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TRANSCRIPT
I
Government of Karnataka
MATHEMATICS
66th Standard
SECOND SEMESTER 2015
KARNATAKA TEXT BOOK SOCIETY (R) 100 Feet Ring Road, Banashankari 3rd Stage,
Bengaluru - 560 085
III
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Second Semester
Slno Unit Page No.
1 Playing with numbers 1-32
2 Fractions 33-66
3 Decimals 67-92
4 Introduction to Algebra 93-107
5 Ratio and proportion 108-127
6 Symmetry 128-140
7 Construction 141-155
8 Mensuration 156-176
Answers 177-182
Contents
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UNIT - 1
PlayINg wITh NUmbers
after learning this unit you can : findthenumbersbywhichagivennumberisdivisible, identifytheprimeandcompositenumbers, write the given numbers as the product of prime
numbers, distinguishbetweenfactorsandmultiples, findHCFandLCMofgivennumbersandusethemto
solvetheproblems.1.1 rules of Divisibility
Givenanumber26,whicharethenumbersbywhichitisdivisible?Isitdivisibleby2?divisibleby3?divisibleby4?divisibleby5 ?andsoon.Byactualdivisionyouwillseethatitisdivisibleby2butnotby3,4and5.
Withoutactuallyperformingdivisioncanwefindoutarulebywhichwecantestwhetheranumberisdivisiblebyagivennumberlike2,3,4,5?Letusverify.
Divisibility by 2:Gracyhaspreparedalistofmultiplesof2.2,4,6,8,10,12,14,16,18,20,22,24,26............
In thesemultiples, she recognised a common (general)pattern(rule).Whatisthatcommonpattern?
Shenoticedthatthemultiplesof2willhave0,2,4,6or8intheirunit'splace.
Shewrotesomemoreevennumbers.
1st Set BW Proof : 04-09-2012 : 000 to 000 (Total 000 Pages) : Print Given : Komala
2nd Set BW Proof : 21-09-2012 : 01 to 32 (Total 32 Pages) : Print Given : Rajeshwari
3rd Set Clr Proof : 31-10-2012 : 1 to 33 (Total 000 Pages) : Print Given : komala
4th Set BW Proof : 15-12-2012 : 000 to 000 (Total 000 Pages) : Print Given : komala
5th Set BW Proof : 12-01-2013 : 1 to 32 (Total 000 Pages) : Print Given : Pallavi
6th BW / Clr Proof : 06-02-2013 : 1 to 32 (Total 000 Pages) : Print Given : Operator
Finalised Draft Print : 00-00-2012 : 000 to 000 (Total 000 Pages) : Print Given : Operator
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Example : 312, 624, 6118, 3090, 5556,............When thesenumbersaredividedby2,theremainderiszero.Shestatedtheruleofdivisibilityby2asfollows.rule for divisibility by 2 : Ifthenumbersareeven,thentheyaredivisibleby2.
Try :Identifythenumbersdivisibleby2:671,586,394,5798,7320,4441
Divisibility by 3:Naveenwrotethemultiplesof3andtriedtofindthegeneral
pattern.Helistedthemultiplesof3 andobserved.3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,......90,
96,....... 102, ............126,............144, ............He couldnotfindanygeneralrulebasedonthedigitsintheunit'splaceandotherplaces.Thenheselectedafewandstartedaddingthedigitsofthenumbersasgivenbelow.
272+7=9
333+3=6
393+9=12
909+0=9
969+6=15
144 1+4+4=9
......
Somepatternisobservedinthesenumbers.Whatisthat?
rule for divisibility by 3 : Ifthesumofthedigitsofanumberis amultiple of three, then that number is completelydivisibleby3.Try :Whichofthesefollowingnumbersaredivisibleby3?
(1)382 (2)692 (3)1086 (4)2367
Divisibility by 6 :Wehavealreadylearnttherulesfordivisibilityby2and
3.Letusknowarulefordivisibilityby6.Letuslistafewmultiplesof6.
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Theyare6,12,18,24,30,36,42,48,54,60,66,72,............ All thesemultiples are completely divisible bothby2 and3.Soifanumberistobedivisibleby6,ithastobedivisibleby2and3.
58isdivisibleonlyby2andnotby3.Isitdivisibleby6?Weseethat58isnotdivisibleby6.
Similarly,27isdivisibleby3andnotby2.Also27isnotdivisibleby6.
Nowstateageneralrulefordivisibilityby6.
rule for divisibility by 6 : Allthenumbersdivisiblebothby2and3aredivisibleby6.
Try :Identifythenumberswhicharedivisibleby6.(1) 84 (2) 95 (3) 378 (4) 6534
Divisibility by 4:
Dividethefollowingnumbersby4.Aretheydivisibleby 4? 100,200,300,.....900,1000,1100,....2500,....3800,.....Yestheseare divisible by 4. These are all themultiples of 100 anddivisibleby4.
• Is 1356 divisible by 4?
We know that themultiples of 100 are divisible by 4.Now1356=1300+56. 1300isdivisibleby4.Is56divisibleby4?Soitisenoughtocheckthedivisibilityby4forthenumber56 whichisformedbyten'sandunit'splaceof1356.
56isdivisibleby4.
∴ 1356isdivisibleby4.
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• Is 5734 divisible by 4?5734 = 5700 + 34. We know that 5700 is divisible by 4.
Now34isthenumberformedbydigitsintheten'sandunit'splaceof5734.Is34isdivisibleby4?No.
So 5734 isnotdivisibleby4.Sowestatethedivisibilityruleby4asfollows.rule for divisibility by 4 : Incaseofthenumbershavingmorethantwodigits,ifthenumberformedbydigitsfromten'sandunit'splacesisdivisibleby4,thenthegivennumberisdivisibleby4.
Try : (i) Isthenumber 6921 divisibleby 4? (ii) Whatistheleastnumbertobeaddedto6921 to makeitdivisibleby4?
Divisibility by 5:Divyalistedafewmultiplesof5asfollows.5,10,15,20,25,.....50,55,60,65,....105,......230,.....Everymultipleof5 willhaveeither0or5initsunit'splace.
So,westatetherulefordivisibilityby5asfollows.
rule for divisibility by 5 : Ifanumberhas0or5initsunit'splacethenthatnumberisdivisibleby5.
Try :Writeafourdigitnumberwhichisdivisibleby5.
Divisibility by 10:Srilathahaslistedafewmultiplesof10asfollows.10,20,30,40,50,60,......100,110,120,130,....By observing these numbers, she found the general
pattern.Canyouguessthat?Allthemultiplesof10have‘0’intheirunit’splace.Thenshedivided230,470,3020,6890...by10andframed
therulefordivisibilityby10asfollows.
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rule for divisibility by 10 : Thenumbershaving'0'intheirunit'splacearedivisibleby10.
Try : 1)Definetherulefordivisibilityby100. 2)Definetherulefordivisibilityby1000.
Divisibility by 8:Notethat100isnotdivisibleby8.But1000isdivisibleby8.Lookatthesemultiplesof1000. 1000,2000,3000,4000,....25000,....
Aretheydivisibleby8?Yes.Allthemultiplesof1000aredivisibleby8.Nowconsiderthenumbers2520,3518.Arethesedivisible
by8?Weknow that, themultiplesof 1000aredivisibleby 8.
Thereforeinthenumbers2520and3518,weneedtoverifythedivisibilityofthenumbersformedfromthedigitsexcludingthousand’splaceofthesenumbers.
Wehavetoseethedivisibilityofthenumberformedbyhundred's,ten'sandunit'splacesby8.2520=2000+520
520÷8=65 8 520
040
0040
48
65g
∴ 2520isdivisibleby8.In3518,3518=3000+518518÷8=64+6istheremainder.In3518thenumber518isnotdivisibleby8. 8 518
038
0632
48
64g
∴ 3518isnotdivisibleby8.So,westatetherulefordivisibilityby8asfollows.
rule for divisibility by 8 : A numberhaving 4 digitsandmorethan 4 digitsisdivisibleby 8, ifthenumberformedfromthedigitsinhundred's,ten'sandunit'splacesisdivisibleby8.
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Divisibilityofnumbershavinglessthan4digitsby8isverifieddirectlybydividingthegivennumberby8.
Try: 1) Testthedivisibilityby8for38532. 2) Whatistheleastnumbertobeaddedto54074to
makedivisibleby8?
Divisibility by 9 : Observethesemultiplesof9:9,18,27,36,45,54,63,72,81,90,99,108,...............261
270,351,567,3618,5895,9756.
Thesemultipleshaveacommonpattern.Whatisthat?
181+8=9
363+6=9
5675+6+7=18
36183+6+1+8=18
58955+8+9+5=27
Observethatthesumofthedigitsofthemultiplesof9isdivisibleby9.
Is479divisibleby9?
4794+7+9=20
Letusverifythesumofthedigitsofthisnumber.
Here, sum of the digits is 20 and this is notdivisibleby9.Alsobydirectdivisionweseethat
479isnotdivisibleby9.
Observetheseexamplesandstatethe"rulefordivisibilityby9".
rule for divisibility by 9 : If the sumof all thedigits of agivennumber isdivisibleby9, then thegivennumber isdivisibleby9.
Try : Oneofthefollowingnumbersisnotdivisibleby9.Identifythatnumber.2853,6003,8279,5976.
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Divisibility by 11:Someofthemultiplesof11are517,2959,3883,and40425.
Observethecommonpatterninthesenumbers.
Multiplesof11
Sumofthedigitsinoddplacefrom
right
Sumofthedigitsinevenplacesfromright
Differenceofsum
517 7 + 5 = 12 1 12 - 1 = 112,959 9+9=18 5 + 2 = 7 18-7=113,883 3+8=11 8+3=11 11-11=0
40,425 5 + 4 + 4 = 13 2+0=2 13 - 2 = 11
Byobservingthelist,trytostatetherulefordivisibilityby11.
rule for divisibility by 11 : Anumberisdivisibleby11,ifthedifferencebetweenthesumofdigitsinoddplacesanddigitsinevenplacesiseither11or0.
Try : Whichofthefollowingnumbersaredivisibleby11?1)6,5562)1,2373)1,3974)1,748
exercise 1.1
I. Whichofthefollowingnumbersaredivisibleby A)2 and B)3
a) 256 b)394 c)618 d)708 e)692
f) 846 g)3,955 h)6,852 i)3,051 j)6,872
II. Whichofthenumbersaredivisibleby(A)4and(B)8
a)692 b)376 c)5,872 d)8,000 e)12,579
f)36,420 g)58,628 h)7,741 i)30,148 j)20,928
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III.Onthebasisofdivisibilityrule,recognisethenumbersdivisibleby6.
a)3,474 b)6,234 c)4,576 d)3,870 e)6,252
f)12,741 g)59,052 h)82,766 i)40,008 j)51,206
IV.Writetheleastnumberintheblanksoastomakeitdivisibleby3.
a)__7,450 b)34,__52c)56,4__3d)47,32__
V. Applythedivisibilityruleandfindthenumberswhicharedivisibleby9.
a)5,876 b)9,486 c)5,670 d)1,572 e)4,653
f)40,926 g)42,531 h)58,673 i)47,320 j)50,985
VI.Applythedivisibilityruleandidentifythenumberswhicharedivisibleby11.
a)4,719 b)8,228 c)9,211 d)2,926 e)8,987
f)38,798 g)42,163 h)80,564 i)39,119 j)68,035
VII.Completethefollowingtableasgiveninthemodel.
NumbersDivisiblebythefollowing?ornot!
2 3 4 5 6 8 9 10 11
356 Yes No Yes No No No No No No
870
945
3,256
30,438
51,720
609
7,690
91,548
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1.2(a)PrimeandCompositenumbers
Teacher gave 7 and 8metal plates toRazia andGopalrespectivelyandaskedthemtoarrangetheplatesindifferentways.
rule : Thenumberofplatesineachrowmustbeequaltothoseinthecolumn.Accordingtothisrule,theyarrangedtheplatesasfollows:
Taking7inarow.7×1=7
8×1=8
Taking8inarow.
Taking1ineachrow,7rows.
1×7=7
4×2=8
Taking4ineachrow,2rows.
2×4=8
Taking2ineachrow,4rows.
Thereareonlytwopossiblearrangements
TheFactorsof7are1and7.
1×8=8
Taking1ineachrow,8rows.
Thereareonly4possiblearrangements
Thefactorsof8are1,2,4and8.
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Observe the hanging chart of factors:
1,31,21
1 2 3 4 5 6 7 8 9 10 11 12 13
1,5 1,7 1,11 1,13
1,2,3,61,2,4 1,2,4,8
1,2,5,10
1,2,34,6,12
1,3,9
Numbershavingonlyonefactor
Numbershavingonlytwofactors
Numbershavingmorethantwofactors
1 2,3,5,7,11,13 4,6,8,9,10,12
Observe the numbers circled in the given numbers.1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12, 13,............ Thesenumbershaveonlytwofactorsandsuchnumbersarecalled'Prime numbers'. Primenumbersaredivisibleonlyby1andthemselves.
The numbers havingmore than two factors are called'Composite numbers'.
Example:4,6,8,9,10,12,14,15,............
Know This:- '1' has only one factor. Therefore '1' is neither a prime number nor a composite number.
Observe the hanging chart given earlier and answer the following :
Whichistheleastprimenumber?Whichistheleastcompositenumber? ‘1’isregardedasneitherprimenorcomposite.Why?
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recognising the prime numbers between 1 and 100.Werecogniseprimenumbersbyusingamethodwhich
wasusedbytheGreekmathematicianEratosthensduring3rdcentury.
1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19 2021 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 4041 42 43 44 45 46 47 48 49 5051 52 53 54 55 56 57 58 59 6061 62 63 64 65 66 67 68 69 7071 72 73 74 75 76 77 78 79 8081 82 83 84 85 86 87 88 89 9091 92 93 94 95 96 97 98 99 100
Activity :- steps : Prepareatableforthenumbers form1to100asshown.Strikeout1(why)?Encircle 2 and strike out all themultiplesof2.Encircle 3 and strike out all themultiplesof3.Encircle5and7,continuethe same.Continue this operation until all the numbers are either circledorstruckoff.
Finally,thenumbersencircledinthetableare
primenumbers.
Observe the table of eratosthens and answer :• Whicharetheprimenumbersbetween1and20?• Howmanyprimenumbersaretherebetween21and50? • Amongtheprimenumbers,howmanyofthemareeven
numbers?Whicharethose?
exercise 1.2 (a)
I. Completebychoosingthecorrectanswer: 1) Smallestprimenumberis a)1 b)2 c)9 d)0 2) Thenumberofprimenumbersbetween1 and10is a)4 b)7 c)2 d)5 3) Thenumberofprimenumbersbetween21and30is a)1 b)2 c)3 d)4 4) Thenumberofprimenumbersbetween91and100is a)4 b)3 c)2 d)1
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II. 1)Findtwoprimenumberslessthan20whosesum isdivisibleby5.
Example:2 + 3 = 5 2)Findtwoprimenumberslessthan20whosesum
isdivisibleby3. Example:7 + 2 = 9
1.2 (b) Co-prime numbers
example 1 : What are the numbers which divide both 10and7?
Factorsof10=1,2,5,10.Factorsof7=1,7.i.e.,1istheonlynumberwhichdividesboth10and7.
example 2 :What are the numbers which divide both 12and13?
Factorsof12=1,2,3,4,6and12.Factorsof13 = 1and13.i.e.,'1'istheonlynumberwhichdividesboth12and13.Thepairofnumbersintheaboveexamples10and7; 12
and13aredivisibleby1only.Thereforesuchpairsarecalledco-primenumbers.
∴ 10and7areco-primenumbers. 12and13areco-primenumbers.Observe this example :Are12and3co-primenumbers?Factorsof12are1,2,3,4,6and12.Factorsof3are1and3.12and3aredivisibleby1and3.∴12and3arenotco-primenumbers.
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Try : VeritywhethertheseoreCo-primes1) 7,21 2) 9,13
exercise 1.2 (b)
I. Identifytheco-primenumbersinthefollowing. 1)4,7 2)12,15 3)15,8 4)21,20
5)12,20 6)2,9 7)14,81 8)14,49
II. 1)Inacowshed,thereare12calvesand16cows.Ifthey areseparated,isitpossibletogroupthemwithequal numbersineachgroup?
a) Ifpossible,howmanycows/calvesarethereineach group?
b) Ifnot,givereasons. 2) In sixth standard, there are 16 boys and 13 girls.
Is it possible to group them in equivalent groups separately?
a) Ifpossible,howmanychildrenarethereineachgroup? b) Ifnot,givereasons.
1.3 To express numbers as the product of prime numbersYouhavealreadylearntabouttheprimenumbers.Observe
thefollowing4 = 2 ×2,6=3×2,10=2× 5andsoonInfactthenumberscanbeexpressedastheproductofprimenumbers.Letuslearnhowtodoit.method 1 : Factorisation methodexample 1 :Write18astheproductofprimenumbers.
18
3
23 9
∗ Goondividingthegivennumberbyprimenumbersandcontinuetheoperationtillyouobtainaprimenumber
∗ Productofprimenumberssoobtainedis 18=2× 3 × 3
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example 2 :Write96astheproductofprimenumbers.96
12
48
6
24
3
2
2
2
2
2
i.e.,96=2×2×2×2×2×3
Try : Expressthefollowingastheproductofprime numbers.1)75 2)100
method 2 : by constructing factor treeexample 1 :Write30astheproductofprimenumbers.
Method1:
30
15 2
3 5
Method2:
30
6 5
3 2
∗Whenthefactortreeiscompleted,recognisetheprimenumbersbyputtingsquarearoundthem.Inthisexample,theprimenumbersobtainedare3,5and2.
∴ 30 = 3 × 5 × 2.example 2 :Write48astheproductofprimenumbers
48
12
2 6
2 3
2 2
4
∴48=2×2×3×2×2
TrytoSolvethisinadifferentway.
Try : Constructfactortreeforthefollowing. 1)802)120
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example 3 : Writetheprimefactorsof250bybothfactorisationmethodandfactortreemethod.
∴250=2×5×5×5
method : 12 250
125255
55
method : 2
25
5 5
250
10
2 5
Try : Constructafactortreeandwriteastheproduct ofprimefactors.1)282)75
exercise 1.3
I. Writethefollowingnumbersastheproductofprimenumbersbyfactormethod.a)20 b)26 c)40 d)80 e)300
f)570 g)680 h)144 i)500 j)1000II. Completethefollowingfactortrees:
68
34
2
1) 150
5 2
10
2) ?
35
5
2 2
5
3)
?
2
3 2
2 5
4)
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III.Constructfactortreeforthesenumbers.
1)70 2)96 3)160 4)200
1.4 Factors and multiples
Inyourpreviousclassyouhavelearntaboutfactorsandmultiples.Recallthem.Factors :
Reetawrites8astheproductofafewnumbersasfollows:8 =8×1
= 4 × 2
= 2 × 4
=1×8
Reetadivides8by1,2,4and8then
quotient=8
remainder=0
1 88
08g 2 8
08
4g
quotient= 4
remainder=0
4 8
08
2g
quotient= 2
remainder=0
8 8
08
1g
quotient= 1
remainder=0
i.e.,8isdivisibleby1,2,4and8.∴ 1,2,4and8arefactorsof8.
A)Ravitriestowrite13astheproductoftwonumbers.
13 = 13 × 1
= 1 × 13
Thisispossibleinthesetwowaysonly.
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B)Ravidivides13by1and13
1 1313
031
03
g
Quotient= 13Remainder=0
13 131
0013g
Quotient= 1 Remainder=0
Thenumber13isdivisibleby1and13.Therefore,1and13arethefactorsof13.
Factorsmeanthenumberswhichdividethegivennumber.
Observethefactorsof8:Factorsof8are1,2,4and8.1,2and4 arelessthan8.8 = 8Also,observethefactorsof13:Factorsof13are1and13.1 < 13
13 = 13
So,thefactorsofagivennumberarelessthanorequaltothegivennumber.
Do this : Find the factors of 12, 20, 30 and compare them with the given numbers.
Common factors example 1 :Whatarethecommonfactorsof12and16?
Factorsof12 = { 1,2,3,4,6,12}
{ 1,2,4,8,16 } Factorsof16 = ∴Thecommonfactorsof12 and 16 are{1,2,4}.
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example 2 :Whatarethecommonfactorsof56and42?Factorsof56 = {1,2,4,7,8,14,28,56 } Factorsof42 = {1,2,3,6,7,14,42 } Commonfactors= {1,2,7,14}
Do yourself: Listthecommonfactorsof 1)16,202)40,50
multiples example : Shylawasaskedtolistthenumbersdivisible
by4.Sherepresentedthosenumbersbyputtingsquareasfollows:
4 × 1 = 4 4 × 6 = 24 4 × 11 = 44
4 × 2 = 8 4 × 7 = 28 4 ×12= 48
4 × 3 = 12 4 × 8 = 32 ........................
4 × 4 = 16 4 × 9 =36 ........................
4 × 5 = 20 4 ×10 = 40 .........................
4 41
g
04
4 287
g
0028
4 369
g
0036
4 4812
g40880
remainder=0 remainder=0 remainder=0 remainder=0
Thenumbersofthissequence4,8,12,16,20,..........aredivisibleby4.
Therefore, the numbers of this sequence 4, 8, 12,16,20,......................arecalledthemultiplesof'4'.which is the highest multiple of 4 ?
Ifweadd4toanymultipleof4wegetamultiplewhichisbiggerthanthegivenmultipleof4.So,therewillbeagreatermultiplethananygivenmultiple.Observehowthemultiplesof4arecomparedwith4.
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example 1 : Multiplesof4 are4,8,12,16,20.........8,12,16,..........>4
4 = 4
ie.,Multiplesof4aregreaterthanorequalto4.example 2 :Multiplesof15=15,30,45,60,75,.........
30,45,60,75,..........>15
15 = 15
ie.,Multiplesof15aregreaterthanorequalto15.Remember
● Multiplesofagivennumberareinfinite.● Multiplesofagivennumberaregreaterthanorequal
tothatnumber.● Allthemultiplesofagivennumberaredivisibleby
thegivennumberitself.
Common multiples :1) Writethecommonmultiplesof9and12.
Multiplesof9={9,18,27,36,45,54,63,72,81,90,---}
Multiplesof12 = {12,24,36,48,60,72,84,96,108,120,---}
Commonmultiplesof9and12 are{36,72,............}
2) Listthecommonmultiplesof5and10.
Multiplesof5 = { 5,10,15,20,25,30,----}
Multiplesof10={ 10,20,30,40,50,--- }
Commonmultiplesof5and10are{10,20,30,......}
Try : Findthefirst3commonmultiplesofthefollowing. 1)4,6 2)8,10
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method to write the multiples :1. Writethefirst5multiplesof7. 7×1=7 7×2=14 7×3=21 7×4=28 7×5=35
∴First5multiplesof7are7,14,21,28,35.2. Writethefirst4multiplesof25. 25×1=2525×2=5025×3=75 25×4=100
∴First4multiplesof25are25,50,75,100.
Try :Writethenextmultiplesofthefollowing.
Multiplesof9 are9,18,27,36,45,____,____,____,____
Multiplesof14 are14,28,42,____,____,____,____,____
exercise 1.4
I. Listthemultiplesofthefollowingnumbers:
a)8 b)25 c)21 d)31 e)42
f)60 g)67 h)100 i)96 j)75
II. Writefirst5multiplesofthefollowing:a)6 b)11 c)15 d)24 e)30
III. Matchthefollowing:
A B
1)1 a)Multipleof 5
2)18 b)Factorofallthenumbers
3)20 c)Multipleof 7
4)49 d)Multipleof 6
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1.5 highest Common Factor (h C F)Youhavealreadylearntthemethodoffindingcommon
factorsofnumbers.Observethefollowing:examples 1:-Whatarethecommonfactorsof8and12?
Factorsof8= { 1, 2, 4, 8 }Factorsof12= { 1, 2, 3, 4, 6, 12, }Commonfactorsof8and12={1,2,4} Amongthesecommonfactorswhichisthehighestfactor?4 i.e.,Thegreatestnumberwhichdividesboth8and12is4.Wesaythat,4isthehighest Common Factorof8and12
(wewriteitasHCF).∴HCFof8and12is4.
example 2:- Findthehighestcommonfactorof24and36
Factorsof24are{1,2,3,4,6,8,12,24 }Factorsof36are{1,2,3,4,6,9,12,18,36 }So,commonfactorsof24and36are{1,2,3,4,6,12}.Whichisthehighestfactoramongthesecommonfactors?12
∴HCFof24and36is12.
∴HCFofgivennumbersistheonlyonehighestnumberthatdividesthegivennumbers.To find the HCF by factor method :example 1 :-WhatistheHCFof8and20?
8,204,102,5
22
steps :• Factoriseusing least common factorof the
givennumbers.• Continuetheoperationtillco-primenumbers
areleft.• Theproductofthenumbersusedinvertical
linecolumnistheHCFofthegivennumbers.HCFof8and20is4.
= 2 × 2 = 4
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example 2:-WhatistheHCFof18and24?18,249,123,4
23
Thenumbers available in the verticallineare2and3.HCFof18and24 = 2 × 3 = 6.
example 3:-WhatistheHCFof18,72and54?Reorganiseandwritethecommonfactorsandidentify.18=2×3×3}72 = 2 × 2 × 2 × 3 × 354 = 2 × 3 × 3 × 3
18 = 2 × 3 × 372 = 2 × 3 × 3 × 2 × 254 = 2 × 3 × 3 × 3
Commonfactorsare 2,3 and 3.Theproductof2,3and3istheHCFofthegivennumbers.∴ HCF =2×3×3=18. To find the HCF using the product of prime factors :example 1 :-WhatistheHCFof16and20?
steps :● Factorise each number using prime
numbers.● Writetheprimefactorsofeachnumber.● Recognisecommonfactorsinthosetwo
groups.● Here,commonfactoris2andisrepeated
twice.● HCF= 2 × 2 = 4
∴HCFof16and20=4
168 4 2
222
2010 5
22
16 = 2 × 2 × 2 × 2 20= 2 × 2 × 5
example 2 :- WhatistheHCFof24and60?
24= 2 × 2 × 2 × 3
60= 2 × 2 × 3 × 5
24 12 6 3
222
6030 15 5
223
Thecommonfactorsof24and60=2,2and3 ∴HCFof24 and60=2× 2 × 3 = 12.
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example 3 :-WhatistheHCFof280and350?280 = 2 × 2 × 2 × 5 × 7
350 = 2 × 5 × 5 × 7
Commonfactorsof280and350are2,5and7.
∴HCFof280and350=2×5×7=70
Do yourself:FindtheHCFofthefollowing. 1)10,15 2)40,60 3)32,48 4)250,175
hCF of Co-prime numbers :example 1 :-WhatistheHCFof8and25?
8=2×2×2 25 = 5 × 5
Observetheprimefactorsofgivennumbers.Whatarethecommonprimefactorsofgivennumbers?
8and25donothaveanycommonprimefactors.ThenwhatistheHCFof8and25?
1istheonlynumberthatdividesboth8and25.∴HCFof8and25is1.
example 2 :-WhatistheHCFof23and48?23 = 23
48=2×2×2×2×3
23and48don'thaveanycommonprimefactors.Whichistheonlynumberthatdividesboththenumbers?1istheonlynumberthatdividesboththenumbers.∴HCFof23and48is1.
TheHCFoftwoco-primenumbersis1.
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hCF of two numbers, when one number is the multiple of the other :
Observe these examples.example 1:WhatistheHCFof6and42?
6 = 2 × 3
42 = 2 × 3 × 7
H.C.F. = 2 × 3
= 6example 2 :WhatistheHCFof18and72?18=2 × 3 × 3
72 = 2 × 2 × 2 × 3 × 3 H.C.F. = 2 × 3 ×3=18.Fromtheaboveexamples,Ifanumberisthemultipleofthe
othernumber,theleastnumberistheHCFofgivennumbers.Try : Find theHCF of the following by observing the
numberscarefully:- 1)8,32 2)17,35 3)48,16 4)30,90 5)91,97 6)7,49,35 7)14,21,23
Problems involving hCF1) Acylindricalvesselhas60litresofmilkandanotherhas
40litresofmilk.Whatisthemaximumvolumeofavesselthatcanmeasurethemilkofboththevesselsinfull.
Factorsof60={ 1,2,3,4,5,6,10,12,15,20,30,60}
Factorsof40={ 1,2,4,5,10,20,40}Commonfactors={1,2,4,5,10,20}
∴HighestCommonFactorof60and40is20. i.e.,20isthehighestnumberthatcandivideboth40and60. ∴Volumeoftherequiredvessel=20litres.
Try : Solvetheaboveproblembyfactorisationmethod.
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2)Rangaiahhastwogroupsofsheep.Inthefirstgroup16 lambsandinsecondgroup28lambsarethere.Hewantedto keep those lambs in their respective sheds in equalnumbers. Find the requirednumber of sheds for eachgroup.(Thereshouldbeleastnumberofsheds.)
16 = 2 × 2 × 2 × 2
28 = 2 × 2 × 7
168 4 2
222
28 14 7
22
Commonfactors= 2 × 2
HCF= 4
i.e.,Hecanaccommodate4sheepsineachshed.
∴Forfirstgroup,numberofshedsrequired= 16 ÷ 4
= 4 sheds
Forsecondgroup,numberofshedsrequired=28÷ 4
= 7 sheds.
Try:Solvetheaboveproblembylistingthefactors.
Think...Answer :- ShreyasisfindingtheHCFof9and16byfactorisationmethod.9=3×3and16=2×2×2×2
Thereisnocommonfactor,thereforehethoughtthatHCFof9and16is‘0’.
1) Isittherightanswer? 2) Ifnot,givereasons. 3) WhatistheHCFof9and16?
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exercise 1.5
I. Findthecommonfactorsofthefollowing: a)15,18 b)24,36 c)40,60 d)56,25
e)6,8,10 f)12,15,24 g)4,16,17 h)50,20,70 II. FindtheHCFbylistingthefactors: a)6,24 b)18,35 c)48,120 d)96,40
e)10,40,16 f)9,36,72 g)24,48,60 h)51,34,68
III.FindtheHCFbyfactormethod: a) 15,20 b) 35,28 c) 12,21 d) 27,63
e) 12,8,16 f) 18,54,81 g) 90,30,120 h) 35,49,112
IV.MentiontheHCFbyobservation. 1)a)15,29 b)7,11 c)31,17 d)51,53
2)a)5,25 b)14,98 c)96,24 d)18,72
V. Solvethefollowing:1) Lengthandbreadthofaschoolauditoriumare20m
and8mrespectively.Bothlengthandbreadtharetobemeasuredcompletelyusingasamemeasuringstick.Whatisthelengthofthelongestmeasuringstickthatcanmeasurethelengthandbreadthoftheauditorium?
2) Onebaghas56kgoftoordalandanotherbaghas96 kgofBengalgramdal.Thesearetobefilledequallyinsmallbagsusingpossibleleastnumberofbags.Thenwhatisthemaximumquantityofdalthatistobefilledineachbag?Whatistheleastnumberofbagsrequired?
3) Therearetwovessels.Firstvesselcontains18litresofmilkandthesecondvesselcontains24 litres ofjuice.Iftheliquidsaretobemeasuredseparately,usingasmallercommonmeasurewhosevolumeis in litres,whatisthevolumeofsuchabiggestmeasure?
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1.6 least Common multiple (lCm)
WhatistheLeastCommonMultipleofus?
I'm"6"
6 8I'm"8"
Observe the method by which Joseph and shivaram have found the answer.
Multiplesof6 = {6,12,18,24,30,36,42,48,54,--------}Multiplesof8={8,16,24,32,40,48,56,64,------------}Commonmultiplesof6and8={24,48,---------------}∴LeastCommonmultipleof6and8=24.
Observe:LeastCommonMultipleof6and8is24.24istheleastnumberwhichisdivisiblebyboth6and8.So,24istheleast Common multiple (lCm)of6and8.LCM of given numbers is the least number which is
divisiblebythesenumbers.
Finding lCm of given numbers by prime factorisation : example 1:FindtheLCMof6and8.step :Findtheprimefactorsof6and8.
6 = 2 × 38=2× 2 × 2
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● 2and3aretheprimefactorsofthesetwonumbers.● Selectafactorwhichisrepeatedmorenumberoftimes
ineachnumberandfinditsproduct.● Here,thefactor2isrepeatedthricein8and3ispresent
oncein6.Findtheproductof(2×2×2)×3.● Sothisproduct(2× 2 ×2)× 3istheLCMof6and8 ∴ LCMof6and8=(2× 2 ×2)× 3 = 24.
example 2 :FindtheLCMof24and60
24 = 2 × 2 × 2 × 3
60=2×2×3×5
24 12 6 3
222
6030 15 5
223
● Theprimefactorsobtainedare2,3and5.● Theprimefactor2hasappearedmaximumthreetimes
in24.i.e,(2×2×2)● Theprimefactor3hasappearedmaximumoncein24
and60.i.e,(3)● Theprimefactor5hasappearedmaximumoncein60
i.e,(5)● Find the product of all the prime factors repeated
maximumnumberoftimes.● i.e.,(2×2×2)×3×5=120 ∴ LCMof24and60=120
example 3 :Whichistheleastnumberthatisdivisibleby6,10and18?Solution : The least number that is divisible by the givennumbermeansthatistheLeastCommonMultipleofthosenumbers.LetusfindtheLCMofthesenumberstofindtheleastnumberwhichisdivisiblebythesethreenumbers.
6 = 2 × 3 10 =2×5 18 =2×3×3 LCM =(2)×(3×3)×(5) =90
∴Theleastnumberdivisibleby6,10and18=90
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example 4: FindtheLCMof32,36and40.• LCMcanbefoundbyusingthefollowingmethod.
Thegivennumbersarewritteninarowandfactorisationisdoneasfollows.2 32,36,40 Dividebyleastprimefactor2
2 16,18,20 Divideagainby2
2 8,9,10 Divideagainby2,carry9whichisnotdivisibleby2.
2 4,9,5 Divide again by 2, carry 9 and5which are notdivisibleby2.
2 2,9,5 Divideagainby2. Carrythenumberswhicharenotdivisible.
3 1,9,5 Divideby3.
3 1,3,5 Divideagainby3.
5 1,1,5 Divideby5.
1,1,1
So,LCM=2×2×2×2×2×3×3×5=1440
Know this :- Factorisethegivennumberstillco-primefactorsareobtainedandfindtheLCM.
2 32,36,40
2 16,18,20
2 8,9,10
Here4,9and5areco-primefactors.4,9,5
So,LCM=2×2×2×4×9×5=1440
Try : FindtheLCMofthefollowing.
1)14,20 2)16,20 3)18,20,24 4)15,20,30,35
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lCm of co-prime numbers :example :
LCM=2×3×7 = 42
LCM=3×3×5 = 45
LCM=2×2×2×2×3=48
1) 6and7 2) 9and5 3) 16and3
3 9,5 3 3,5 5 1,5
1,1
2 16,3 2 8,3 2 4,3
2,3
2 6,7 3 3,7 7 1,7
1,1
In the above problems find the product of the givennumbers and comparewith their LCM.What do you inferfromthis?Explain.LCMofco-primenumbersistheproductofthosenumbers.
Answerinstantly:WhatistheLCMofthefollowing? 1)8,72)4,113)15,44)2,3,5
lCm of the given numbers if one number is the multiple of the other number : example :
3 21,7 7 7,7
1,1
3 9,45 3 3,15
1,5
2 4,8 2 2,4
1,2
LCM=2×2×2 =8
= 3 × 7= 21
= 3 × 3 × 5= 45
1)4and8 2) 21and7 3)9and45
IntheaboveproblemscomparethegivennumberswiththeirLCM,whatdoyouinferfromthis?In thegivennumbers, if onenumber is themultipleofanotherthen,thatmultiple(greaternumber)istheirLCM.
Answerinstantly:WhatistheLCMof? 1)12,42)5,203)30,90
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Problems involving lCm:example 1 :Thestudentswhoare ready for thedrill canbearrangedin8or12or14ineachline.Thenfindtheleastnumberofstudentsreadyforthedrill.Solution : According to the problem, the least number ofstudentswhoarereadyfordrillisdivisibleby8,12and14. ∴ItistheLeastCommonMultipleofthesenumbers.
2 8,16,142 4,8,72 2,4,71,2,7
LCM=2×2×2×2×7
= 112 ∴Noofstudentsreadyforthedrill= 112
example 2 :Afarmerhassomequantityoftoordhal.Ifhefills4or5or9or12kgofdhalineachbagthen3kgofdalwillstillbeleft.Findtheleastquantityofdhalwithhim?Solution:Accordingtotheproblem,thequantityoftoordalwiththefarmerwhendividedby4,5,9,12leavesbehindtheremainder3.So,thequantityofdalis3kgmorethantheLCM of4,5,9and12.Therefore,findtheLCMof4,5,9and12andadd3togettherequiredanswer.
2 4,5,9,122 2,5,9,63 1,5,9,31,5,3,1
LCM =2×2×3×5×3 =180∴Thequantityoftoordalwiththefarmeris 3kg.morethantheLCM =180+3kg.∴Quantityofdhalwiththefarmer=183kg.
Can you solve this puzzle ? Duringsummer,Neethawenttoher grandfather’s house.Aneighbourwas running a dairy.Neethaaskedtheneighbour,“Howmanycowsdoyouhave?Theneighbouransweredintheformofapuzzle.
Neighbour:Thenumberofcowswithmeisinsuchawaythat,Itiethemto10polesequally,letthemtograzeunder5treesinequalnumbersandletthemtodrinkwaterin4tanksinequalnumbers.Suchaleastnumberofcowsarewithme.Then,howmanycowsarerearedbytheneighbour?
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exercise 1.6
I. FindtheLCMandidentifythecommonpropertyinthefollowing.
1) a)7,12 b)15,11 c)5,13 d)2,17
2) a)6,18 b)20,80 c)40,8 d)60,12
II. FindtheLCM. 1) a)8,20 b)12,16 c)24,60 d)35,100
2) a)6,8,20 b)16,24,32 c)36,18,45 d)20,30,50
III. Solvethefollowing.
1) Findtheleastnumberdivisibleby8,6and12completely.
2) Whenmilkinacertainvesselismeasuredusingjarsof3litresor5litresor6litres,eachtime2litresofmilkisleftinthevessel.Thenwhatisthequantityofmilkfoundinthatvessel?
3) Threechildrenbegintowalktogetherinaschoolfield.Eachofthemcoversadistanceof60cm,65cmand70cmrespectivelyinonestep.Theydecidedtostopaftercoveringthesameleastdistance.Thenwhatistheleastdistancecoveredbythem.
4) Threewall clocksare ina room.All the clocks ringtogetherat6 'o'clockinthemorning.Afterthat,1stclockringsonceforevery20minutes2ndclockringsonceforevery30minutesand3rd clockringsforevery40minutes.Whatwouldbethetimeatwhichalltheclocksringtogetherforthesecondtime?
l l l l l
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Unit- 2
Fractions
after learning this unit you can : understand the meaning of a fraction, classify the fractions into different types, represent different types of fractions through figures and
also on a number line, write fractions in ascending and descending order, do the addition and subtraction operations of like and
unlike fractions.
2.1 FractionYou know that in our daily life we use fractional quantities
like half, one forth and three forth. You have already learnt to read and write such fractional quantities (fractions) in your previous classes. Let us try to recollect them.
Observe the fractions representing the figures given below. The shaded portion is written in the form of fractions.
41
21
83
32
32
43
125
1st Set BW Proof : 04-09-2012 : 000 to 000 (Total 000 Pages) : Print Given : Komala
2nd Set BW Proof : 21-09-2012 : 01 to 32 (Total 32 Pages) : Print Given : Rajeshwari
3rd Set Clr Proof : 31-10-2012 : 1 to 33 (Total 000 Pages) : Print Given : komala
4th Set BW Proof : 15-12-2012 : 000 to 000 (Total 000 Pages) : Print Given : komala
5th Set BW Proof : 12-01-2013 : 33 to 66 (Total 000 Pages) : Print Given : Pallavi
6th Set BW Proof : 6-02-2013 : 33 to 66 (Total 000 Pages) : Print Given :
Extra BW / Clr Proof : 00-00-2012 : 000 to 000 (Total 000 Pages) : Print Given : Operator
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Fraction is a selected part out of equal number of parts of an object or a group.
a fraction consists of two parts.Example:-
41 →numerator numerator → selected number of parts
→denominator denominator → total number of parts
125 →numerator
→denominator
In an object or in a group the total number of equal parts are indicated by the denominator. The number of selected equal parts from the total equal parts is indicated by the numerator.
try. 1) In 53 , which is the denominator ?
2) In 87 , which is the numerator ?
to represent fractions on a number line :We represent the whole numbers 0, 1, 2, 3............on a
number line. In the same way we can represent fractions also on a number line.Example 1 :- Locate 2
1 on number line.
21 is greater than '0' and less than '1'. So it is in between
'0' and '1'. So it is represented like this.
21 10
Example 2 : Locate 32 on number line.
32 is greater than '0' and less than '1'.We can represent 3
2 after dividing one whole into 3 equal parts.
031
32
33
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Example 3 : Locate 50 and 5
4 on a number line.
observe
50 is same as '0'
51
50` j 5
253
54
55
1
0
try : 1) Show 65 on a number line.
2) Show 85 , 80 , 82 , 87 on a number line.
Exercise 2.1
I. Write a fraction which represents the shaded part of the following.
a) b) c) d)
II. Shade the parts of the following figures equal to the fractions given.
41
82
96
31
87
41
82
96
31
87
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2.2 types of a Fraction Proper fraction :
is the shaded portion bigger than the whole in the following figures ?
41
62
847
4
Here each fraction is less than 1. Observe the numerator
and denominator in the fractions 41 , 62 , 74 and 8
4 . Here the
numerator is less than the denominator.
Observe the representation of 41 and 7
4 on a number line.
41
42
43 10
71
72
73
74
75
76 10
The fractions 41 and 7
4 are less than '1'. Such fractions
are called "Proper fractions".
Proper fraction : If the fraction is less than '1', then such a fraction is called "Proper fraction". In proper fractions, the numerator is less than the denominator.
Example : 53 , 32 , 10
7 , 125 , 9
4
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try : 1) Show 62 and 8
4 on a number line.
2) Write any four proper fractions whose denominator is 8.
3) Write any four proper fractions whose numerator is 7.
improper fraction
Example 1 : In this picture given below, how many 41 parts
are there?
41
41
41
41
41
There are 5 parts of 41 .
We write this as 45 .
Is 45 greater or smaller than '1'? (It is greater than '1').
Example 2 : Here, how many 21 parts are there?
There are 7 parts
of 21 . This can
be written as 27 .
21
21
21
21
21
21
21
Is 27 greater than '1' or less than '1'?
It is greater than '1'.
Example 3 : How many 31 parts are there in this figure ?
31
31
31
There are 3 parts of 31 . This can be written as 3
3 .
Is the fraction 33 greater than '1' or less than '1'?
(It is equal to '1'.)
So the fractions 45 , 27 , 33 are greater than '1' or equal to '1'.
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Let us represent the above numbers on a number line.
45 :
41
42
43
44
45
460
33 :
31
32
330
1
27 :
21
22
24
26
23
25
27
280
1 2 3 4
45 greater than '1'.
33 is equal to '1'.
27 is greater than '1'.
Observe the numerator and denominator of 45,27,33 . Here
the numerator is greater than the denominator or equal to the
denominator. Such fractions are called "improper fractions".
try : 1) Write any 7 improper fractions whose denominator is 4. 2) How many improper fractions can be written whose
numerator is 5 ? Write them.
Mixed Fraction: Give the following circles in the form of a fraction.
Here how many whole circles are there? (2)How many fractional parts of the circle are there? one half 2
1` j.
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This is written as 2 21 . It is read as "two and one by two".
2 21
Whole number Proper fraction
Express the following squares in fractions. Total squares = 3
Fractional part = 43
This is written as 3 43 .
It is read as 'three and three by four'.
3 43
Whole number Proper fraction
Let us locate 2 21 and 3 4
3 on a number line.
2 21 :
2 210 1 2 3 4
3 43 :
3 430 1 2 3 4
In this way fraction which has both non zero whole number and proper fraction is called mixed fraction. The mixed fraction is always more than '1'. try : 1) Identify the whole number and proper fraction in 8 5
2 .
2) Write any four mixed fractions.
3) Locate 1 43 on a number line.
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Activity:- The teacher prepares different flash cards of different types of fractions and whole numbers and organizes the activity as given below.
43 2
31
58
72
65 3
21 6
41 5
105
59
95
65 2
43
912
86 3
• Arrange the children in a circle.• Keep all the flash cards at the centre and ask each
one of them to pick up a flash card.• When the teacher says ''Improper fraction'', then the
children having flash cards of improper fraction will come forward.
• The teacher verifies the flash cards and gives '5' marks to those who have correct flash cards.
• Again the students will stand in a circle.• The teacher follows the same method for proper
fraction, mixed fraction, whole number and continues the game.
• After completing the game once, the teacher repeats the game.
2.4 converting improper fraction into mixed fraction and vice versa
converting improper fraction to Mixed fraction :
Example 1: Convert 25 into a mixed fraction.
21
21
21
21
21 In 2
5 , there are 5 equal parts of 21 .
How many wholes can be made from the above? What is the remaining fractional part?
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1 121
2 wholes and one 21 part.
This is written as 2 21 2 2
1+ = .
Briefly it is written as
25
22
22
21= + +
2 21 2 2
1= + =
∴ 25 = 2 2
1
Improper Fraction
MixedFraction
Know the alternate method :Write 2
5 as a mixed fraction.
Denominator → 2
Numerator
Whole
52
14 2
5 2 21=`
Example 2 : Convert 314 into a mixed fraction.
314 = 3
12 + 32
= 4 32
Method - 1
Denominator→ 3
Numerator
Whole
314 4 3
2=`
144
212
Method - 2
Writing mixed fraction into improper fraction :Example 1: Convert 2 4
3 into an improper fraction.
41
41
41
What is the smallest fractional
part here ? 41` j
Divide the whole into the same
fractional parts.
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41
41
41
41
41
41
41
41
41
41
41
∴8 fractional parts of 41 3+
fractional parts of 41 .
∴ 2 43 = 4
8 + 43
= 48 3+ = 4
11
∴ 2 43 = 4
11
alternate method (whole number × denominator) + numerator
denominator
2 43 = 4
(2 4) 3+# = 48 3
411+ =
Example 2 :- Convert 3 85 into improper fraction.
8181
81
81
81
111 What is the smallest
fractional part here ? 81` j
Divide the total objects into the same fractional part.In total how many fractional parts do we get?
8181
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
81
824
85
824 5
829
3 85
829
+ = + =
=`
Alternate method (Whole number × denominator)
+ numerator
denominator
3 85
8(3 8) 5
824 5
3 85
829
= + = +
=
#
`
Example 3 :Convert 5 7
4 into improper fraction (use alternate method)
5 74
denominatorwhole number denominator numerator
5 74
75 7 4
735 4 5 7
4739
= +
= +
= + =
#
#
`
^
^
h
h
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try :
a) Convert the following into mixed fractions. 49 , 512 , 7
22
b) Convert the following into improper fractions. 3 52 ,4 6
5
Exercise : 2.2I. Draw a number line and mark the points to represent the
following fractions.
a) 41 , 42 , 43 , 44 b) 3
1 , 32 , 33 c) 6
1 , 63 , 65 , 66
II. Write the following as mixed fractions.
a) 310 b) 4
17 c) 518 d) 2
19 e) 925 f) 7
26
III. Write the following as improper fractions.
a) 4 53 b) 3 2
1 c) 2 85 d) 76
5 e) 9 32 f) 11 7
3
Activity :- Draw diagrams to represent these numbers.
2 43a) Example: 2
5b)Example:
1) 47 2) 3 2
1 3) 2 42 4) 8
3 5) 2 85
2.3 Equivalent fractionsExample 1: Observe the shaded portion representing fractions in the following figures.
42
21
84
21
42
84= =
When the shaded part 21 , 42 , 84
are arranged one above the other, they are found to be equal. So they are equivalent fractions.
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Method of getting equivalent fractions :
We know that, 21
42=
From 21 , how do we get 4
2 ? Think.
Multiply the numerator and denominator of 21 by 2.
21
2 21 2
42= =
##
In the same way multiply both numerator and denominator of the given fraction by the same number to get a few other equivalent fractions.
2 31 3
63=
## 2 4
1 484=
## 2 5
1 5105=
## 2 6
1 6126=
##
21
63
84
105
126= = = =
By multiplying numerator and denominator of the given fraction (except 0) by the same number, we get equivalent fractions.
Observe the equivalent fractions of 43 written by Dinesh.
They are.
4 23 2
86=
## 4 3
3 3129=
## 4 4
3 41612=
## 4 5
3 52015=
##
43
86
129
1612
2015= = = = ..........
Many more equivalent fractions can be written in the same way.Method of identifying equivalent fractions :Example 1: 2
1 , 42 are these fractions equivalent?
Observe the cross multiplication of the numerator and denominator of these two fractions.
21
42
1 4 42 2 4
21
42;
== =
## `
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Example 2:- 53 , 15
9 are these fractions equivalent?
53
159
3 15 455 9 45
53
159;
== =
## `
Example 3:- Are the fractions 43 and 8
7 equivalent ?
43
87
3 8 244 7 28= =
=##
The products obtained are not equal. They are not equivalent fractions. 4
387!
alternate method of identifying equivalent fractions :By dividing the numerator and denominator of the given
fraction by the same number, (except zero) we get equivalent fraction.
Example 1:- 168
16 28 2
84=
''
16 48 4
42=
''
16 88 8
21=
''
168
84
42
21= = =`
168
84 16 4 64 8 8 64
168
84
; ,= =
=
# #
`
Verification
Example 2 :- 129
129
129
33
43= =
''
129
43=`
129
43 12 3 36 9 4 36
129
43
; ,= =
=
# #
`
Verification
Example 3 : Sahana has written a few equivalent fractions
for 4020 . Observe them.
4020
22
2010=
'' 40
2044
105=
'' 40
2055
84=
'' 40
202020
21=
''
Verify by cross multiplication whether the fractions written by Sahana are equivalent fractions.
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When the numerator and denominator are multiplied or divided by the same number (except zero), the fraction will not differ. By this process we get equivalent fractions for the given fraction.
try : 1) Find a few equivalent fractions of the following. 1) 7
2 2) 1512 3) 60
40
2) Write the correct number in the blank space so as to become equivalent fractions.
96
3=d , 7
4 12= d , 1812
6=d
simplest form of a fractionWrite equivalent fractions for 36
24 .
3624
22
1812 , 36 4
24 496 , 36 12
24 1232
3624
1812
96
32
= =
= = =
''
''
'' =
Which is the least form in these fractions? Why?1 is the only factor of numerator and denominator of the
fraction 32 (They are co-prime numbers). 3
2 is the lowest form. (Both are not divisible by each other). If the common factor of both the numerator and denominator of a fraction is '1'. Then it is considered as the lowest form of the fraction. the method of getting the lowest form of a fractionExample 1 : What is the lowest form of 48
36 ? Let us list the common factors of the numerator and
denominator of this fraction.They are 1, 2, 3, 4, 6 and 12. Among these common factors, by
which number do you divide the numerator and denominator of 48
36 , to get the lowest form ?
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4836
22
2418=
'' (Not the lowest form)
4836
66
86=
'' (Not the lowest form)
The HCF of 48 and 36 = 12.
So divide the numerator and denominator by 12.
4836
1212
43='
('1' is the common factor of numerator and denominator of these fractions.)
∴ 43 is the lowest form of the given fraction.
Example 2 : What is the lowest form of 6050 ?
Find the HCF of 60 and 50.The HCF of 60 and 50 is 10.
Then divide the numerator and the denominator by the HCF.
6050 = 60 10
50 1065=
'' ∴ 6
5 is the lowest form.
Example 3 : What is the lowest form of 4530 ?
The HCF of the numerator and denominator of the fraction is 15. Divide the numerator and the denominator by 15.
4530 = 45
301515
32=
''
∴ 32 is the lowest form
try : 1) Write the lowest form of. a) 20
16 b) 2821 c) 63
35
2) To get the lowest form of a fraction, Why should the
numerator and the denominator be divided by the HCF ?
Think and answer.
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Exercise 2.3I. Write the next four equivalent fractions for the following
by multiplying both numerator and denominator by the same number.
Example : 32
64
96
128
1510= = = =
1) 53 2) 8
5 3) 127 4) 9
4
II. Write two equivalent fractions for the following. (By dividing both numerator and the denominator by
the same number) Example : 2016 = 10
8 = 54
1)126 2) 24
20 3) 6030 4) 40
20
III. Fill in the blanks with numbers so as to make equivalent fractions.
1) 53
10=d 2) 41
8=d 3) 65 15=d
4) 73 12=d
5) 2515
5=d 6) 3018
10=d 7) 2114 2=d
8) 6036 3=d
IV. Are the two given fractions equivalent? Verify.
1) 72 ,14
4 2) 83 , 24
8 3) 65 , 1815 4) 9
5 , 1810
V. Find the lowest form of the following fractions.
1) 2510 2) 12
10 3) 2613 4) 45
18
5) 10075 6) 30
6 7) 4025 8) 200
50
2.4 comparison of fractions
Sridhara ate 2 41 chapatis and Veena ate 3 4
1 chapatis. Among them who ate more?
Sridhara ate 2 complete chapatis and little more Veena ate 3 complete chapatis and little more. Naturally 3 is greater than 2. So here Veena ate more chapatis. ∴ 3 4
1 2 412
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Observe the fractions which represent the coloured parts in the following figures.
• Which is greater in 21
and 31 ? 2
1` j
• Which is greater in 31
and 41 ? 3
1` j
• Which is greater in 61
and 41 ? 4
1` j
21
31
41
61
By representing 21 , 31 , 41 , 61 in the form of picture we learnt
to identify greater fraction among the given fractions. But it is not easy every time to represent and compare the fractions through diagrams. So let us learn the alternate methods to compare them.
2.4 (a) comparison of Like fractionsObserve the following fractions.
43 , 54 , 32 , 21 , 53 , 42 , 31
Let us form the groups of fractions having same denominator.
43 , 4
2 54 , 5
3 32 , 3
1
Example 1: Which is greater between 54 and 5
3 ?
Here the total object has been divided into 5 equal parts. Then 4 and 3 equal parts are selected and coloured. Here 5
4 is greater than 5
3 . ∴ 54
532 .
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Example 2:Which is greater between 8
3 and 87 ?
These are like fractions and 7 > 3.
So 87 is greater. ∴ 8
7832
So, in like fractions, the fraction with greater numerator will be greater. The fraction with smaller numerator is smaller.Example 3:
Which is smaller between 158 and 15
3 ?
These are like fractions and 3 < 8
So, 153 is smaller ∴ 15
31581
ascending order and Descending order of the fractions Example 1: Write the following in ascending order.
73 , 76 , 72 , 74
Solution : These are like fractions. So observe their numerators and write them in ascending order. Among the above fractions,
72 is the smallest. Next bigger is 7
3 , next 74 and 7
6 comes to the last as it is the biggest. So the ascending order of the given fractions is 7
2 , 73 , 74 , 76
Example 2 : Write the following in descending order.
125 ,12
9 ,124 ,12
8 ,127
Solution : Among the above fractions, 129 is the greatest. So
start from 129 and write the next largest among the remaining
and so on. We get the descending order as 129 ,12
8 ,127 ,12
5 ,124 ,
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try : 1. Write > or< sign in between these fractions.
1) 107
109d 2) 23
152312d 3) 18
5184d 4) 75
247529d
2. Write the following in ascending and descending order. 1) 6
5 , 61 , 64 , 63 , 68 2) 16
9 ,167 ,1612 ,16
4 ,162
2.4 (b) comparison of unlike fractions Observe the groups of the following fractions.
21 , 32 5
4 , 41 , 63 8
7 ,107 , 71
The denominators of the fractions in each group are different. So we call them unlike fractions. Let us learn the method of comparing them.
Example 1: Which is greater between 52 and 3
2 ?
52
32
Here in the first figure, the whole is divided into 5 equal
parts and two parts are coloured 52` j.
In the second figure, the whole is divided into 3 equal parts
and two parts are coloured 32` j.
In 52 and 3
2 the numerators are same but denominators
are different. In 52 there are 2 parts of 5
1 , where as in 32 there
are 2 parts of 31 . We know that 3
1 > 51 . So 3
2522 .
Example 2 : Which is smaller between 73 and 11
3 ?
Among 73 and 11
3 , the fraction 73 is 3 parts of 7
1 and 113 is
3 parts of 111 . We know that 11
1 < 71 . So 11
3 < 73 .
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From the above two examples, we conclude that the unlike fractions whose numerators are same, then the fraction with greater denominator will be smaller and the fraction with lesser denominator will be greater.
85
45 ;1 6
141 ;1 8
7107 ;2 20
162516 ;2
ascending order and Descending order of fractions.
Example 1: Write 83 , 53 ,12
343and in ascending order.
Solution : These are fractions having same numerators but different denominators. So the fraction with greater denominator will be smaller. Based on this we write the
ascending order of the given fractions as 123 , 8
3 , 53 , 43 .
Example 2: Write 159 ,10
9 ,119
209and in descending order.
Solution : These are fractions with same numerators but different denominators. So the fraction with lesser denominator will be bigger. Based on this, we write the descending order
of the given fractions as 109 ,11
9 ,159 , 20
9 .
Activity :- Write the following in the ascending order.
3 ,106 ,10
2 ,105 ,10
4 ,107
10
try : Write the following in ascending order first and then in descending order.
1) 137 , 9
7 ,157 , 6
7 2) 2016 ,16
16 , 2416 ,18
16
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2.4 (c) comparison of unlike fractions
Example 1: Compare 41 and 6
2 .
These two fractions have different numerators as well as
different denominators. We do not know how to compare such
fractions. But we know how to compare like fractions. Can
we write 41 and 6
2 as fractions with same denominator? We
can do this by writing equivalant fractions of 41 and 6
2 .
41
82
123
164
205
246= = = = = ...........
62
124
186
248
3010= = = = .............
The equivalent fractions which have the same denominators
for 41 and 6
2 are 123 and 12
4 . But 123
1241 ∴ 4
1621 .
alternate method : Compare 4
1 and 62 and say which is greater.
We can compare by getting equivalent fractions by multiplying numerator and denominator of one fraction with the denominator of the other fraction.
41
66
246 , 6
244
248
246
248
41
62
= =##
##
`1 1
Example 2 : Compare 53 and 8
4 and say which is greater.
Solution : Write equivalent fractions for 53 and 8
4 . Among them select the fractions which have the same denominator and compare them.
53
106
159
2012
2515
3018
3521
4024
4527= = = = = = = = ............
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84
168
2412
3216
4020
4824
5628= = = = = = ..................
The equivalent fractions of 53 and 8
4 which have the same
denominator are 4024 and 40
20 .
4024
40202 5
384` 2 .
alternate method • We get equivalent fractions by finding LCM of denominators
and then compare the fractions and say which is greater. 1) Compare 5
3 , 84 and say which is greater.
53
88 , 8
455=
##
##
4024 , 40
20=
4024
4020= 2
53
84` 2
• The LCM of 5 and 8 is 40.
• Write equivalent fractions so that the denominator of the fractions is 40.
• Then they are compared.
Example 3 : Compare 106 and 12
8
Solution : Find the LCM of the denominator of the two given fractions Find their equivalent fractions and then compare them.
The LCM of denominators 10,12 = 60.
106
10 66 6
6036
128
12 58 5
6040= = = =
##
##
6036
60401 10
6128` 1
try : Compare the following fractions and use the correct symbol in between them. (< , >)
1) a) 85
65d b) 15
9209d c) 12
8108d d) 17
122012d
2) a) 54
106d b) 12
81512d
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Activity: Write the following fractions in the descending order in the given rectangular bars.
153 , 7
3 , 123 , 4
3 , 93 , 83
Exercise 2.4
I. Write the correct symbol between these fractions after comparing them. (<, > or =)
a) 1) 63
65d 2) 10
7105d 3) 12
10128d 4) 15
3157d
b) 1) 75
65d 2) 8
393d 3) 6
2124d 4) 13
111211d
II. Write in the ascending order:
a) 1) 127 ,12
5 ,121 ,12
9 2) 153 ,1510 ,15
7 ,156
3) 86 , 82 , 85 , 88 4) 20
18 , 2011 , 20
6 , 2019
b) 1) 72 , 52 , 32 , 42 2) 9
7 ,167 ,10
7 , 87
3) 103 ,12
3 , 53 ,14
3 4) 209 ,15
9 ,129 ,16
9
III. Write in the descending order :
1) 121 , 23
1 , 51 , 7
1
2) 51 , 5
3 , 54 , 5
2
3) 71 , 7
4 , 72 , 7
6
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IV. Observe the following figures. Use the sign < or > = in between the fractions given.
10
11
20
21
22
40
41
42
43
44
80
81
82
83
84
85
86
87
88
30 3
132
33
a) 21
42d b) 4
131d c) 2
185d d) 8
743d
e) 30
33d f) 8
121d g) 8
332d
V. Solve the following problems.
1) Mary studies mathematics for 54 hour and Shilpa
studies for 43 hour. Who studies mathamatics for more
time?
2) In a field, the farmers have grown 85 part of dicotyledons
and 93 part of sugarcane. Which crop is grown in lesser
part of the field?VI. Compare the following fractions and use the correct sign
between them (< or >).
1) 52
43d 2) 6
281d 3) 10
6125d 4) 9
552d
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2.5 Addition and subtraction of fractions.Already you have learnt the addition and subtraction of
whole numbers. Now let us learn to add and subtract the fractions. Observe the following figures. Find the total quantity in fractions from each figures.
fig 1
41
41
41
41
41
31
41
fig 2
In figure-1, there are 2 parts of 41 and 3 parts of 4
1 . So the total quantity is 5 parts of 4
1 i.e. 45 .
In figure-2, what is the total quantity? Can you tell immediately? (No) Why?
In figure 1, all the parts are equal. That means they are fractions with equal denominators. 4
243
45+ =
In figure-2, there are different parts. That means they are fractions with different denominators. Let us learn to add such fractions.
2.5.(a) addition of fractions with equal denominators
Example 1: Johnson read 51 part of children's story book on
Saturday and 52 part on Sunday. What part of the book did
he read in these days all together?
Part of the book read on Saturday = 51
Part of the book read on Sunday = 52
Total part of the book read in two days = 51
52+
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These fractions are with equal denominators. So add the numerators and write the answer.
So 51 2
53+ = . ∴ The total part of the book that he read is 5
3 .
Example 2 : A farmer grows 123 part of coconuts, 12
5 part of
paddy and 122 part of malbary leaves in his field. What is the
total part of his field in which he has grown these crops?
Part of the field for growing coconut = 123
Part of the field for growing paddy = 125
Part of the field for growing malbary = 122
Total part of the field used for irrigation = 123
125
122+ +
These fractions are with equal denominators.
So, 123 5 2+ + 12
10= 6
5
1210
65= =
∴ The total part of the field in which he has grown the
crops = 65 .
Example 3 : Hussain paid ` 5 207 to purchase a note book
and ` 2016 to purchase a pencil. What is their total cost?
5 207
2016+ (One of these is a mixed fraction.)
20107
2016= + (It is converted into improper fraction.)
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20107 16= + (We know how to add these two fractions
as they have equal denominators.)
20133= (This is an improper fraction.)
6 2013= (So the answer is converted into mixed
fraction.)
∴ Total cost is ` 6 2013 .
try : 1) 165
163+ 2) 8
582
83+ + 3) 45
153+ 4) 2 3
2 2 31+
2.5 (b) subtraction of fractions with equal denominatorsExample 1: Subtract 8
5 from 86 .
86
85- (These are fractions with equal denominators.)
= 86 5- (So find the difference between the numerators.)
= 81 (Write the answer.)
Example 2: Raju ate 102 part out of 10
8 part of the fruit. What
is the remaining part of the fruit?
108
102- (Fractions with equal denominators.)
= 108 2- (So find the difference between the
numerators.)
= 106 (Write the answer.)
= 106
53=
3
5
(The fraction is reduced to its lowest form.)
∴ Remaining part of the fruit is 53 .
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Example 3: Nivedita has got ̀ 8 41 . She bought same bangles
by paying ` 5 42 . So what is the amount remaining with her?
8 41 5 4
2- (These are mixed fractions.)
433
422= - (They are converted into improper fractions.)
433 22= - (The difference of the numerators are found as
they are fractions with equal denominators.)
411= (This is an improper fraction)
411 2 4
3= = (It is converted into mixed fraction.)
∴ Remaining amount = ` 2 43 .
Try : 1) 159
157- 2) 24
182413- 3) 4 8
385- 4) 5 3
1 2 32-
Exercise 2.5I. Simplify .
A) 1) 73
72+ 2) 15
7154+ 3) 10
7102
103+ + 4) 12
5123
121+ +
B) 1) 4 32 2 3
1+ 2) 1 62 26
1+ 3) 3107
104+ 4) 212
5 1127+
II. Complete the following addition and subtraction square.
53
54 (a)
52
52 (b)
(c) (d) (e)
(+)
(-)
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III. SimplifyA) 1) 8
681- 2) 13
9132- 3) 20
18207- 4) 30
28305-
B) 1) 2 41
43- 2) 3 7
5 1 76- 3) 2 4
3 1 41- 4) 112
7126-
IV. Solve the following problems.A. 1) When a thread is cut into two parts, the threads of
lengths 84 m and 8
3 m were obtained. What is the total length of the thread?
2) Ravi ate 123 part of the fruit and Razia ate 12
5 part of the fruit. What is the total fruit they ate?.
3) The cost of one ball is ̀ 3 53 and the cost of a doll
is ` 2 53 . What is their total cost?
B. 1) There is 87 kilograms of sugar in a box. To
prepare the juice 83 kilograms of sugar was used.
So what is the remaining amount of sugar? 2) Sreyan has got ` 510
3 . In this he spent ` 2105 .
So what is the remaining amount?
2.6 (a) addition of fractions having different denominators
Example 1:
41
31
In this figure, we have two different parts. What is their sum ? Here 3
1 and 41 parts are there. They have
different denominators. So they are not equal parts. We don't know how to add them directly. But we know how to add two fractions with same denominators. So if we convert 3
141and into fractions with equal denominators we
will be able to add them. How to get fractions with equal denominators of 3
1 and 41 ?
Think.
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31
41+ (Find the LCM of denominators LCM of 3
and 4 =12.)
3 41 4
4 31 3= +
##
## (Multiply both numerator and denominator
of both the fractions by the LCM to get equivalent fractions.)
124
123= + (Now we have fractions
with equal denominators.)Briefly1) 3
141
(1 4) (1 3)
124 3
127
3 4
+
+
= + =#
# #
124 3= + (Add the
numerators.)
127= ∴ Total sum = 12
7=
Example 2: There is 43 kilograms of sugar, 2
1 kilogram of soji
and 54 kilograms of dhal in a bag. What is their total weight?
To find the total weight we have to find, the sum of those
fractions.
i.e., 43
21
54+ + (We have to convert them into
fractions with same denominators.)
So find the LCM of denominators 4, 2, 5. LCM of 4, 2, 5 = 20
4 53 5
2 101 10
5 44 4= + +
##
##
## (Form equivalent fractions so that
the denominator is 20.)
2015
2010
2016= + +
2015 10 16= + + (Add the numerators)
2041= (This is an improper fraction)
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2041 2 20
1= = (Convert it into mixed fraction)
∴ Total weight = 2 201 kilogram. Briefly
43
21
54
20(3 5) (1 10) (4 4)
2015 10 16
2041 2 20
1
+ +
= + +
= + + = =
# # #
Example 3 :
Sharanya travelled 1 81 kilometres from her place bicycle
and 2103 kilometres by bus and reached Taluk kendra. So
What is the distance of 'Taluk kendra' from her place?
1 81 210
3+ (These are mixed fractions)
89
1023= + (Convert them into improper fractions)
So construct equivalent fractions so that the denominators are 40.
8 59 5
10 423 4= +
##
## (The LCM of denominators 8 and 10=40)
4045
4092= + (These are fractions with same
denominators)Briefly1 81 210
3
89
1023
40(9 5) (23 4)
4045 92
40137 3 40
17
+
= +
+
= +
= =
# #
4045 92= + (Their numerators are added)
40137= (This is an improper fraction. )
3 4017= (Converted into mixed fraction)
∴ Distance to the Taluk Kendra is 3 4017 km.
try : 1) 6
581+ 2) 10
7154+ 3) 5
341
65+ + 4) 26
1 3 41+
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2.6 (b) subtraction of fractions having different denominators
Example 1: Sharada has ̀ 54 , out of this money she spent ̀ 10
3 to buy a rubber. So what is the remaining amount with her?
54 and 10
3 are the fractions with different denominators. If we convert them into fractions with the same denominators, subtraction can be done easily. Observe these steps.
54
103- These are fractions with different denominators.
So find the LCM of their denominators. LCM of 5, 10 = 10
Form equivalent fractions so that the denominator is 10.
5 24 2
10 13 1=
##
##- Briefly
54
103
10(4 2) (3 1)
108 3
105
21
21
2
1
-
=-
= -
=
=
# #
=Y
108
103-
108 3= - (Find the difference of the
numerators.)
105= (Write the answer in the lowest
form.)
21= ∴Remaining amount = ` 2
1
Example 2: A wire of length 261 metre is cut into two parts.
The length of one piece is 43 metre. What is the length of the
other piece?
261
43- (Here the first fraction is mixed fraction. So
convert this into improper fraction.)
613
43= -
In 613 and 4
3 the LCM of denominators of 6 and 4 is 12. (Convert 6
13 and 43 into equivalent fractions so
that the denominator is 12.)6 213 2
4 33 3= -
##
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1226
129= - Briefly
2 61
43
613
43
12(13 2) (3 3)
1226 9
1217 112
5
-
= -
=-
= -
= =
# #
1226 9= - (Find the difference of the
numerators)
1217= (This is an improper fraction.
Write the answer in mixed fraction)∴ The length of the remaining wire = 112
5 metre.
1125=
try : 1) 65
122- 2) 25
19105- 3) 2 8
3 1101-
Exercise 2.6I Simplify : 1. a) 6
531+ b) 12
11152+ c) 8
7125+ d) 15
142013
103+ +
2. a) 1 43
21+ b) 2 5
4 3101+ c) 46
183+ d) 2 3
1127+
II Complete this square of addition and subtraction.
54
43 a)
41
21 b)
c) d) e)
(+)
(-)
III Simplify.
1. a) 65
31- b) 8
762- c) 12
11107- d) 15
13155-
2. a) 1 43
81- b) 2 5
4 1103- c) 3 8
5 2103- d) 3 8
5 265-
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IV Solve the following problems.
1) The workers have cleared the weeds in the field.
Morning they cleared 85 part and in the afternoon they
cleared 61 part. What is the total part the weeds they
cleared from the field?
2) Raghu has filled water to 103 part of a tank and Rafiq
to 63 part of a tank. So what is the quantity of water
filled by both of them?
3) Sushrutha has bought a pen for ` 3107 and a book for
` 4 53 . What is their total cost?
V. Solve the following problems.
1) Teachers and children of a school jointly prepared a
big poster to be pasted to the wall. In that 3019 part is
prepared by the children. What is the part of the poster
prepared by the teachers ? (Hint : Here, consider one
full picture is equal to 3030 .)
2) Huvappa had Reserved 83 part of his income for his
children's education. In this 51 part was spent for the
education. What is the remaining amount?
3) A farmer has 3 85 acres of land. Out of this, he has grown
paddy in 2 43 acres and banana in the remaining part of
the land. So in how many acres of land did the farmer
grow banana?
l l l l l
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Unit - 3
DECiMALs
After learning this unit you can :• knowthemethodtoreadandwritedecimalnumbers,•writethedecimalnumbersontheplacevaluetableand
identifytheplacevalueofeachdigit,•locatethedecimalnumbersonthenumberline,•addandsubtractthedecimalnumbers,•solvetheproblemsrelatingtoadditionandsubtraction
of decimal numbers with respect to length, weightandmoney.
3.1 Decimal numberYouhavealreadylearntthemethodofreadingandwriting
decimalnumbers inyourpreviousclasses. In thisunit letus learnmore about addition and subtraction of decimalnumbers.
Observe this example.AfterReshmaandGopalbroughtgroceriesfromtheshop,
theirmotheraskedastohowmuchmoneyisremaining.Reshmasaid` 5and75paiseisremaining.Gopalwritesinachit̀ 5.75andshowed.Outofthesetwo,
whichiscorrect?(Botharecorrect)Gopalhasshowntheremainingamountindecimalform
byusingthedecimalpoint. 5.75 .
decimalpoint
In5.75,thepointrepresentsthedecimalpoint.
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3.1 tenthsExample1:-Observethisfigure.
Herehowmanyobjectsarecoloured?
2completeobjectsand103 partofthethirdobject
Totalcolouredpart=2+103 .Thus2completeobjects+3
Tenths(3by10).Thisiswrittenas2.3(twopointthree)byusing
decimal point. This is written in place value table.
Observe.• In2.3,2isinonesplace. • Wewritethetenthsplaceontherightside
Ones tenths
1 101
2 3=2.3
oftheonesplace.• Write3inthetenthsplace. • Thevalueofthetenthsplaceis10
1 oftheonesplace.
Example 2 : Whatisthelengthofthiscomb?
Observe: In the scaleeachunitof1cmlengthis divided into 10 equalparts. So the length ofeach smaller unit is 10
1
cm.
Thelengthofthecombismorethan6cmandlessthan7cm.Whenweobservethescale,wefindthatitslengthis6cm+10
7 cm. Thatmeans,6cm+7tenthsofacm.Wewritethisas6.7cm(sixpointseven).
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Sothelengthofthecombis6.7cm.Seehow6.7 iswritten in theaboveplace
Ones Tenths
1 101
6 7= 6.7
valuetable. 6iswritteninonesplace,7iswrittenintenthsplace.
Thenumberinthetenthsplaceiswrittenafterthedecimalpointas6.7(wereadassixpointseven).Example 3:Observethesetiles.
Totaltiles=1groupoftens+3ones+4partsoftenths10+3+10
4 =13+104 =13.4
In13.4,1isintensplace,3isintens Ones tenths
10 1 101
1 3 4onesplaceand4isintenthsplace.Thisiswrittenintheadjacentplacevaluetable.Observe.
Methodofwriting:13.4Methodofreading:Thirteenpointfour.
Know this: After ones place, before writing the number in the tenths place, we put the point and then write the number.
Observethefollowingnumberswrittenindecimals. 1) 5 10
6
5.6
+
=
2) 20 4 109
24 109
24.9
+ +
= +
=
3) 300 8 105
308 105
308.5
+ +
= +
=
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try : 1)Writethefollowingnumbersgiveninplacevaluetablebyusingdecimalpoint.
Hundreds100
Tens10
Ones1
Tenths
101
8 3
3 6 4
7 9 1
4 8 5 7
2)Writeanyfournumbersas given in the adjacentplace value table andwritetheminthedecimalform.3)Measure the length ofyour and your friend’spencil and write theirlengthinthedecimalform.
to denote decimal numbers on a number lineWealreadyknowtodenotefractionsonthenumberline.
Nowletusdenotedecimalnumbersonthenumberline.1) Denote 0.6 on the number line.
0.6isgreaterthan‘0’andlessthan1.0.6means6tenths.Sothedistancebetween0and1isdividedintotenequalparts.Thenmark0.6asshownbelow.
0 10.6 2 3
2) Locate 2.5 on the number line.2.5 isgreater than2and lessthan3.Thatmeansafter
2,5tenths.Sothedistancebetween2and3isdividedintotenequalpartsandmarkasshown.
0 1 2.52 3
try :1)Writeanyfourdecimalnumbersbetweenzero andoneandlocatethemonthenumberline. 2)Writeanytwodecimalnumbersbetween3and4 andlocatethemonthenumberline.
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to write the fractions in decimal form :
We have already learnt to write the fractions whosedenominatorsareteninthedecimalform.
Example :106 0.6= , 10
1 0.1,= 108 0.8,= 10
7 0.7=
Howtowrite1023 indecimalform?
1023
1020
103 2 10
3 2.3= + = + =
1023 2.3=`
Nowletuslearntowritethefollowingfractionsindecimalform.
a)53 b)5
7 c) 21 d) 2
9
Writeequivalentfractionssothatthedenominatorsofthegivenfractionsbecome10.Thenwriteindecimalform.
a) 53
5 23 2
106 0.6 5
3 0.6= = = =## `
b) 57
5 27 2
1014
1010
104 1 10
4 1.4 57 1.4= = = + = + = =
## `
c) 21
2 51 5
105 0.5 2
1 0.5= = = =## `
d) 29
29
55
1045
1040
105 4 10
5 4.5 29 4.5= = = + = + = =
## `
try :Writethefollowingfractionsindecimalform.
a) 25 b)5
3 c) 524 d) 2
17 e)1031
to write decimals in the form of fractions :Asyouwritefractionsindecimal,canyouwritedecimals
infraction?
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Observe the following examples.
a) 0.6→6Tenths∴106 =5
3
b) 2.4→2Completeobjects+4tenths∴2104 =2 5
2
c) 32.7→32Completeobjects+7tenths∴32107
d) 7.3→7103
try :Writeintheformoffraction. a)0.3 b)6.5 c)15.9 d)250.4
Activity:- Group the children in pairs. Let themexchangeoneeachoftheirpen,pencil,notebookandtextbook.Letthemcalculatethelengthofeachoftheseobjectstheygotandwritethelengthsinatableindecimalform.Letthemexchangethetableandobjectsandverify.
Exercise - 3.1
I.(1)Writethenumbersrepresentedbythefollowingfiguresinplacevaluetable.
a)
Tens Ones Tenths
b)
Hundreds Tens Ones
c)
TensOnes Tenths
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2)Writethefollowingdecimalnumbersinplacevaluetable.
a)0.7 b)2.8 c)26.7 d)165.4
3)Writethefollowingnumbersasdecimalsinnumerals.
a)fourtenths d)Sixhundredpointseven.
b)ninebyten e)Thirtytwotenths
c)Sixtysevenpointsix
4) Writethefollowingnumbersindecimalform.
a)3 107+ b)20 5 10
9+ + c)200 7 103+ +
d)600 40 102+ + e)20 10
1+
II.1)WritethedecimalnumberswhichrepresentthepointsA,B,CandDonthenumberline.
0 1 2 3
A= C=
B= D=
2) Identifythefollowingnumbersonanumberline. a)0.8 b)1.4 c)2.3 d)3.53) Writethefollowingdecimalsintheformoffractions. a)0.6 b)3.5 c)4.7 d)62.5 e)740.64)Writethefollowingfractionsindecimalform. a)10
7 b)1021 c)5
4 d) 511 e) 2
17
5) Writethefollowingincentimeters.(Hint:1cm=10mm) Example:32mm.=3.2cm. a)7mm b)27mm c)30mm d)4cm5mm e)6cm8mm
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3.2 Hundredths
Activity: Takeasquarecardboardanddrawlinessoastogettenequalparts.
Noweachpartis101 ofthetotalobject.
Inthis,whatisthecolouredpart?(101
means,1tenth,0.1).Now divide another complete squareinto10equalparts.Againdivideeachpartintotenequalpartsasshowninthefiguregivenbelow.Nowintotalhowmanysmallsquaresareobtained?(100)Each small square is 100
1 part of thecompleteobject.Notethat
1001 is10
1 partof10
1 . Usingdecimalpointwewritethisas0.01.Howtorepresent itontheplacevaluetable?Itiswrittenas0.01usingdecimal
point.Sowewritehundredthplaceontherightsideoftenthsplace.Howmanyhundredthsaretherein0.01?Only1.Sowewrite0.01intheplacevaluetableasinthetablebelow.
Ones Tenths Hundredths
1 101
1001
0 0 1
Wewrite1001 0.01= .
That means each smallsquareonthecardboardaboveis one hundredths. This iswritten as 0.01 by using the
decimalpoint.
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Observe these examples.Example 1:Ineveryfigure,tellthecolouredpartinfraction
andthenindecimalform.
a) Colouredpart=1005 .5hundredths=0.05
[Readingmethod:"Zeropointzerofive"]
Ones Tenths Hundredths0 0 5
b) Colouredpart=10012
12 hundredths = 10 hundredths+2hundredths=1tenth+2hundredths
Sowewritethisas10012 0.12= .Intheplace
valuetableitiswrittenas
Ones Tenths Hundredths0 1 2
Wereadthisaszeropointonetwo.
c) Colouredpart=10057 .
57 hundredths = 50 hundredths +
7hundredths=5tenths+7hundredths.
Sowewriteas10057 =0.57.Wewriteinthe
placevaluetableas
Ones Tenths Hundredths0 5 7
Wereaditaszeropointfiveseven.
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d)Colouredpart=10096 =96hundredths=0.96
90Hundredths+6Hundredths
9Tenths+6Hundredths.
10096 =0.96
(Wewritethisintheplacevaluetableas)Ones Tenths Hundredths
0 9 6
Wereadthisaszeropointninesix.
e)Colouredpart1 10015+
=1and15hundredths
=110015 1.15= [onepointonefive]
Ones Tenths Hundredths1 1 5
Example 2:Hereafewsmallcubesaregiven.Writethedecimalnumber
thatrepresentsthem.
Hundreds Tens Ones Tenths Hundredths
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Here1hundred+3tens+4ones+2tenths+6hundredthsarethere. Hundreds Tens Ones Tenths Hundredths
100 10 1 101
1001
1 3 4 2 6
Inthisnumber,wehave100+30+4+102 +
1006 .Thedecimal
formis134.26.Thewayofreadingisonehundredthirtyfourpointtwosix.Example 3 : Writethefollowingnumbergivenintheplacevaluetableasadecimalnumber.Ones Tenths Hundredths
1 101
1001
3 5 7
Here 3 ones + 5 tenths + 7hundredthsarethere.∴Number=3.57(Threepointfiveseven)
Example 4 :- Writethefollowingnumbergivenintheplacevaluetableasdecimalnumber.Tens Ones Tenths Hundredths
10 1 101
1001
5 4 3 7
101
1001
103
1007
5 10 4 1 3 7
50 4 54.37
+ + +
= + + + =
# # # #
In the place value table (among the numbers)as wemovefromlefttorighteveryplacewillhave10
1 ofitsvalueofleftsideplace.Example: Therightsideplaceof100thplaceistensplace.
Itsvalueis100of101 =10.
Therightsideplaceofthe10thplaceisonesplace.Itsvalueis10of10
1 =1.
Therightsideplaceofonesplaceistenthsplace.Itsvalueis1of10
1 =101 .
Therightsideplaceoftenthsplaceishundredthsplace.Itsvalueis10
1 of101 =100
1 .
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try :1)Writethenumbersintheplacevaluetablebyusingdecimalpoint.
Hundreds Tens Ones Tenths Hundredths100 10 1 10
11001
a) 7 4 2b) 6 5 0 3c) 2 4 2 7 9d) 7 6 4 3
2)Readthefollowingnumbersandwriteinwords
a) 9.73 b) 16.49 c) 3.05 d) 245.43to write the fractions in the form of decimals :Example 1: Write the following fractions in decimal form. a)5
3 b) 43 c)50
7 d)10005
a) 53
5 23 2
106 0.6= = =
##
b) How towrite 43 indecimalnumber form?There isno
completemultiplewhich can convert 4 to 10. Sowriteequivalentfractionsothatthedenominatorof 4
3 shouldbecome100. 4 25
3 2510075 75= =
## Hundredths=0.75
c) Towrite 507 asdecimal,writeequivalentfractionsothat
thedenominatorof507 shouldbecome100.50
750 27 2
10014= =
##
=14hundredths=0.14.
thousandth : Observe1000
5 .Inthefraction,ifthedenominatoris10,itistenths.Ifthedenominatoris100,itishundredths.Ifthedenominatoris1000,..................?Ifthedenominatoris1000,itisthousandths.
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Nowletuswrite10005 inplacevaluetable.Thatmeansin
theplacevaluetableafterhundredthplacetotherightside
oneplaceshouldbeincreased.
Thatplacevalueisequalto101 ofhundredthsplace
Thatmeans101 of100
1 =1001 ×10
1 =10001
Sotherightsideplaceafterhundredthsplace iscalledthousandthplace.
10005 →
Tens Ones Tenths Hundredths Thousandths
10 1 101 100
110001
0 0 0 510005 =0.005
Example 1: Write the following fractions in decimal form.
a)510038 b)425100
27 c)401000375
solution :
a)510038 5.38= (Fivepointthreeeight)
b)42510027 425.27= (Fourhundredandtwentyfivepointtwoseven]
c)401000375 40.375= (Fortypointthreesevenfive)
Example 2 :Write this fraction in decimal form: 81000725
8 107
1002
10005
8 1000725
8.725
= + + +
= +
=
107
1002
10005
1000700 20 5
1000725
+ +
= + +
=
p
r
qqqqqqqqq
t
v
uuuuuuuuu
81000725
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to write decimal numbers in fractional form Examples :Writethefollowinginfractionalform.a)0.05 b)3.54 c)2.430
Solution:a)0.05 1005= Lowestform100
555
201 0.05 20
1= ='
' `
b)3.54 3 10054
3.54 3 5027 3 50
27
= +
= + =
100 254 2
5027=
''c m
c)2.430 2 1000430
210043
= +
=
1000 10430 10
10043=
''` j
try :1.Writethefollowingnumbersindecimalform.
a)310025 b)2100
3 c)15 1000347+ d)75 10
51004
10002+ + +
2.Writethefollowinginplacevaluetable.
a)5.43 b)26.275 c)16.34 d)8.564
3.Writethefollowingdecimalinfractionalform.
a)0.75 b)2.56 c)32.45 d)5.75
Exercise 3.2
I. Writethenumbersrepresentedbythesefiguresinplacevaluetable.
a)
Ones Tenths Hundredths
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b)
Tenths Hundredthsc)
Ones Hundredths
II. Writethenumbersintheplacevaluetableindecimalform.Hundreds Tens Ones Tenths Hundredths Thousandths
100 10 1 101
1001
10001
5 4 2 0
1 8 0 5 4
2 3 2 7 9
1 0 7 3 5 0
III.1)Write the following decimal numbers in place valuetable.
a)0.35 b)2.43 c)25.027 d)256.49 e)8.756
2)Writethefollowingasdecimalnumbers.
a)310052 b)56100
9 c)251000754 d)181000
54 e)62810007
3)Writethefollowinginthelowestformoffractions.
a)0.50 b)0.450 c)0.85 d)0.124 e)2.550
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3.3 Usage of decimals3.4(a) Money :Example:Convert these to rupees in decimal form.
a)75paise b)5paise c)145paise d)` 2and8paise
Solution:Weknowthat`1=100paise.
So1paise =` 1001 or ` 0.01
a)75paise =`10075 = `0.75
b)5paise =`1005 = `0.05
c)145paise=`100145 = ` 1100
145 =`1.45
d)`2,8paise=`21008 = `2.08
try : (1)Write` 3and47paiseand` 3and7paiseas decimalnumbersinrupees.
(2)Write` 8and6paiseand` 8and26paiseas decimalnumbersinrupees.
3.3(b) Measurement of length : Millimetre (mm) and Centimetre (cm)
Weknowthat1cm=10mm.1mm=101 cm.Observe
thatmmhasbeenwrittenincmindecimal.6mm
106 cm
0.6 cm
=
=
28mm
1028 cm
2108 cm
2.8 cm
=
=
=
16mm
1016 cm
1106 cm
1.6 cm
=
=
=
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Metre and Centimetre
Weknowthat1m=100cmand1cm=1001 mObserve
the followingexamples.Knowthemethod toconvert them
intodecimalform.140 cm
100140 m
110040 m
1.40m
=
=
=
205 cm
100205 m
21005 m
2.05 m
=
=
=
6m8 cm61008 m
6.08m
=
=
Kilometre (Km) and metre
1km =1000m.So1m=10001 km.Observethefollowing
examplesindecimalform.328m
1000328 km
0.328 km
=
=
28m
100028 km
0.028 km
=
=
2 km75m2100075 km
2.075 km
=
=
try : 1)Writethefollowingincmasdecimalnumber. a)56mm b)5mm c)25mm
2)Writethefollowinginmetresasdecimal number.
a)64cm b)135cm c)205cm 3)Writeinkilometresasdecimalnumber. a)475m b)3475m c)5km254m
3.3(c) Measurement of weight :
Weknowthat1kg=1000grams.∴1gram=10001 kg.
Basedonthisobservationthatthemeasurementofweight
iswrittenintheformofdecimals.
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750 grams
1000750 kg
0.750 kg
=
=
3625 grams
10003625 kg
31000625 kg
3.625 kg
=
=
=
5025 kg
10005025 kg
5100025 kg
5.025 kg
=
=
=
try :Writetheseinkgasdecimalnumbers. (a) 658g (b)85g (c)6018g
3.3 (d) Measurement of Liquid :Weknowthat1Litre=1000ml
∴1ml=10001 l.
Basedonthisobservethatthemeasurementofliquidsinlitresarewrittenindecimalform.
320m
1000320
0.320
l
l
l
=
=
25m
100025
0.025
l
l
l
=
=
3 574m3 1000
574
3.574
l l
l
l
= +
=
try :Writethefollowinginlitres(asdecimalnumber)(a)684ml (b)84ml (c)4ml(d)2l75ml (e)3l5ml
3.3(e) time :Thoughtimecanbeexpressedindecimalform,likeother
measurementsinourdailylifewerarelywritetimebyusingdecimalpoint.Observethefollowingexample.
1hour=60minutes.
6045 h
6045
43
2525
10075 0.75 h
4
3= =
=
= =
#
6030 h
6030
21
21 h
21
55
105 0.5 h
= = =
= = =#
30minutes 45minutes
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Exercise - 3.3
I. Solvetheproblems.
1) Writethefollowinginrupees(indecimalform).
a)50paise b)5paise c)40paise
d)`4,80paise e)325paise
2) Writethefollowingincentimetres.(indecimalform)
a)8mm b)25mm c)7cm5mm
d)84mm e)175mm
3) Writeinmetres.(indecimalform)
a)60cm b)6cm c)2m30cm
d)2m3cm e)378cm
4) WriteinKilometres.(indecimalform)
a)876m b)76m c)6m
d)2km68m e)3005m
5) WriteinKg.(indecimalform)
a)763gram b)63gram c)3gram
d)3kg54g e)2825g
6) Writeinlitres.(indecimalform)
a)675ml b)75ml c)5ml
d)2l38ml e)5l40ml
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3.4 Comparison of decimals You have learnt in your previous classes about the
methodofcomparingwholenumbers.Nowletuslearnaboutcomparingthedecimalnumbers.
Activity:- Between0.1and0.07,which is greater ?• Taketwosquarepapersofthesamemeasurement.• Make100 equal parts in each square as shown in
thefigure.• In the first square, colour 0.1 squares and in the
second0.07squares.
0.1 101
10010= =
0.07 1007=
Observe the squareswhichyouhavecoloured.
Tellwhichnumberisgreater?Inthese,0.1isgreaterthan0.07ie,0.1>0.07.
Example 1:Compare 5.4 and 5.43. Which is greater among them?
Compareinordereachdigitofthegivennumberbystartingfromthegreatestplace.Between5.4and5.43.
• Thevalueofthedigitsinonesplaceareequal.
• Thevalueofthedigitsintenthsplacearealsoequal.
• Sonextwehavetocompareandseethehundredthsplace.
• Then,5.4=5+104
1000+ 5.43=5+10
41003+
Inhundredthsplaceof5.4thereiszeroandin5.43thereis'3'.Now3>0.So,5.43isgreaterthan5.4.∴5.43>5.4
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Example 2 : Which is greater ?a)2or0.99 b)3.094or3.5Solution:2or0.99(a)In2,thereare2completeobjects In 0.99, there is nocomplete object. So 2 isgreaterthan0.99.∴2>0.99
(b)3.094or3.5• In these numbers, thedigits in ones place areequal.(3=3)• Among the digits in thetenths place, 5 is greaterthan0,So3.5isgreaterthan3.094.∴3.5>3.094
Observe another method of comparing decimal numbers.
3.094 3 100
1009
10004= + + +
3.5 3 105
1000
10000= + + +
Whenwecomparethevalueofeachdigitfromlefttoright,thegreatervaluecanbeknown.∴3.5>3.094try:Usethecorrectsigninbetweenthesenumbers(>or<) a)3.7 3.07 b)4.35 4.53 c)5 0.624
d)3 3.002 e)4.603 4.630
Exercise 3.4 I. Comparethefollowingandwriteusingcorrectsign(>
or<).a)8.2 8.5 b)3.715 3.157 c)0.43 0.4
d)2.5 5.473 e)5.075 5.7
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II. Tellwhichissmallerinthefollowing. a) 8.3cmand8.2cm b)0.85mand0.89m c) 2.047kmand2.74km d)6.509kgand6.9kg e) 3.5land3.425l
3.5 (a) Addition of decimal numbersObserve these examples.1) Asrafboughtapenfor` 5.75andapencilfor`3.5.What
isthetotalvalueofthosetwoobjects?5.75+3.5=?
Thesenumbersarewritteninplacevaluetable.Observe.Asintheadditionofwholenumbersgoonaddingthenumberineachplacevalue.
Ones tenths Hundredths53
75
5
9 2 5
Priceof1pen =` 5.75
Priceof1pencil=` 3.50 Total=` 9.25
∴Theirtotalvalue=`9.25.2) Ramappagoestohisfieldswhichareindifferentdirections
fromhisplacewalking.Hewalks5.74kmonMonday,8.268kmonTuesdayand6.050kmonWednesday.Whatisthetotaldistancehehaswalkedinthosethreedays?
tens Ones tenths Hundredths thousandths586
720
465
080
2 0 0 5 8 DistancemovedonMonday =5.740km DistancemovedonTuesday =8.268km DistancemovedonWednesday =6.050km Total=20.058km ∴Totaldistancehehasmoved=20.058km
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3)Inabagtherearepacketsof4.25kgofsugar,0.750kgofdaland8.5kgofrice.Whatisthetotalweightofprovisioninthatbag?
Tens Ones Tenths Hundredths Thousandths
4
8
275
550
000
1 3 5 0 0
Sugar = 4.250 kgDal = 0.750 kgRice = 8.500 kgTotalweight=13.500 kg
∴ Totalweightoftheprovisionis13.500kg.
(4) Kamalahasacowinherhouse.Afterusingthemilkforherhouse,shesendstheremainingmilktothemilk(dairy)kendra.Duringthefirstweekshesends20.75litres,secondweek18.4litres,thirdweek22.850litresandfourthweek20litres.Whatisthetotalquantityofmilkthatshesendstothemilkdiaryduringthatmonth?
1stweek = 20.750 litres• All the numbers arewritten in equal decimalplacesandthentheyareadded.
2ndweek = 18.400 litres
3rdweek = 22.850 litres
4thweek = 20.000 litres
Total = 82.000 litres
∴Totalmilksenttothedairyis82litres.try :Findthesumofthefollowing. 1)7.3+6.25+3.43 2)5.325+2.45+1.6
3)36.40+8.23+3.452 4)6.5+0.9+0.07
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3.5(b) subtraction of decimal numbersObserve the following examples.Example 1 : Thereis8.46mlengthofribboninaribbonroll.Inthatalengthof2.35miscutandtakenout.Whatisthelengthoftheribbonremaininginthatroll?•Asthesubtractionofwholenumbersisdone,thissubtractionisalsodone.
Ones Tenths Hundredths82
43
65
6 1 1Theribbonintheroll =8.46mCutandtakenoutribbon=2.35mRemainingribbon =6.11m
∴Remainingribbon=6.11metres.
Example 2 : Sharvari broughthome3.5 l ofmilk. In this1.76lofmilkwasspent.Whatistheremainingquantityofmilk?
Ones Tenths Hundredths Thousandths31
57
06
00
1 7 4 0
Milkwhichwasbrought = 3.500lMilkwhichwasspent = 1.760lRemainingmilk = 1.740l
Example 3 : Josephisgoingtohisofficewhichisat16.7km.Alreadyhehascoveredadistanceof12.525km.Whatistheremainingdistancehehastocover?
Distancetotheoffice=16.7 0 0 km6 9 10Y Y Y
Distancealreadycovered=12.525km Distancetobecovered =04.175km
• Asinwholenumberssubtraction,10hasbeenborrowedfromthenextnumberandsubtractionisdone.
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try :1)Subtract
a)8.54−3.13 b)20.3−12.28
c)36.75−15.293 d)8.605−3.462)a)Subtract3.75from8.64 b)Subtract0.59from2.3
Exercise 3.5
I A.Findthesumofthefollowing. 1) 0.754+0.23+0.5 2)5.6+3.75+4.321
3) 0.65+5.437+2.5 4)5.4+3.279+6.3+4
5) 8.753+26.49+156.7
B.Solvethefollowing.
1) Nidhi purchased blue coloured ribbon of length3.5mandwhitecolouredribbonoflength2.7mforherexercisedisplay.Whatisthetotallengthoftheribbonshehaspurchased?
2) Johnson'smothergavehim`35.5,hisfather`42.75andgrandfather`60.20.WhatisthetotalamountJohnsongotfromallofthem?
3) Amoghapurchased2kg300gofapple,3kg250gofbananaand1kg500gofmangofruits.Whatisthetotalweightoffruitsshehaspurchased?
4) Mohammedtravelledfromhisplace2.570kmincycle,8.43kminbusand1.3kminrickshawandreachedtheTalukoffice.Whatisthetotaldistancehehastravelled?
5) InRajesh'shouse,onafestivaldaythemilkspentduringmorning,afternoonandeveningwere2l350ml,1l250mland2l25mlrespectively.Whatwasthetotalquantityofmilkspentonthatday?
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II A.Subtract
1) 0.43from0.75 2)0.95from2.57
3) 2.34from5.2 4)28.7from36.45
5) 3.738from5.954 6)1.467from3.05
B.Answerthefollowing.
1)Desouzahad`35.40.Inthat`21.35wasspent.Whatistheamountremainingwithhim?
2)Inabundletherewas18.76mofcloth.Inthatsomeclothwassoldout.Theninthebundle12.90mofclothwasremaining.Thenhowmuchclothwassoldout?
3)Therewas10m.50cmlengthofwireinaelectriccoilofwire.Inthat2m78cmlengthofwirewascutandtakenout.Sowhatistheremaininglengthofwireinthecoil?
4)Inthehousetherewas5.720kgofsugar.Inthat2.570kgofsugarwasusedtopreparesweets.Thenwhatistheremainingquantityofsugar?
5)Ina can therewas3l 452ml ofhoney.After selling1l750mlofhoneywasleft.Sohowmuchhoneywassold?
6)In a city themaximum temperature of the day is30.40celsiusandminimumtemperatureis26.80celsius.Whatisthedifferenceintemperatureonthatday?
l l l l l
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Unit - 4intRODUCtiOn tO ALGEBRA
After Learning this unit you can : recognise the variables, write the expression with variable, identify algebraic expressions,convert the statement into an algebraic expression and
algebraic expression into a statement.4.1 introduction
You have already learnt about numbers, their properties and operations on them. The branch of mathematics that deals with numbers is Arithmetic. You also learnt about figures in two and three dimensions and their properties. The branch of mathematics that deals with shapes is Geometry.
Now we study another branch of mathematics called Algebra.
Beginning of AlgebraAround 300 BC, use of letters to denote unknowns and forming
expressions from them was quite common in India. Many great Indian Mathematicians Aryabhata (476 AD), Brahmagupta (598 AD), Mahavira (who lived around 850 AD) and Bhaskara II (born 1114 AD)and others contributed a lot to the study of Algebra. They gave names such as Beeja, Varna etc to unknowns (variables). The Indian name for Algebra is Beejaganit.
The word Algebra originated from the word 'AL-JABR' from the book titled “iLM – AL- JABR –’ALMUGABALAH” written by an Arabian Mathematician named AL-KHOORiZMi. In 825 AD.
Algebra is a branch of mathematics in which principles of Arithmetic are generalized by using letters as symbols for representing numbers.
Let us begin our study with simple examples.
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ice Candy Stick patterns :
Example 1: Rashmi and Ramya would like to make some patterns with the ice candy sticks as shown in the figure.
To make one t how many ice candy sticks are used? (Two)To make two ts how many ice candy sticks are used? (Four)They prepared a table as shown below.
Number of Ts formed 1 2 3 4 5 6 ...
Number of ice candy sticks required
2 4 6 8 10 12 ...
2×1 2×2 2×3 2×4 2×5 2×6 ...
Observe the table carefully What is the relation between number of ts formed and the
number of ice candy sticks required? The number of ice candy sticks required is twice the
number of ts formed. That is the number of ice candy sticks required = 2 × Number of ts.
For convenience, let us consider the letter ‘n’ for the number of ts.
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If one t is made, then n = 1. If two ts are made, then n=2 and so on; Here n takes the values 1, 2, 3, 4, 5…..
Then we write the number of ice candy sticks required = 2 × n and write it as 2n.
Note : 2n is same as 2 × nThus, from the above table, you found a rule that the
number of ice candy sticks required for making any number 'n' of Ts is 2n.Example 2 : Rashmi and Ramya made the following triangular patterns as shown in the figure. How many ice candy sticks are used to make one triangle? (Three ice candy sticks) How many ice candy sticks are used to make two triangles? (Six ice candy sticks) They prepared a table as given below for the above triangular pattern. But some entries in the table are left blank. Complete them.
Number of triangles 1 2 3 5 6 .......
Number of ice candy sticks used
3 6 12 18 .......3×1 3×2 3×3 3×4 .......
What is the relation between number of triangles and number of ice candy sticks?
The number of ice candy sticks required is thrice the number of triangles formed.
That is number of ice candy sticks required = 3 × number of triangles.
Let 'y' be the number of triangles.Then number of ice candy sticks required =3 × y. This is
written as 3y. Here ‘y’ takes values 1,2,3,4, 5…..Note : 3y is same as 3 × y
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In example 1, ‘n’ does not have a fixed value but can take any value among 1,2,3,4,5...... and similarly in example 2. ‘y’ also does not have a fixed value but can take any value among 1,2,3,4,5...... So we call ‘n’ and ‘y’ as variables.
The word ‘variable’ means something that can vary i.e. change. The value of variable is not fixed. It can take different values.
In practice the variables (unknown quantities) are denoted by English small letters like a, b, c, ….n, ….x, y, z. Thus the quantities expressed by the English alphabet are called ‘Literal numbers’ or ‘Algebraic variables’.
The literal numbers used for representing the unknown quantities are called ‘variables’.Example 3 : Rakesh and Rohan who were carefully observing the patterns made by Rashmi and Ramya also made the following pattern.
They make one ‘W’ using 4 ice candy sticks. Now write the rule for the number of ice candy sticks
required for making patterns of ‘W’.More examples of variables :
Book Store
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Example 1 : Some students went to buy notebooks from the school bookstore. Price of one note book is ̀ 8. Chandru wants to buy 6 notebooks, Suraj wants to buy 7 notebooks, Meera wants to buy 5 note books and so on. How much money should student carry when he or she goes to the book store to buy notebooks? This will depend on the number of notebooks that the student wants to buy. The students work together to prepare a table.
Number of notebooks required
1 2 3 4 5 6 7 ... m
Total cost in ` 8 16 24 32 40 48 56 ... 8m
The letter ‘m’ stands for the number of notebooks astudent wants to buy; ‘m’ is a variable which can take any value 1,2,3,4,5,……The total cost of ‘m’ notebooks is given by the rule.
The total cost of ‘m’ note books (in `) = 8 × number of notebooks required = ` 8 m.
If Chandru wants to buy 6 notebooks then taking m= 6, we say that Chandru should carry ` 8 × 6 or ` 48 with him to the school book store. Now you find out how much money should Suraj and Meera carry?
Let us take another example. Example 2 : For the Independence Day celebration in the school, children are going to perform mass drill in the presence of the chief guest. They stand 15 in every row .How many children are there in the drill?
The number of children will depend upon the number of rows. If there is one row there will be 15 children, If there are two rows there will be 15 × 2 = 30 children and so on. If there are ‘n’ rows there will be 15 × n which we write as 15n. children in the drill; Here ‘n’ is a variable which stands for the number of rows and so ‘n’ takes on values 1,2,3,4,5,6……
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Example 3:
Ramanna thought of planting coconut saplings in his field. If he plants 10 saplings in each row, the number of saplings will depend upon the number of rows. If there is one row he can plant 10 saplings, If there are two rows he can plant 10×2 or 20 saplings and so on. If there are ‘t’ rows he can plant 10 ‘t’ saplings. Here ‘t’ is a variable which stands for the number of rows and ‘t’ it takes values 1, 2, 3, 4, 5,......etc.
there can be different situations as well in which numbers are added to or subtracted from the variable as
seen below.Example 4 : Nancy and Joycy are sisters. Joycy is younger to Nancy by 4 years. When Nancy is 12 years old, Joycy is 8 years old. When Nancy is 17 years old Joycy is 13 years old. We do not know Nancy’s age exactly. Let ‘p’ denote Nancy’s age in years, then ‘p’ is a variable. If Nancy’s age in years is ‘p’ then Joycy’s age in years is p-4. The expression p-4 is read as ‘p minus four’. As you would expect when ‘p’ is 12, (p-4) is 8 and when ‘p’ is 17, (p-4) is 13. Example 5 : Rani and Sonali have the hobby of collecting stamps. Rani says that she has collected 5 more stamps in her collection than Sonali. If Sonali has 25 stamps then Rani has 30. If Sonali has 30 stamps then Rani will have 35 and so on. We do not know exactly how many stamps Sonali has.
But we know that, number of stamps with Rani and number of stamps with Sonali is equal + 5.
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We shall denote number of stamps with Sonali’s stamps by the letter ‘k’. Here ‘k’ is a variable which can take any value 1,2,3,4,…..etc. Using ‘k’ we write number of stamps with Rani = k+5. The expression (k+5) is read as ‘k plus five’. It means 5 added to k. If ‘k’ is 25 then (k+5) is 30. If ‘k’ is 30 then (k+5) is 35 and so on.
Know this : The expression k+5 cannot be simplified further. Do not confuse k+5 with 5k, they are different. In 5k ‘k’ is multiplied by 5. In (k+5), 5 is added to ‘k’. We may check this for some values of ‘k’. For example if k=2, 5k = 5 × 2 = 10 and k + 5 = 2 + 5 = 7.If k = 10 , 10k = 10×10=100 and k + 10 =10+10=20.
Exercise 4.1I. Find the rule which gives the number of match sticks
required to make the following patterns or shapes. Use a variable to write the rule.
1. A pattern of letter V as
2. A pattern of letter C as
3. A pattern of letter M as
4. A pattern of letter E as
5. A pattern of letter L as
6. A pattern of letter Z as
7. A pattern of letter S as
8. A pattern of letter R as
9 A pattern of the shape ⌂10. A pattern of the shape
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II. Answer the following.1) Make a list of letters which have the same rule when they
are arranged using match sticks. Eg: The letters T,V,L etc give us the rule 2n 2) N.C.C. cadets are marching in a parade with 6 cadets in
each row. What is the rule which gives the number of cadets, given the number of rows ? ( use ‘n’ for the number of rows)
3) The teacher distributes 4 chocolates per student. How many chocolates are needed if there are 's' number of students ?
4) Rashmi’s mother has made certain number of laddus. After she gave some laddus to her friends and family members, still 8 laddus remain. If she distributes 'x' number of laddus, how many laddus did she make?
5) Rohan is Kumar’s younger brother. Rohan is 3 years younger to Kumar. Write Rohan’s age in terms of Kumar’s age (Let Kumar’s age be ‘l’ years)
6) If there are 75 mangoes in a box, how will you write the total number of mangoes in terms of number of boxes? (use ‘b’ for the number of boxes)
7) There are a certain number of cement bags in a godown. Some cement bags were loaded to a lorry. If 20 cement bags are remaining after loading, find the total number of cement bags? (Use ‘c’ for the number of cement bags)
8) There are 25 rows in a two wheeler parking lot. How many vehicles can be parked if the number of vehicles in each row is ‘v’ ?
9) Give an example for each of the following rule. 1) 4m 2) d+6 3) r-7
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4.2 Use of variables in common formulaePerimeter of a square:
We know that perimeter of any polygon (A closed figure made up of three or more line segments) is the sum of the lengths of its sides. A square has 4 sides and they are equal in length.
l
l
l
l Therefore the perimeter of a square= sum of the lengths of the sides of the square = 4 times the length of a side of the square = 4 × l. (Here 'l' is the length of the side of the square). ∴ the perimeter of a square = 4l
Thus we get the rule for the perimeter of a square. The use of the variable ‘l’ allows us to write the general rule in a way that is concise and easy to remember.Perimeter of a rectangle: A rectangle has four sides.
Pb b
l
lS R
Q In the adjoining figure PQRS is a rectangle, the four sides are PQ, QR, RS, and SP. The opposite sides of any rectangle are always equal in length. In rectangle
PQRS let us denote length by ‘l’ and the breadth by ‘b’ then Perimeter of a rectangle
= Length of PQ+ Length of QR+ Length of RS + Length of SP= 2 × length of PQ + 2 × breadth of QR= 2 × l + 2 × b = 2l + 2bThe formula for the perimeter of a rectangle = 2l + 2b (where ‘l’ and ‘b’ are length and breadth of the rectangle
respectively).note: Here both ‘l’ and ‘b’ are variables and they take values independent of each other that is the value of one variable does not depend on other.
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Exercise 4.2
I. Answer the following:
1) The side of an equilateral triangle is shown by ‘a’. Express the perimeter of the equilateral triangle using ‘a’.
2) The side of a regular pentagon is denoted by ‘a’. Express the perimeter of the pentagon using ‘a’. (Hint: A regular pentagon has all its 5 sides equal in length).
a
a
a
a
a
3) A cube is a three dimensional figure as shown in the figure. It has six faces and all of them are identical squares. If the length of an edge of the cube is denoted by ‘a’, find the formula for the total length of the edges of the cube.
aa
a
4) The diameter of a circle is a line which joins two points on the circle and passes through the centre of the circle. In the figure. CD is the diameter of a circle and ‘O’ is its centre. Express the diameter of the circle d in terms of its radius r.
E
C Dor r
4.3 Expressions with variables
In Arithmetic we have come across expressions like (2×8) + 4, 10-(5-3), 3 + (4×3) etc. These expressions are formed from numbers. Expressions can be formed from variables too. In fact we have already seen expressions with variables for example 2n, 3y, x +10 etc.
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An expression made up of variables and numerals connected by arithmetic operations is called an algebraic expression.
These expressions with variables are obtained by operations of addition, subtraction, multiplication and division on variables. For example the expression 2n is formed by multiplying the variable ‘n’ by 2, the expression (x+10) is formed by adding 10 to the variable ‘x’.
We know that variables can take different values as they have no fixed value. But the value they take are numbers. That is why in the case of numbers operations of addition, subtraction, multiplication and division can be done on them.
One important point must be noted regarding the expressions containing variables. The value of number expression like (2×3)+4 can be immediately found as (2×3)+4=6+4=10. But the value of an expression like 2x + 3 which contains the variable ‘x’ cannot be determined unless the value of x is given.
Activity :- Take one red and one green box. Put some marbles into them. Let the number of marbles in red box be 'x'. and in green box be 'y'. Ask your friend to pick up 3 marbles from red box and ask him, to tell the number of remaining marbles in red box, then ask him to put one marble into green box and now ask him to tell the number of marbles in the green box.
Continue the above process of adding and removing of marbles to both the boxes and list the results in the table given below.
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Colour of the box
Number of marbles in
the box
No. of marbles
removed from the box
No. of marbles added to the box
No. of marbles
remaining in the box
Red x 3 x - 3
Green y - 1 y + 1
Red x 2 x + 2
Green y 4 - y - 4
Red x 10 - x - 10
Green y - 8 y + 8
the Following are a few examples of the statement and their expression.
Statement Expression1) 5 is added to x x + 52) 7 is subtracted from k k - 7 3) 'y' is divided by 7 7
y 4) 'q' is multiplied by -3 -3q5) 'y' is multiplied by 2 and then 5 is subtracted from the product 2y – 5
Activity :- Write ten more such simple expressions.A few examples of the expression and their statements. Expressions Statements
1) x - 12 12 is subtracted from x.2) m + 25 25 is added to m.3) 14p p is multiplied by 14.4)
2y y is divided by 2.
5) -9z z is multiplied by -9.6) 10r + 7 r is multiplied by 10 and then 7 is added to the product.7) 2a -1 a is multiplied by 2 and 1 is subtracted from the product.
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Know this :- Fundamental operations of variables.1. Addition : Let ‘x’ and ‘y’ be two variables. Then their
sum is written as x + y or y + x.2. Subtraction : Let ‘x’ and ‘y’ be two variables then their
difference is written as x - y or y - x. Note that x - y ≠ y - x (when x ≠ y).3. Multiplication : Let ‘x’ and ‘y’ be two variables, then
their multiplication or product is written as xy or x × y, Also x.y = y.x.
4. Division : Let ‘x’ and ‘y’ be two variables, then their division is written as x ÷ y or x
y where y ≠ o.
Note that xy x
y! .
Exercise 4.3
I. Which out of the following are expressions with numbers only?
a) x + 3 b) (3 × 25) − 10 y c) 16 − 2 d) 9p
e) 8 − 8r f) (15 × 20) − (6 × 8) − 25 + n g) 6(23 − 5) + 8 × 3
II. Write the expressions for the following statements.a) 5 is added to y.b) 7 is subtracted from p.c) x is multiplied by 5.
d) y is divided by 8.
e) 6 is subtracted from –m.f) -z is multiplied by 5.
g) -z is divided by 7.
h) m is divided by -10.
i) r is multiplied by -15.
j) y is multiplied by 5 and 3 is added.
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k) y is multiplied by 5 and 3 is subtracted.l) p is multiplied by -8 and 6 is added to the product.m) p is multiplied by 4 and 15 is subtracted from the
product.n) p is multiplied by -4 and 15 is added to the product.o) If Ramya’s present age is ‘y’ years, then answer the
following.1) What will be her age after 4 years?2) What was her age 5 years back?3) Age of Ramya’s grandfather is 6 times her age . What is the age of her grandfather?4) Ramya's grandmother is 2 years younger to her grand father. What is her grandmother’s age?5) Ramya’s father’s age is 5 years more than 3 times Ramya’s age. What is her father’s age?
III. Write the following expressions in words.
1) x + 2 2) 8a − 3 3) 41y +
4) 7p + 3 5) (10m + n)y
Activity :- Ask your friend to think of a number in mind and follow the steps given below.
1) Multiply it by 2.
2) Add 4 to the product obtained.
3) Multiply the sum by 5
4) Subtract 20 from the product.
5) Divide the resulting number by 10
The number obtained is the number kept in mind by your friend.
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Example : Let 'y' be the number kept in mind of your friend. 1) When 'y' is multiplied by 2, product is 2y 2) When 4 is added to 2y, sum obtained is 2y + 43) When this sum is multiplied by 5, the product is equal to
(2y + 4) 5 = 10y + 20.4) When 20 is subtracted from this product, we get10y+20 - 20=10y
5) When 10 y is divided by 10, 1010y y=
y is the number kept in your friend's mind. Let 8 be the number kept in your friend's mind and then
the above steps are followed we get.1) 2 × 8 = 16
2) 16 + 4 = 20
3) 20 × 5 = 100
4) 100 − 20 = 80
5) 1080 8=
It is true for all values of 'y'.
Remember :• The letters of the English alphabet use in algebraic
expressions are called literal numbers.• Quantities that can change are called variables • Algebra is a branch of mathematics that involves
letters of the alphabet and numbers. • Using the basic operations of arithmetic in algebra we
obtain, sum x + y, Difference x − y, Product xy and Quotient y
x
l l l l l
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UNIT - 5
RaTIo aNd PRoPoRTIoN
after Learning this unit you can : understand the meaning of ratio, express the given quantities in ratio, recognise the terms of ratio, understand the concept of proportion, recognise the terms of proportion, write the proportion using symbol. solve the problems by using proportionality law. understand the scope of ratio and proportion in daily
life.
5.1 RaTIoIn our daily life, we come across many situations involving
comparison of two things in respect of their magnitudes using numbers. For example, number of boys and girls in a class, number of different types of toys, price of items etc, In these situations we use ratio for comparing quantities conveniently. Then what is a ratio? To understand this, observe the following example.Example 1: Vijay has 8 toys out of which 5 are dolls and 3 are balls. What is the ratio of dolls to balls?
Observe the figure. How many dolls are there? There are 5 dolls.How many balls are there?There are 3 balls.How do you compare the number of dolls to number of balls?
dolls
balls
It is compared as number of dolls to number of balls is 35 .
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35 is also written as 5:3.
5:3 is a ratio and it is read as ‘5 is to 3’.Example 2: John wants to compare the number of boys in his class to number of girls. There are 40 students in his class. Among them 24 are girls. How does he compare the number of boys to number of girls?Solution :- To find the ratio of the above, you need to know the number of boys and girls in the class.
How many students are there in the class? There are 40.How many girls are there in the class? There are 24.How many boys are there in the class?(No. of boys = 40-24). There are 16 boys.Then what is the ratio of number of boys to the
number of girls? The ratio of boys to girls is16 is to 24 = 24
16 or 16 : 24
But both the two numbers are divisible by 8.(8 is the HCF of 16 and 24)Divide both the numbers by 8.We get, 24
16 = 32 or 2 : 3
Hence, John found that the ratio of number of boys and girls is 2:3. Here 2 is the antecedent and 3 is the consequent.Try : Now from example 2, you try to compare
1) number of girls to the number of boys. 2) number of boys to the total number of students.
Example 3: Nadeem is studying in 6th standard. His height is 4.5 feet and weight is 40 kg. What is the ratio of his height to weight ?Solution : What is the height of Nadeem ? = 4.5 feet. What is the weight of Nadeem ? = 40 kg.
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Observe, Whether height and weight are same kind of quantities or not ?
Height and Weight are not same kind of quantities. ∴ You cannot compare the height to the weight.
While comparing the two quantities, both the quantities should be of the same kind. Activity :- Answer the questions given below by writing the ratio for each.
1) Ratio of apples to oranges
2) Ratio of oranges to apples
1) Ratio of biscuits to chocolates
2) Ratio of chocolates to biscuits
1) Ratio of tables to chairs
2) Ratio of chairs to tables
1) Ratio of cars to buses
2) Ratio of buses to cars
1) Ratio of pencils to pens
2) Ratio of pens to pencils
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Now, can you define a ratio?
Comparison of two quantities of the same kind as a quotient is called ratio.
If two quantities a and b are to be compared then, their ratio is written as ‘ b
a ’ or a:b.
The symbol ‘:’ is used to express ratio and is read as ‘is to’. Hence a:b is read as ‘a is to b’.
In the ratio a:b, a and b are called the two terms of the ratio where ‘a’ is called the antecedent and ‘b’ is called the consequent.
Note : A ratio a:b can be written in three notations:
i) Word notation : a is to b
ii) Symbolic notation : a:b
iii) Fraction notation : ba
order of the terms :
In example 2 what is the ratio of number of boys to girls? 2:3
In this statement, which quantity is stated first?
Number of boys is stated first.
Which number represents the number of boys in 2:3 ?
The number 2 represents the number of boys.
What is the position of 2 in 2:3 ie, 2 is in first place
Hence, which quantity is to be written first in the ratio?
The first number in a ratio is always the first quantity stated to compare.
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Activity :-1) Find the ratio of total number of students to the
teachers of your school.2) Find the ratio of number of windows to the doors in
your house. 3) Find the ratio of number of notebooks to textbooks
in your bag.4) Visit your nearby resource centre to collect the
population of your village or town or city, and find the ratio of number of men to women.
Worked Problems1) Write the antecedent and consequent in the following
ratio 4:9 In 4:9, antecedent is 4 and consequent is 9.2) Write the following in fraction. Fraction form of 8:15 is 15
8
3) Express the following fraction in the form of ratio 136 .
Ratio form of 136 is 6 :13
4) Write the following statements using ratio symbol (:) a) 4 is to 7 = 4 : 7 b) 19 is to 12 = 19 : 125) Write the ratio for the following in fraction.
ratio of pencils to erasers is 23
ratio of balls to tenycoit rings is 35
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Exercise 5.1
I. Read the following ratios. a) 4 : 3 b) 9 : 17 c) x : y d) 125 : 20
II. Write the following statements using ratio symbol (:)
1) One is to one 2) Three is to two
3) Five is to sixteen 4) Twelve is to seventeen
5) Twenty is to six 6) Seventy five is to sixty five. 7) 'a' is to 'b' 8) 'm' is to 'n'
III. Write antecedent and consequent of the following.
Ratio antecedent consequent1:2 1 2
2:7
8:5
4:6
11:12
50:75
13:1
m:n
IV. Write the following in fraction.
1) 5:7 2) 3:1 3) 12:25 4) 43:55 5) 128:98
V. Write the following in ratio.
1) 32 2) 4
7 3) 1512 4) 65
35 5) 8476 6) 101
20
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5.2 To compare the quantities having different units :Example 1: Observe the length and breadth of a scale
given below and find the ratio of length to breadth.
20 mm
10 cm
What is the length of the given scale ? = 10 cm.What is the breadth of the given scale ? = 20 mm.If we do not take the units into consideration, the ratio of
length to breadth seems to be, 20
10 = 10:20 = 21
That means, the length of the scale seems to be half of its breadth.
But, is the length really half of its breadth in the figure? Certainly not, why? Think.
Observe the units of length and breadth of the scale carefully.
Are they in the same unit? No.Since, length and breadth are in different units, these
cannot be expressed in ratio.Hence the above ratio of length to breadth 1:2 is wrong.
Then how do you compare these two quantities?Now, express both the quantities in the same unit i.e.,
Length of the scale = 10cm = 10 × 10 mm (1cm = 10mm) = 100mm.
Breadth of the scale = 20mm. ∴ Ratio of length to the breadth 20
10015 5:1= = =
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ie, The length of the scale is 5 times of its breadth. Similarly, find the ratio of breadth to the length.
10020
51 1 5length
breadth := = =
Example 2: Find the ratio of 50 paise to ` 4. Can you compare if the units are in paise and rupees? No.How do you compare 50 paise and ` 4?Both of them must be converted into the same unit and
then we compare.` 1 = 100 paise
` 4 = 400 paise
∴ 50 paise to ` 4 = 50 : 400 = 40050 (Divide by their HCF ie., 50)
81 1:8or=
Remember :• To compare the quantities having different units both
the quantities must be converted to the same units.• While expressing in ratio, units are not indicated, only
numbers are written.
Model Problems
1. Find ratio of the following and write them in their simplest form:
a) ratio of 4 km to 500m
1 km =1000 m
4 km =4000 m
∴ ratio of 4 km to 500 m = 5004000
540
18 8:1or
1
8
= =
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b) ratio of 250 ml. to 2l
1 l = 1000 ml
2 l = 2000 ml ratio of 250 ml to 2l 2000
25081 1:8= = =
c) ratio of 2 hr 45 min to 75 min
1hr = 60 min
2hr 45 min =120 + 45 = 165 min
∴ ratio of 2hr 45 min to 75 min 75165
1533
33:15
= =
=
Note : Ratios are to be expressed in simplest possible form.
Exercise 5.2I. Find the ratio of the following :
a) ratio of ` 2 to 150 paise b) ratio of 3 month to 1 year 3 month
c) ratio of 500 ml to 2 litre d) ratio of 40 cm to 1.5 m e) ratio of 5km to 750 m f) ratio of 200 gram to 2 kilogram
II. The length of an agricultural field is 50m and breadth is 30m. Find the ratio of length to breadth.
III. The population of a village is 9,500 out of which 4,500 are women. Then find the following.
a) Ratio of men to women. b) Ratio of total population to men.
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IV. Kaleem earns ` 30,000 every month. He saves ` 5,000 to his childrens' education and he spends remaining to lead his life. Then find the following.
a) Ratio of savings to income. b) Ratio of his savings to expenditure. c) Ratio of expenditure to income.V. Present age of Malini is 11 years and her father's age is
33 years. Then find the following. a) Ratio of Malini’s age to her father’s age. b) When Malini’s father’s age is 30 years ratio of her age
to her father’s age. c) After 10 years, ratio of Malini’s father’s age to her age.
5.3 Proportion In our daily life, we come across many situations where
we compare the two ratios. For example price and quantity of objects, speed and distance travelled by vehicles etc.
In such situations we need the concept of proportion. To understand the concept of proportion, observe the
following examples:Example 1: Navya bought 2 chocolates for ` 5 and Nageena bought 4 chocolates for `10. Then whose chocolate’s price is costly?
Name of the girl
No. of chocolates bought
Amount paid (in ` )
Navya 2 5
Nageena 4 10
What is the ratio between the number of chocolates bought by Navya to Nageena ?
2:4 42
21 1:2= = =
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What is the ratio of the amounts paid by Navya and Nageena?
5:10 105
21 1:2= = =
Now compare the ratios obtained.
ie, 2 : 4 = 1 : 2, 5 : 10 = 1 : 2
∴ 2 : 4 = 5 : 10Whose chocolates is more expensive ?The price of the chocolates bought by both is the same.
Why ? Because, the ratio between the number of chocolates is equal to the ratio between the amounts paid by them.
Example 2: A car travels a distance of 50 km in an hour. If it travels at the same speed then the distance covered by the car at regular intervals of time is listed as follows:
Time duration (in hours) 1 2 3 4 5 6
Distance covered (in kms) 50 100 150 200 250 300
Find the ratio between any two successive time durations and ratio between their respective distances covered and compare them.observe the table, What is the distance covered by the car in 2 hours ? 100 km.What is the distance covered by the car in 3 hours? 150 km.The ratio of 2 hr to 3 hr is 2:3
What is the ratio of distances covered in 2 hour to 3 hour ? 100:150 150
10032 2:3= = = =
Compare the two ratios obtained i.e., 2:3 = 100:150What do you conclude now?
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The ratios 2 : 3 and 100 : 150 are equal ratios.Similarly try for other time durations and their respective
distances travelled. Now, the ratios in example (1) and (2) are said to be in proportion.
What is a proportion ?• If two ratios are equal, then it is called proportion.• If two ratios a:b and c:d are equal, then the four terms
a, b, c and d in that order are said to be in proportion and we express it as a : b :: c : d.
• Here :: is the proportionality symbol. a : b :: c : d can also be written as a : b = c : d and read as 'a is to b' is proportional to 'c is to d'. (:: is as to)
• In a : b :: c : d, a, b, c and d are called four terms of the proportion.
• In a:b:: c:d, a and d are called extremes b and c are called means.
Relationship among the four terms of a proportion (proportionality rule)
To understand the relationship among the four terms of a proportion, observe the following examples.Example 1 : Find the product of extremes and means of the proportion given below 2 : 4:: 3: 6
What is the other form of this proportion ? What is the product of extremes ? 2 × 6 =12
What is the product of means ? 4 × 3 = 12
What do you conclude from this ? 2 × 6 = 4 × 3
extremes
means2:4=3:6
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Example 2 : Compare the product of extremes and means of the proportion given below. 5 : 3 :: 25 : 15
How do you write this proportion in another way?
5:3=25:15What is the product of extremes? 5 × 15 = 75What is the product of means? 3 × 25 = 75What do you conclude from this? 5×15= 3×25=75
From the above two examples we see that, product of extremes = product of means.
In fact this is true for all proportions. Remember : Now, can you tell the relationship among the four terms of a proportion?
• In proportion, ''product of the extremes is equal to the product of the means". This is called the proportionality law.
• If a: b : : c : d then ad = bc.• Proportionality of any four terms can be verified by using
proportionality law.• In a proportion, if any three terms are known, the remaining
term can be found out using proportionality law.
Model Problems1) Verify the proportionality of the following by using proportionality rule. a) 3:5 :: 6:10
Solution: According to proportionality law, Product of the extremes = product of the means. Identify the extremes and means.
3 × 10 = 5 × 630 =30∴3, 5, 6 and 10 are in proportion.So, 3 : 5 :: 6 : 10 is a proportion.
extremes
means
3 : 5 = 6 : 10
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b) 1 : 5 :: 3 : 15 Solution : In the proportion, identify the extremes and
means
extremes
means
1 : 5 = 3 : 15
1 × 15 = 5 × 3
15 =15
∴ Product of the extremes = product of the means.
∴ 1,5,3 and 15 are in proportion.
So, 1 : 5 :: 3 : 15 is a proportion.
2) Check whether the following terms are in proportion.
a) 6, 10, 15, 20
Solution : Ratios formed are 6 : 10 and 15 : 20
6 : 10 = 3 : 5 (Dividing by 2) (106
53 )
5
3
=
15 : 20 = 3 : 4 (Dividing by 5) (2015
43 )
4
3
=
But 6 : 10 ≠ 15 : 20
∴6, 10, 15, 20 are not in proportion.
b) 3, 12, 4, 16
Solution : Ratios formed are 3 : 12 and 4 : 16
3 : 12 = 1 : 4 (Dividing by 3) (123
41 )
4
1
=
4 : 16 = 1 : 4 (Dividing by 4)
3 : 12 = 4 : 16
(164
41 )
4
1
=
∴3, 12, 4, 16 are in proportion.
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3 a) Find the value of x if 6 : 3 :: 8 : x Solution : If 6 : 3 :: 8 : x then
6 3 8
63 8
1 44
x
x
xx
12
1 4=
=
==
# #
#
` #
Extremes
Means
6 : 3 :: 8 : x
b) Find the value of ‘x’ if 4 : x :: 16: 48 Solution : If 4: x : :16 : 48 then
4 : x = 16 : 48
16 4 48
164 48
1 1212
x
x
x
x
4
1 12
1
# #
#
#
`
=
=
=
=
Note : Generally in proportion the product of the extremes is written first. But, while finding value of an unknown term, it is written first irrespective of its position (extreme or mean).
4) Cost price of 15 m cloth is ` 825. what is the cost price of 10 m of cloth ?
Solution: Cost price of 15m cloth = ` 825
Let the cost price of 10m cloth be ` x Write the given quantities as below :
Length of cloth Cost price (in `)
15 82510 x
Now the required proportion is 15 : 10 :: 825 : x.
∴ Cost price of 10 m cloth is ` 550.
15 : 10 = 825 : x
3
15 × x = 10 × 825
x = = 2 × 27510 × 825
15
2
1
275
∴ x = ` 550
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5) a labourer works for 5 days and earns `500. How many days should he work to earn `2000?
Solution : Let the number of days he worked to earn ̀ 2000 be 'x'
The required proportion is
5 : x = 5 0 0 : 2 0 0 0
500 5 2000
5005 2000 1 20
20
x
x
x1
1=
= =
=
# #
`#
#
`
Days Earning (in `)
5 500x 2000
∴ The labourer worked for 20 days to earn ` 2000.Exercise 5.3
I. Write the extremes and means of the following proportions.
a) 3 : 6 :: 4 : 8 b) 5 : 3 :: 20 : 12 c) 2 : 3 :: 8 : 12 d) 6 :12 :: 12 : 24 e) 7 : 21 :: 9 : 27 f) 20 : 10 :: 10 : 5II. Verify, whether the following numbers are in proportion
or not. a) 2, 3, 4 and 6 b) 15, 8, 6 and 3 c) 3, 6, 15 and 45 d) 24, 28, 36 and 48 III. Find the value of ‘x’ in the following. a) x : 3 :: 15:5 b) 4 : x :: 5 : 15 c) 8 : 64 :: x : 24 d) 21 : 42 :: 3 : xIV. Solve the following. 1) Price of 5 kg of sugar is `160. Find the price of 8 kg
of sugar? 2) A motor bike uses 3 litre of petrol to travel 180 km.
How much petrol is required to travel 240 km at the same speed?
3) 6 dozen of banana costs `150. What is the cost of 2 dozen of banana ?
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5.4 Unitary MethodIf you are given the cost of one object, you can find the
cost of many objects by multiplying the cost of one object with the given number of objects. But, If you are given the cost of certain number of objects, how do you find the cost of required number of objects? Let us consider two such situations.Situation 1 : Pranathi bought 2 note books for ` 28 from a book shop. Praveen also wants to buy 5 note books of the same kind. How much he has to pay ?
Solution : The amount paid for 2 note books ` 28.
What is the price of one note book ? 2
28 = `14 Price of one note book is ` 14 , What is the cost price of 5 note books ? Price of 5 note books = 14×5 = ` 70
The amount paid by Praveen = ` 70 Situation 2: Manjappa sold 2 quintal of paddy for ̀ 2,300. How many quintals of paddy is sold for ` 9,200 ?
Solution : What is the selling price of 2 quintal of paddy? = ` 2,300
What is the selling price of 1 quintal of paddy? = ,
22 300 = ` 1,150
∴ The price of 1 quintal of paddy = ` 1,150
How do you find the quantity of paddy sold for ` 9,200
Divide the given value by 1 quintal value =
,,1 1509 200 = 8 quintals
8 quintals of paddy is sold for ` 9,200.In the above situations, the problems are solved by unitary
method.
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What is unitary method?The method in which finding the value of one unit
and then finding the value of required number of units is known as unitary method.
Note : By Knowing the value of one unit we can find the value of any number of required units.
Activity :- Complete the following table
Time (in hr)
Distance travelled (in km)Camel Bicycle Car Bus
1 5 30 70
2 4 10
3 90 210
4 8 120
5 25
6
7 14 210
8 560
9 45 270
Model problems1) Cost of 4 pens is ` 80. What is the cost of 7 pens?
Solution : Cost of 4 pens = ` 80
∴ Cost of 1 pen = 480 = ` 20
∴ Hence the cost of 7 pens = ` 20 × 7
= ` 140
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2) Austin pays ` 4,500 house rent for 3 months Calculate
the rent paid by him for 1 year.
Solution : House rent for 3 month = ` 4,500
∴Rent for 1 month = 3
45001
1500
= ` 1,500
∴ Rent paid for 1 year (12 months) = 1,500 × 12 = ̀ 18,000
∴ Rent paid for 1 year = ` 18,000.
Exercise 5.4
1) The Price of 4 cakes is ̀ 20. What is the price of 5 cakes?
2) Narasimha earns ` 1,225 in a week. How much did he earn in 30 days?
3) Weight of 32 biscuit packets is 4 kg. What is the weight of 10 biscuit packets of the same kind?
4) A lorry covers a distance of 1,740 km using 87l of diesel. Then how many litres of diesel is required to cover a distance of 620 km ?
5) The price of 25 kg of wheat flour is ` 800. Then,
a) What is the price of 10 kg of wheat flour?
b) How many kg of wheat flour can be bought for ` 2,400?
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Try :Four workers work for 30 days and earn total salary of ` 12,000. Then,
i) How many days do they work to get a total salary of ` 9,000 ?ii) What is the salary of a) four workers for 6 days ? b) a worker in a month ? c) a worker for a day ? d) two workers for a week ?
Know this :-
• The ratio of length and breadth of our national flag is always 3 : 2
• In the construction of a building, the concrete used has
i) the ratio of sand to gravel is 4 : 2
ii) the ratio of cement to sand is 1 : 4
iii) the ratio of cement to gravel is 1 : 2
• Idly can be softened by rice and urud dhal mixing in the ratio 3 : 1
• In the extraction of iron, iron ore, lime stone and coke are used in the ratio 8 : 1 : 4 respectively.
• In water, hydrogen and oxygen are in the ratio 2:1.
l l l l l
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Unit - 6SYMMEtRY
After Learning this unit you can :• explainthemeaningofsymmetry,• identify2Dsymmetricalobjects,• understandreflectionsymmetryof2Dobjectstakingtheir
mirrorimages,• identifyaxesofreflectionsymmetryandknowreasons
forreflectionsymmetry.Inyourpreviousclassyouhavestudiedbasicelementsof
symmetricalfigures.Theconceptofsymmetryhelpstheartists,designersof
jewelleryorclothing,architectsetc.intheirworktopreparebeautifulartpieces,designsandornaments.Knowledgeofsymmetryhelpsustostudythepatternsinvarioustypesofmathematicalconceptsandingeometricalshapes.What is symmetry?
Observethefiguresgivenbelow.
EFoldthesefiguresalongthedottedline.Whatdoyou
observe?Wefindthatlines,curvesandanyotherelementsofone
halfwillexactlycoincidewiththeotherhalf. (over laps) Insuchacasewesaythatthepictureissymmetrical.So,whatissymmetry?
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“Thefigureswhichhavebalancedportionsareknownassymmetricalfiguresorafigureissaidtobesymmetricalwhenanimaginarylineseemstodividethefigureintotwoidenticalpartssuchthatiftheyaresuperimposedoneachothertheycoincide”.
Activity 1 : Herearesomefiguresgivenbelow.Drawsuchtypeoffiguresinasheetofpaperandfoldthefigurealongthelinedrawnonitsothattwohalvesareoneovertheother.
A
B
A
B
A
B
A B
A
B
What you observe ?
Wefindthatbothhalvesfiteachotherexactly.Sowecansaythattheabovefiguresaresymmetrical.
Activity 2 : -Takeapieceofpaper.Folditatthemiddleandopenitout.Putadropofinkononehalf sideofthesheet. Fold the paper carefully andpressthetwosidestogether,openoutthe fold and you wil l f ind aninterestingsymmetricaldesignformedonthepaper.Oneofsuchdesignsisgivenhere.
In naturewe can see symmetry in the flowers, leaves,butterflies etc.Manyman-madestructuresormonumentslike Tajmahal, Vidhana Soudha, IndiaGate are beautifulbecauseofthesymmetry.Tounderstandmoreaboutvarioussymmetricalobjectsobservethefollowingandexperiencethebeautyoftheirsymmetry.
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1)
2)
3)
4)
H I O X5) B C D E Activity 1 :- Select the letters of your namewhichhavesymmetryandwritethelineofsymmetryforthoseletters.
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Reflection SymmetryObservation and identification of 2D symmetrical objects for reflection symmetry :
Haveyouseenyourreflectioninamirror?Yes,wehaveseenourreflectioninamirror.Wecansee
reflectionofanyobjectinamirror.Doyouknowwhethertheobjectanditsimagearesymmetricalornot?Nodoubt,theobjectanditsreflectionaresymmetrical.
LineofSymmetry
Mirror
ImageObjecty
x
E E MMLineof
Symmetry
Mirror
ImageObject
x
y
Object
Image
Mirror(lineofsymmetry)yx
Observe the above figures.Intheabovepicturesthelineinbetweentwofiguresisthe
positionofthemirror.Ifwefoldthispicturealongthislinetheobjectandimagecoincidewitheachother.Thenwesaythattheobjectanditsreflectionaresymmetrical.TheobjectanditsimageareequidistantfromthelineXY(mirror).
the line which divides the picture into two halves is known as axis of symmetry or line of symmetry.
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Observethesesymmetricalfigures.Howmanyhalvesarethereineachofthefollowing?
Wecanseethattherearetwohalvesineachpicture.Ifwefoldthesepicturesalongthelineofsymmetrytheleftandrighthalvesmatchexactlyeachother.Ineachfigurethetwohalvesaremirrorimagesofeachother.Inthesepicturesthedottedlineisthepositionofthemirror.Thisdottedlineistheaxisofsymmetry.
Then what is reflection symmetry?Apictureissaidtohavereflectionsymmetryifitsimage
coincideswithitsreflectionalongaparticularline.Thislineiscalledasaxisofsymmetry.
Some pictures of living beings which show reflectionsymmetryaregivenbelow.
ManylettersoftheEnglishalphabetalsoshowreflectionsymmetry.Fewexamplesofsuchalphabetaregivenbelow.Ineachfigurethedottedlineisthelineofsymmetry.
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XO U A H MMany 2D geometrical figures show reflection symmetry.
Fewexamplesaregivenbelow.Thedottedlineineachfigureisthelineofsymmetryoraxisofsymmetry.
Rhombus Parallelogram Trapezium Rectangle
HexagonOctagon Pentagon Circle
Activity :- Collect somemore examples for reflectionsymmetryofEnglishalphabet,2Dgeometricalfiguresandliving beings. Draw the figures andmark the axis ofsymmetry.
Operation of reflection (Taking mirror images)
Wehavealreadystudiedaboutreflectionsymmetry.Weknowtheobjectanditsmirrorimagesaresymmetricalandthelinebetweenthemirrorandobjectformstheaxisofsymmetry.
Whatchangesdoyouobservewhenyouseeyourreflectioninamirror?
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Observe the figures and their mirror images given below. What do you observe?
Mirror(lineofsymmetry)
BPC
B
PC
Mirror(lineofsymmetry)
E EMirror
(lineofsymmetry)
Mirror(lineofsymmetry)
A A
Mirror(lineofsymmetry)
MM
Duringtheformationofmirrorimagesthereisalateralinversion in amirror. Thatmeans ifwe lift our lefthand,thereflectionwillshowthattherighthandislifted.Also,weobservethefollowingpointsduringthereflectioninamirror.
1) The distance of the imagefromthemirrorisequaltothedistanceoftheobjectfromthemirror.
2) The height of the image andtheheightoftheobjectarethesame.
3)Thereflectionofanobjectlookslaterallyinverted.
Lateral inversion = Lateral change.
Object
lineofsymmetry(mirror)
Image
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Activity :- Collect themirror images of all EnglishalphabetfromAtoZ.e.g:- B→
C→BC
Identifying axis of reflection symmetryWehavealreadystudiedaboutreflectionsymmetryand
axisofreflectionsymmetry.Thenwhatistheaxisofreflectionsymmetry?
Thelinebetweentheobjectandtheimageiscalledaxisofreflectionsymmetry(oritisamirror)orthelinewhichdividesthepictureintotwohalvesisknownasaxisofsymmetry.Observe the following figures and identify the axis of symmetry.
A AA B C H M
Ineachoftheabovefiguresthedottedlinerepresentstheaxisofsymmetryorlineofsymmetry.
Note:Thelineofsymmetrymaybehorizontalorverticalorneitherhorizontalnorvertical.
What type of line of symmetry is there in each of the fol-lowing figures ?
Intheabovefiguresthelineofsymmetryisvertical.
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What type of line of symmetry do the following figures have?
H E DIntheabovefiguresthelineofsymmetryishorizontal.
What type of line of symmetry do the following figures have?
HIn the above figures the line of symmetry is both
verticalandhorizontal.What is the type of line of symmetry in the following figures?
In the above figures the line of symmetry is neitherhorizontalnorvertical.Observe the following figures. How many lines of symmetry do they have?
1lineofsymmetry2linesofsymmetry
(Equilateraltriangle)3linesofsymmetry
4linesofsymmetry
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5linesofsymmetry
6linesofsymmetry
Multiplelinesofsymmetry
A symmetrical figure may have one or more than one line of symmetry.
e.g: Isosceles triangle is having only one line ofsymmetry.
How many lines of symmetry are there in a rectangle?Inarectanglethereare2linesofsymmetry.
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How many lines of symmetry are there in a square?
Inasquarethereare4linesofsymmetry.
a1
a2
a3
a4 a5a6
Count the number of lines of sym-metry In the given figure?
In the given regular hexagonthereare6linesofsymmetry.
How many lines of symmetry are there in a circle? Inacirclethereareuncountable
linesofsymmetry.Socircleishavingmultiplelines
ofsymmetry.
Activity 1 :- Takedifferentsheetsofpaperandcutitintoashapeofsquare,rectangle,regularpentagon,regularhexa-gonandfoldittogetcorrecthalvesandhencefindthelineofsymmetryforalltheaboveplanefigures.
Activity 2 :- WritetheaxesofsymmetryforallthelettersoftheEnglishalphabetwhicharehavingsymmetry.
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Worked Examples
I. WriteanyfivelettersoftheEnglishalphabetthathavenolinesofsymmetry?Ans:1)F 2)G 3)J 4)L 5)P
II.WriteanyfivelettersoftheEnglishalphabethavingonelineofsymmetry?Ans:1)A 2)E 3)C 4)D 5)M
III.Drawthelinesofsymmetrytothefollowingfigures.
Figure LinesofSymmetry
1)
a4 a1 a3
a2
2)
3)
4)
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Exercise - 6
I Draw the line of symmetry for the following letters of
Englishalphabet.
a) b) c) d)
II. Drawthefiguresforthefollowinganddrawthepossible
numberoflinesofsymmetryforeach.
1)Circle 2)Parallelogram
3)Scalenetriangle 4)RegularPentagon
5)Equilateraltriangle 6)Isoscelestriangle
III.Write two letters of English alphabet for each of the
following.
1)Havingtwoaxesofsymmetry
2)Havingoneaxisofsymmetry
3)Withnoaxisofsymmetry
IV.Writethedigitsfrom0to9havinglinesofsymmetry.
l l l l l
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Unit- 7COnStRUCtiOnS
After Learning this unit you can : draw a line segment of given length, construct a circle of given radius using scale and
compasses, draw perpendicular bisector of a given line, construct an angle of given measure using protractor, construct angles measuring 600 and 1200 using scale and
compasses only.
For geometrical construction, the ruler and the compasses are only two geometrical instruments recommended by Euclid. Therefore, the construction, which is done with the help of ruler and compasses, is called Euclidean construction. Protractor and set squares are also used for construction. However, for accuracy and geometrical logic, ruler and compasses are recommended by Euclid.
In geometry, construction means drawing a correct and accurate figure from the given data. The skill of construction is useful not only for an architect, but also in all walks of life. In science and engineering it is an important step for professional training.
7.1 How to draw a line segment of a given length?A line segment is a part of a line having two end points.
Therefore, it has a definite length. We can draw a line segment of given length by using i) A scale, ii) ruler and compasses. Let us know how to draw a line segment.
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i) Drawing a line segment using a scale :Example 1 :
Draw a line segment of length 5.2 cm.Steps of construction:
Step 1 : Take the scale and place it on a page of the notebook.
Step 2 : Mark two points on the paper which are 5.2 cm apart and very close to the graduated edge of the scale. Place the ruler in such a way that the zero (0) mark of the ruler coincides with 'P' and the point at 5.2 cm co-incides with Q.
P Q
Step 3 : Join the points P and Q. Now PQ is the required line segment.
P Q 5.2 cm.
ii) Using ruler and compasses : Steps of construction:Step 1 : Draw a line l, mark a point on it and name it as ‘P’.
P l
Step 2 : Place the metallic end of the compasses on the zero point of a scale and open out the compasses such that pencil end is on the 5.2 cm mark and thus set your compasses to have the radius equal to 5.2 cm.
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Step 3 : Draw an arc of radius 5.2 cm. with ‘P’ as centre which
cuts the line l at a point right of point P. Name the point of intersection as ‘Q’.
Step 4 : Now, the length of line segment PQ = 5.2 cm.
5.2 cm.P Ql
(iii) to draw a line segment equal to a given length of line segment
Given:- PQ is a line segment.
P Q
Draw a line segment AB equal to the length of PQ.
Step 1: Draw a line ‘l’ and mark a point ‘A’ on it.
A l
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Step 2 : Measure the length of PQ using a compasses.
Step 3 : Keep the needle of the compasses at A on the line ‘l’ and draw an arc at B with the same measure.
A B CStep 4 : AB is the required line segment which is equal
to length of PQ i.e., AB = PQ. Verify the measurement of PQ and AB using scale.
Exercise 7.1
1) Draw a line segment of length 6.5 cm using a ruler.
2) Construct a line segment of length 7.4 cm using ruler and compasses.
3) Construct AB of length 7.8 cm. Cut off AC = 4.3 cm from 'A'. Measure BC.
4) Given AB of length 8.5 cm. and CD of length 3.8 cm construct a line segment XY , such that the length of XY is equal to the difference of lengths of AB and CD. Verify by measuring.
5) Draw any line segment PQ. Without measuring PQ construct a copy of PQ.
6) Construct a line segment AB of length 12 cm. Cut off a segment AC of length 4.5 cm. Measure the remaining segment CB.
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7) Draw a line segment AB whose length is 3 cm. Construct PQ such that the length of PQ is twice that of AB, using the compasses.
7.2 Construction of a circle
Example 1 :Construct a circle of a given radius with the help of
compasses. Steps of construction:
Step 1: Mark a point to be taken as centre on your notebook with a sharp pencil and name it as ‘O’.
O
Step 2: Open the arms of the compasses for the required radius.
Step 3: Place the needle of the compasses at ‘O’.
Step 4: Slowly turn the
arm carrying pencil around ‘O’ and complete one round of movement in one instant.
Geometrical figure thus obtained is a circle.
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Example 2 :
Construct a circle of radius 4.1 cm using compasses. Steps of construction :Step 1: Mark a point ‘O’ in the middle of the page. This is taken as the centre of circle.
O
Step 2: Place the metallic point of the compasses on the initial point of the scale. Open the pencil end and adjust in such a manner that the distance between the metallic point and the pencil is equal to 4.1 cm.Step 3: Place the metallic point of the compasses on point ‘O’. Now slowly turn the arm carrying pencil around ‘O’ and complete one round of movement in one instance without changing the radius.Now you will get a circle of radius 4.1 cm.
Think:- How many circles can you draw with a given centre?
note: • Every circle has a centre and a definite radius.• Diameter is two times the radius. ie., d = 2r.
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Exercise 7.2
1) Draw a circle of radius 3 cm.
2) Draw 2 circles of radius 4 cm and 2.5 cm with the same centre ‘O’.
3) Draw a circle of radius 4 cm. Mark the following.
i) Centre ‘O’ ii) Radius OA iii) Diameter XY
4) Mark any two points A and B on a sheet of paper. Using compasses draw a circle with centre ‘O’ and passing through A and B.
5) Draw a circle of radius 3.5 cm. Draw any two of its diameters. And name the figure obtained by joining the end points of the diameters.
6) Draw a circle with centre ‘O’ and diameter
o
C3
C2
C1
8 cm. Measure its radius.
7) In the adjoining figure measure the radii of circles C1, C2, and C3.
8) Measure the radius and diameter of the following circles
o
C1 C2
o oo
C3
What do you observe ?
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7.3 to draw the perpendicular bisector to a given line segment.
Example 1: Draw the perpendicular bisector of the line segment EF.
Steps of construction:
Step 1: Observe the line segment EF given below.
E F
Step 2: Using compasses with E as centre and radius equal to more than half of EF, draw two arcs one above EF and the other below as shown. FE
Step 3: With the same radius, with 'F' as centre draw two more arcs to cut the previous arcs at 'M' and 'N'. FE
Step 4: Join M and N. MN bisects the line segment EF at 'O'. MN is the perpendicular bisector of EF.
F
M
N
E O
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Example 2:
Draw the perpendicular bisector of a line segment of length 6 cm.
Steps of construction:Step 1: Draw the line segment AB = 6 cm with the help of a scale and pencil.
6 cmA B
Step 2: Using compass with 'A' as centre and radius equal to more than half of AB, draw two arcs one above AB and the other below as shown. BA
Step 3: With the same radius with 'B' as centre draw two more arcs to cut the previous arcs. BA
Step 4: Name the points of intersection of the arcs as P and Q.
Step 5: Join P and Q. PQ bisects AB at 'R'.
B
P
Q
A R
Step 6: PQ is the required perpendicular bisector of AB.
Verify: Measure AR and RB line segments. Also, measure ∠PRA and ∠PRB. What do you infer?
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Note : The line that is perpendicular to a given line segment and divides it into two equal parts is called the perpendicular bisector.
Exercise 7.3
1) Identify the perpendicular bisector in the following figures.
Q
X
P
Y
SA
R B
F
N
E
M
R
L
K
S
2) Draw a line segment PQ of length 8.4 cm. and construct a perpendicular bisector to it.
3) Draw a line segment AB of length 7.4 cm and construct the perpendicular bisector and verify whether two parts are of equal measure using a ruler.
4) Draw a perpendicular bisector of XY of length 9.6 cm. and mark point of bisector as 'P'. Examine whether PX = PY.
5) Draw a line segment AB=10 cm. using ruler and compasses, divide it into 4 equal parts.
6) Draw the perpendicular bisectors of the following line segments.
A
B
a)P
Q
b)
X Y
c)
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7.4 (a) Construction of angles using scale and protractor.
In your previous class you have studied about angles. Now you will learn to construct the angles using a protractor. A protractor is an instrument in geometry used to measure and construct a given angle. Example 1: Construction of an angle of 500.
The following steps are followed to construct an angle of measure 500.Steps of construction:
Step 1: Draw a line segment AB of any length. A B
Step 2: Place the protractor on AB so that the centre of the protractor coincides with the initial point A of the line segment AB and base line of the protractor coincides with AB.
A B
Step 3: Start from 00 of the protractor move upto 500 and mark a point 'C' at 500.
A B
C
Step 4: Remove the protractor and join points A and C using scale.
Now BAC is the required angle of measure 500.
A B
C
500
Try : Draw a line segment AB and construct an angle of measure 500 at B.
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Example 2 : Construction of an angle of 900.
The following steps are to be followed to construct an angle of 900.
Step 1: Draw a line segment PQ of any length.
P Q
Step 2: Place the centre of the protractor at ‘P’ and let its base line coincide with PQ. P Q
Step 3 : Start from 00 and move upto 900 and mark a point 'R' at 900.
P Q
R90°
Step 4 : Join PR. ∠RPQ= 900 which is the required angle.
P Q
R
90°
Try : Draw a line segment PQ and construct an angle of 900 at Q.
Example 3: Construction of 1300.Follow these steps to construct an angle of 1300.
Steps of construction:
Step 1 : Draw a line segment RS of any length. R S
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Step 2: Place the protractor on RS so that the centre of the protractor coincides with the initial point R of the line segment RS and base line of protractor coincides with RS.
R S
Step 3: Start from 00 of the protractor, move upto 1300 and mark a point 'T' at 1300.
R
T
S
Step 4: Remove the protractor and join the points R and T using a scale. Now ∠TRS is the required angle of measure 1300. R
T
S
1300
7.4(b) to construct an angle using scale and compasses.There are some elegant and accurate methods to construct
some angles of special size which do not require the use of protractor. We shall learn this.Example 1: Construction of 600.Steps of construction:Step 1: Draw a line segment OB of any length.
O B
Step 2: With O as centre and with any suitable radius draw an arc which cuts OB at 'P'.
PO B
Step 3: With P as centre and with the same radius draw another arc to cut the previous arc at Q.
PO
Q
B
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Step 4: Draw OA through Q.
Thus formed ∠AOB measures 600.
Verification:Using protractor, measure ∠AOB. PO
Q
A
B
Example 2 : Construction of 1200.Steps of construction:
Step 1: Draw a line segment OB. O B
Step 2: With 'O' as centre with any suitable radius draw an arc to cut the line segment OB at 'P'.
PO B
Step 3: With P as centre and with the same radius, draw another arc to cut the previous arc at Q.
PO B
Q
Step 4: With Q as centre and with same radius, draw one more arc to cut the original arc at 'R'.
PO B
QR
Step 5: Join the points O and A through 'R' and extend it to A. Thus formed ∠AOB measures 120o. Verify ∠AOB =1200 using protractor. PO B
QR
A
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Exercise 7.4
1) Construct the following angles with the help of a protractor.
a) 300 b) 450 c) 550 d)1300 e) 900
2) Using a ruler and compasses construct the following angles:
i) 600 ii) 1200
3) Construct an angle of 600 at point ‘B’ on the line given below. (Use compasses)
A B
4) Construct an angle of 1200 at the point M on the given line. (Use compasses)
S
M
5) Construct an angle of 900 at the points R, B and L on the following lines.
Q
R
B
A
K
La) b) c)
l l l l l
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unit - 8
mensuration
after Learning this unit you can :• knowtheperimeterandareaofclosedplanefigure,• calculatetheperimeterofclosedplanefigures,• calculatethesurfaceareaofclosedplanefigures,• distinguishbetweenperimeterandareaofclosedplane
figures,• solveproblemsonperimeterandareaofclosedfigures
usingformula.
Wearefamiliarwiththetermslikeboundaryandregionsofclosedplanefigures.So,whatisaboundaryofaplanefigure?
Tounderstandthis,letusdothefollowingactivity.Activity 1:- Take some match sticks and arrange them as shown in the figure.
(i) (ii) (iii) (iv)
What type of figures are i), ii), iii) and iv) ? Figures i) and ii) are closed figures where as iii) and iv) are open figures.To understand the boundary of a plane figure we need to select closed figures.Boundaryismadeupoflinesegmentsoftheclosedplanefigure.
Insomecaseswerequirethetotallengthoftheboundaryoftheclosedplanefigures.Afewsuchexamplesaregivenbelow.
i) Inordertofencetheboundaryofafieldafarmertakesthetotallengthoftheboundary.
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ii)WhilebuildingacompoundaroundahouseanEngineermeasurestheboundaryofthatcompound.
Hence,thetotallengthoftheboundaryisnothingbuttheperimeter.
So,whatistheperimeterofaplanefigure?
Perimeterofaclosedplanefigureisthetotallengthoftheboundaryofthatfigure.
Inthepreviousactivity(1),whatis Doyouknow?Perimeter isaGreek
word:Perimeansaroundmetronmeansmeasure.
the perimeter of the closed planefiguresi)andii)?
The perimeter of the closedplanefigure infigure i) is the totallengthofthethreematchsticksandinfigureii)isthetotallengthofthefourmatchsticks.How to measure the perimeter of a closed plane figure?Activity 1 :-
A B
Ca)
b)
A B
CD Thisclosedplanefigurealsohasalinear
boundary.Theperimeterofthefigure(b)=AB BC CD DA+ + + .
Lookatthisfigure.Whattypeofboundaryithas?Ithasalinearboundary.Perimeterofthefigure(a)=AB BC CA.+ +
Perimeterofanyclosedplanefigurewithlinearboundaryisthesumofthemeasuresofallitssides.
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Activity 2 : Findtheperimeterofthefollowingfiguresbymeasuringtheirsidesusingtheruler.
E F
D B
Q
P R
ST
A
D C(i) (ii) (iii)
Perimeter of regular polygons Let us know the method of finding the perimeter of regular polygons.
A
B C P
Q R
S B C
A E F
GH3cm2.5cm3c
m
3cm
3cm
3.5cm
2cm 3cm
3cm
3cm 3cm
3cm
3cm
3cm
fig (a) fig (b) fig (c) fig (d)
In the above figures (c) and (d) are regular polygons,because their sides and angles are equal. For finding theperimeterofaregularpolygon,wemeasurethelengthofasideandmultiplyitbythenumberofsidesofthepolygon.example 1:Calculatetheperimeterofaregular(equilateral)triangleofside5cm.
A
CB5cm
5cm
5cm
Inanequilateraltriangleallsidesareequal.Perimeter of the equilateral triangleABC AB BC CA
5 5 515 cm
= + += + +=
Thisisasgoodas,Perimeterof∆ABC=No.ofsides×measure ofoneside =3×5 =15cm.
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example 2 :
D
A B
C2cm
2cm
2cm2cm
Calculate the perimeter of a regularquadrilateral(Square)ofside2cm.
Inasquareallsidesareequal.
PerimeterofasquareABCD AB BC CD DA2 2 2 28 cm
= + + += + + +=
PerimeterofasquareABCD=8cm
Or,Perimeterofasquare =4×measureofoneside
=4×2 =8cm
example 3 :Calculatetheperimeterofregularpentagonofside4cm.Perimeterofregularpentagon=5×measureofoneside
=5×4
A B
C
D
E
4cm
4cm
4cm4cm
4cm
=20
Perimeterofaregularpentagon=20cmexample 4 :
Calculate theperimeterof regularhexagonofside5cm.
Perimeterofregularhexagon=6×measureofoneside. =6×5
=30cm
∴Perimeterofregularhexagon=30cm
S
PQ
RT
U5cm
5cm
5cm5cm
5cm
5cm
Hence, perimeter of any regular polygon=number ofsides×measureofoneside
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Perimeter of a rectangle:Name the equal sides of the rectangle given in the figure
A
D
b b
C
Bl
l
Theequalsidesarei) AB DCii) AD BC
and
and
Inarectanglethetwolengthsareequalandtwobreadthsareequal.
Activity: Measure the sides of the rectangle given below and calculate its perimeter.D
A
b b
B
Cl
l
Measuresofsides: ABBCCDDA
_
_
_
_
_
cm
cm
cm
cm
cmTotal
=====
∴PerimeterofrectangleABCD
AB BC CD DAb b
2 2b2( b)
l lll
= + + += + + += += +Hereperimeterofrectangle
Inanyrectanglewhoselengthisl andbreadthisb,thenitsperimeteris=2(l+b)units
example :Calculatetheperimeterofarectangleoflength8cmandbreadth4cm.
Givenl=8cm, b=4cm
Perimeterofrectangle=2(l+b)=2(8+4)=2(12)=24
think it! The perimeter of a
rectangleP=2(l+b).Frameformulaforfindinglength(l ) andbreadth(b).
∴Perimeterofarectangle=24cm
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Perimeter of a square:
S P
Ql
l
l
l
R What do you know about the sides ofasquare?Weknowthatinasquareallsidesareequal.Squareisaregularpolygon.Perimeterofasquare=4×measureofoneside. =4×l
∴Perimeterofasquare=4×lwherel =measureofoneside.
example : Calculatetheperimeterofthesquareofside5cm. Given:l=5Perimeterofthesquare=4×l=4×5=20∴ Perimeterofthesquare=20cm.
Points to remember :* Perimeterofanylinearboundaryofplanefigureisthe
sumofallsides.* Perimeterofanyregularplanefigureistheproductof
numberofsidesandmeasureofoneside.* Perimeter of a rectangle = 2(l+b), Where l = length,
b =breadth.* Perimeterofasquare=4× l,where l =measureofa
side.
Worked examples1) Calculatetheperimeterofthefollowingfigure.
2cm
3cm
P
W
U T
SQ
R
V
3cm1cm
2cm
5cm
3cm
3cm
Perimeter=PQ+QR+RS+ST+TU+UV+VW+WP =2+1+3+5+2+3+3+3
=22
∴Perimeter=22cm
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2)Arectangularplotoflength20mandbreadth10mtobefenced.Whatisthetotallengthofwirerequiredtofencetheplotthreetimes?Given :l=20m b=10m.Lengthofwireneededtofenceonce=perimeterofthe
rectangularplot =2(l+b)=2(20+10)=2(30)=60m
∴Lengthofwireneededtofenceitthrice=60× 3=180m
3) Perimeterofarectangleis20cm.Findthebreadthifitslengthis8cm?Given:Perimeterofarectangle=20cml =8cm,b=?Perimeterofrectangle=2(l +b)
4) Asquaresheetofcardboardhasaperimeterof40cm.Findthelengthofeachside?Given:Perimeter=40cm,l=?Perimeterofasquare=4xl
40 =4xl
440 l
10
=
∴ l=10cm.
20 2 220 2(8) 220 16 220 16 24 2
24
2
l bb
bb
bb
b cm
= += += +- ==
=
=`
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5) Findthecostofwirerequiredtofencearectangularparkoncehavingthelength40mandbreadth20mattherateof`10permeter.
Given:l=40m, b=20m.
Lengthofthewireforfencingtherectangularpark= Perimeteroftherectangle.
Perimeteroftherectangle=2(l +b )=2(40+20)=2(60)=120m.
∴Costofwirepermeter =`10
Costof120mwire =120×10=` 1200
Group activity :- With a group of 5 students try tomeasuretheperimeterofthefloorofyourclassroomandblackboard.
exercise 8.1
I Answerthefollowingquestionsasperinstructions:
1)Findtheperimetersofthefollowingfigures.
a) b)
L
A B
C D
F
GH
E
K
IJ
3cm
3cm
3cm
3cm
3cm
3cm
2cm 2cm
2cm
2cm
2cm2cm
3cm
3cm
3cm
3cm
4cm
4cm
A B
DE
F C
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2) Findtheperimeteroftherectangleswhoselengthandbreadtharegiven:(allmeasurementsareincm)
l 5 6 10 12b 4 2 5 4
3) Findtheperimeterofthesquareswhosesidesaregiven(allmeasurementsareincm)
measureofeachside 6 3.5 5 25
4) Find the length of each side of the square whoseperimeteris80cm.
5) Find the breadth of a rectanglewhose perimeter is40cmandlength11cm.
6) Findthelengthofarectanglewhoseperimeteris30cmandbreadthis5cm.
7) Findtheperimeterof the followingregularpolygonswhosesideis5cm.
a)regulartriangle b)regularpentagon
c)regularhexagon d)regularoctagon
8) Rajujoggstwotimesaroundarectangularfieldoflength40mandbreadth25m.FindthedistancecoveredbyRaju.
9) Asquareparkoflength80mtobefenced.Findthelengthofthewirerequiredtofencearounditfourtimes.
10)Find the cost of wire needed to fence around therectangularplotoflength20mandbreadth15mattherateof`10permeter.
Activity:- Find the perimeter of cover page of yourmathematicstextbookandnotebook.
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8.2 Shapes of different figures with same perimeterActivity 1 : -Observethefollowingfiguresandcalculatetheperimeterofeachfigure.
6
6
6
6
3 3
3
3
3
3
336
(i) (ii) (iii)In fig (i) the perimeter = 3×6=18 units.In fig (ii) the perimeter = 2(l +b)=2(6+3)=2(9)=18 units.In fig (iii) the perimeter = 6×3=18 Units.Activity 2 :- Observe the following figures and find the perimeters of fig (i) and fig (ii).
8
8
2 5
5
5
52
(i) (ii)
Infigure(i)theperimeter=2(l+b)=2(8+2)=2×10=20unitsInfig(ii)theperimeter=4×5=20units.
Activity 3 : Observe the following figures. Find the perimeter.
3
6
6 8 7
7
22
8
113
(i) (ii) (iii)
Perimeteroffig(i)=2(l + b)=2(6+3)=2(9)=18units.Perimeteroffig(ii)=2(l + b)=2(8+1)=2(9)=18units.Perimeteroffig(iii)=2(l + b)=2(7+2)=2(9)=18unitsWhatdoyouconcludefromtheabovethreeactivities?
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Fromthepreviousthreeactivitiesweconcludethat,eventhoughthefiguresareofdifferentshapes,theyhavesameperimeterandviceversa.
think it : Isitpossibletodrawdifferentsquareswithsameperimeter?
Worked examples1) Drawanytworoughfiguresshowingregularpolygons
ofperimeter24units,sothatthemeasureofeachsideisawholenumber.
6
6
66
A
B C
8 8
8
2)Write the rough sketches of any four rectangles ofdifferentmeasureswithperimeter20unitseach.
a)
9
9
11 b)
8
8
2 2
c)
4
6
6
4 d)
7
7
3 3
think it : How many different types of polygons arepossiblewithperimeter24unitseach,sketchit.Ex.:
6
8
1044
75 3
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exercise 8.2
1) Sketchfourrectanglesofdifferentmeasureseachhavingperimeter16units.
2) Theperimeterofarectangleandasquareareequal.Thelengthandbreadthoftherectangleare20cmand12cmrespectively.Findthelengthofthesideofthesquare.
3) Perimeterofasquareandarectangleareequalwhichisequalto20units.Findthelengthofeachsideofsquareandlengthofrectangleifitsbreadthis4units.
8.3 areaThe interior and exterior regions and boundary of the
followingfiguresareshown.Observe.
boundary
Exteriorregion
Interiorregion
Interiorregion
Exteriorre
gion
boundaryA B
CD
What is the interior region of a figure?Itisthetotalregionboundedbytheboundaryofthefigure.
Observethegivenfigures. Shadedregioniscalledtheareaofthefigure.Thenwhatisthearea?
The total surface enclosedby the boundary in a closedfigureiscalleditsarea.
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Do you Know? : Areaisalatinwordwhichmeanspieceoflevelground.
Activity :- Observeeachpairoffiguresgivenbelow:
a)
1 2
b)
1 2
c)
1 2
d)
21
e)
1 2
f)
21
Comparetheshadedregionofeachfigureinapairwhichhas thegreaterarea ineachpair.Tabulate the resultasfollows.
Figureofgreaterarea
Figureofsmallerarea
a) fig(1) fig(2)
b)
c)
d)
e)
f)
Wecan'tconcludeabouttheareainfigurese)andf)withoutmeasuringit.
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How to measure the area of plane figures?
D
A B
C
1cm
1cm
1cm
1cm
Lookatthisfigure.Whatistheareaofthissquare?Areaofthissquareis1cm×1cm=1cm2.
(Becausethefigurehas2dimensions).
Note:Whilemultiplyingtheunitsintwodimensionswegetsquareunit.
The area of any figure is measured by taking unitsquaresasstandardunit.
Inthecaseofthesquareof1mside,theareais1m2.Incaseofthesquareof1kmside,theareais1km2.Look at the figures drawn on the graph sheet
What is themeasure of each side of thesquareABCD?
Itis1cm.WhatistheareaoffigureABCD?
Area=1cm×1cm=1cm2.
ABCDisaunitsquareofarea1cm2.HowmanyunitsquaresarethereinPQRS?
Thereare4unitsquares.
∴TheareaofPQRSis=4cm2
Howmanyunit squaresare there in thefigureKLMN?Thereare8unitsquares inthefigureKLMN.
∴TheareaofKLMNis8cm2.
Know it: Whilemeasuringtheareaofanyplanefigurewetaketheareaofaunitsidedsquareasstandardsize.
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area of a rectangleActivity : -
D
A B
C
3
2 2
35
5
S R
QP
3 3
6
6
4 4
Draw these rectangles on a graph paper of having squares of 1 cm × 1 cm.
ABCD Rectangle covers unit squaresinagraphpaper.∴AreaofarectangleABCD=3cm.×2cm.=6sq.cm.=3×2=l×bRectanglePQRScovers15unitsquaresonagraphpaper.∴AreaofrectanglePQRS=5cm.×3cm.=15sq.cm=5×3=l×b
KLMNrectanglecovers24unitsquaresinagraphpaper.∴AreaofrectangleKLMN=24sq.cm=6×4= l×b
Observing the above three examplesAreaofrectangleEFGH=l ×bsquareunit.Where l=length,b=breadth.
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Whatdoyouconcludefromthisactivity?Weconcludethattheareaofarectangleistheproductofitslengthandbreadth.∴AreaofarectangleA=l×b,wherel=lengthb=breadth.
area of the squareActivity : Draw these squares on a graph paper having 1 cm × 1cm square 4
N
K L
M
4
44
3
3
3 3
R
QP
S2D
A B
C
2
2 2
Square ABCD covers 4 unit squares in thegraphpaper.∴AreaofABCD=4sq.cm=2×2=a×a=a2
SquarePQRScovers9unitsquareinthegraphpaper∴AreaofPQRS=9sq.cm=3×3=a×a=a2
SquareKLMNcovers16unit squares in thegraphpaper∴AreaofKLMN=16sq.cm=4×4=a×a=a2
Byobservingtheaboveexample;AreaofthesquareEFGH=a×a=a2units.Wherea=lengthofeachsideofthesquare.Whatdoyouconcludefromtheaboveactivity?
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From the above activity we conclude that area of asquareofeachside'a',thatisA=a×a=a2cm.Areaofanysquare=a2units,where'a'isthemeasureofeachside.
How to measure the area of irregular shapes?Wecanestimatetheareaofirregularshapesbydrawing
theirregularfigureonagraphsheet.Countthenumberofunitsquaresinsidethefigure.Numberoffullsquares=17.Numberofsquaresmorethanhalf=12.Numberofhalfsquares=0.∴Totalnumberofunitsquares=29.∴Areaofthisirregularshape=29cm2.
Note:Neglectthesquareswhicharehavingarealessthanhalfunitandconsidertheareaofthesquarewithmorethanhalfunitandhalfunitasoneeach.
Activity 1 :- Findtheareaofthefollowingfigures.
1) 2)
Activity 2 :-1) Findtheareaofthefloorofyourclassroom.(insq.ft)2) Find the area of the black board of your class room
(insq.ft).3) Find the area of any one door of your class room
(insq.ft).4) Find the area of coverpage of yourmathematics text
book(insq.cm).
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Worked examples1) Findtheareaofarectanglewhoselengthandbreadthare
10cmand6cm.method :l =10cm,b=6cmAreaoftherectangleA= l×b =10×6=60Areaoftherectangle=60squarecentimetre=60cm2
2)Findtheareaofasquareofside8cm. method :a=8cmAreaofasquare=a2=a×a
=82=8×8=64Areaofthesquare=64sq.cm.
3) Theareaofarectangularflagis72sq.cm.Ifitslengthis9cm.Whatisthewidthoftheflag?method :A=72sq.cm,l=9cm,b=?AreaofrectangleA=l×b 72=l×b72 9 b
972 b
b 8
8=
=
=
#
Thewidthoftheflag=8cm4) A rectangle and a square are of equal areas. Area of
thesquareis144sq.cm.Findthelengthofarectangleofbreadthis8cm?method :AreaofsquareA=144sq.cm.Breadthoftherectangleb=8cmlengthoftherectanglel=? Areaofrectangle=Areaofthesquare l×b=144 l×8=144
l= 8144
l=18cm ∴lengthoftherectangle=18cm.
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5) Findtheareaofthesquarewhoseperimeteris20cmmethod:P=20cm. ∴AreaofthesquareA=a2
P=4a 20 420 4
420
5
a
a
a
a cm
5
==
=
=
#
52=5×5 A=25sq.cm
6) Findtheareaofthefollowingfiguresbydividingthemintoseveralrectanglesorsquares.
1) 2cm
3cm
3cm 4cm
4cm
1cm 1cm1 1
11 1D C
A
E F
B K J
H2
2G
44
AreaofrectangleABCD=l ×b=4×1=4sqcm
AreaofrectangleGHJK=4×1=4sqcm
AreaofrectangleCEFG=2×1=2sqcm
∴ Totalareaoftherectangle=4+4+2=10sqcm
method:
2)
method:
AreaofRectangleABKL=l ×b=2×1=2AreaofsquareBCDE=1×1=1
AreaofRectangleEFGH=2×1=2AreaofSquareHIJK=1×1=1AreaofSquareKHEB=1×1=1TotalArea=2+1+2+1+1=7
TotalAreaofthefigure=7sq.cm
1
1 12
2 2
1
1 1
J I
DC
A
KL GH
B E F
2cm1cm
2
2
21
1
1 1
1
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exercise 8.3
I. Findtheareaofthefollowingrectangleswhosesidesare
a) 5cmand4cm
b) 6.5cmand2cm
c) 7cmand4cm
d) 1mand20cm
II. Findtheareaofthesquarewhosesidesare
a.15cm b.12cm
c.8cm d.20cm
III.Solvethefollowingproblems.
1) Findthebreadthofarectangulargardenwhoseareais400sq.mandlengthis25m.
2) Find the cost of carpet required to spread over arectangularfloorof a roomof length12metersandbreadth4metersattherateof̀ 240persquaremeter?
3) Findhowmanysquaremetersofcarpetisrequiredtocoverasquareshapedfloorofsidewhichmeasures2.5meter.
4) Findthewidthofarectangularfieldofarea4400sqmtsandwhoselengthis110m.
5) Findtheareaofasquarewhoseperimeteris40cm.
6) Findthebreadthofarectangleof length20cmandperimeter50cm.
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7) Findtheareaofthefollowingfiguresbydividingintoseveralrectanglesandsquares.(Measurementsareincms)
A
BC
D E
51 1
2 2
3
1
FG
HG 6
42
7
62
4
7
A
B C
H
D E
Fa) b)
Try:
4
4
4A
D C
B
4
ABCDisasquare.Numericalvalueofareaandperimeterofthissquareisequal.Itis16.Trytofindafewotherexamplesofthistype.Whether this ispossible in caseofrectangle!Think!
Activity :- 1) Write a few examples of pairs of rectangles and squares whose perimeters are equal and areas are different.
Eg: 8
8
4 46
6
6
6
2)Iftheperimeterofarectangleandsquareareequalthenwhichhasmorearea?Knowitbycollectingsomeexamplesofthistype.
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AnswersUNIT - 1 PLAYING WITH NUMBERS -EXERCISE 1.1
I A) Numbers divisible by 2:- a) 256 b) 394 c) 618 d) 708 e) 692 f) 846 h) 6,852 j)6,872 B) Numbers divisible by 3:- c) 618 d)708 f) 846 h) 6,852I. 3,051 II. A) Numbers divisible by 4 a) 692 b) 376 c) 5,872 d) 8,000 f) 36,420 g) 58,628 i) 30,148 j) 20,928B) Numbers divisible by 8 c) 5,872 d) 8,000 j) 20,928III. a) 3,474 b) 6,234 d) 3,870 e) 6,252 g) 59,052 i) 40,008IV. a) 2 b)1 c) 0 d)2 V. b)9,486 c) 5,670 e) 4,653 j) 50,985VI. a) 4,719 b) 8,228 d) 2,926 e) 8,987 g) 42,163 h)80,564 j) 68,035EXERCISE 1.2 (a) I.1. b) 2 2. a) 4 3. b) 2 4. d) 1II. 1. (2,13) (3,7) (7,13) (11,19) (13,17) 2. (2,13) (2,19) (5,7) (5,13) (5,19) (7,11) (7,17) (11,19) (13,17)EXERCISE 1.2 (b) I,1) 4,7 3) 15,8 4) 21,20 6) 2,9 7) 14,81II 1. a) Yes, Number of cows = 4, Number of calves = 32. a) No, b) 16 and 13 are co-prime numbersEXERCISE 1.3 I. a) 20 = 2 × 2 × 5 b) 26 = 2 × 13,c) 40 = 2 × 2 × 2 × 5 d) 80 = 2 × 2 × 2 × 2 × 5 e) 300 = 2 × 2 × 3 × 5 × 5 f) 570 = 2 × 3 × 5 × 19 g) 680 = 2 × 2 × 2 × 5 × 17 h) 144 = 2 × 2 × 2 × 2 × 3 × 3 i) 500 = 2 × 2 × 5 × 5 × 5j) 1000 = 2 × 2 × 2 × 5 × 5 × 5 EXERCISE 1.4 - I. a) 14, 28, 42, 56, ..... b) 25, 50, 75, 100, .....c) 21, 42, 63, 84, ..... d) 31, 62, 93, 124, .....e) 42, 84, 126, 168, ..... f) 60, 120, 180, 240, .....g) 67, 134, 201, 268, ..... h) 100, 200, 300, 400, .....i) 96, 192, 288, 384, ..... j) 75, 150, 225, 300, .....II. a) 6, 12, 18, 24, 30 b) 11, 22, 33, 44, 55 c) 15, 30, 45, 60, 75d) 24, 48, 72, 96, 120 e) 30, 60, 90, 120, 150III. A B (answers)1. 1 Factors of all the numbers2. 18 Multiple of 63. 20 Multiple of 54. 49 Multiple of 7EXERCISE 1.5 I a) 1,3 b)1, 2, 3, 4, 6, 12 c)1, 2, 4, 5, 10 d) 1 e)1, 2 f) 1 g) 1 h) 1, 2, 5, 10
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II. a) Factors of 6 = { 1, 2, 3, 6} Factors of 24 = {1, 2, 3, 4, 6, 12, 24} Common Factors of 6 and 24 = {1, 2, 3, 6} ∴HCF of 6 and 24 = 6 b)1 c) 24 d) 8 e) 2 f) 9 g) 12 h) 17 III. a) 5 b) 7 c) 3 d) 9 e) 4 f) 9 g) 30 h) 7 IV.1 a) 1 b) 1c) 1d)1[Co-Prime numbers] 2. a)5 b)14 c)24 d)18 [ Multi-ples of a numbers] V. 1) Length of longest measuring tape = HCF of 20 and 8 = 4m, 2)a) Maximum quantity of dal in each bag = HCF of 56 and 96,= 8 kgs b) Number of bags required= 193) Volume of the biggest measure = HCF of 18 and 24 = 6 ltr.EXERCISE 1.6 - I. 1. a)84 b)165 c) 65 d) 34 Common Property:-LCM of mutually prime numbers is the product of those numbers. 2. a) 18 b) 80 c) 40 d) 60 Common Property:- If one number is the multiple of another, then that multiple (greater number) is their LCMII. 1) a) 40 b) 48 c) 120 d) 700 2) a) 120 b) 96 c) 180 d) 300III. 1) 72 2) 32ltr 3) 5460 cms 4) 8'o'clock in the morning.
UNIT-2 FRACTIONS - EXERCISE 2.1 I. a) b) c) d) EXERCISE 2.2- II. a) .b) c). e) f)
III. a) b) c) d) e) f)
EXERCISE 2.3 - I. 1)
2) 3)
4)
II.1) 2)
3) 4) III. 1) 6 2) 2 3) 18
4) 28 5) 3 6) 6 7) 3 8) 5 IV. 1) Yes 2) No 3) Yes 4) Yes
V.1) 2) 3) 4) 5) 6) 7) 8)
EXERCISE 2.4 I A. 1) < 2) > 3) > 4) < B. 1) <2) > 3) = 4) <II A) 1) 2)
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3) 4)
B) 1) 2)
3) 4)
III. 1) 2) 3) IV. a) = b)<c)<d)>e)<f)<g) <
V. 1) Mary 2) Sugar Cane.
VI. 1) < 2) > 3) > 4) >EXERCISE 2.5- I. A. 1) 2) 3) 4) , B. 1) 7 2) 3) 4) 4II. a) b) c) d) e)
III. A. 1) 2) 3) 4) B. 1) 2) 3) 4)
IV. A. 1) 2) 3) B. 1) 2)
EXERCISE 2.6- I 1) a) b) c) d) 2) a) b) c)
d) II a) b) c) d) e) III 1) a) b) c) d)
2) a) b) c) d) IV 1) 2) 3) V 1)
2) 3)
UNIT-3 DECIMAL - EXERCISE 3.1
abc
H T U Tenths5 1 3
1 3 72 4 3
I 1) H T U Tenths0 72 8
2 6 71 6 5 4
abcd
2)
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3) a) 0.4 b) 0.9 c) 67.6 d) 600.7 e) 3.2 4) a) 3.7 b) 25.9 c) 207.3
d) 640.2 e) 20.1 II. 1) A→ 0.2 B→ 1.6 C→ 2.4 D→ 3.6
3) a) b) c) d) e) 4) a) 0.7 b) 2.1 c) 0.8
d) 2.2 e) 8.5 5) a) 0.7 cm b) 2.7 cm c) 3 cm d) 4.5 cm e) 6.8 cm
EXERCISE 3.2
I.
abc
ones Tenths Hundreths2 5 3
3 83 0 6
II. 5.420, 18.05423.279,107.350
III. 1)Hun-dreds (100)
Tens (10)
Ones (1)
Tenths ( )
Hundreths ( )
Thou-sandths
2) a) 3.52 b) 56.09 c) 25.754 d) 18.054 e) 628.0073) a) b)
c) d)
e)
0 3 52 4 3
2 5 0 2 72 5 6 4 9
8 7 5 6
abcde
EXERCISE 3.3- I 1. a) ` 0.5 b) ` 0.05 c) ` 0.4 d) ` 4.8 e) ` 3.252. a) 0.8 cm b) 2.5 cm c) 7.5 cm d) 8.4 cm e) 17.5 cm3. a) 0.6 m b) 0.06 m c) 2.3 m d) 2.03 m e) 3.78 m4. a) 0.876 km b) 0.076 km c) 0.006 km d) 2.068 km e) 3.005 km5. a) 0.763 kg b) 0.063 kg c) 0.003 kg d) 3.054 kg e) 2.825 kg6. a) 0.675 l b) 0.075 l c) 0.005 l d) 2.380 l e) 5.040 lEXERCISE 3.4 - I a) < b) > c) > d) < e) < II a) 8.2 cm b) 0.85m c)2.047 km d) 6.509 kg e) 3.425 lEXERCISE 3.5 I. A) 1)1.484 2) 13.671 3) 8.587 4) 18.979 5) 191.943B)1) 6.2m 2) `138.45 3) 7.050 4)12.300km 5)5.850 II. A)1)0.32 2) 1.62 3) 2.86 4) 7.75 5) 2.216 6) 1.583 B)1) ` 14.05 2)5.86 m 3) 7.72 4) 3.150 kg 5) 1l.702ml 6) 3.60c
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UNIT-4 INTRODUCTION TO ALGEBRA - EXERCISE 4.1
I 1) 2x 2) 3x 3) 4x 4) 5x 5) 2x 6) 3x 7) 5x 8) 6x 9) 5x 10) 6x
II 1) The Letters T, V, L, Y, give us the rule 2n.
The Letters A, C, F, N, U, Z, give us the rule 3n.
The Letters D, J, K, M, O, W, X, give us the rule 4n.
The Letters E, P, Q, S, give us the rule 5n.
The Letters G, R, give us the rule 6n.
2) 6n 3) 4s 4) l + 8 5) x - 3 6) 75b 7) c + 20 8) 25v
EXERCISE 4.2- I 1) 3a 2) 5a 3) 12a 4) d = 2r
EXERCISE 4.3- I. c) 16-2 g) 6(23 - 5) + 8 × 3 II. a) y + 5 b) p -7 c) 5 x d)
e) -m-6 f) -5z g) h) i) -15r j) 5y+3 k) 5y-3 l) -8p+6 m) 4p-15
n) -4p+15 o) 1)y+4 2) y-5 3) 6y 4) 6y-2 5) 3y+5
III 1) 2 is added to ‘x’. 2) ‘a’ is multiplied by 8 then 3 is subtracted.
from the product. 3) 1 is added to y then the result is divided by 4.
4) ‘P’ is multiplied by 7 then 3 is added to the product.
5) ‘m’ is multiplied by 10 then ‘n’ is added to the product then result
is multiplied by ‘y’.
UNIT - 5 RATIO AND PROPORTION-EXERCISE 5.1
I. a) 4 is to 3 b) 9 is to 17 c) x is to y d) 125 is to 20
II.1) 1 : 1 2) 3 : 2 3) 5 : 16 4) 12 : 17 5) 20 : 6 6) 75 : 65 7) a : b
8) m : n
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III Antecedent 1 2 8 4 11 50 13 mIV 1) , 2) , 3)
4) 5) Consequent 2 7 5 6 12 75 1 n
V.1) 2 : 3 2) 7 : 4 3) 12 : 15 4) 35 : 65 5) 76 : 84 6) 20 : 101
EXERCISE 5.2 I. a) 4 : 3 b) 1 : 5 c) 1 : 4 d) 4 : 15 e) 20 : 3 II.5 : 3
III a) 10 : 9 b) 19 : 10 IV a) 1 : 6 b) 1 : 5 c) 5 : 6EXERCISE 5.3 I
a b c d e fExtremes 3,8 5,12 2,12 6,24 7,27 20,5
Means 6,4 3,20 3,8 12,12 21,9 10,10
II.a) is in proportion. b) , c) and d) are not in proportion.
III.a) 9 b) 12 c) 3 d) 6 IV. 1) `256 2) 4litre 3) `50
EXERCISE 5.4 1) `25 2) `5,250 3) 0.8kg 4) 31 litre
5) a) `320 b) 75kg
UNIT - 6 SYMMETRY- EXERCISE 6
III .1)H, X 2) K, M 3) P, R IV 0,1,3,8
UNIT - 8 MENSURATION-EXERCISE 8.1
1) a) 30 cm b) 20 cm. 2) 18 cm, 16 cm, 30 cm, 32 cm.3) 24 cm,
14 cm, 20 cm, 100 cm.4) 20 cm. 5) 9 cm.6) 10 cm.7) a)15 cmb)25
cmc)30 cm d) 40 cm 8) 260 m.9) 1,280 m. 10) ` 700
EXERCISE 8.2 2) 16 cm 3) 5 cm, 6cm
EXERCISE 8.3 I. a)20 sq.cm b)13 sq.cm c) 28 sq.cm d) 20 sq.cm
II. a) 225 sq.cm b) 144 sq.cm c) 64 sq.cm d)400 sq.cm
III. 1)16 cm. 2) `11,520 3) 6.25 sq.m 4) 40 m 5) 100 sq.cm 6) 5 cm
7) a) 34 sq.cm b) 8 sq.cm
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