inverse is the negative of the given number.So, additive inverse of −11−25 is −1125.

Page No 1.14:

Question 4:

Write the negative (additive inverse) of each of the following:
(i) −25

(ii) 7−9
(iii) −1613
(iv) −51

(v) 0
(vi) 1
(vii) −1

Answer:

(i) Negative of −25 is 25.(ii) Negative of 7−9 is 79.(iii) Negative of −1613 is 1613.(iv) Negative of −51 is 51.(v) Negative value of 0 is 0.(vi) Negative of 1 is −1.(vi) Negative of −1 is 1.

Page No 1.14:

Question 5:

Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:
(i) 25+73+−45+−13

(ii) 37+−49+−117+79
(iii) 25+83+−1115+45+−23
(iv) 47+0+−89+−137+1721

Answer:

(i) We have:25+73+−45+−13=(25+−45)+(73+−13)=(2−45)+(7−13)=−25+63=−6+3015=2415=85(ii)We have:37+−49+−117+79=(37+−117)+(−49+79)=(3−117)+(−4+79)=−87+39=−72+2163=−5163=−1721(iii) We have:25+83+−1115+45+−23=(25+45)+(+83+−23)+−1115=(2+45)+(8−23)+−1115=65+63+−1115=18+30−1115=3715(iv)We have: 47+0+−89+−137+1721=(47+−137)+(−89)+1721=(4−137)+(−89)+1721=−97+−89+1721=−81−56+5163=−8663

Page No 1.14:

Question 6:

Re-arrange suitably and find the sum in each of the following:
(i) 1112+−173+112+−252

(ii) −67+−56+−49+−157
(iii) 35+73+95+−1315+−73
(iv) 413+−58+−813+913
(v) 23+−45+13+25
(vi) 18+512+27+712+97+−516

Answer:

(i) (112+−252)+−173+1112=(11−252)+−173+1112=−142+−173+1112=−7+−173+1112=−84−68+1112=−14112=−474(ii) (−67+−157)+−56+−49=(−6−157)+−56+−49=−217+−56+−49=−378−105−56126=−539126=−7718(iii) (35+95)+(−73+73)+−1315=(3+95)+(−7+73)+−1315=125+−1315=36−1315=2315(iv) (413−813+913)+−58=(4−8+913)+−58=513+−58=40−65104=−25104(v) (23+13)+(−45+25)=33+−25=15−615=915=35(vi) (18+−516)+(512+712)+(27+97)=(2−516)+(5+712)+(2+97)=−316+1212+117=−63+336+528336=801336=267112

Page No 1.18:

Question 1:

Subtract the first rational number from the second in each of the following:
(i) 38, 58

(ii) −79, 49
(iii) −211, −911
(iv) 1113, −413
(v) 14, −38
(vi) −23, 56
(vii) −67, −1314
(viii) −833, −722

Answer:

(i) 58−38=5−38=28=14(ii) 49−−79=4−(−7)9=4+79=119(iii) −911−−211=(−9)−(−2)11=−9+211=−711(iv) −413−1113=−4−1113=−1513(v) −38−14=−3−28=−58(vi) 56−−23=56−−46=5−(−4)6=5+46=96=32(vii) −1314−−67=−13−(−12)14=−13+1214=−114(viii) −722−−833=−2166−−1666=(−21)−(−16)66=−21+1666=−566

Page No 1.18:

Question 2:

Evaluate each of the following:
(i) 23−35

(ii) −47−2−3
(iii) 47−−5−7
(iv) −2−59
(v) −3−8−−27
(vi) −413−−526
(vii) −514−−27
(viii) 1315−1225
(ix) −613−−713
(x) 724−1936
(xi) 563−−821

Answer:

(i) 23−35=1015−915=10−915=115(ii) −47−2−3=−1221−−1421=(−12)−(−14)21=−12+1421=221(iii) 47−−5−7=4−57=−17(iv) −2−59=−189−59=−18−59=−239(v) −3−8−−27=2156−−1656=21−(−16)56=21+1656=3756(vi) −413−−526=−826−−526=−8−(−5)26=−8+526=−326(vii) −514−−27=−514−−414=−5−(−4)14=−5+414=−114(viii) 1315−1225=6575−3675=65−3675=2975(ix) −613−−713=−6−(−7)13=−6+713=113(x) 724−1936=2172−3872=21−3872=−1772(xi) 563−−821=563−−2463=5−(−24)63=5+2463=2963

Page No 1.18:

Question 3:

The sum of the two numbers is 59.

If one of the numbers is 13,

find the other.

Answer:

It is given that the sum of two numbers is 59, where one of the numbers is 13.Let the other number be x.∴  x+13=59⇒ x=59−13⇒x=59−39⇒ x=5−39⇒ x=29

Page No 1.18:

Question 4:

The sum of two numbers is −13.

If one of the numbers is −123,

find the other.

Answer:

It is given that the sum of two numbers is −13, where one of the numbers is −123.Let the other number be x.∴ x+−123=−13⇒ x=−13−−123⇒ x=−1−(−12)3=−1+123=113

Page No 1.18:

Question 5:

The sum of two numbers is −43.

If one of the numbers is −5, find the other.

Answer:

It is given that the sum of two numbers is −43, where one of the numbers is −5.Let the other number be x.∴ x+(−5)=−43⇒ x=−43−(−51)⇒ x=−43−−153 ⇒x=−4−(−15)3⇒x=−4+153=113

Page No 1.18:

Question 6:

The sum of two rational numbers is −8. If one of the numbers is −157,

find the other.

Answer:

It is given that the sum of two rational numbers is −8, where one of the numbers is −157.Let the other rational number be x.∴ x + (−157)=−8⇒x=−81−−157⇒x=−567−−157=−56−(−15)7=−56+157=−417Therefore, the other rational number is −417.

Page No 1.18:

Question 7:

What should be added to −78

so as to get 59?

Answer:

Let x be added to −78 so as to get 59.∴ x+−78=59⇒x=59−−78⇒x=4072−−6372 ⇒x=40−(−63)72⇒x=40+6372=10372

Page No 1.18:

Question 8:

What number should be added to −511

so as to get 2633?

Answer:

Let x be added.∴ x+−511=2633 ⇒x=2633−−511⇒x=2633−−1533⇒ x=26−(−15)33⇒ x=26+1533⇒ x=4133

Page No 1.18:

Question 9:

What number should be added to −57

to get −23?

Answer:

Let x be added.∴ x+−57=−23 ⇒x=−23−−57⇒x=−1421−−1521⇒ x=−14−(−15)21⇒x=−14+1521⇒x=121

Page No 1.18:

Question 10:

What number should be subtracted from −53

to get 56?

Answer:

Let x be subtracted.∴ −53−x=56⇒ x=−53−56⇒ x=−106−56⇒ x=−10−56=−156=−52

Page No 1.19:

Question 11:

What number should be subtracted from 37

to get 54?

Answer:

Let, x be subtracted.∴ 37−x=54 ⇒x=37−54⇒x=1228−3528 ⇒x=12−3528=−2328

Page No 1.19:

Question 12:

What should be added to (23+35)

to get −215?

Answer:

Let x be added.∴ x+(23+35)=−215⇒x+(1015+915)=−215⇒x+(10+915)=−215⇒x=−215−1915⇒ x=−2−1915⇒ x=−2115⇒ x=−75

Page No 1.19:

Question 13:

What should be added to (12+13+15)

to get 3?

Answer:

Let x be added.∴ x+(12+13+15)=3⇒x+(1530+1030+630)=3⇒ x+(15+10+630)=3 ⇒x+3130=3⇒x=31−3130⇒ x=9030−3130 ⇒x=90−3130⇒x=5930

Page No 1.19:

Question 14:

What should be subtracted from (34−23)

to get −16?

Answer:

Let x be subtracted.∴ (34−23)−x=−16⇒ (912−812)−x=−16⇒(9−812)−x=−16⇒x=112−−16⇒x=112−−212 ⇒x=1−(−2)12⇒x=1+212⇒x=312⇒x=14

Page No 1.19:

Question 15:

Fill in the blanks:
(i) −413−−326=. . .

(ii) −914+. . .=−1
(iii) −79+. . .=3
(iv) . . .+1523=4

Answer:

(i) −413−−326=−826−−326 =−8−(−3)26=−8+326=−526(ii) −914+x=−1⇒x=−1−−914 ⇒x=−1414−−914⇒x=−14−(−9)14⇒x=−14+914⇒x=−514(iii) −79+x=3⇒x=3−−79⇒x=279−−79⇒x=27−(−7)9⇒x=27+79⇒x= 349(iv) x+1523=4⇒x=4− 1523⇒x=9223− 1523⇒x=92−1523⇒x=7723

Page No 1.22:

Question 1:

Simplify each of the following and write as a rational number of the form pq:

(i) 34+56+−78
(ii) 23+−56+−79
(iii) −112+76+−58
(iv) −45+−710+−815
(v) −910+2215+13−20
(vi) 53+3−2+−73+3

Answer:

(i) 34+56+−78=1824+2024+−2124=18+20+(−21)24=18+20−2124=1724(ii) 23+−56+−79=1218+−1518+−1418=12+(−15)+(−14)18=12−15−1418=−1718(iii) −112+76+−58=−13224+2824+−1524=(−132)+28+(−15)24=−132+28−1524=−11924(iv) −45+−710+−815=−2430+−2130+−1630=(−24)+(−21)+(−16)30=−24−21−1630=−6130(v) −910+2215+13−20=−5460+8860+−3960=(−54)+88+(−39)60=−54+88−3960=−560=−112(vi) 53+3−2+−73+3=106+−96+−146+186=10+(−9)+(−14)+186=10−9−14+186=56

Page No 1.23:

Question 2:

Express each of the following as a rational number of the form pq:

(i) −83+−14+−116+38−3
(ii) 67+1+−79+1921+−127
(iii) 152+98+−113+6+−76
(iv) −74+0+−95+1910+1114
(v) −74+53+−12+−56+2

Answer:

(i) −83+−14+−116+38−3=−6424+−624+−4424+924−7224=(−64)+(−6)+(−44)+9+(−72)24=−64−6−44+9−7224=−17724=−598(ii) 67+1+−79+1921+−127=5463+6363+−4963+5763+−10863=54+63+(−49)+57+(−108)63=54+63−49+57−10863=1763(iii) 152+98+−113+6+−76=18024+2724+−8824+14424+−2824=180+27+(−88)+144+(−28)24=180+27−88+144−2824=23524(iv) −74+0+−95+1910+1114=−245140+−252140+266140+110140=(−245)+(−252)+266+110140=−245−252+266+110140=−121140(v) −74+53+−12+−56+2=−2112+2012+−612+−1012+2412=(−21)+20+(−6)+(−10)+2412=−21+20−6−10+2412=712

Page No 1.23:

Question 3:

Simplify:
(i) −32+54−74

(ii) 53−76+−23

(iii) 54−76−−23

(iv) −25−−310−−47

(v) 56+−25−−215

(vi) 38−−29+−536

Answer:

(i) −32+54−74Taking the L.C.M. of the denominators:−64+54−74=−6+5−74=−84=− 2(ii) 53−76+−23Taking the L.C.M. of the denominators:106−76+−46=10−7+(−4)6=10−7−46=−16(iii) 54−76−−23Taking the L.C.M. of the denominators:1512−1412−−812=15−14−(−8)12=15−14+812=912=34(iv) −25−−310−−47Taking the L.C.M. of the denominators:−2870−−2170−−4070=(−28)−(−21)−(−40)70=−28+21+4070=3370(v) 56+−25−−215Taking the L.C.M. of the denominators:2530+−1230−−430=25+(−12)−(−4)30=25−12+430=1730(vi) 38−−29+−536Taking the L.C.M. of the denominators:2772−−1672+−1072=27−(−16)+(−10)72=27+16−1072=3372=1124

Page No 1.25:

Question 1:

Multiply:
(i) 711 by 54

(ii) 57 by −34
(iii) −29 by 511
(iv) −317 by −5−4
(v) 9−7 by 36−11
(vi) −1113 by −217
(vii) −35 by −47
(viii) −1511 by 7

Answer:

(i) 711×54=7×511×4=3544(ii) 57×−34=5×(−3)7×4=−1528(iii) −29×511=(−2)×59×11=−1099(iv) −317×−5−4=(−3)×(−5)17×(−4)=15−68(v) 9−7×36−11=9×36(−7)×(−11)=32477(vi) −1113×−217=(−11)×(−21)13×7=23191=3313(vii) −35×−47=(−3)×(−4)5×7=1235(viii) −1511×7=−1511×71=(−15)×711×1=−10511

Page No 1.25:

Question 2:

Multiply:
(i) −517 by 51−60

(ii) −611 by −5536

(iii) −825 by −516

(iv) 67 by −4936

(v) 8−9 by −7−16

(vi) −89 by 364

Answer:

(i) −517×51−60=−517×3×17−5×3×4=−5×3×17−17×5×3×4=14(ii) −611×−5536=−611×−5×116×6=56(iii) −825×−516=−15×5×−52=110(iv) 67×−4936=67×−7×76×6=−76(v) 8−9×−7−16=1−9×−7−2=−718(vi) −89×364=−83×3×38×8=−124

Page No 1.26:

Question 3:

Simplify each of the following and express the result as a rational number in standard form:
(i) −1621×145

(ii) 76×−328

(iii) −1936×16

(iv) −139×27−26

(v) −916×−64−27

(vi) −507×143

(vii) −119×−81−88

(viii) −59×72−25

Answer:

(i)−1621×145=−2×2×2×23×7×2×75=−3215(ii) 76×−328=72×3×−32×2×7=−18(iii) −1936×16=−192×2×3×3×2×2×2×2=−769(iv) −139×27−26=−133×3×3×3×3−2×13=32(v) −916×−64−27=−916×−4×16−3×9=−43(vi) −507×143=−507×2×73=−1003(vii) −119×−81−88=−119×−9×9−8×11=−98(viii) −59×72−25=−59×8×9−5×5=85

Page No 1.26:

Question 4:

Simplify:
(i) (258×25)−(35×−109)

(ii) (12×14)+(12×6)

(iii) (−5×215)−(−6×29)

(iv) (−94×53)+(132×56)

(v) (−43×12−5)+(37×2115)

(vi) (135×83)−(−52×113)

(vii) (137×1126)−(−43×56)

(viii) (85×−32)+(−310×1116)

Answer:

(i) (258×25)−(35×−109)=54−−23=5×3+2×412=2312(ii) (12×14)+(12×6)=18+3=1+3×88=258(iii) (−5×215)−(−6×29)=−23−−43=−2+43=23(iv) (−94×53)+(132×56)=−3×54+13×512=−154+6512=−15×3+6512=2012=53(v) (−43×12−5)+(37×2115)=4×45+1×35=165+35=16+35=195(vi) (135×83)−(−52×113)=13×815−−5×116=10415−−556=104×2+55×530=48330(vii) (137×1126)−(−43×56)=1×117×2−−2×53×3=1114−−109=11×9+10×14126=239126(viii) (85×−32)+(−310×1116)=4×(−3)5+−3×1110×16=−125+−33160=−12×32−33160=−384−33160=−417160

Page No 1.26:

Question 5:

Simplify:
(i) (32×16)+(53×72)−(138×43)

(ii) (14×27)−(514×−23)+(37×92)

(iii) (139×−152)+(73×85)+(35×12)

(iv) (311×56)−(912×43)+(513×615)

Answer:

(i) (32×16)+(53×72)−(138×43)=14+356−136=1×3+35×2−13×212=3+70−2612=4712(ii)(14×27)−(514×−23)+(37×92)=114−−521+2714=1×3+5×2+27×342=9442=4721(iii) (139×−152)+(73×85)+(35×12)=−13×56+7×815+310=−656+5615+310=−65×5+56×2+3×330=−20430=−345(iv) (311×56)−(912×43)+(513×615)=522−1+213=5×13−286+2×22286=65−286+44286=−177286

Page No 1.31:

Question 1:

Verify the property: x × y = y × x by taking:
(i) x=−13, y=27

(ii) x=−35, y=−1113
(iii) x=2, y=7−8
(iv) x=0, y=−158

Answer:

We have to verify that x×y=y×x.(i) x=−13, y=27x×y=−13×27=−221y×x=27×−13=−221∴ −13×27=27×−13Hence verified.(ii) x=−35, y=−1113x×y=−35×−1113=3365y×x=−1113×−35=3365∴ −35×−1113=−1113×−35 Hence verified.(iii) x=2, y=7−8 x×y=2×7−8= 7−4y×x=7−8×2= 7−4∴ 2×7−8=7−8×2Hence verified.(iv) x=0, y=−158 x×y=0×−158=0y×x=−158×0=0∴ 0×−158=−158×0Hence verified.

Page No 1.31:

Question 2:

Verify the property: x × (y × z) = (x × y) × z by taking:

(i) x=−73, y=125, z=49

(ii) x=0, y=−35, z=−94

(iii) x=12, y=5−4, z=−75

(iv) x=57, y=−1213, z=−718

Answer:

We have to verify that x×(y×z)=(x×y)×z.(i) x=−73, y=125, z=49x×(y×z)=−73×(125×49)=−73×1615=−11245(x×y)×z=(−73×125)×49=−285×49=−11245∴−78×(155×49)=(−78×155)×49Hence verified.(ii) x=0, y=−35, z=−94x×(y×z)= 0×(−35×−94)= 0×2720=0(x×y)×z=(0×−35)×−94= 0×−94=0∴ 0×(−35×−94)=(0×−35)×−94Hence verified.(iii) x=12, y=5−4, z=−74x×(y×z)= 12×(5−4×−74)=12×3516=3532(x×y)×z=(12×5−4)×−74=5−8×−74=3532∴ 12×(5−4×−74)=(12×5−4)×−74Hence verified. (iv) x=57, y=−1213, z=−718x×(y×z)=57×(−1213×−718)=57×1439=1039(x×y)×z=57×−1213)×−718=−6091×−718=1039∴ 57×(−1213×−718)=(57×−1213)×−718Hence verified.

Page No 1.32:

Question 3:

Verify the property: x × (y + z) = x × y + x × z by taking:

(i) x=−37, y=1213, z=−56

(ii) x=−125, y=−154, z=83

(iii) x=−83, y=56, z=−1312

(iv) x=−34, y=−52, z=76

Answer:

We have to verify that  x×(y+z)=x×y+x×z.(i) x=−37, y=1213, z=−56x×(y+z)=−37×(1213+−56)=−37×72−6578=−37×778= −126x×y+x×z=−37×1213+−37×−56=−3691+514=−36×2+5×13182=−72+65182=−126∴ −37×(1213+−56)=−37×1213+−37×−56Hence verified.(ii) x=−125, y=−154, z=83x×(y+z)=−125×(−154+83)=−125×−45+3212=−125×−1312=135x×y+x×z=−125×−154+−125×83=91+−325=45−325=135∴−125×(−154+83)=−125×−154+−125×83Hence verified.(iii) x=−83, y=56, z=−1312x×(y+z)=−83×(56+−1312)=−83×10−1312=−83×−312=23x×y+x×z=−83×56+−83×−1312=−209+269=−20+269=23∴ −83×(56+−1312)=−83×56+−83×−1312Hence verified . (iv) x=−34, y=−52, z=76x×(y+z)=−34×(−52+76)=−34×−15+76=−34×−86=1x×y+x×z=−34×−52+−34×76=158+−78=15−78=1∴−34×(−52+76)=−34×−52+−34×76Hence verified.

Page No 1.32:

Question 4:

Use the distributivity of multiplication of rational numbers over their addition to simplify:
(i) 35×(3524+101)

(ii) −54×(85+165)

(iii) 27×(716−214)

(iv) 34×(89−40)

Answer:

(i) 35×(3524+101)=35×3524+35×101=78+61=7+488=558(ii) −54×(85+165)=−54×85+−54×165=−21+−41=−6(iii) 27×(716−214)=27×716−27×214=18−32=1−128=−118(iv) 34×(89−40)=34×89−34×40=23−30=2−903=−883

Page No 1.32:

Question 5:

Find the multiplicative inverse (reciprocal) of each of the following rational numbers:
(i) 9
(ii) −7
(iii) 125

(iv) −79
(v) −3−5
(vi) 23×94
(vii) −58×1615
(viii) −2×−35
(ix) −1
(x) 03

(xi) 1

Answer:

(i) Multiplicative inverse (reciprocal) of 9 = 19(ii) Multiplicative inverse (reciprocal) of −7 = −17(iii) Multiplicative inverse (reciprocal) of 125 = 512(iv) Multiplicative inverse (reciprocal) of −79 = −97(v) Multiplicative inverse (reciprocal) of −3−5 = −5−3 or 53(vi) Multiplicative inverse (reciprocal) of 23×94 = 32×49=23(vii) Multiplicative inverse (reciprocal) of −58×1615 = 8−5×1516=−32(viii) Multiplicative inverse (reciprocal) of −2×−35 = 1−2×5−3=56(ix) Multiplicative inverse (reciprocal) of −1 = 1−1=−1(x) Multiplicative inverse (reciprocal) of 03 = 30=undefined (ix) Multiplicative inverse (reciprocal) of 1 = 11=1

Page No 1.32:

Question 6:

Name the property of multiplication of rational numbers illustrated by the following statements:
(i) −516×815=815×−516

(ii) −175×9=9×−175

(iii) 74×(−83+−1312)=74×−83+74×−1312

(iv) −59×(415×−98)=(−59×415)×−98

(v) 13−17×1=13−17=1×13−17

(vi) −1116×16−11=1

(vii) 213×0=0=0×213

(viii) −32×54+−32×−76=−32×(54+−76)

Answer:

(i) Commutative property
(ii) Commutative property
(iii) Distributivity of multiplication over addition
(iv) Associativity of multiplication
(v) Existence of identity for multiplication
(vi) Existence of multiplicative inverse
(vii) Multiplication by 0
(viii) Distributive property

Page No 1.32:

Question 7:

Fill in the blanks:
(i) The product of two positive rational numbers is always .....
(ii) The product of a positive rational number and a negative rational number is always .....
(iii) The product of two negative rational numbers is always .....
(iv) The reciprocal of a positive rational number is .....
(v) The reciprocal of a negative rational number is .....
(vi) Zero has ..... reciprocal.
(vii) The product of a rational number and its reciprocal is .....
(viii) The numbers ..... and ..... are their own reciprocals.
(ix) If a is reciprocal of b, then the reciprocal of b is .....
(x) The number 0 is ..... the reciprocal of any number.
(xi) Reciprocal of 1a, a≠0

is .....
(xii) (17 × 12)⁻¹ = 17⁻¹ × .....

Answer:

(i) Positive
(ii) Negative
(iii) Positive
(iv) Positive
(v) Negative
(vi) No
(vii) 1
(viii) -1 and 1
(ix) a
(x) not
(xi) a
(xii) 12−1

Page No 1.33:

Question 8:

Fill in the blanks:
(i) −4×79=79× ......

(ii) 511×−38=−38× ......
(iii) 12×(34+−512)=12×......+......×−512
(iv) −45×(57+−89)=(−45×.....)×−89

Answer:

(i) −4x×y=y×x  (commutativity)(ii) 511x×y=y×x  (commutativity)(iii) 34;12x×(y+z)=x×y+x×z  (distributivity of multiplication over addition)(iv) 57x×(y×z)=(x×y)×z  (associativity of multiplication )

Page No 1.35:

Question 1:

Divide:
(i) 1 by 12

(ii) 5 by −57

(iii) −34 by 9−16

(iv) −78 by −2116

(v) 7−4 by 6364

(vi) 0 by −75

(vii) −34 by −6

(viii) 23 by −712

(ix) −4 by −35

(x) −313 by −465

Answer:

(i) 1÷12=1×21=2(ii) 5÷−57=5×7−5=−7(iii) −34÷9−16=−34×−169=43(iv) −78÷−2116=−78×16−21=23(v) 7−4÷6364=7−4×6463=−169(vi) 0÷−75=0×5−7=0(vii) −34÷−6=−34×1−6=18(viii) 23÷−712=23×12−7=−87(ix) −4÷−35=−4×5−3=203(x) −313÷−465=−313×65−4=154

Page No 1.36:

Question 2:

Find the value and express as a rational number in standard form:
(i) 25÷2615

(ii) 103÷−3512

(iii) −6÷(−817)

(iv) −4099÷(−20)

(v) −2227÷−11018

(vi) −36125÷−375

Answer:

(i) 25÷2615=25×1526=313(ii) 103÷−3512=103×12−35=−87(iii) −6÷−817=−6×17−8=514(iv) −4099÷(−20)=−4099×1−20=299(v) −2227÷−11018=−2227×18−110=215(vi) −36125÷−375=−36125×75−3=365

Page No 1.36:

Question 3:

The product of two rational numbers is 15. If one of the numbers is −10, find the other.

Answer:

Let the other number be x.∴ x×(−10)=15or x=15−10=3−2So, the other number is −32.

Page No 1.36:

Question 4:

The product of two rational numbers is −89.

If one of the numbers is −415,

find the other.

Answer:

Let the other number be x.∴ x×−415=−89or x=−89÷−415or x=−89×15−4or x=103Thus, the other number is 103.

Page No 1.36:

Question 5:

By what number should we multiply −16

so that the product may be −239?

Answer:

Let the number be x.∴ x×−16=−239x=−239÷−16 x=−239×6−1=463Therefore, the other number is 463.

Page No 1.36:

Question 6:

By what number should we multiply −1528

so that the product may be −57?

Answer:

Let the other number be x.∴ x×−1528=−57or x=−57÷−1528or x=−57×28−15or x=43Thus, the other number is 43.

Page No 1.36:

Question 7:

By what number should we multiply −813

so that the product may be 24?

Answer:

Let the number be x.∴ x×−813=24or x=24÷−813or x=24×13−8or x=−39Thus, the number is −39.

Page No 1.36:

Question 8:

By what number should −34

be multiplied in order to produce 23?

Answer:

Let the other number that should be multiplied with −34 to produce 23 be x.∴ x×−34=23or x=23÷−34or x=23×4−3or x=−89Thus, the number is −89.

Page No 1.36:

Question 9:

Find (x + y) ÷ (x − y), if
(i) x=23, y=32

(ii) x=25, y=12
(iii) x=54, y=−13
(iv) x=27, y=43
(v) x=14, y=32

Answer:

Page No 1.36:

Question 10:

The cost of 723

metres of rope is Rs 1234.

Find its cost per metre.

Answer:

The cost of 723 metres of rope is Rs 1234.∴ Cost per metre =1234÷723=514÷233=514×323=15392=Rs 16192                                     

Page No 1.36:

Question 11:

The cost of 213

metres of cloth is Rs 7514.

Find the cost of cloth per metre.

Answer:

The cost of 213 metres of cloth is Rs 7514.∴ Cost per metre=7514÷213=3014÷73=3014×37=1294=Rs 3214 Thus, Rs 3214 or Rs 32.25 is the cost of cloth per metre.                                      

Page No 1.36:

Question 12:

By what number should −3316

be divided to get −114?

Answer:

Let the number be x.∴ −3316÷x=−114or −3316×1x=−114or 1x=−114×16−33or 1x=43or  x=34Thus, the number is 34.

Page No 1.36:

Question 13:

Divide the sum of −135

and 127 by the product of −317 and −12.

Answer:

(−135+127)÷(−317×−12)=−13×7+12×535÷3114=−91+6035÷3114=−3135×1431=−25

Page No 1.36:

Question 14:

Divide the sum of 6512 and 127

by their difference.

Answer:

(6512+127)÷(6512−127)=65×7+12×1284÷65×7−12×1284=455+14484÷455−14484=59984÷31184=59984×84311=599311

Page No 1.36:

Question 15:

If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser?

Answer:

 Cloth needed to prepare 24 trousers=54 m∴ Length of the cloth required for each trousers= 54÷24=5424=94 m = 214 metres

Page No 1.43:

Question 1:

Find a rational number between −3 and 1.

Answer:

Rational number between −3 and 1 = −3+12=−22=−1

Page No 1.43:

Question 2:

 Find any five rational numbers less than 2.

Answer:

We can write:2=21=2×51×5=105Integers less than 10 are 0, 1, 2, 3, 4, 5 ... 9.Hence, five rational numbers less than 2 are 05,15,25,35 and45 .

Page No 1.43:

Question 3:

Find two rational numbers between −29 and 59.

Answer:

Since both the fractions (−29 and 59) have the same denominator, the integers between the numerators(−2 and 5) are −1, 0, 1, 2, 3, 4.Hence, two rational numbers between −29 and 59 are 09 or 0 and 19.

Page No 1.43:

Question 4:

Find two rational numbers between 15 and 12.

Answer:

Rational number between 15 and 12=(15+12)2=2+5102=720Rational number between 15 and 720=(15+720)2=4+7202=1140Therefore, two rational numbers between15 and 12 are 720 and 1140.

Page No 1.43:

Question 5:

Find ten rational numbers between 14 and 12.

Answer:

The L.C.M of the denominators (2 and 4) is 4.So, we can write  14 as it is.Also,  12=1×22×2=24As the integers between the numerators 1 and 2 of both the fractions are not sufficient, we will multiply the fractions by 20.∴14=1×204×20=208024=2×204×20=4080Between 20 and 40, there are 19 integers. They are 21, 22, 23, 24, 25, 26, 27....39, 40. Thus, 2140,2240,2340,2440,2540,...................3840 and 3940 are the fractions.We can take any 10 of these.

Page No 1.43:

Question 6:

Find ten rational numbers between −25 and 12.

Answer:

L.C.M of the denominators (2 and  5) is 10.We can write: −25=−2×25×2=−410 and 12=1×52×5=510Since the integers between the numerators (−4 and 5 ) of both the fractions are not sufficient, we will multiply the fractions by 2.∴−410=−4×210×2=−820510=5×210×2=1020There are 17 integers between −8 and 10, which are −7,−6,−5,−4...................8, 9.These can be written as:−720,−620,−520,−420,−320,...................820 and 920We can take any 10 of these.

Page No 1.43:

Question 7:

Find ten rational numbers between 35 and 34.

Answer:

The L.C.M of the denominators 5 and 4 of both the fractions is 20.We can write:35=3×45×4=122034=3×54×5=1520Since the integers between the numerators 12 and 15 are not sufficient, we will multiply both the fractions by 5.1220=12×520×5=601001520=15×520×5=75100There are 14 integers between 60 and 75. They are 61, 62, 63.......73 and 74.Therefore, 60100,61100,62100..........73100 and74100 are the 14 fractions.We can take any 10 of these.

Page No 1.5:

Question 1:

Add the following rational numbers.
(i) −57 and 37

(ii) −154 and 74
(iii) −811 and −411
(iv) 613 and −913

Answer:

(i) −57 + 37 = −5+37 = −27(ii) −154 + 74 = −15+74 = −84 = −2(iii) −811 + −411 = −8−411 = −1211(iv) 613 + −913 = 6−913 = −313

Page No 1.6:

Question 2:

Add the following rational numbers:
(i) 34 and −58

(ii) 5−9 and 73
(iii) −3 and 35
(iv) −727 and 1118
(v) 31−4 and −58
(vi) 536 and −712
(vii) −516 and 724
(viii) 7−18 and 827

Answer:

(i) Clearly, denominators of the given numbers are positive. The L.C.M. of denominators 4 and 8 is 8. Now, we will express 34in the form in which it takes the denominator is 8.3×24×2=6834 + −58=68+−58=6+(−5)8=6−58= 18(ii) 5−9 + 73=−59 + 73The L.C.M. of denominators 9 and 3 is 9. Now, we will express 73 in the form in which it takes the denominator is 9.7×33×3=219−59 + 73=−59 + 219=(−5)+219=−5+219= 169(iii) −3 +35=−31 +35The L.C.M. of denominators 1 and 5 is 5. Now, we will express −31 in the form in which it takes the denominator 5.−3×51×5=−155  So−31 +35=−155 +35= −15+35= −125(iv) The L.C.M. of denominators 27 and 18 is 54. Now, we will express −727 and 1118 in the form in which they take the denominator 54.−7×227×2=−1454  11×318×3=3354−727 + 1118=−1454+3354= −14+3354= 1954(v)We have 31−4 + −58=−314 + −58The L.C.M. of denominators 4 and 8 is 8. Now, we will express −314 in the form in which it takes the denominator 8.−31×24×2=−628So−314 + −58=−628 + −58= (−62)+(−5)8= −62−58=−678(vi) The L.C.M. of denominators 36 and 12 is 36. Now, we will express −712 in the form in which it takes the denominator 36.−7×312×3=−2136So536 + −712=536 + −2136= 5+(−21)36=5−2136=−1636=−49(vii) The L.C.M. of denominators 16 and 24 is 48. Now, we will express −516 and 724 in the form in which they take the denominator 48.−5×316×3=−1548  7×224×2=1448So−516 + 724=−1548+1448= (−15)+1448= −15+1448= −148(viii) 7−18 + 827=−718 + 827The L.C.M. of denominators 18 and 27 is 54. Now, we will express −718 and 827 in the form in which they take the denominator 54.−7×318×3=−2154  8×227×2=1654So−718 + 827=−2154+1654= −21+1654= −554=−554

Page No 1.6:

Question 3:

Simplify:
(i) 89+−116

(ii) 3+5−7
(iii) 1−12+2−15
(iv) −819+−457
(v) 79+3−4
(vi) 526+11−39
(vii) −169+−512
(viii) −138+536
(ix) 0+−35
(x) 1+−45

Answer:

(i) 89 + −116L.C.M. of the denominators 9 and 6 is 18. Now, we will express 89and −116 in the form in which they take the denominator 18.8×29×2=1618−11×36×3=−331889 + −116=1618 + −3318= 16+(−33)18= 16−3318=−1718(ii) 3+5−7=31+−57L.C.M. of the denominators 1 and 7 is 7. Now, we will express 31 in the form in which it takes the  denominator 7.3×71×7=21731+−57=217+−57=21+(−5)7=21−57=167(iii) 1−12+2−15=−112+−215L.C.M. of the denominators 12 and 15 is 60. Now, we will  express −112and −215in the form in which they take the denominator 60.−1×512×5=−560−2×415×4=−860−112+−215=−560+−860=(−5)+(−8)60=−5−860=−1360(iv) −819 + −457L.C.M. of the denominators 19 and 57 is 57. Now, we will express −819in the form in which it takes the denominator 57.−8×319×3=−2457−819 + −457=−2457 + −457=−24−457=−2857(v) 79 + 3−4=79 + −34L.C.M. of the denominators 9 and 4 is 36. Now, we will express 79and −34in the form in which they take the  denominator 36.7×49×4=2836−3×94×9=−2736So79 + −34=2836+−2736=28+(−27)36=28−2736= 136(vi) 526 + 11−39=526 + −1139L.C.M. of the denominators 26 and 39 is 78. Now, we will  express 526and −1139in the form in which they take the  denominator 78.5×326×3=1578−11×239×2=−2278So526 + −1139=1578+−2278=15−22)78= −778(vii) −169 + −512L.C.M. of the denominators 9 and 12 is 36. Now, we will express −169and −512in the form in which they take the  denominator 36.−16×49×4=−6436−5×312×3=−1536−169 + −512=−6436+−1536= (−64)+(−15)36= −64−1536=−7936(viii) −138+ 536L.C.M. of the denominators 8 and 36 is 72. Now, we will express −138and 536 in the form in which they take the  denominator 72.−13×98×9=−117725×236×2=1072−138+ 536=−11772+1072= −117+1072= −10772(ix) 0 + −35Taking L.C.M. of the denominator:= 0×5−35= −35(x) 1 + −45= 11 + −45L.C.M. of the denominators 1 and 5 is 5. Now, we will express 11in the form in which it takes the  denominator 5.1×51×5=5511 + −45=55 + −45= 5−45= 15

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Question 4:

Add and express the sum as a mixed fraction:
(i) −125 and 4310

(ii) 247 and −114

(iii) −316 and −278

(iv) 1016 and 78

Answer:

(i) We have −125 + 4310. L.C.M. of the denominators 5 and 10 is 10. Now, we will express −125 in the form in which it takes the denominator 10.−12×25×2=−2410∴−125 + 4310=−2410+4310= −24+4310= 1910 = 1910(ii) We have 247 + −114.L.C.M. of the denominators 7 and 4 is 28. Now, we will express 247and −114 in the form in which they take the denominator 28.24×47×4=9628−11×74×7=−7728∴247 + −114=9628+−7728= 96−7728= 1928(iii) We have −316+ −278.L.C.M. of the denominators 6 and 8 is 24. Now, we will express −316and −278 in the form in which they take the denominator 24.−31×46×4=−12424−27×38×3=−8124∴−316+ −278=−12424+−8124= −124−8124= −20524 = −81324(iv) We have 1016 + 78.L.C.M. of the denominators 6 and 8 is 24. Now, we will express 1016and 78 in the form in which they take the denominator  24.101×46×4=404247×38×3=2124∴ 1016 + 78=40424+2124= 404+2124= 42524= 171724