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1.1 Introduction

1.2 Revision-Profit,LossandSimpleInterest

1.3 ApplicationofPercentage,ProfitandLoss,OverheadExpenses,DiscountandTax

1.4 CompoundInterest

1.5 DifferencebetweenS.I.andC.I.

1.6 FixedDepositsandRecurringDeposits

1.7 CompoundVariation

1.8 TimeandWork

1.1 Introduction

Every human being wants to reach the height of ‘WIN-WIN’situation throughouthis life.Toachieve iteffectively,heallocateshistimetowork,toearnmorewealthandfame.

Fromstoneagetopresentworld,frommaterialexchangetomoneytransaction,forhisproduceandland,manhasbeenapplyingtheideaofratioandproportion.ThemonumentalbuildingsliketheTajMahalandtheTanjoreBrihadisvaraTemple,knownfortheiraestheticlooks,alsodemonstrateourancestors’knowledgeandskillofusingrightkindofratiotokeepthemstrongandwonderful.

Manyoftheexistingthingsintheworldareconnectedbycauseandeffectrelationshipasinrainandharvest,nutritionandhealth,incomeandexpenditure,etc.andhencecompoundvariationarises.

Inourefforttosurviveandambitiontogrow,weborrowordepositmoneyandcompensatetheprocesspreferablybymeansofcompoundinterest. The government bears the responsibility of the sectors likesecurity,health,educationandotheramenities.Todeliver these toallcitizens,wepayvarioustaxesfromourincometothegovernment.

Thischaptercoversthetopicswhichareinterwoveninourlives.

Roger Bacon[1214-1294]

Englishphilosopher,Wonderfulteacheremphasisedonempiricalmethods.Hebecameamaster at Oxford.

He stated: “Neglectofmathematics worksinjurytoallknowledge”.Hesaid,“The

importanceofmathematics for a common man tounderpinnedwheneverhevisitsbanks,shoppingmalls,railways,postoffices,insurancecompanies,ordealswithtransport,businesstransaction,importsandexports,tradeandcommerce”.

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1.2 Revision: Profit, Loss and Simple Interest

Wehavealreadylearntaboutprofitandlossandsimpleinterestinourpreviousclass.Letusrecallthefollowingresults:

ReSuLtS on PRofIt, LoSS and SImPLe InteReSt

(i) ProfitorGain = Sellingprice–Costprice

(ii) Loss = Costprice–Sellingprice

(iii) Profit% = 100.C.P.Profit

#

(iv) Loss% = 100C.P.Loss

#

(v)Simpleinterest(I) =100

Principal Time Rate# # Pnr100=

(vi) Amount = Principal+Interest

1.3 application of Percentage, Profit and Loss, overhead expenses, discount and tax

1.3.1. application of Percentage

Wehavealreadylearntpercentagesinthepreviousclasses.Wepresenttheseideasasfollows:

(i) Twopercent = 2%=1002

(ii) 8%of600kg =1008 ×600=48kg

(ii) 125% =100125 =

45 =1

41

Now,welearntoapplypercentagesinsomeproblems.Example 1.1

Whatpercentis15paiseof2rupees70paise?Solution 2rupees70paise = (2×100paise+70paise) = 200paise+70paise = 270paise Requiredpercentage =

27015 100# =

950 =5 %

95 .

=21 =50%

=41 =25%

=43 =75%

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Example 1.2Findthetotalamountif12%ofitis`1080.SolutionLetthetotalamountbex .Given : 12%ofthetotalamount = `1080 x

10012

# = 1080 x = ×

12108 1000 =`9000

` Thetotalamount = `9000.Example 1.3

72% of 25 students are good in Mathematics. How many are not good inMathematics?

Solution PercentageofstudentsgoodinMathematics = 72% NumberofstudentsgoodinMathematics = 72%of25students =

10072 25# = 18 students

NumberofstudentsnotgoodinMathematics = 25–18=7.Example 1.4

Findthenumberwhichis15%lessthan240.Solution 15%of240 =

10015 240# =36

` Therequirednumber = 240–36=204.Example 1.5

ThepriceofahouseisdecreasedfromRupeesFifteenlakhstoRupeesTwelvelakhs.Findthepercentageofdecrease.

Solution Originalprice = `15,00,000 Changeinprice = `12,00,000 Decreaseinprice = 15,00,000–12,00,000=3,00,000 ̀ Percentageofdecrease= 100

1500000300000

# =20%Remember

Percentageofincrease = Original amountIncrease in amount ×100

Percentageofdecrease = Original amountDecrease in amount ×100

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exeRcISe 1.1

1. Choosethecorrectanswer.

(i) Thereare5orangesinabasketof25fruits.Thepercentageoforangesis___ (A)5% (B)25% (C)10% (D)20%

(ii) 252 =_______%.

(iii) 15%ofthetotalnumberofbiscuitsinabottleis30. Thetotalnumberofbiscuitsis_______.

(A)100 (B)200 (C)150 (D)300

(iv) Thepriceofascooterwas` 34,000lastyear.Ithasincreasedby25%thisyear.Thentheincreaseinpriceis_______.

(A) `6,500 (B)`8,500 (C)`8,000 (D)`7,000

(v) A man saves `3,000permonthfromhistotalsalaryof`20,000.Thepercentageofhissavingsis_______.

(A)15% (B)5% (C)10% (D)20%

2. (i) 20%ofthetotalquantityofoilis40litres. Findthetotalquantityofoilinlitres.

(ii) 25%ofajourneycovers5,000km.Howlongisthewholejourney? (iii) 3.5%ofanamountis` 54.25. Find the amount. (iv) 60%ofthetotaltimeis30minutes.Findthetotaltime. (v) 4%salestaxonthesaleofanarticleis`2.Whatistheamountofsale?

3. Meenuspends`2000fromhersalaryforrecreationwhichis5%ofhersalary.Whatishersalary?

4. 25%ofthetotalmangoeswhicharerottenis1,250.Findthetotalnumberofmangoesinthebasket.Also,findthenumberofgoodmangoes.

15sweetsaredividedbetweenSharathandBharath,sothattheyget20%and80%ofthemrespectively.Findthenumberofsweetsgotbyeach.

MyGrandmasays,inherchildhood,goldwas`100pergram.Readanewspaper toknow thepriceofgoldandnotedownthepriceon thefirstofeverymonth.Findthepercentageofincrease every month.

(A)25 (B)4 (C)8 (D)15

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5. ThemarksobtainedbyRaniinhertwelfthstandardexamsaretabulatedbelow.Expressthesemarksaspercentages.

Subjects maximum marks marks obtained Percentage of

marks (out of 100)(i)English 200 180(ii)Tamil 200 188(iii) Mathematics 200 195(iv) Physics 150 132(v) Chemistry 150 142(vi)Biology 150 140

6. Aschoolcricketteamplayed20matchesagainstanotherschool.Thefirstschoolwon25%ofthem.Howmanymatchesdidthefirstschoolwin?

7. Rahimdeposited`10,000inacompanywhichpays18%simpleinterestp.a.Findtheinteresthegetsforaperiodof5years.

8. Themarkedpriceofatoyis`1,200.Theshopkeepergaveadiscountof15%.Whatisthesellingpriceofthetoy?

9. Inaninterviewforacomputerfirm1,500applicantswereinterviewed.If12%ofthemwereselected,howmanyapplicantswereselected?Alsofindthenumberofapplicantswhowerenotselected.

10. Analloyconsistsof30%copperand40%zincandtheremainingisnickel.Findtheamountofnickelin20kilogramsofthealloy.

11. PandianandThamaraicontestedfortheelectiontothePanchayatcommitteefromtheirvillage.Pandiansecured11,484voteswhichwas44%ofthetotalvotes.Thamaraisecured36%ofthevotes.Calculate(i)thenumberofvotescastinthevillageand(ii)thenumberofvoterswhodidnotvoteforboththecontestants.

12. Amanspends40%ofhisincomeforfood,15%forclothesand20%forhouserentandsavestherest.Whatisthepercentageofhissaving?Ifhisincomeis `34,400,findtheamountofhissavings.

13. Jyothikasecured35marksoutof50inEnglishand27marksoutof30inMathematics.Inwhichsubjectdidshegetmoremarksandhowmuch?

14. Aworkerreceives`11,250asbonus,whichis15%ofhisannualsalary.Whatishismonthlysalary?

15. Thepriceofasuitisincreasedfrom`2,100to`2,520.Findthepercentageofincrease.

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1.3.2 applications of Profit and Loss Inthissection,welearntosolveproblemsonapplicationsofprofitandloss.

(i) Illustration of the formula for S.P.Considerthefollowingsituation:Rajesh buysapenfor`80andsellsittohisfriend.

Ifhewantstomakeaprofitof5%,canyousaythepriceforwhichhewouldhavesold?

Rajesh bought the pen for ` 80 which is the CostPrice(C.P.).Whenhesold,hemakesaprofitof5%whichiscalculatedontheC.P.

` Profit = 5%ofC.P.=1005 ×80=` 4

Sincethereisaprofit,S.P.>C.P. S.P. = c.P. + Profit = 80+4=` 84.` ThepriceforwhichRajeshwouldhavesold=` 84.Thesameproblemcanbedoneusingtheformula.

Sellingprice(S.P.) = 100100 Profit% C.P#

+^ h .

=100

100 580#

+^ h =100105 80# =` 84.

1. 40%=100%–_____% 2. If25%ofstudentsinaclasscometoschoolbywalk,65%ofstudents

come by bicycle and the remaining percentage by school bus, whatpercentageofstudentscomebyschoolbus?

3. Inaparticularclass,30%ofthestudentstakeHindi,50%ofthemtakeTamilandtheremainingstudentstakeFrenchastheirsecondlanguage.WhatisthepercentageofstudentswhotakeFrenchastheirsecondlanguage?

4. Inacity,30%arefemales,40%aremalesandtheremainingarechildren.Whatisthepercentageofthechildren?

AmuthabuyssilksareesfromtwodifferentmerchantsGanesanandGovindan.Ganesanweaves200gmofsilverthreadwith100gmofbronzethreadwhereasGovindanweaves300gmofsilverthreadwith200gmofbronzethreadforthesarees.Calculatethepercentageofsilverthreadineachandfindwhogivesabetterquality.[Note:Morethesilverthreadbetterthequality.]

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(ii) Illustration of the formula for c.P.Considerthefollowingsituation:Supposeashopkeepersellsawristwatchfor`540

makingaprofitof5%,thenwhatwouldhavebeenthecostofthewatch?

Theshopkeeperhadsoldthewatchataprofitof5%ontheC.P.SinceC.P.isnotknown,letustakeitas`100.

Profitof5%ismadeontheC.P. ` Profit = 5%ofC.P. =

1005 ×100

Weknow, S.P. = c.P. + Profit = 100+5 = `105.Here,ifS.P.is`105,thenC.P.is`100.WhenS.P.ofthewatchis`540,C.P. =

105540 100# = ` 514.29

` Thewatchwouldhavecost`514.29fortheshopkeeper.Theaboveproblemcanalsobesolvedbyusingtheformulamethod.

c.P. = %100100profit S.P.#

+c m

=100 5100+

×540

=105100 ×540

= ` 514.29.

WenowsummarizetheformulaetocalculateS.P.andC.P.asfollows:

1. Whenthereisaprofit 1. Whenthereisaloss

(i)C.P.= 100 profit%100 S.P.#

+c m (ii)C.P.= 100 loss%

100 S.P.#-

` j

2.Whenthereisaprofit 2. Whenthereisaloss,

(i)S.P.= %

100

100 profit C.P.#+

c m (ii)S.P.= %100

100 loss C.P.#-` j

=` 5.

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Revathiusedtheformulamethod: Loss = 15%. S.P. = `13,600

C.P. = 100 Loss%100-

×S.P.

=100 15

100-

×13600

=85100 ×13600

= `16,000

Example 1.6 HameedbuysacolourT.Vsetfor`15,200andsellsitatalossof20%.Whatis

thesellingpriceoftheT.Vset?SolutionRaghulusedthismethod:Lossis20%oftheC.P.

= 1520010020

#

= `3040

S.P. = C.P.–Loss

= 15,200–3,040

= `12,160

BothRaghulandRoshancameoutwiththesameanswerthatthesellingpriceoftheT.V.setis`12,160.

Example 1.7 Ascootyissoldfor`13,600andfetchesalossof15%.Findthecostpriceof

the scooty.Deviusedthismethod:Lossof15%means,if C.P. is `100,Loss=` 15Therefore,S.P.wouldbe (100–15) = ` 85IfS.P.is`85,C.P.is`100WhenS.P.is`13,600,then C.P. =

85100 13600#

= `16,000Hencethecostpriceofthescootyis`16,000.

Items cost price in ` Profit/LossSelling

price in `WashingMachine 16,000 9%Profit

MicrowaveOven 13,500 12%LossWoodenShelf 13%Loss 6,786

Sofaset 12½%Profit 7,000Air Conditioner 32,400 7%Profit

Roshanusedtheformulamethod: C.P. = `15,200 Loss = 20% S.P. = 100

100 Loss%- × C.P.

= 15200100

100 20#-

= 1520010080

#

= `12,160

OR

OR

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Example 1.8 Thecostpriceof11pensisequaltothesellingpriceof10pens.Findtheloss

orgainpercent.SolutionLetS.P.ofeachpenbex. S.P.of10pens = `10x S.P.of11pens = ` 11xGiven: C.P.of11pens = S.P.of10pens=`10xHere,S.P.>C.P. ` Profit = S.P.–C.P. = 11x–10x=x Profit% = C.P.

Profit ×100=x

x10

×100=10%.Example 1.9

Amansellstwowristwatchesat`594each.Ononehegains10%andontheotherheloses10%.Findhisgainorlosspercentonthewhole.

SolutionGiven :S.P.ofthefirstwristwatch =`594,Gain%=10%

` C.P.ofthefirstwristwatch =%100 profit

100 S.P.#+

= 594100 10

100#

+^ h

= 594110100

# =`540.

Similarly,C.P.ofthesecondwatchonwhichheloses10%is

= 100 Loss%100 S.P.#-^ h

= 594100 10

100#

-^ h =90100 594# =` 660.

TosaywhethertherewasanoverallProfitorLoss,weneedtofindthecombinedC.P.andS.P.

TotalC.P.ofthetwowatches = 540+660=`1,200. TotalS.P.ofthetwowatches = 594+594=`1,188. NetLoss = 1,200–1,188=` 12.

Loss% = 100C.P.Loss

#

=120012 100# =1%.

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1.3.3. application of overhead expenses

Maya went with her father to purchase an Aircooler.Theybought it for` 18,000.The shopwhereinthey bought was not closer to their residence. Sothey had to arrange a vehicle to take the air cooler totheir residence. They paid conveyance charges of ` 500. Hence the C.P. of the air cooler is not only` 18,000 but it also includes the conveyance charges (Transportation charges) `500whichiscalledasoverhead expenses .

Now, C.P.oftheaircooler = Realcost+Conveyancecharges = 18,000+500=`18,500Consideranothersituation,whereKishore’sfatherbuysanoldMaruticarfrom

aChennaidealerfor`2,75,000andspends`25,000forpaintingthecar.Andthenhetransportsthecartohisnativevillageforwhichhespendsagain`2,000.Canyouanswerthefollowingquestions:

(i) Whatisthetheoverallcostpriceofthecar? (ii) Whatistherealcostpriceofthecar? (iii) Whataretheoverheadexpensesreferredhere?In the aboveexamplethepaintingchargesandthetransportationchargesare

theoverheadexpenses.

\Costpriceofthecar = Realcostprice+Overheadexpenses

= 2,75,000+(25,000+2,000) = 2,75,000+27,000=`3,02,000.Thus,wecometotheconclusionthat,

Sometimeswhen an article is bought or sold, some additional expenses occurwhilebuyingorbeforesellingit.Theseexpenseshavetobeincludedinthecostprice.Theseexpensesarereferredtoasoverhead expenses.Thesemayincludeexpenseslikeamountspentonrepairs,labourcharges,transportation,etc.,

Example 1.10Raju bought amotorcycle for` 36,000 and then bought some extra fittings

tomakeitperfectandgoodlooking.Hesoldthebikeataprofitof10%andhegot `44,000.Howmuchdidhespendtobuytheextrafittingsmadeforthemotorcycle?

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SolutionLettheC.P.be`100. Profit = 10%,S.P.=`110IfS.P.is`110,thenC.P.is`100.WhenS.P.is`44,000 C.P. =

11044000 100# =`40,000

\Amountspentonextrafittings=40,000–36,000=`4,000.

exeRcISe 1.2

1. FindtheCostprice/Sellingprice.

cost price Selling price Profit Loss

(i) `7,282 `208

(ii) ` 572 ` 72

(iii) `9,684 ` 684

(iv) `1,973 ` 273

(v) `6,76,000 `18,500

2. Filluptheappropriateboxesandleavetherest.

C.P. S.P. Profit & Profit % Loss & Loss%

(i) `320 ` 384

(ii) `2,500 `2,700

(iii) `380 ` 361

(iv) `40 ` 2 Loss

(v) `5,000 `500Profit.

3. FindtheS.P.ifaprofitof5%ismadeon

(i) abicycleof`700with`50asoverheadcharges.

(ii) acomputertableboughtat`1,150with`50astransportationcharges.

(iii) atable-topwetgrinderboughtfor`2,560andanexpenseof`140onrepaircharges.

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4. Bysellingatablefor`1,320,atradergains10%.FindtheC.P.ofthetable. 5. Thecostpriceof16notebooksisequaltothesellingpriceof12notebooks.Find

thegainpercent.

6. Amansoldtwoarticlesat`375each.Onthefirstarticle,hegains25%andontheother,heloses25%.Howmuchdoeshegainorloseinthewholetransaction?Also,findthegainorlosspercentinthewholetransaction.

7. Anbarasanpurchasedahousefor̀ 17,75,000andspent̀ 1,25,000onitsinteriordecoration.He sold thehouse tomakeaprofitof20%.Find theS.P.of thehouse.

8. AfterspendingRupeessixtythousandforremodellingahouse,Amlasoldahouseataprofitof20%.IfthesellingpricewasRupeesfortytwolakhs,howmuchdidshespendtobuythehouse?

9. Jaikumarboughtaplotof land in theoutskirtsof thecity for`21,00,000.Hebuiltawallarounditforwhichhespent`1,45,000.Andthenhewantstosellit at `25,00,000bymakinganadvertisementinthenewspaperwhichcostshim `5,000.Now,findhisprofitpercent.

10. Amansoldtwovarietiesofhisdogfor`3,605each.Ononehemadeagainof15%andontheotheralossof9%.Findhisoverallgainorloss.

[Hint: FindC.P.ofeach]

1.3.4 application of discounts

YesterdayPoojawent to a shopwith her parents topurchaseadressforPongalfestival.Shesawmanybannersintheshop.Thecontentofwhichwasnotunderstandbyher.

With an unclear mind, she entered the shop andpurchasedafrock.

Thepricelabelledonthefrockwas`550.ItiscalledasMarkedPrice(abbreviatedasM.P.)andshegavetheshopkeeper` 550.Buttheshopkeeperreturnedthebalanceamountandinformedherthattherewasadiscountof20%.

Here,20%discountmeans,20%discountontheMarkedPrice.

Discount=10020 ×550=`110.

discount is the reduction in value on themarked Price or List Price of the article.

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AmountpaidbyPoojatotheshopkeeperis`440

= `550- `110

= MarkedPrice-DiscountHenceweconcludethefollowing:

Discount = MarkedPrice-SellingPrice SellingPrice = MarkedPrice-Discount MarkedPrice = SellingPrice+Discount

Example 1.11A bicyclemarked at ` 1,500 is sold for ` 1,350.What is the percentage of

discount?SolutionGiven :MarkedPrice=`1500,SellingPrice=`1350 Amountofdiscount = MarkedPrice–SellingPrice = 1500–1350 = `150 Discountfor`1500 = `150

Discountfor`100 =1500150 100#

Percentageofdiscount = 10%.Example 1.12

Thelistpriceofafrockis`220.Adiscountof20%onsalesisannounced.Whatistheamountofdiscountonitanditssellingprice?

Solution Given : List(Marked)Priceofthefrock=`220,Rateofdiscount=20% Amountofdiscount = 100

20 220#

= ` 44 \SellingPriceofthefrock = MarkedPrice–Discount = 220–44 = ` 176.

During festival seasonsand in the Tamil monthof ‘‘Aadi’’, discounts orrebates of 10%, 20%,30%, etc., are offered toattract customers byCo-optex, Khadi and othershopsforvariousitemstopromotesales.

SinceDiscountisonMarkedPrice,wewillhavecalculatediscount on M.P.

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Example 1.13Analmirahis sold at ` 5,225 after allowing a discount of 5%. Find its

markedprice.SolutionKrishnausedthismethod:Thediscountisgiveninpercentage.Hence,theM.P.istakenas`100.

Rateofdiscount = 5% Amountofdiscount =

1005 ×100

= ` 5. SellingPrice = M.P.–Discount = 100–5=` 95

IfS.P.is`95,thenM.P.is`100.WhenS.P.is`5225,

M.P. =95100 × 5225

` M.P.ofthealmirah=`5,500.Example 1.14

Ashopkeeperallowsadiscountof10%tohiscustomersandstillgains20%.Findthemarkedpriceofanarticlewhichcosts`450totheshopkeeper.

SolutionVanithausedthismethod:LetM.P.be`100.Discount=10%ofM.P. =

10010 ofM.P.= 100

10010

#

=`10S.P. =M.P.-Discount =100- 10=`90Gain =20%ofC.P. = 450

10020

# =` 90

S.P. =C.P.+Gain =450+90=`540.IfS.P.is` 90,thenM.P.is`100.WhenS.P.is`540,

M.P.=90

540 100# =`600

` TheM.P.ofanarticle=`600

Vigneshusedtheformulamethod: S.P. = Rs5225Discount = 5% M.P. = ?

M.P.=100 Discount%

100 S.P.#-

` j

= 5225100 5100

#-

` j

= 522595100

#

= `5,500.

Vimalusedtheformulamethod:Discount=10%,Gain=20%,C.P.=`450,M.P.=?

M.P. =100 Discount%100 Gain% C.P#-+ .

=100 10

100 20450#

-

+

^^

hh

=90120 450#

=`600

[OR]

[OR]

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Example 1.15 Adealerallowsadiscountof10%andstillgains10%.Whatisthecostprice

ofthebookwhichismarkedat`220?SolutionSugandanusedthismethod:M.P. =`220.Discount =10%ofM.P.

= 22010010

# = ` 22

S.P. =M.P.–Discount =220–22=` 198LetC.P.be`100.Gain =10%ofC.P.

= 10010010

# = ` 10

S.P. =C.P.+Gain =100+10 =`110.

IfS.P.is`110,thenC.P.is`100.WhenS.P.is`198, C.P. =

110198 100#

=`180.Example 1.16

Atelevisionsetwassoldfor`14,400aftergivingsuccessivediscountsof10%and20%respectively.Whatwasthemarkedprice?

Solution SellingPrice = `14,400LettheM.P.be`100. Firstdiscount = 10%= 100

10010

# =`10S.P.afterthefirstdiscount= 100–10=`90 Seconddiscount = 20%= 90

10020

# =` 18SellingPriceaftertheseconddiscount =90 – 18=` 72IfS.P.is`72,thenM.P.is`100.WhenS.P.is`14,400, M.P. =

7214400 100# =`20,000

M.P. = `20,000.

Mukundanusedtheformulamethod:

Discount=10%

Gain =10%

M.P. =`220

C.P. = 100 Gain%100 Discount% M.P.#

+-

=100 10100 10 220#

+-

=11090 220# =`180.

[OR]

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Example 1.17Atraderbuysanarticlefor`1,200andmarksit30%abovetheC.P.Hethen

sellsitafterallowingadiscountof20%.FindtheS.P.andprofitpercent.Solution:LetC.P.ofthearticlebe`100M.P.=30%aboveC.P.=`130If C.P. is `100,thenM.P.is`130.

WhenC.P.is`1200, M.P. =100

1200 130# =`1560

Discount=20%of1560 =10020 1560# =` 312

S.P. =M.P.–Discount

= 1560–312=` 1248

Profit = S.P.–C.P.

= 1248–1200=` 48.

\Profit% = 100C.P.Profit

#

=120048 100# =4%

Ashopgives20%discount.WhatwillbetheS.P.ofthefollowing? (i) Adressmarkedat`120 (ii) Abagmarkedat`250 (iii) Apairofshoesmarkedat`750.

1.3.5 application of tax

Very oftenwe find advertisements in newspapersandon televisionrequestingpeople topay their taxes intime.Whatisthistax?WhydoestheGovernmentcollectthetaxfromthecommonpeople?

Governmentneedsfundstoprovideinfrastructurefacilitieslikeroads,railways,hospitals,schoolsetc.,forthepeople.TheGovernmentcollectsthefundsrequiredbyimposingvarioustaxes.

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Taxesareoftwotypes:

1. direct tax

TaxwhichiscollectedintheformofIncomeTax,PropertyTax,ProfessionalTax,WaterTax,etc.,iscalledasDirectTax.Thesearepaiddirectlytothegovernmentbythepublic.

2. Indirect tax

SomeofthetaxeswhicharenotpaiddirectlytothegovernmentareIndirectTaxesandareexplainedbelow.

excise tax

Thistaxischargedonsomeitemswhicharemanufacturedinthecountry.ThisiscollectedbytheGovernmentofIndia.

Service tax

Tax which is charged in Hotels, Cinema theatres, for service of CharteredAccountants,TelephoneBills,etc.,comeunderServiceTax.ThistaxiscollectedbytheserviceproviderfromtheuseranddepositedtotheGovernment.

Income taxThis is themost important source of revenue for theGovernmentwhich

iscollectedfromeverycitizenwhoisearningmore thanaminimumstipulatedincomeannually.Astruecitizensofourcountry,weshouldbeawareofourdutyandpaythetaxontime.

Sales tax / Value added tax

Sales taxSalesTaxisthetaxleviedonthesalesmade

byaselleratthetimeofsellingtheproduct.Whenthebuyerbuysthecommoditythesalestaxispaidbyhimtogetherwiththepriceofthecommodity.

ThissalestaxischargedbytheGovernmenton thesellingpriceofan itemand isadded to thevalueofthebill.

These days, however, the prices include thetaxknownasValue added tax (Vat).ThismeansthatthepricewepayforanitemisaddedwithVAT.

SalesTaxischargedbytheGovernmentontheSalesofanItem.

7777 7777

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CanyoufindtheprevailingrateofSalesTaxforsomecommoditiesintheyear2011.

1.Electricalinstruments_____%2.Petrol_____% 3.Diesel___%4.Domesticappliances_____%5.Chemicals_____%

calculation of Sales tax

AmountofSalestax = 100Rate of Sales tax Cost of the item#

RateofSalestax = Cost of the itemAmount of Sales tax 100#

Billamount = Costoftheitem+AmountofSalestax

Example 1.18 Vinodhpurchasedmusicalinstrumentsfor`12,000.Iftherateofsalestaxis

8%,findthesalestaxandthetotalamountpaidbyhim.Solution Valueofthemusicalinstruments = `12,000 RateofSalesTax = 8% AmountofSalesTax = 12000

1008

#

= `960 TotalamountpaidbyVinodhincludingSalesTax = 12,000+960 =`12,960

Example 1.19 Arefrigeratorispurchasedfor`14,355,includingsalestax.Iftheactualcost

priceoftherefrigeratoris`13,050,findtherateofsalestax.SolutionGiven:Fortherefrigerator,billamount=`14,355,Costprice=`13,050. Salestax = Billamount–Costoftheitem = 14,355–13,050=`1,305

RateofSalesTax = Cost of the itemAmount of Sales Tax 100#

= 100130501305

# =10%

TheGovernmentgivesexemptionofSalesTaxforsomecommoditieslike rice, sugar, milk, salt, pen,pencilsandbooks.

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Example 1.20 Priyaboughtasuitcasefor`2,730.TheVATforthisitemis5%.Whatwasthe

priceofthesuitcasebeforeVATwasadded?AlsostatehowmuchistheVAT.SolutionGiven :VATis5%.IfthepricewithoutVATis`100,thenthepriceincludingVATis`105.Now,whenpriceincludingVATis`105,originalpriceis`100.WhenpriceincludingVATis`2,730,theoriginalpriceofthesuitcase =

105100 2730# =`2,600

Theoriginalpriceofthesuitcase=` 2,600 \VAT = 2,730–2,600=`130

1. Findthebuyingpriceofeachofthefollowingwhen5%SalesTaxisaddedonthepurchaseof:

(i)Apillowfor`60(ii)Twobarsofsoapat` 25 each. 2. If8%VATisincludedintheprices,findtheoriginalpriceof: (i) An electric water heater bought for ` 14,500 (ii) A crockery set

boughtfor`200.

exeRcISe 1.3 1. Choosethecorrectanswer: (i) Thediscountisalwaysonthe_______. (A)MarkedPrice (B)CostPrice (C)SellingPrice (D)Interest

(ii) IfM.P.=`140,S.P.=`105,thenDiscount=_______. (A) `245 (B)` 25 (C) `30 (D)` 35

(iii) ______=MarkedPrice–Discount. (A)CostPrice (B)SellingPrice (C)ListPrice (D)Marketprice

(iv) Thetaxaddedtothevalueoftheproductiscalled______ Tax. (A)SalesTax (B)VAT (C)ExciseTax (D)ServiceTax

(v) IftheS.P.ofanarticleis`240andthediscountgivenonitis`28,thentheM.P.is_______.

(A) `212 (B)` 228 (C) `268 (D)` 258

2. Thepricemarkedonabookis`450.Theshopkeepergives20%discountonitainbookexhibition.WhatistheSellingPrice?

3. Atelevisionsetwassoldfor`5,760aftergivingsuccessivediscountsof10%and20%respectively.WhatwastheMarkedPrice?

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4. Sekarboughtacomputerfor`38,000andaprinterfor`8,000.Iftherateofsalestaxis7%fortheseitems,findthepricehehastopaytobuythesetwoitems.

5. ThesellingpricewithVAT,onacookingrangeis`19,610.IftheVATis6%,whatistheoriginalpriceofthecookingrange?

6. Richardgotadiscountof10%onthesuithebought.Themarkedpricewas `5,000forthesuit.Ifhehadtopaysalestaxof10%onthepriceatwhichhebought,howmuchdidhepay?

7. Thesalestaxonarefrigeratorattherateof9%is`1,170.Findtheactualsaleprice.

8. Atradermarkshisgoods40%abovethecostprice.Hesellsthematadiscountof5%.Whatishislossorgainpercentage?

9. AT.V.withmarkedprice`11,500issoldat10%discount.Duetofestivalseason,theshopkeeperallowsafurtherdiscountof5%.FindthenetsellingpriceoftheT.V.

10. Apersonpays`2,800foracoolerlistedat`3,500.Findthediscountpercentoffered.

11. Deepapurchased15shirtsattherateof`1,200eachandsoldthemataprofitof5%.Ifthecustomerhastopaysalestaxattherateof4%,howmuchwilloneshirtcosttothecustomer?

12. Findthediscount,discountpercent,sellingpriceandthemarkedprice.

Sl. no Items m. P Rate of

discountamount of discount

S. P

(i) Saree `2,300 20%

(ii) Pen set `140 `105

(iii) Diningtable 20% `16,000

(iv)WashingMachine `14,500 `13,775

(v) Crockeryset `3,224 12½%

Whichisabetteroffer?Twosuccessivediscountsof20%and5%orasinglediscountof25%.Giveappropriatereasons.

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1.4. compound Interest

InclassVII,wehavelearntaboutSimpleInterestandtheformulaforcalculatingSimpleInterestandAmount.Inthischapter,weshalldiscusstheconceptofCompoundInterestand themethodofcalculatingCompoundInterestandAmountattheendofacertainspecifiedperiod.

Vinay borrowed ` 50,000 from a bank for a fixedtimeperiodof2years.attherateof4%perannum.

Vinayhastopayforthefirstyear,

Simpleinterest = n r100

P # #

=100

50000 1 4# # =`2,000

Supposehefailstopaythesimpleinterest`2,000attheendoffirstyear,thenthe interest `2,000isaddedtotheoldPrincipal`50,000andnowthesum=P+I=`52,000becomesthenewPrincipalforthesecondyearforwhichtheinterestiscalculated.

Nowinthesecondyearhewillhavetopayaninterestof

S.I. = n r100

P # #

=100

52000 1 4# # =`2,080

Therefore Vinay will have to pay moreinterest for the second year.

Thisway of calculating interest is calledcompound Interest.

Generallyinbanks,insurancecompanies,postoffices and inother companieswhich lendmoneyandacceptdeposits,compoundinterestisfollowedtofindtheinterest.

Example 1.21Ramlaldeposited`8,000withafinancecompanyfor3yearsataninterestof

15%perannum. WhatisthecompoundinterestthatRamlalgetsafter3years?SolutionStep 1: Principalforthefirstyear = `8,000

Interestforthefirstyear = n r100

P # #

=100

8000 1 15# # =`1,200 Amount at the end of first year = P+I=8,000+1,200=`9,200

When the interest is paidon the Principal only, it iscalledSimple Interest.ButiftheinterestispaidonthePrincipalaswellason theaccruedinterest,itiscalledcompound Interest.

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Step 2: TheamountattheendofthefirstyearbecomesthePrincipalforthe second year.

Principalforthesecondyear = `9,200

Interestforthesecondyear = n r100

P # #

=100

9200 1 15# # =`1,380

Amount at the end of second year = P+I =9,200+1,380=`10,580Step 3: TheamountattheendofthesecondyearbecomesthePrincipalforthe

third year.

Principalforthethirdyear =`10,580

Interestforthethirdyear = n r100

P # #

=100

10580 1 15# # =`1,587

Amount at the end of third year = P+I

= 10,580+1,587=`12,167Hence,theCompoundInterestthatRamlalgetsafterthreeyearsis

A–P = 12,167–8,000=`4,167.

deduction of formula for compound Interest

The above method which we have used for the calculation of CompoundInterestisquitelengthyandcumbersome,especiallywhentheperiodoftimeisverylarge.HenceweshallobtainaformulaforthecomputationofAmountandCompoundInterest.

IfthePrincipalisP,Rateofinterestperannumisr %andtheperiodoftimeorthenumberofyearsisn,thenwededucethecompoundinterestformulaasfollows:

Step 1 : Principalforthefirstyear = P

Interestforthefirstyear = n r100

P # #

= 100P 1 r# # =100

Pr

Amount at the end of first year = P+I

= P 100Pr+

= rP 1 100+` j

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Step 2 : Principalforthesecondyear = P 1 100r+` j

Interestforthesecondyear =1

1001r r

100

P # #+` j

(usingtheS.I.formula)

= r rP 1 100 100#+` j

Amount at the end of second year = P+I

= r r rP 1 100 P 1 100 100#+ + +` `j j

= r rP 1 100 1 100+ +` `j j

= rP 1 1002

+` j

Step 3 : Principalforthethirdyear = rP 1 1002

+` j

Interestforthethirdyear =1

1001P r r

100

2

# #+` j

(usingtheS.I.formula)

= r rP 1 100 1002#+` j

Amount at the end of third year = P+I

= r r rP 1 100 P 1 100 1002 2

#+ + +` `j j

= r rP 1 100 1 1002

+ +` `j j

= rP 1 1003

+` j

Similarly, Amount at the end of nth year is A = 1P 100r n

+` j

and C.I.attheendof‘n’yearsisgivenbyA–P

(i. e.) C. I. = 1100

P Pr n

+ -` j

to compute compound Interest

case 1: compounded annually

WhentheinterestisaddedtothePrincipalattheendofeachyear,wesaythattheinterestiscompoundedannually.

Here 1A P 100r n

= +` j andC.I.=A–P

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case 2: compounded Half - Yearly (Semi - annually)

WhentheinterestiscompoundedHalf-Yearly,therearetwoconversionperiodsinayeareachafter6months.Insuchsituations,theHalf-Yearlyratewillbehalfoftheannualrate,thatis

2r` j.

Inthiscase, 1 r21100A P

n2

= + ` j8 B andC.I.=A–P

case 3: compounded Quarterly

Whentheinterestiscompoundedquarterly,therearefourconversionperiodsinayearandthequarterlyratewillbeone-fourthoftheannualrate,thatis r

4` j.

Inthiscase, 1 r41100A P

n4

= + ` j8 B andC.I.=A–P

case 4: compounded when time being fraction of a year

When interest is compounded annually but time being a fraction.Inthiscase,wheninterestiscompoundedannuallybuttimebeingafractionof

ayear,say5 41 years,thenamountAisgivenby

A = 1r r41

100P 1 100

5

. .

+ +` `j j8 BandC.I.=A–P

for 5 years for ¼ of yearExample 1.22

Find the C.I. on `15,625at8%p.a.for3yearscompoundedannually.SolutionWeknow,

Amountafter3years = rP 1 1003

+` j

= 15625 11008 3

+` j

= 15625 1 252 3

+` j

= 15625 2527 3

` j

= 156252527

2527

2527

# # #

= `19,683Now,Compoundinterest = A–P=19,683–15,625 = `4,058

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to find the c.I. when the interest is compounded annually or half-yearlyLetusseewhathappensto`100overaperiodofoneyearifaninterestis

compoundedannuallyorhalf-yearly.

S.No annually Half-yearly

1P=`100at10%perannumcompoundedannually

P=`100at10%perannumcompoundedhalf-yearly

2 Thetimeperiodtakenis1year Thetimeperiodis6monthsor½year.

3 I100

100 10 1# #= =`10 12I 100

100 10# #= =` 5

4 A=100+10=`110 A=100+5=`105Forthenext6months,P=`105

So,12I 100

105 10# #= =` 5.25

andA=105+5.25=`110.25

5 A=`110 A=`110.25

If interest is compounded half - yearly, the amount is more than whencompoundedannually.Wecomputetheinteresttwotimesandrateistakenashalfoftheannualrate.

Example 1.23 Find the compound interest on` 1000 at the rate of 10%per annum for 18

monthswheninterestiscompoundedhalf-yearly.SolutionHere,P=`1000,r=10%perannumand n=18months=

1218 years= 2

3 years=1 21 years

\Amountafter18months = r21

100P 1

2n+ ` j8 B

= 1000 121

10010 2

23

+#

` j8 B

= 1000 1 20010 3

+` j

= 1000 2021 3

` j

= 10002021

2021

2021

# # #

= ` 1157.625 = ` 1157.63 C.I. = A–P =1157.63–1000=` 157.63

A sum is taken foroneyear at 8%p. a. Ifinterest is compoundedafter every three months, how manytimes will interest bechargedinoneyear?

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Example 1.24 Findthecompoundintereston`20,000at15%perannumfor2

31 years.

Solution

Here,P=`20,000,r=15%p.a.andn =231 years.

Amount after 2 31 years=A = r r

100P 1 100 1 3

12+ +` `j j8 B

=1001520000 1 100

15 1 312

+ +` `j j8 B

= 20000 1203 1

2012

+ +` `j j

= 200002023

20212

` `j j

= 200002023

2023

2021

# # #

= `27,772.50 C.I. = A–P = 27,772.50–20,000 = `7,772.50

Inverse Problems on compound InterestWehavealreadylearnttheformula,A= rP 1 100 ,n+` j

wherefourvariablesA,P,r and nareinvolved.Outofthesefourvariables,ifanythreevariablesareknown,thenwecancalculatethefourthvariable.

Example 1.25 Atwhatrateperannumwill`640amountto`774.40in2years,wheninterest

isbeingcompoundedannually?Solution:Given:P=`640,A=`774.40,n=2years,r=?

Weknow, A = rP 1 100n

+` j

774.40 = r640 1 1002

+` j

.640

774 40 = r1 1002

+` j

6400077440 = r1 100

2

+` j

100121 = r1 100

2

+` j

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1011 2

` j = r1 1002

+` j

1011 = r1 100+

r100 = 10

11 1-

r100 = 10

11 10-

r100 = 10

1

r =10100

Rate r = 10%perannum.

Example 1.26 Inhowmuchtimewillasumof`1600amountto`1852.20at5%perannum

compoundinterest.SolutionGiven:P=`1600,A=`1852.20,r=5%perannum,n=?

Weknow, A = rP 1 100n

+` j

1852.20 = 1600 11005 n

+` j

.

16001852 20

= 100105 n

` j

160000185220

= 2021 n

` j

80009261 =

2021 n

` j

2021 3

` j = 2021 n

` j

\ n =3years

1.5 difference between Simple Interest and compound Interest

WhenPisthePrincipal,n=2yearsandristheRateofinterest,

DifferencebetweenC.I.andS.I.for2years= rP 1002

` j

Example 1.27 FindthedifferencebetweenSimpleInterestandCompoundInterestforasum

of `8,000lentat10%p.a.in2years.SolutionHere,P=`8000, n=2years,r=10%p.a.

Findthetimeperiodand rate for each of the cases givenbelow:1.Asumtakenfor2yearsat

8%p.a.compounded half-yearly.

2.Asumtakenfor1½yearsat4%p.a.compoundedhalf-yearly.

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DifferencebetweenCompoundInterestandSimpleInterestfor2years= rP 1002

` j

= 8000 10010 2

` j

= 8000 101 2

` j

= 8000101

101

# # =`80

exeRcISe 1.4

1. FindtheAmountandCompoundInterestinthefollowingcases:Sl.No. PrincipalinRupees Rate%perannum Timeinyears

(i) 1000 5% 3(ii) 4000 10% 2

(iii) 18,000 10% 221

2.Sangeethaborrowed`8,000fromAlexfor2yearsat12½%perannum.WhatinterestdidSangeethapaytoAlexiftheinterestiscompoundedannually?

3. Maria invested `80,000inabusiness.Shewouldbepaidinterestat5%perannumcompoundedannually.Find(i)theamountstandingtohercreditattheend of second year and (ii) the interest for the third year.

4. Findthecompoundintereston`24,000compoundedhalf-yearlyfor1½yearsattherateof10%perannum.

5. FindtheamountthatDravidwouldreceiveifheinvests`8,192for18monthsat12½%perannum,theinterestbeingcompoundedhalf-yearly.

6. Findthecompoundintereston`15,625for9months,at16%perannumcompoundedquarterly.

7. FindthePrincipalthatwillyieldacompoundinterestof`1,632in2yearsat4%rateofinterestperannum.

8.Vickyborrowed`26,400fromabanktobuyascooterattherateof15%p.a.compoundedyearly.Whatamountwillhepayattheendof2yearsand4monthstocleartheloan?

9. Ariftookaloanof`80,000fromabank.Iftherateofinterestis10%p.a.,findthedifferenceinamountshewouldbepayingafter1½yearsiftheinterestis (i)compoundedannuallyand(ii)compoundedhalf-yearly.

10. Findthedifferencebetweensimpleinterestandcompoundintereston`2,400at 2yearsat5%perannumcompoundedannually.

11. Findthedifferencebetweensimpleinterestandcompoundintereston`6,400 for2yearsat6¼%p.a.compoundedannually.

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12. The difference betweenC. I. and S. I. for 2 years on a sum ofmoney lent at 5%p.a.is`5.Findthesumofmoneylent.

13. Sujathaborrows`12,500at12%perannumfor3yearsat simple interestandRadhika borrows the same amount for the same period at 10% per annumcompoundedannually.Whopaysmoreinterestandbyhowmuch?

14. Whatsumisinvestedfor1½yearsattherateof4%p.a.compoundedhalf-yearlywhichamountsto`1,32,651?

15. Gayathriinvestedasumof`12,000at5%p.a.atcompoundinterest.Shereceivedan amount of `13,230after‘n’years.Findthevalueof‘n’.

16. Atwhatratepercentcompoundinterestperannumwill`640amountto

`774.40in2years?

17. Findtheratepercentperannum,if`2,000amountto`2,315.25inanyearandahalf,interestbeingcompoundedhalf-yearly.

1.5.1 appreciation and depreciation

a) appreciation

In situations like growth of population, growth ofbacteria,increaseinthevalueofanasset,increaseinpriceofcertainvaluablearticles,etc.,thefollowingformulaisused.

A= rP 1 100n

+` j

b) depreciationIncertaincaseswherethecostofmachines,vehicles,

valueofsomearticles,buildings,etc.,decreases,thefollowingformulacanbeused.

A= rP 1 100n

-` j

Example 1.28Thepopulationofavillageincreasesattherateof7%everyyear.Ifthepresent

populationis90,000,whatwillbethepopulationafter2years?SolutionPresentpopulationP=90,000,Rateofincreaser=7%,Numberofyearsn=2. Thepopulationafter‘n’years = rP 1 100

n+` j

\Thepopulationaftertwoyears = 90000 11007 2

+` j

World Population Year Population 1700 600,000,000 1800 900,000,000 1900 1,500,000,000 2000 6,000,000,000In3centuries,populationhasmultiplied10fold.

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= 90000100107 2

` j

= 90000100107

100107

# #

= 103041

Thepopulationaftertwoyears = 1,03,041Example 1.29

Thevalueofamachinedepreciatesby5%eachyear.Amanpays`30,000forthemachine.Finditsvalueafterthreeyears.

SolutionPresentvalueofthemachineP=`30,000,Rateofdepreciationr =5%, Numberofyearsn = 3

Thevalueofthemachineafter‘n’years = rP 1 100n

-` j

\Thevalueofthemachineafterthreeyears = 30000 11005 3

-` j

= 3000010095 3

` j

= 3000010095

10095

10095

# # #

= 25721.25 Thevalueofthemachineafterthreeyears = `25,721.25

Example 1.30Thepopulationofavillagehasaconstantgrowthof5%everyyear.Ifitspresent

populationis1,04,832,whatwasthepopulationtwoyearsago?SolutionLetPbethepopulationtwoyearsago.

` P 11005 2

+` j = 104832

P100105 2

` j = 104832

P100105

100105

# # = 104832

P =105 105

104832 100 100#

# #

= 95085.71 = 95,086(roundingofftothenearestwholenumber)\Twoyearsagothepopulationwas95,086.

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exeRcISe 1.5

1. Thenumberofstudentsenrolledinaschoolis2000.Iftheenrollmentincreasesby5%everyyear,howmanystudentswillbethereaftertwoyears?

2. Acarwhichcosts`3,50,000depreciatesby10%everyyear.Whatwillbetheworthofthecarafterthreeyears?

3. Amotorcyclewasboughtat`50,000.Thevaluedepreciatedattherateof8%perannum.Findthevalueafteroneyear.

4. InaLaboratory,thecountofbacteriainacertainexperimentwasincreasingattherateof2.5%perhour.Findthebacteriaattheendof2hoursifthecountwasinitially5,06,000.

5. Fromavillagepeoplestartedmigratingtonearbycitiesduetounemploymentproblem.Thepopulationofthevillagetwoyearsagowas6,000.Themigrationistakingplaceattherateof5%perannum.Findthepresentpopulation.

6. Thepresentvalueofanoilengineis`14,580.Whatwastheworthoftheengine3yearsbeforeifthevaluedepreciatesattherateof10%everyyear?

7. Thepopulationofavillageincreasesby9%everyyearwhichisduetothejobopportunitiesavailableinthatvillage.Ifthepresentpopulationofthevillageis11,881,whatwasthepopulationtwoyearsago?

1.6 fixed deposits and Recurring deposits

Banks,postofficesandmanyotherfinancialinstitutionsacceptdepositsfrompublicatvaryingratesofinterest.Peoplesaveintheseinstitutionstogetregularperiodicalincome.

Differentsavingschemesareofferedbythesefinancialinstitutions.Fewofthoseschemesare

(i)FixedDepositand(ii)RecurringDeposit

(i) fixed depositInthistypeofdeposit,peopleinvestaquantumofmoneyforspecificperiods.

SuchadepositiscalledFixedDeposit(inshortform,F.D.)note:Depositscaneitherbeforashorttermorlongterm.Dependingonthe

periodofdeposits,theyofferahigherrateofinterest.

(ii) Recurring depositRecurringDeposit(inshortform,R.D.)isentirelydifferentfromFixedDeposit.Inthisscheme,thedepositorhasthefreedomtochooseanamountaccording

tohissavingcapacity,tobedepositedregularlyeverymonthoveraperiodofyearsinthebankorinthepostoffice.

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Thebankorpostofficerepaysthetotalamountdepositedtogetherwiththeinterestattheendoftheperiod.ThistypeofDepositisknownasRecurringDeposit.note: TheinterestonRecurringDepositiscalculated usingsimpleinterestmethod.

to find the formula for calculating interest and the maturity amount for R.d :

Let r %betherateofinterestpaidand‘P’bethemonthlyinstalmentpaidfor‘n’months.

Interest= 100PN ,r where

2

1n nN 121 +

=^ h; E years

TotalAmountdueatmaturityisA=P 100PNn r+

Example 1.31TharunmakesadepositofRupeestwolakhsinabankfor5years.Iftherateof

interestis8%perannum,findthematurityvalue.SolutionPrincipaldepositedP =`2,00,000,n=5years,r=8%p.a. Interest = 100

Pnr =200000 51008

# #

= `80,000 \Maturityvalueafter5years = 2,00,000+80,000=`2,80,000.

Example 1.32Vaideeshdeposits`500atthebeginningofeverymonthfor5yearsinapost

office.Iftherateofinterestis7.5%,findtheamounthewillreceiveattheendof5years.

Solution Amountdepositedeverymonth,P =`500 Numberofmonths,n =5×12=60months Rateofinterest,r = 7 % %

21

215=

Totaldepositmade =Pn=500×60 =`30,000 Periodforrecurringdeposit,N =

2

1n n

121 +^ h; E years

=241 60 61# # = 2

305 years

Themonthly instalmentscan be paid on any daywithinthemonthforR.D.

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Interest,I = 100PNr

=5002

3052 100

15# #

#

=`5,718.75

Totalamountdue = nP 100PNr+

=30,000+5,718.75 =`35,718.75

Example 1.33Vishaldeposited`200permonthfor5yearsinarecurringdepositaccountin

apostoffice.Ifhereceived`13,830findtherateofinterest.SolutionMaturityAmount,A=`13,830,P=`200,n=5×12=60months

Period,N =2

1n n

121 +^ h; E years

= 121 60

261

2305

# # = years

AmountDeposited=Pn =200×60=`12,000

MaturityAmount = Pn100PNr+

13830 =12000 200 r2

305100

# #+

13830–12000 =305×r

1830 =305×r

\ r =3051830 =6%

1.6.1 Hire Purchase and Instalments

Banksandfinancialinstitutionshaveintroducedaschemecalledhirepurchaseandinstalmenttosatisfytheneedsoftoday’sconsumers.

Hire purchase: Underthisscheme,thearticlewillnotbeownedbythebuyerforacertainperiodoftime.Onlywhenthebuyerhaspaidthecompletepriceofthearticlepurchased,he/shewillbecomeitsowner.

Instalment: Thecostofthearticlealongwithinterestandcertainotherchargesisdividedbythenumberofmonthsoftheloanperiod.Theamountthusgotisknownastheinstalment.

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equated monthly Instalment ( e.m.I. )

EquatedMonthlyInstalmentisalsoasequivalentastheinstalmentschemebutwithadimnishingconcept.Wehavetorepaythecostofthingswiththeinterestalongwithcertaincharges.Thetotalamountshouldbedividedbytheperiodofmonths.TheamountthusarrivedisknownasEquatedMonthlyInstalment.

E.M.I Number of monthsPrincipal Interest

=+

different schemes of Hire purchase and Instalment scheme

1. 0% interest scheme: Companiestakeprocessingchargeand4or5monthsinstalmentsinadvance.

2. 100% finance:Companiesaddinterestandtheprocessingchargestothecostprice.

3. discount Sale:Topromotesales,discountisgivenintheinstalmentschemes.4. Initial Payment:Acertainpartofthepriceofthearticleispaidtowardsthe

purchaseinadvance.ItisalsoknownasCashdownpayment.Example 1.34

Thecostpriceofawashingmachineis̀ 18,940.Thetablegivenbelowillustratesvariousschemestopurchasethewashingmachinethroughinstalments.Choosethebestschemetopurchase.

Sl. no

different schemes

S. P. in `

Initial payments

Rate of interest

Processing fee Period

(i) 75%Finance 18,940 25% 12% 1% 24

months

(ii) 100%Finance 18,940 Nil 16% 2% 24

months

(iii) 0%Finance 18,940

4 E. M. I. in

advanceNil 2% 24

months

CalculatetheE.M.I.andthetotalamountfortheaboveschemes.Solution(i) 75% finance P = `18,940,Initialpayment=25%,Rate=12%,Processingfee=1% Processingfee = 1%of`18,940 = 18940

1001

# =`189.40- ` 189 Initialpayment = 25%of`18,940 = 18940

10025

# =`4,735

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Loanamount = 18,940–4,735=`14,205

Interest =100

14205 12 2# #

= `3,409.20- `3,409

E.M.I. = Number of monthsLoan amount Interest+

=24

14205 3409+ =24

17614

= ` 733.92 - ` 734\Totalamounttobepaid= 4,735+14,205+3,409+189 = `22,538(ii) 100% finance Processingfee = 2%of`18,940 = 18940

1002

# =`378.80- ` 379 RateofInterest = 16% Interest = 18940 100

16 2# #

= `6060.80- `6,061

E.M.I. = Number of monthsLoan amount Interest+

=24

18940 6061+ =24

25001

= `1,041.708- `1,041.71 = `1,042 Totalamounttobepaid = 6,061+18,940+379=`25,380(iii) 0% interest scheme Processingfee = 2%of`18,940

= 189401002

# =`378.80- ` 379

E.M.I. = Number of monthsLoan amount Interest+

=24

18940 0+ =24

18940

= ` 789.166 - ` 789

Totalamounttobepaid = 18,940+3,156+379=`22,475

AdvanceE.M.I.paid = `789×4=`3,156Hence,0%interestschemeisthebestscheme.

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Example 1.35Thecostofacomputeris`20,000.Thecompanyoffersitin36months,but

charges10%interest.Findthemonthlyinstalmentthepurchaserhastopay.SolutionCostofcomputer=`20,000,Interest=10%p.a.,Period=36months(3years)

TotalInterest = 2000010010 3# #

= `6,000

\Totalamounttobepaid = 20,000+6,000

= `26,000

MonthlyInstalment = Number of monthsTotal amount

=36

26000

= ` 722. 22

- ` 722

exeRcISe 1.6

1. Ponmanimakesafixeddepositof`25,000inabankfor2years.Iftherateofinterestis4%perannum,findthematurityvalue.

2. Devamakesafixeddepositof`75,000inabankfor3years.Iftherateofinterestis5%perannum,findthematurityvalue.

3. Imrandeposits`400permonthinapostofficeasR.D.for2years.Iftherateofinterestis12%,findtheamounthewillreceiveattheendof2years.

4. Thecostofamicrowaveovenis`6,000.Pooraniwantstobuyitin5instalments.Ifthecompanyoffersitattherateof10%p.a.SimpleInterest,findtheE.M.I.andthetotalamountpaidbyher.

5. Thecostpriceofarefrigeratoris`16,800.Ranjithwantstobuytherefrigeratorat0%financeschemepaying3E.M.I.inadvance.Aprocessingfeeof3%isalsocollectedfromRanjith.FindtheE.M.I.andthetotalamountpaidbyhimforaperiodof24months.

6. Thecostofadiningtableis`8,400.Venkatwantstobuyitin10instalments.IfthecompanyoffersitforaS.I.of5%p.a.,findtheE.M.I.andthetotalamountpaidbyhim.

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1.7 compound Variation

IntheearlierclasseswehavealreadylearntaboutDirectandInverseVariation.Letusrecallthem.

direct Variation

If two quantities are such that an increase or decrease in one leads to acorresponding increase or decrease in the other, we say they vary directly or thevariationisDirect.

examples for direct Variation:

1.Distance andTime are inDirectVariation, becausemore the distancetravelled,thetimetakenwillbemore(ifspeedremainsthesame).

2.PrincipalandInterestareinDirectVariation,becauseifthePrincipalismoretheinterestearnedwillalsobemore.

3.PurchaseofArticlesandtheamountspentareinDirectVariation,becausepurchaseofmorearticleswillcostmoremoney.

Indirect Variation or Inverse Variation:

If two quantities are such that an increase or decrease in one leads to acorrespondingdecreaseor increase in theother,wesay theyvary indirectlyor thevariation is inverse.

examples for Indirect Variation:

1. WorkandtimeareinInverseVariation,becausemorethenumberoftheworkers,lesserwillbethetimerequiredtocompleteajob.

2. SpeedandtimeareinInverseVariation,becausehigherthespeed, theloweristhetimetakentocoveradistance.

3. PopulationandquantityoffoodareinInverseVariation,becauseifthepopulationincreasesthefoodavailabilitydecreases.

compound Variation

Certainproblemsinvolveachainoftwoormorevariations,whichiscalledasCompoundVariation.

Thedifferentpossibilitiesofvariationsinvolvingtwovariationsareshowninthefollowingtable:

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Variation I Variation IIDirect DirectInverse InverseDirect InverseInverse Direct

Letusworkoutsomeproblemstoillustratecompoundvariation.Example 1.36

If20mencanbuildawall112meterslongin6days,whatlengthofasimilarwallcanbebuiltby25menin3days?

Solution:method 1: Theprobleminvolvessetof3variables,namely-Numberofmen,

Numberofdaysandlengthofthewall.number of men number of days Length of the wall in metres

20 6 11225 3 x

Step 1 : Considerthenumberofmenandthelengthofthewall.Asthenumberofmenincreasesfrom20to25,thelengthofthewallalsoincreases.SoitisinDirectVariation.

Therefore,theproportionis20:25::112:x ...... (1)

Step 2: Considerthenumberofdaysandthelengthofthewall.Asthenumberofdaysdecreasesfrom6to3,thelengthofthewallalsodecreases.So,itisinDirectVariation.

Therefore,theproportionis6:3::112:x ..... (2)

Combining(1)and(2),wecanwrite

:

:: x

20 25

6 3112S1

Weknow,Product of extremes = Product of means.

extremes means extremes

20 : 25 ::112 : x 6 : 3

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So, x20 6# # = 25 3 112# #

x =20 6

25 3 112#

# # =70meters.method 2

number of men number of days Length of the wall in metres20 6 11225 3 x

Step 1:Consider thenumberofmenand lengthof thewall.As thenumberofmenincreasesfrom20to25,thelengthofthewallalsoincreases.Itisindirect variation.

Themultiplyingfactor2025=

Step 2: Considerthenumberofdaysandthelengthofthewall.Asthenumberofdaysdecreases from6 to3, the lengthof thewallalsodecreases. It is indirect variation.

Themultiplyingfactor 63= .

\ x = 1122025

63

# # =70meters

Example 1.37Sixmenworking10hoursadaycandoapieceofworkin24days.Inhowmany

dayswill9menworkingfor8hoursadaydothesamework?Solutionmethod 1: Theprobleminvolves3setsofvariables,namely-Numberofmen,

WorkinghoursperdayandNumberofdays.number of men number of hours per day number of days

6 10 249 8 x

Step 1: Considerthenumberofmenandthenumberofdays.Asthenumberofmenincreasesfrom6to9,thenumberofdaysdecreases.SoitisinInverseVariation.

Thereforetheproportionis9:6::24: x ..... (1)Step 2: Considerthenumberofhoursworkedperdayandthenumberofdays.

Asthenumberofhoursworkingperdaydecreasesfrom10to8,thenumberofdaysincreases.Soitisinverse variation.

Thereforetheproportionis8:10::24:x ..... (2)Combining(1)and(2),wecanwriteas

:

:: : : x

9 6

8 10241

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Weknow,Product of extremes = Product of means. extremes means extremes 9 : 6 : : 24 : x 8 :10So, 9×8×x = 6×10×24 x =

9 86 10 24

## # =20days

note: 1. DenotetheDirectvariationas. (Downwardarrow) 2. DenotetheIndirectvariationas- (Upwardarrow) 3. Multiplying Factors can bewritten based on the arrows. Take the

numberontheheadofthearrowinthenumeratorandthenumberonthetailofthearrowinthedenominator.

Formethodtwo,usetheinstructionsgiveninthenoteabove.method 2 : (using arrow marks)

number of men number of hours per day number of days6 10 249 8 x

Step 1 : Considermenanddays.Asthenumberofmenincreasesfrom6to9,thenumberofdaysdecreases.Itisininverse variation.

Themultiplyingfactor=96

Step 2 : Considerthenumberofhoursperdayandthenumberofdays.Asthenumberofhoursperdaydecreasesfrom10to8,thenumberofdaysincreases.Itisalsoininverse variation.

Themultiplyingfactor=810

\ x = 24 2096

810

# # = days.

exeRcISe 1.7 1. Twelvecarpentersworking10hoursadaycompleteafurnitureworkin18days.

Howlongwouldittakefor15carpentersworkingfor6hoursperdaytocompletethesamepieceofwork?

2. Eightymachinescanproduce4,800identicalmobilesin6hours.Howmanymobilescanonemachineproduceinonehour?Howmanymobileswould25machinesproducein5hours?

3. If14compositorscancompose70pagesofabookin5hours,howmanycompositorswillcompose100pagesofthisbookin10hours?

4. If2,400sq.m.oflandcanbetilledby12workersin10days,howmanyworkersareneededtotill5,400sq.m.oflandin18days?

5. Working4hoursdaily,Swaticanembroid5sareesin18days.Howmanydayswillittakeforhertoembroid10sareesworking6hoursdaily?

6. A sum of ` 2,500depositedinabankgivesaninterestof` 100in6months.Whatwillbetheintereston`3,200for9monthsatthesamerateofinterest?

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1.8 time and Work

When we have to compare the work of several persons, it is necessary toascertaintheamountofworkeachpersoncancompleteinoneday.Astimeandworkareofinversevariationandifmorepeoplearejoinedtodoawork,theworkwillbecompletedwithinashortertime.

Insolvingproblemshere,thefollowingpointsshouldberemembered:

1. Ifamanfinishestotalworkin‘n’days,theninonedayhedoes‘n1 ’of

thetotalwork.Forexample,ifamanfinishesaworkin4days,theninone day he does

41 ofthework.

2. If the quantity ofwork doneby aman in oneday is given, then thetotalnumberofdaystakentofinishthework=1/(oneday’swork).Forexample,ifamandoes

101 oftheworkin1day,thenthenumberofdays

takentofinishthework

101

11110

#= =` j

=10days.

Example 1.38Acandoapieceofworkin20daysandBcandoitin30days.Howlongwill

theytaketodotheworktogether?SolutionWorkdonebyAin1day= 20

1 ,WorkdonebyBin1day= 301

WorkdonebyAandBin1day =201

301+

= 603 2+ =

605

121= ofthework

TotalnumberofdaysrequiredtofinishtheworkbyAandB= 12112

1 = days.Example 1.39

AandBtogethercandoapieceofworkin8days,butAalonecandoit12days.HowmanydayswouldBalonetaketodothesamework?

Solution WorkdonebyAandBtogetherin1day = 8

1 ofthework

WorkdonebyAin1day = 121 ofthework

WorkdonebyBin1day =81

121- =

243 2

241- =

NumberofdaystakenbyBalonetodothesamework= 24124

1 = days.

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Example 1.40

TwopersonsAandBareengagedinawork.Acandoapieceofworkin12daysandBcandothesameworkin20days.Theyworktogetherfor3daysandthenAgoesaway.InhowmanydayswillBfinishthework?

Solution

WorkdonebyAin1day =121

WorkdonebyBin1day =201

WorkdonebyAandBtogetherin1day = 121

201+

=60

5 3608

152+ = =

WorkdonebyAandBtogetherin3days =152 3

52

# =

RemainingWork = 152

53- =

NumberofdaystakenbyBtofinishtheremainingwork=20153

53

120

#=

= 12days.

Example 1.41AandBcandoapieceofworkin12days,BandCin15days,CandAin20

days.Inhowmanydayswilltheyfinishittogetherandseparately?Solution WorkdonebyAandBin1day =

121

WorkdonebyBandCin1day =151

WorkdonebyCandAin1day = 201

Workdoneby(A+B)+(B+C)+(C+A)in1day = 121

151

201+ +

Workdoneby(2A+2B+2C)in1day = 605 4 3+ +

Workdoneby2(A+B+C)in1day = 6012

WorkdonebyA,BandCtogetherin1day =21

6012

# =101

WhileA,BandCworkingindividually can complete ajobin20,5,4daysrespectively.Ifall jointogetherandwork,find in howmany days theywillfinishthejob?

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` A,BandCwillfinishtheworkin10days.

WorkdonebyAin1day

(i.e.) [(A+B+C)’swork–(B+C)’swork] =101

151

303 2

301- = - =

` Awillfinishtheworkin30days.

WorkdonebyBin1day

(i.e.) [(A+B+C)’swork–(C+A)’swork] =101

201

202 1

201- = - =

` Bwillfinishtheworkin20days.

WorkdonebyCin1day

(i.e.) [(A+B+C)’swork–(A+B)’swork] =101

121

606 5

601- = - =

` Cwillfinishtheworkin60days.

Example 1.42Acandoapieceofworkin10daysandBcandoitin15days.Howmuchdoes

eachofthemgetiftheyfinishtheworkandearn` 1500?Solution

WorkdonebyAin1day = 101

WorkdonebyBin1day = 151

Ratiooftheirwork = :101

151 =3:2

TotalShare = `1500 A’sshare = 1500

53# =`900

B’sshare = 15052 0# =`600

Example 1.43Twotapscanfillatankin30minutesand40minutes.Anothertapcanempty

itin24minutes.Ifthetankisemptyandallthethreetapsarekeptopen,inhowmuchtimethetankwillbefilled?

SolutionQuantityofwaterfilledbythefirsttapinoneminute=

301

Quantityofwaterfilledbythesecondtapinoneminute=401

Quantityofwateremptiedbythethirdtapinoneminute=241

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Quantityofwaterfilledinoneminute,whenallthe3tapsareopened=30

1401

241+ -

=120

4 3 5+ - =1207 5-

=1202 =

601

Timetakentofillthetank =160

1 =60minutes

=1hour

exeRcISe 1.8

1. Amancancompleteaworkin4days,whereasawomancancompleteitinonly12days.Iftheyworktogether,inhowmanydays,cantheworkbecompleted?

2. Twoboyscanfinishaworkin10dayswhentheyworktogether.Thefirstboycandoitalonein15days.Findinhowmanydayswillthesecondboydoit allbyhimself?

3. ThreemenA,BandCcancompleteajobin8,12and16daysrespectively.

AandBworktogetherfor3days;thenBleavesandCjoins.Inhowmanydays,canAandCfinishthework?

4. AtapAcanfilladrumin10minutes.AsecondtapBcanfillin20minutes.

AthirdtapCcanemptyin15minutes.Ifinitiallythedrumisempty,findwhenitwillbefullifalltapsareopenedtogether?

5. Acanfinishajobin20daysandBcancompleteitin30days.Theyworktogetherandfinishthejob.If`600ispaidaswages,findtheshareofeach.

6. A,BandCcandoaworkin12,24and8daysrespectively.Theyallworkforoneday.ThenCleavesthegroup.InhowmanydayswillAandBcompletetherestofthework?

7. Atapcanfillatankin15minutes.Anothertapcanemptyitin20minutes.Initiallythetankisempty.Ifboththetapsstartfunctioning,whenwillthetankbecomefull?

abbreviation: C.P.=CostPrice,S.P.=SellingPrice,M.P.=MarkedPrice,

P=Principal,r=Rateofinterest,n =timeperiod, A=Amount,C.I.=CompoundInterest.

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Percent means per hundred. A fraction with its denominator 100 is called a percent.

Incaseofprofit,wehave

Profit=S.P.–C.P.;Profitpercent=C.PProfit 100#

S.P.=100

100 Profit% C.P.#+` j ;C.P.=100 Profit%

100 S.P.#+

` j

IncaseofLoss,wehave

Loss=C.P.–S.P.;Losspercent= 100C.P.Loss

#

S.P.=100

100 Loss% C.P.#-` j ;C.P.=100 Loss%

100 S.P.#-

` j

DiscountisthereductiongivenontheMarkedPrice.

Selling Price is the price payable after reducing theDiscount from theMarkedPrice.

Discount=M.P.–S.P.

M.P.=100 Discount%

100 S.P.#-

; S.P.=100

100 Discount% M.P.#-

C.P.=100 Profit%100 Discount% M.P.#

+- ; M.P.=

100 Discount%100 Profit% C.P#

-+ .

DiscountPercent= 100.M.P.

Discount#

When the interest is

(i)compoundedannually,A= rP 1100

n+` j

(ii)compoundedhalf-yearly,A= 1 r21

100P

2n+ ` j8 B

(iii)compoundedquarterly,A=100r1

41P

n4+ ` j8 B

Appreciation,A= rP 1100

n+` j ; Depreciation,A= rP 1

100n

-` j

ThedifferencebetweenC.I.andS.I.for2years= r100

P 2

` j

Oneday’sworkofA= 1Number of days taken by A

Workcompletedin‘x’days=Oneday’sworkx x

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2Geometry

2.1 Introduction

2.2 ConcurrencyinTriangle

2.3 PythagorasTheorem

2.4 Circles

2.1 Introduction

Geometrywas developed byEgyptiansmore than 1000 yearsbeforeChrist. Itwas abstracted by theGreeks into logical systemofproofswithnecessarybasicpostulatesoraxioms.

Geometryplaysavitalroleinourday-to-daylifeinmanyways.Especially the triangles are used in severalways.The area of a landwhichisinpolygonshapecanbefoundbydividingthemintotriangles.Thesumoftheareaofthetrianglesistheareaoftheland.Ifthetrianglesareinrightanglesthentheareacanbefoundveryeasily.Otherwise,theareacanbefoundbydrawingtheperpendicularlinefromthebasetothe vertex.

Hereweseetheconcurrencyintrianglesandsomepropertiesofthem.

Pythagoras (582 - 497 BC)

TheGreekMathematician whowasone

of the foremost Mathematicians ofalltimes.Hewasperhapsbestknownfortherightangledtrianglerelationwhichbearshis

name.

47

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2.2 concurrency in trianglesDrawthreeormorelinesinaplane.Whatarethepossibleways?Thepossibilitiesmaybeasfollows:

(a) (b) (c) (d)In fig (a),AB,CDandEFareparallelsotheyarenot intersecting.In fig (b),ABandCDintersectatP,ABandEFintersectatQ.SoP,Qaretwo

points of intersection.In fig (c),P,Q,Rarethree points of intersection.Butinfig (d),Pistheonlypointofintersection.HereAB,CD,EFarethelines

passing through the samepointP.These linesarecalledasconcurrent lines.ThepointP is called the point of concurrency.

Inatriangletherearesomespecialpointsofconcurrence,whichareCentroidofatriangle,Orthocentreofatriangle,IncentreofatriangleandCircumcentreofatriangle.Nowwearegoingtostudyhowtoobtainthesepointsinatriangle.

2.2.1 centroid of a triangleIntheadjacentfigure,ABCisatriangle.DismidpointofBC.JoinAD.HereADisoneofthemediansofDABC.A medianofatriangleisthelinesegmentjoiningavertex

andthemidpointoftheoppositeside.Nowconsidertheadjacentfigure,inwhichAD,BE,CFare

the three medians of 3 ABC.TheyareconcurrentatG.Thispointiscalledascentroid.Thethreemediansofatriangleareconcurrentandthepoint

ofconcurrencyisknownasCentroid.Itisdenotedby‘G’.note : (i) TheCentroiddivideseachofthemedianintheratio2:1 (ii) TheCentroidwouldbethephysicalcentreofgravity.

A

B

F

DQ

PC

E

A

B

C

D

E FQ R

P

E

FC

DA

BP

A B

C D

E F

A

B D C2cm 2cm

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2.2.2 orthocentre of a triangle

Intheadjacentfigure,ABCisatriangle.FromA,drawaperpendiculartoBC,ADisperpendiculartoBC.ADB ADC 900

+ += = . Here D need not be themidpoint.HereADisanaltitudefromvertexA.

Altitude of a triangle is a perpendicular linesegmentdrawnfromavertex to theoppositeside.NowconsiderthetriangleABC,inwhichAD,BE,CFarethethreealtitudes.

TheyareconcurrentatH.ThispointisknownasOrthocentre.

Thethreealtitudesofatriangleareconcurrentandthepointofconcurrencyisknownasorthocentre.

different positions of orthocentre

(a) (b) (c)case (i) : Infig(a),ABCisanacuteangledtriangle. HereorthocentreliesinsidetheDABC.case (ii): Infig(b),ABCisarightangledtriangle. Hereorthocentreliesonthevertexattherightangle.case (iii): Infig(c),ABCisanobtuseangledtriangle. HereorthocentreliesoutsidetheDABC.

2.2.3 Incentre of a triangle

Intheadjacentfigure,ABCisatriangle.The A+ isbisectedintotwoequalpartsbyAD.Therefore BAD DAC+ += .HereADissaidtobetheanglebisectorof+A.

B

A

CD

A

CB

A

BD C

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Anglebisectorofatriangleisalinesegmentwhichbisectsanangleofatriangle.

Nowconsider thefigure inwhichAD,BE,CFarethreeanglebisectorsof3ABC.

TheyareconcurrentatI.Thispointisknownasincentreofthetriangle.Thethreeanglebisectorsofatriangleareconcurrent

andthepointofconcurrenceiscalledtheIncentre .

2.2.4 circumcentre of a triangle

Wehavelearntaboutperpendicularbisectorinpreviousclass.Whatisaperpendicularbisectorinatriangle?Observethefollowingfigures:

(a) (b) (c)Infig(a):ADisperpendicularfromAtoBCbutnotbisectingBC.Infig(b):ADbisectsBC.HenceBD=DCandADisperpendiculartoBC.Infig(c):DXisperpendiculartoBCandDXalsobisectingBC.BD=DCbut

DXneednotpassesthroughthevertex‘A’.Theperpendicularbisectorofthesideofatriangleisthelinethatisperpendicular

toitandalsobisectstheside.

Now,considertheabovefigure.

A

CB

M

Q

R N

SP

O

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HerePQ,RS,andMNarethethreeperpendicularbisectorsofBC,ACandABconcurrentatO.Oisknownasthecircumcentre.

Thethreeperpendicularbisectorsofatriangleareconcurrentandthepointofconcurrenceisknownas circumcentre .

note : (i) In any triangle ABC , Circumcentre (O) ,Centroid(G)andOrthocentre(H)alwayslieononestraightline,whichiscalledaseuler Line,and OG:GH=1:2.

(ii) In particular for an equilateral triangle,Circumcentre(O),Incentre(I),Orthocentre(H)andCentroid(G)willcoincide.

2.3 Pythagoras theorem2.3.1 Pythagoras Theorem

In a right angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let us consider ABCD with C 90o+ = .BC a, CA b= = and AB c= .Then,a b c2 2 2

+ = .This was proved in number of ways by different

Mathematicians.We will see the simple proof of Pythagoras

theorem.Now,weconstructasquareofside(a+b)asshowninthefigure,and using the construction we prove

Pythagorastheorem.Thatis,weprove .a b c2 2 2+ =

WeknowthatAreaofanysquareissquareof its side.

Areaofasquareofside(a+b)=(a b)2+

Fromthefigure,Area of the square of side (a +b)=(a b)2+ =sumoftheareaofthe trianglesI,II,IIIandIV+ theareaofthesquarePQRS

A

CB

hypo

tenuse

a

b

base

c

heig

ht

I II

IIIIV

a P b

b

Q

a

b R a

a

S

bc

c c

c

A

B C

M HE

Mc

Ma

Mb

Hc

Hb

Ha

Euler (1707-1783)Switzerland

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i.e.,(a b)2+ =4(AreaofrightangledD)+(AreaofthesquarePQRS)

(a b)2+ = 4 21 a b c2# # +` j

a b 2ab2 2+ + = 2ab c2+

a b2 2` + = c2

HenceweprovedPythagorastheorem.

Pythagoras theoremDraw a right angled triangle

ABC,such that C 90 ,o+ = AB = 5 cm, AC=4cmandBC=3cm.

Construct squares on the three sidesofthistriangle.

Divide these squares into smallsquares of area one cm

2 each.

By counting the number of smallsquares, pythagoras theorem can beproved.

NumberofsquaresinABPQ=25NumberofsquaresinBCRS=9NumberofsquaresinACMN=16`NumberofsquaresinABPQ= NumberofsquaresinBCRS+ NumberofsquaresinACMN.

ThenumberswhicharesatisfyingthePythagorastheoremarecalledthePythagorean triplets.

Example 2.1In ABC, B 90o+D = ,AB=18cmandBC=24cm.CalculatethelengthofAC.Solution ByPythagorasTheorem, AC2 = AB BC2 2

+

= 18 242 2+

= 324+576 = 900 \AC = 900 =30cm

Example 2.2Asquarehastheperimeter40cm.Whatisthesumofthediagonals?

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Solution Let‘a’bethelengthofthesideofthesquare.ACisadiagonal. PerimeterofsquareABCD = 4a units 4a = 40cm[given] a =

440 =10cm

Weknowthatinasquareeachangleis90o andthediagonalsareequal.

In ABCD ,AC2 = AB BC2 2+

= 10 102 2+ =100+100= 200

AC` = 200

= 2 100# =10 2

= 10 1.414# =14.14cm DiagonalAC = DiagonalBDHence,Sumofthediagonals = 14.14+14.14=28.28cm.

Example 2.3From the figure PT is an altitude

ofthetrianglePQRinwhichPQ=25cm,PR=17cmandPT=15 cm.IfQR=x cm. Calculatex.

Solution Fromthefigure,wehaveQR=QT+TR.

Tofind:QTandTR.In the right angled trianglePTQ ,

PTQ 90o+ = [PTisattitude]ByPythagorasTheorem,PQ2 =PT QT2 2

+ PQ PT2 2` - =QT2

QT2` = 25 152 2- =625-225=400

QT = 400 =20cm .....(1)Similarly,intherightangledtrianglePTR,byPythagorasTheorem, PR2 = PT TR2 2

+

TR2` = PR PT2 2-

= 17 152 2-

= 289- 225=64 TR = 64 =8cm .....(2)Form(1)and(2) QR = QT+TR=20+8=28cm.

xcm

P

Q RT

25cm

17cm

15cm

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Example 2.4Arectangularfield isofdimension40mby30m.Whatdistance issavedby

walkingdiagonallyacrossthefield?Solution Given:ABCDisarectangularfieldofLength=40m,Breadth=30m, B 90o+ = IntherightangledtriangleABC,ByPythagorasTheorem, AC2 = AB BC2 2

+

= 30 402 2+ =900+1600

= 2500 AC` = 2500 =50mDistancefromAtoCthroughBis = 30+40=70m Distancesaved = 70–50=20m.

exeRcISe 2.1

1. Choosethecorrectanswer

(i) Thepointofconcurrencyofthemediansofatriangleisknownas (A)incentre (B)circlecentre (C)orthocentre (D)centroid

(ii) Thepointofconcurrencyofthealtitudesofatriangleisknownas (A)incentre (B)circlecentre (C)orthocentre (D)centroid

(iii) Thepointofconcurrencyoftheanglebisectorsofatriangleisknownas (A)incentre (B)circlecentre (C)orthocentre (D)centroid

(iv)Thepointofconcurrencyoftheperpendicualarbisectorsofatriangleisknownas (A)incentre (B)circumcentre (C)orthocentre (D)centroid

2. InanisoscelestriangleAB=ACand B 65+ = c.Whichistheshortestside?

3. PQRisatrianglerightangledatP.IfPQ=10cmandPR=24cm,findQR.

4. Checkwhetherthefollowingcanbethesidesofarightangledtriangle AB=25cm,BC=24cm,AC=7cm.

5. QandRofatrianglePQRare25°and65°.IsDPQRarightangledtriangle?Moreover PQ is 4cm and PR is 3 cm. Find QR.

6. A15mlongladderreachedawindow12mhighfromtheground.Onplacingitagainstawallatadistancex m. Find x.

7. Findthealtitudeofanequilateraltriangleofside10cm.

8. Arethenumbers12,5and13formaPythagoreanTriplet?

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9. Apaintersetsaladderuptoreachthebottomofasecondstoreywindow16feetabovetheground.Thebaseoftheladderis12feetfromthehouse.Whilethepaintermixesthepaintaneighbour’sdog bumps the ladderwhichmoves the base 2feetfartherawayfromthehouse.Howfarupsideofthehousedoestheladderreach?

2.4 circles

You are familiar with the following objects. Can you say the shape of thefollowing?

(a)Cyclewheel (b)AshokachakrainourNationalEmblem (c)FullmoonSure,youranswerwillbecircle.Youknowthatacircleisdescribedwhena

pointPmovesinaplanesuchthatitsdistancefromafixedpointintheplaneremainsconstant.

definition of circleAcircleisthesetofallpointsinaplaneataconstantdistancefromafixedpoint

inthatplane.Thefixedpointiscalledthecentreofthecircle.Theconstantdistance isknownas the radiusof the

circle.Inthefigure‘O’iscentreandOA,OB,OCareradii

ofthecircle. Here,OA = OB=OC =r

Note: Alltheradiiofthecircleareequal.

chord

Achordisalinesegmentwithitsendpointslyingonacircle.

Infigure,CD,ABandEFarechords.

HereABisaspecialchordthatpassesthroughthecentre O.

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diametera diameter is a chord that passes through the centre of the circle and

diameter is the longest chord of a circle.Inthefigure,AOBisdiameterofthecircle.OisthemidpointofABandOA=OB=radiusofthecircleHence,Diameter=2×radius (or) Radius=(diameter)÷2

note : (i) Themid-pointofeverydiameterofthecircleisthecentreofthecircle.(ii) Thediametersofacircleareconcurrentandthepointofconcurrencyis

thecentreofthecircle.

Secant of a circle

Alinepassingthroughacircleandintersectingthecircleattwopointsiscalledthesecantofthecircle.

Inthegivenfigure,lineABisaSecant.ItcutsthecircleattwopointsAandB.Now, let us move the secant AB

downwards.ThenthenewpositionsareA1B1,A2B2,....etc.,

While secant AB moves down, thepointsAandBaremovingclosertoeachother.

So distance between A and B isgraduallydecreases.

AtonepositionthesecantABtouchesthecircleatonlyonepointL.Atthisposition,thelineLMiscalledastangentandittouchesthecircleatonlyonepoint.

tangent

Tangent is a line that touchesacircleat exactlyonepoint, and thepoint isknownaspointofcontact.

arc of a circle InthefigureABisachord.ThechordABdividesthe

circleintotwoparts.ThecurvedpartsALBandAMBareknownasArcs.

detonarcswillbede bythesymbol‘!

’.

!

Thesmallerarc ALB is the minor arc. Thegreaterarc AMB is the major arc.

A

B A1

B1 A2

B2

L MN

!

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MINOR SEGMENT

A

BMAjOR SEGMENT

L

M

Segment of a circle

Achordofacircledividesthecircularregioninto twoparts.Eachpart iscalledassegmentofthecircle.

Thesegmentcontainingminorarciscalledthe minor segment.

Thesegmentcontainingmajorarciscalledthe major segment.

Sector of a circle

ThecircularregionenclosedbyanarcofacircleandthetworadiiatitsendpointsisknownasSectorofacircle.

ThesmallersectorOALBiscalled the minor sector.ThegreatersectorOAMBiscalledthemajor sector.

exeRcISe 2.2

1. Choosethecorrectanswer:

(i) The_______ofacircleisthedistancefromthecentretothecircumference. (A)sector (B)segment (C)diameters (D)radius

(ii) Therelationbetweenradiusanddiameterofacircleis______ (A)radius=2×diameters (B)radius=diameter+2 (C)diameter=radius+2 (D)diameter=2(radius)

(iii) Thelongestchordofacircleis (A)radius (B)secant (C)diameter (D)tangent

2. Ifthesumofthetwodiametersis200mm,findtheradiusofthecircleincm.

3. Definethecirclesegmentandsectorofacircle.

4. Definethearcofacircle.

5. Definethetangentofacircleandsecantofacircle.

o

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WecanfindmanyPythagoreanTripletsusingtheconditionm n , m n , 2mn2 2 2 2+ - here m>n;m,ne N .

If m = 2 and n = 1 then 2 1 5 2 1 3 2 2 1 4m n , m n , 2mn2 2 2 22 2 2 2# #+ -= + = = - = = =

5 , 3 , 4 isaPythagoreanTriplet.

ThemultipleofPythagoreanTripletisalsoaPythagoreanTriplet.

Eg,Themultiplesof ( 5 , 3 , 4 ) : ( 10, 6, 8 ) , ( 15 , 9, 12 ), ( 20 , 12 , 16 ) , ... arealsothePythagoreanTriplets.

CanyoufindmorePythagoreanTriplets?

Centroid : PointofconcurrencyofthethreeMedians.

Orthocentre :PointofconcurrencyofthethreeAltitudes.

Incentre :PointofconcurrencyofthethreeAngleBisectors.

Circumcentre : Point of concurrency of the Perpendicular Bisectors of thethree sides.

Circle : A circle is the set of all points in a plane at a constant distance from a fixedpointinthatplane.

Chord :Achordisalinesegmentwithitsendpointslyingonacircle.

Diameter : A diameter is a chord that passes through the centre of the circle.

A line passing through a circle and intersecting the circle at two points is called the secant of the circle.

Tangent is a line that touches a circle at exactly onepoint, and the point isknownaspointofcontact.

Segment of a circle :Achordofacircledividesthecircularregionintotwoparts.

Sector of a circle :ThecircularregionenclosedbyanarcofacircleandthetworadiiatitsendpointsisknownasSectorofacircle.

Mathematics Club Activity PYTHAGOREAN TRIPLET

3DataHandling

3.1 Introduction

Everydaywecomeacrossdifferentkindsof information in the

formofnumbersthroughnewspapersandothermediaofcommunication.

Thisinformationmaybeaboutfoodproductioninourcountry,

populationoftheworld,importandexportofdifferentcountries,drop-

outsofchildrenfromtheschoolsinourstate,theaccidentialdeaths,etc.

In all these information, we use numbers. These numbers are

calleddata.Thedatahelpusinmakingdecisions.Theyplayavitalpart

inalmostallwalksoflifeofeverycitizen.Hence,itisveryimportantto

knowhowtogetrelevantandexactinformationfromsuchdata.

Thecalculateddatamaynotbesuitableforreading,understanding

andforanalysing.Thedatashouldbecarefullyhandledsothatitcanbe

presentedinvariousforms.Acommonmanshouldbeabletounderstand

andvisualizeandgetmoreinformationonthedata.

3.1 Introduction

3.2 RecallingtheFormationofFrequencyTable

3.3 DrawingHistogramandFrequencyPolygonforGroupedData

3.4 ConstructionofSimplePiechart

3.5 MeasuresofCentralTendency

R.a. fisher[17th Feb., 1890 - 29th July, 1962]

Fisherwasinterested in the theory of errors thateventuallylethimtoinvestigatestatisticalproblems.Hebecameateacherin Mathematics and Physics between1915and 1919. He studiedthedesignofexperimentsbyintroducingrandomisation andtheanalysisof variance proceduresnowusedthroughouttheworld.Heisknownas “Father of Modern Statistics”.

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3.2 Recalling the formation of frequency table

We have learnt in seventh standard, how to form a frequency table. Let us recallit.

3.2.1 formation of frequency table for an ungrouped data

Example 3.1Considerthefollowingdata: 15,17,17,20,15,18,16,25,16,15, 16,18,20,28,30,27,18,18,20,25, 16,16,20,28,15,18,20,20,20,25.Formafrequencytable.SolutionThefrequencytableisgivenbelow:

number(x) tally mark frequency

(f)

15 4

16 5

17 2

18 5

20 7

25 3

27 1

28 2

30 1

total 30

3.2.2 formation of frequency table for a grouped data

Example: 3.2Themarksobtainedby50studentsinaMathematicstestwithmaximummarks

of100aregivenasfollows:

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43,88,25,93,68,81,29,41,45,87,34,50,61,75,51,96,20,13,18,35, 25,77,62,98,47,36,15,40,49,25,39,60,37,50,19,86,42,29,32,61, 45,68,41,87,61,44,67,30,54,28.Prepareafrequencytablefortheabovedatausingclassinterval.Solution Totalnumberofvalues = 50 Range = Highestvalue-Lowestvalue = 98–8=90Letusdividethegivendatainto10classes.

` LengthoftheClassinterval = Number of class intervalRange

=1090 9=

Thefrequencytableofthemarksobtainedby50studentsinamathematicstestisasfollows:

classInterval (c.I) tally mark frequency (f)

0-10 2

10-20 4

20-30 6

30-40 7

40-50 9

50-60 4

60-70 8

70-80 2

80-90 5

90-100 3

total 50

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Thusthegivendatacanbegroupedandtabulatedasfollows:class

Interval (c.I)

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

frequency(f) 2 4 6 7 9 4 8 2 5 3

3.3 drawing Histogram and frequency Polygon for Grouped data

The statistical data can be represented by means of geometrical figures ordiagramswhicharenormallycalled“graphs’’.Thegraphicalrepresentationofdatamakesitselfinterestingforreading,consuminglesstimeandeasilyunderstandable.Therearemanywaysofrepresentingnumericaldatagraphically.Inthischapter,wewillstudythefollowingtwotypesofdiagrams:

(i) Histogram (ii) frequency Polygon

3.3.1 Histogram

A two dimensional graphical representation of a continuous frequencydistributioniscalledahistogram.

Inhistogram,thebarsareplacedcontinuouslysidebysidewithnogapbetweenadjacentbars.Thatis,inhistogramrectanglesareerectedontheclassintervalsofthedistribution.Theareasofrectangleareproportionaltothefrequencies.

3.3.1 (a) drawing a histogram for continuous frequency distribution

Procedure:

Step 1 :Representthedatainthecontinuous(exclusive)formifitisinthediscontinuous(inclusive)form.

Step 2 :MarktheclassintervalsalongtheX-axisonauniformscale.Step 3 :MarkthefrequenciesalongtheY-axisonauniformscale.Step 4 : Constructrectangleswithclassintervalsasbasesandcorresponding

frequenciesasheights.

Themethodofdrawingahistogramisexplainedinthefollowingexample.Example 3.3

Drawahistogramforthefollowingtablewhichrepresentthemarksobtainedby100studentsinanexamination:

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marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80number of students 5 10 15 20 25 12 8 5

Solution

Theclassintervalsareallequalwithlengthof10marks.LetusdenotetheseclassintervalsalongtheX-axis.DenotethenumberofstudentsalongtheY-axis,withappropriatescale.Thehistogramisgivenbelow.

Fig. 3.1

note:Intheabovediagram,thebarsaredrawncontinuously.Therectanglesareoflengths(heights)proportionaltotherespectivefrequencies.Sincetheclassintervalsareequal,theareasofthebarsareproportionaltotherespectivefrequencies.

3.3.1 (b) drawing a histogram when class intervals are not continuous

Example 3.4:The heights of trees in a forest are given as follows. Draw a histogram to

representthedata.

Heights in metre 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55number of trees 10 15 25 30 45 50 35 20

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SolutionIn thisproblem, thegivenclass intervals arediscontinuous (inclusive) form.

Ifwedrawahistogramasitis,wewillgetgapsbetweentheclassintervals.Butinahistogramthebarsshouldbecontinuouslyplacedwithoutanygap.Thereforeweshouldmaketheclassintervalscontinuous.Forthisweneedanadjustmentfactor.

AdjustmentFactor =21 [(lowerlimitofaclassinterval)–

(upperlimitoftheprecedingclassinterval)]

= 21 (21–20)=0.5

Intheaboveclassinterval,wesubtract0.5fromeachlowerlimitandadd0.5ineachupperlimit.Thereforewerewritethegiventableintothefollowingtable.

Heights in metre 15.5-20.5 20.5-25.5 25.5-30.5 30.5-35.5 35.5-40.5 40.5-45.5 45.5-50.5 50.5-55.5

number of trees 10 15 25 30 45 50 35 20

Nowtheabovetablebecomescontinuousfrequencydistribution.Thehistogramisgivenbelow

Fig. 3.2

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note:Inthehistogram(Fig.3.2)alongtheX-axisthefirstvaluestartsfrom15.5,thereforeabreak(kink)isindicatedneartheorigintosignifythatthegraphisdrawnbeginningat15.5andnotattheorigin.

3.3.2 frequency polygon

FrequencyPolygon is anothermethodof representing frequencydistributiongraphically.

Drawahistogramforthegivencontinuousdata.Markthemiddlepointsofthetopsofadjacentrectangles.Ifwejointhesemiddlepointssuccessivelybylinesegment,wegetapolygon.Thispolygoniscalledthefrequency polygon. It is customary to bring the ends of the polygon down to base level by assuming a lower class of afrequencyandhighestclassofafrequency.

FrequencyPolygoncanbeconstructedintwoways:(i) Usinghistogram(ii) Withoutusinghistogram.

3.3.2 (a) to draw frequency polygon using histogram

Procedure:Step 1 :Obtain the frequencydistribution fromthegivendataanddrawa

histogram.Step 2 : Jointhemidpointsofthetopsofadjacentrectanglesofthehistogram

bymeansoflinesegments.Step 3 :Obtain the mid points of two assumed class intervals of zero

frequency, one adjacent to the first bar on its left and anotheradjacenttothelastbaronitsright.Theseclassintervalsareknownas imagined class interval.

Step 4 :Complete the polygon by joining the mid points of first and thelastclassintervalstothemidpointoftheimaginedclassintervalsadjacenttothem.

Example 3.5Draw a frequency polygon imposed on the histogram for the following

distribution

class interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90frequency 4 6 8 10 12 14 7 5

ThebreakisindicatedbyaZig-Zagcurve.

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SolutionTaketheclass-intervalsalongtheX-axisandfrequenciesalongtheY-axiswith

appropriatescaleasshownintheFig3.3.Drawahistogramforthegivendata.Nowmarkthemidpointsoftheupper

sidesoftheconsecutiverectangles.Wealsomarkthemidpointsoftheassumedclassintervals0-10and90-100.Themidpointsarejoinedwiththehelpofaruler.Theendsof thepolygonare joinedwith themidpointsof0-10and90-100.Now,weget thefrequencypolygon.ReferFig3.3.

Fig. 3.3

Example 3.6Drawafrequencypolygonofthefollowingdatausinghistogram

class interval 0-10 10-20 20-30 30-40 40-50 50-60

frequency 5 10 25 16 12 8

SolutionMarktheclassintervalsalongtheX-axisandthefrequenciesalongtheY-axis

withappropriatescaleshowninFig3.4.

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Drawahistogramforthegivendata.Now,markthemidpointsoftheuppersidesoftheconsecutiverectangles.Alsowetaketheimaginedclassinterval(-10)-0and 60-70.Themidpointsarejoinedwiththehelpofaruler.Theendsofthepolygonarejoinedwiththemidpointsoftheimaginedclassintervals(-10)-0and60-70.Nowwegetthefrequencypolygon.(ReferFig3.4).

Fig. 3.4

note:Sometimesimaginedclassintervalsdonotexist.for example,incaseofmarksobtainedby the students in a test,wecannotgobelowzero andbeyondmaximummarksonthetwosides.Insuchcases,theextremelinesegmentsareonlypartlydrawnandarebroughtdownverticallysothat theymeetat themidpointsoftheverticalleftandrightsidesoffirstandlastrectanglesrespectively.

Usingthisnote,wewilldrawafrequencypolygonforthefollowingexample:Example 3.7

Drawafrequencypolygonforthefollowingdatausinghistogram

marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

number of students 5 4 6 8 5 7 4 9 5 7

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SolutionMarktheclassintervalsalongtheX-axisandthenumberofstudentsalongthe

Y-axis.Drawahistogramforthegivendata.Nowmarkthemidpointsoftheuppersidesoftheconsecutiverectangles.Themidpointsarejoinedwiththehelpofaruler.Notethat,thefirstandlastedgesofthefrequencypolygonmeetatthemidpointoftheverticaledgesofthefirstandlastrectangles.

Fig. 3.5

3.3.2 (b) to draw a frequency polygon without using histogram

Procedure:Step 1 :Obtain the frequency distribution and compute themid points of

eachclassinterval.Step 2 :RepresentthemidpointsalongtheX-axisandthefrequenciesalong

the Y-axis.Step 3 :Plotthepointscorrespondingtothefrequencyateachmidpoint.Step 4 : Jointhesepoints,bystraightlinesinorder.Step 5 :Tocompletethepolygonjointhepointateachendimmediatelytothe

lowerorhigherclassmarks(asthecasemaybeatzerofrequency)ontheX-axis.

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Example 3.8Drawafrequencypolygonforthefollowingdatawithoutusinghistogram.

class interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90

frequency 4 6 8 10 12 14 7 5

Solution:Mark the class intervals along the

X-axisandthefrequencyalongtheY-axis.Wetaketheimagined classes 0-10atthebeginning and 90-100 at the end, eachwith frequency zero. We have tabulatedthedataasshown.

Usingtheadjacenttable,plotthepointsA(5,0),B(15,4),C(25, 6),

D(35,8),E(45,10),F(55,12), G(65,14),H(75,7),I(85,5) andJ(95,0).WedrawthelinesegmentsAB,BC,CD,DE,EF,FG,GH,HI,IJtoobtainthe

requiredfrequencypolygonABCDEFGHIJ,whichisshowninFig3.6.

Fig. 3.6

class interval midpoints frequency0-10 5 010-20 15 420-30 25 630-40 35 840-50 45 1050-60 55 1260-70 65 1470-80 75 780-90 85 590-100 95 0

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exeRcISeS 3.1

1. Drawahistogramtorepresentthefollowingdata

Classintervals 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 8 12 6 14 10 5

2. Drawahistogramwiththehelpofthefollowingtable

Yieldperacre(Quintal) 11-15 16-20 21-25 26-30 31-35 36-40Numberofricefield 3 5 18 15 6 4

3. Drawahistogramtorepresentthefollowingdataofthespectatorsinacricketmatch

Ageinyears 10-19 20-29 30-39 40-49 50-59 60-69Numberofspectators 4 6 12 10 8 2

4. Inastudyofdiabeticpatientsinavillage,thefollowingobservationswerenoted

Age(inyears) 10-20 20-30 30-40 40-50 50-60 60-70

Numberofpatients 3 6 13 20 10 5

Representtheabovedatabyafrequencypolygonusinghistogram.

5. Constructahistogramandfrequencypolygonforthefollowingdata

Classinterval 0-10 10-20 20-30 30-40 40-50 50-60

Frequency 7 10 23 11 8 5

6. Thefollowingtableshowstheperformanceof150candidatesinanIntelligencetest.Drawahistogramandfrequencypolygonforthisdistribution

IntelligentQuotient 55-70 70-85 85-100 100-115 115-130 130-145

Numberofcandidates 20 40 30 35 10 15

7. Constructafrequencypolygonfromthefollowingdatausinghistogram.

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100Noofstudents 9 3 4 6 2 3 4 5 7 8

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8. Drawafrequencypolygonforthefollowingdatawithoutusinghistogram

Age(inyears) 0-10 10-20 20-30 30-40 40-50 50-60 60-70Numberofpersons 6 11 25 35 18 12 6

9. Constructafrequencypolygonforthefollowingdatawithoutusinghistogram

Classinterval 30-34 35-39 40-44 45-49 50-54 55-59

Frequency 12 16 20 8 10 4

10. Thefollowingarethemarksobtainedby40studentsinanEnglishexamination(outof50marks).Drawahistogramandfrequencypolygonforthedata

29,45,23,40,31,11,48,11,30,24,25,29,25,32,31,22,19,49,19,13, 32,39,25,43,27,41,12,13,32,44,27,43,15,35,40,23,12,48,49,18.

3.4 construction of Simple Pie chart

Haveyouevercomeacrossthedatarepresentedinacircularformasshown inFigure3.7andFigure3.8?

The figures similar to the above are called circlegraphs. A circle graph shows the relationship betweenawhole and its parts. Here, thewhole circle is dividedinto sectors. The size of each sector is proportionalto the activity or information it represents. Since,the sectors resemble the slices of a pie, it is called a pie chart.

Thetimespentbyaschoolstudentduringaday(24hours).

ViewerswatchingdifferenttypesofchannelsonTV.

Fig. 3.7 Fig. 3.8

Pie is an American food item

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Forexample,inthepiechart(Fig3.7).

The proportion of the sectorfor hours spent in sleeping

G = whole daynumber of sleeping hours

= 24 hours8 hours =

31

So,thissectorisdrawn31 rdpartofthecircle.

The proportion of the sectorfor hours spent in school

G = Whole daynumber of school hours

= 24 hours6 hours =

41

So,thissectorisdrawn41 thofthecircle.

The proportion of the sectorfor hours spent in play

G = Whole daynumber of play hours

= 24 hours3 hours = 8

1

So,thissectorisdrawn81 thofthecircle.

The proportion of the sectorfor hours spent in homework

G = whole daynumber of home work hours

= 24hours3 hours =

81

So,thissectorisdrawn81 thofthecircle.

The proportion of the sectorfor hours spent in others

G = whole daynumber of others hours

= 24hours4 hours = 6

1

So,thissectorisdrawn61 thofthecircle.

Addingtheabovefractionsforallactivities,

Wegetthetotal = 31

41

81

81

61+ + + +

=24

8 6 3 3 4+ + + + =2424 =1.

the sum of all fractions is equal to one.Here the time spent by a schoolstudentduringadayisrepresentedusingacircleandthewholeareaofthecircleistakenasone.Thedifferentactivitiesoftheschoolstudentarerepresentedinvarioussectorsbycalculatingtheirproportion.Thisproportionalpartcanalsobecalculatedusingthemeasureofangle.Since,thesumofthemeasuresofallanglesatthecentralpointis360°,wecanrepresenteachsectorbyusingthemeasureofangle.

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In the following example,we are going to illustrate how a pie chart can beconstructedbyusingthemeasureofangle.

Example 3.9Thenumberofhoursspentbyaschoolstudentonvariousactivitiesonaworking

day,isgivenbelow.Constructapiechartusingtheanglemeasurment.

activity Sleep School Play Homework Others

number of hours 8 6 3 3 4

Solution

Numberofhoursspentindifferentactivitiesinadayof24hoursareconverted

into component parts of 360°. Since the duration of sleep is 8 hours, it should be

representedby248 3600

# =120°.Therefore,thesectorofthecirclerepresentingsleephoursshouldhaveacentral

angleof120°.Similarly,thesectorrepresentingotheractivitiessuchasschool,play,homework,

andothersarecalculatedinthesamemannerintermsofdegree,whichisgiveninthefollowingtable:

activity duration in hours central angle

Sleep 8248 3600

# =1200

School 6246 3600

# =900

Play 3 360243 0

# =450

Homework 3 360243 0

# =450

Others 4 360244 0

# =600

Total 24 3600

drawing pie chart

Nowwedrawacircleofanyconvenientradius.Inthatcircle,startwithanyradiusandmeasure120°atthecentreofthecircleandmarkthesecondarmofthisangle.Thissectorrepresentsthehoursspentinsleeping.

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From this armnowmarkoffasecondsector,bymeasuringanangleof90°inthesamesenseasbefore.Thissectorrepresentsthe school hours. Proceedinglikethis,weconstructthesectorsfor play and home work. Theremaining sector will representthelastclass(i.e.others).

The sectors may beshaded or coloured differentlyto distinguish one from theother.Thecompletedpiechartisrepresentedinthefigureasshownabove.

note:Inapiechart, thevariousobservationsorcomponentsarerepresentedbythesectorsofacircleandthewholecirclerepresentsthesumofthevalueofallthecomponents.Clearly,thetotalangleof360°atthecentreofthecircleisdividedaccordingtothevaluesofthecomponents.

Thecentralangleofacomponent= Total valueValue of the component 3600#; E

Sometimes,thevalueofthecomponentsareexpressedinpercentages.Insuchcases,thecentralangleofacomponent= 100

Percentage value of the component 3600#; E.

Steps for construction of pie chart of a given data:

1. Calculatethecentralangleforeachcomponent,usingtheaboveformula. 2. Drawacircleofconvenientradius. 3. Withinthiscircle,drawahorizontalradius. 4. Draw radiusmaking central angle of first componentwith horizontal

radius;thissectorrepresentsthefirstcomponent.Fromthisradius,drawnextradiuswithcentralangleofsecondcomponent;thissectorrepresentssecondcomponentandsoon,untilweexhaustallcomponents.

5. Shadeeachsectordifferentlyandmarkthecomponentitrepresents. 6. Giveakey. 7. Givetheheading.

Thus,weobtaintherequiredpiechartforthegivendata.

Fig. 3.9

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Example 3.10Thefollowingtableshowsthemonthlybudgetofafamily

Particulars Food House Rent Clothing Education Savings Miscella-

neousExpenses

(in `) 4800 2400 1600 800 1000 1400

Drawapiechartusingtheanglemeasurement.SolutionThecentralangleforvariouscomponentsmaybecalculatedasfollows:

Particulars Expenses(`) Centralangle

Food 4800 360°120004800

# =144°

House rent 2400 360°120002400

# =72°

Clothing 1600 360°120001600

# =48°

Education 800 360°12000800

# =24°

Savings 1000 360°120001000

# =30°

Miscellaneous 1400 360°120001400

# =42°

total 12000 360°

Clearly,weobtaintherequiredpiechartasshownbelow.MonthlyBudgetofaFamily

Fig. 3.10

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Example 3.11TheS.S.L.CPublicExaminationresultofaschoolisasfollows:

Result Passed in firstclass

Passed in secondclass

Passed inthirdclass Failed

Percentageofstudents 25% 35% 30% 10%

Drawapiecharttorepresenttheaboveinformation.Solution

Centralangleforacomponent= 100Percentage value of the component 3600#

Wemaycalculatethecentralanglesforvariouscomponentsasfollows:

Result Percentage of students central angle

Passed in first class 25% 10025 3600

# =900

Passed in second class 35% 36010035 0

# =1260

Passed in third class 30% 36010030 0

# =1080

failed 10% 36010010 0

# =36°

total 100% 3600

Clearly,weobtaintherequiredpiechartasshownbelow:

SSLCPublicExaminationResults

Fig. 3.11

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exeRcISe 3.2

1. Yugendran’sprogressreportcardshowshismarksasfollows:

Subject Tamil English Mathematics Science SocialscienceMarks 72 60 84 70 74

Drawapiechartexhibitinghismarkinvarioussubjects.

2. Thereare36studentsinclassVIII.Theyaremembersofdifferentclubs:

Clubs Mathematics N.C.C j.R.C ScoutNumberofstudents 12 6 10 8

Representthedatabymeansofapiechart. 3. Thenumberofstudentsinahostelspeakingdifferentlanguagesisgivenbelow:

Languages Tamil Telugu Malayalam Kannada English Others

Numberofstudents 36 12 9 6 5 4

Representthedatainapiechart.

4. Inaschool,thenumberofstudentsinterestedintakingpartinvarioushobbiesfromclassVIIIisgivenbelow.Drawapiechart.

Hobby Music Pottery Dance Drama SocialserviceNumberofstudents 20 25 27 28 20

5. Ametalalloycontainsthefollowingmetals.Representthedatabyapiechart.

Metal Gold Lead Silver Copper ZincWeights(gm) 60 100 80 150 60

6. Onaparticularday,thesales(in`)ofdifferentitemsofabaker’sshoparegivenbelow.Drawapiechartforthisdata.

Item OrdinaryBread FruitBread Cakes Biscuits OthersCost (`) 320 80 160 120 40

7. Themoneyspentonabookbyapublisherisgivenbelow:

Item Paper Printing Binding Publicity RoyaltyMoneyspent(`) 25 12 6 9 8

Representtheabovedatabyapiechart.

8. Expenditureofafarmerforcultivationisgivenasfollows:

Item Ploughing Fertilizer Seeds Pesticides IrrigationAmount (`) 2000 1600 1500 1000 1100

Representthedatainapiechart.

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9. Thereare900creaturesinazooasperlistbelow:

Creatures Wildanimal Birds Otherlandanimals Wateranimals ReptilesNumberofCreatures 400 120 135 170 75

Representtheabovedatabyapiechart.

10. Inafactory,fivevarietiesofvehiclesweremanufacturedinayear,whosebreakupisgivenbelow.Representthisdatawiththehelpofapiechart.

Vehicle Scooter Motorbike Car Jeep VanNumber 3000 4000 1500 1000 500

11. Afoodcontainsthefollowingnutrients.Drawapiechartrepresentingthedata.

Nutrients Protein Fat Carbohydrates Vitamins MineralsPercentage 30% 10% 40% 15% 5%

12. Thefavoriteflavoursoficecreamforstudentsofaschoolisgiveninpercentagesasfollows

Flavours Chocolate Vanilla Strawberry OtherflavoursPercentageofStudentspreferringtheflavours 40% 30% 20% 10%

Representthisdatabyapiechart.

13. Thedataonmodesoftransportusedbystudentstocometoschoolaregivenbelow:

Modeoftransport Bus Cycle Walking Scooter CarPercentageofstudents 40% 30% 15% 10% 5%

Representthedatawiththehelpofapiechart.

14. Mr.RajanBabuspends20%ofhisincomeonhouserent,30%onfoodand10%forhischildren’seducation.Hesaves25%,whiletheremainingisusedforotherexpenses.Makeapiechartexhibitingtheaboveinformation.

15. Thepercentageofvariouscategoriesofworkersinastatearegiveninfollowingtable

Categoryofworkers

Cultivators Agriculturallabours

IndustrialWorkers

CommercialWorkers Others

Percentage 40% 25% 12.5% 10% 12.5%

Representtheinformationgivenaboveintheformofapiechart.

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3.5 measures of central tendency

Evenaftertabulatingthecollectedmassofdata,wegetonlyahazygeneralpictureofthedistribution.Toobtainamoreclearpicture,itwouldbeidealifwecandescribethewholemassofdatabyasinglenumberorrepresentativenumber.Togetmoreinformationaboutthetendencyofthedatatodeviateaboutaparticularvalue,there are certainmeasureswhich characterise the entire data.Thesemeasures arecalledthemeasures of central tendency.Suchmeasuresare

(i)ArithmeticMean, (ii)Median and (iii) Mode

3.5.1 arithmetic mean (a.m)

Thearithmeticmeanistheratioofthesumofalltheobservationstothetotalnumberofobservations.

3.5.1. (a) arithmetic mean for ungrouped data

If there are nobservations , ,, ,x x x xn1 2 3 g forthevariablex then their arithmetic

meanisdenotedbyx anditisgivenbyx =n

x x x xn1 2 3

g+ + + + .

In Mathematics, the symbolin Greek letter R , is called Sigma. This notation is used to represent thesummation.Withthissymbol,thesumof

, ,, ,x x x xn1 2 3 g is denoted as x

ii

n

1=

/ orsimply

as xiR .Thenwehave xnxiR= .

note:Arithmeticmeanisalsoknownas average or mean.

Example 3.12Themarksobtainedby10studentsinatestare15,75,33,67,76,54,39,12,78,

11. Find the arithmetic mean.SolutionHere,thenumberofobservations,n=10 A.M=x =

1015 75 33 67 76 54 39 12 78 11+ + + + + + + + +

x =10460 =46.

more about notation : S

kk 1

3

=

/ = 1+2+3=6

nn 3

6

=

/ = 3+4+5+6=18

n2n 2

4

=

/ = 2×2+2×3+2×4=18

5k 1

3

=

/ = 5k 1

3

=

/ × k0

= 5×10+5×20+5×30

= 5+5+5=15

1k2

4

K-

=

^ h/ = (2–1)+(3–1)+(4–1)=6

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Example 3.13Iftheaverageofthevalues9,6,7,8,5andxis8.Findthevalueofx.SolutionHere,thegivenvaluesare9,6,7,8,5andx,alson=6. Byformula,A.M.=x = x

69 6 7 8 5+ + + + + = x

635 +

Bydata,x = 8 So, x

635 + = 8

i.e.35+x = 48 x = 48–35=13.

Example 3.14The average height of 10 students in a class was calculated as 166 cm. On

verificationitwasfoundthatonereadingwaswronglyrecordedas160cminsteadof150cm.Findthecorrectmeanheight.

Solution Here,x = 166cmandn=10

Wehavex =nxR = x

10R

i.e.166 = x10R or Sx=1660

TheincorrectSx = 1660 ThecorrectSx = incorrectSx–thewrongvalue+correctvalue = 1660–160+150=1650Hence,thecorrectA.M. =

101650 =165cm.

3.5.1 (b) arithmetic mean for grouped dataArithmeticmeanforgroupeddatacanbeobtainedintwomethodswhichare (i)DirectMethod and (ii) Assumed Mean Method

(i) to calculate arithmetic mean (direct method)Supposewehavethefollowingfrequencydistribution

Variable x1 x2 x3 g xn Frequency f1 f2 f3 g fn

Thenthistableistobeinterpretedinthefollowingmanner:Thevalue: x1 occurs f1 times x2 occurs f2 times x3 occurs f3 times ggggg

ggggg

xn occurs fn times.

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Here , ,, ,x x x xn1 2 3

g arethedistinctvaluesofthevariablex.Inthiscase,thetotalnumberofobservationsisusuallydenotedbyN.

(i.e.,) f f f fn1 2 3

g+ + + + =N(or)i

n

1=

/ fi=NThenthetotalvaluesobserved

= x x x f x x x f x x x ftimes times timesn n n n1 1 1 1 2 2 2 2

g g g g+ + + + + + + + + + + +^ ^ ^h h h

= f x f x f xn n1 1 2 2# # #g+ + + = f x

i iR

Hence, x = Total number of observationsTotal values observed =

f

f x

i

i i

R

R

Usually,itiswrittenasx =ffxR

R =NfxR , whereN= .fR

Example 3.15CalculatetheArithmeticmeanofthefollowingdatabydirectmethod

x 5 10 15 20 25 30 f 4 5 7 4 3 2

Solution

x f f x

5 4 20

10 5 50

15 7 105

20 4 80

25 3 75

30 2 60

Total N=25 fxR =390

ArithmeticMean,x =Nf xR

=25390 =15.6.

(ii) to calculate arithmetic mean (assumed mean method)

In the above example multiplication looks very simple, since the numbersaresmall .Whenthenumbersarehuge, theirmultiplicationsarealways tediousoruninterestingandleadstoerrors.

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Toovercomethisdifficultyanothersimplermethodisdevised.Inthismethodweassumeoneofthevaluesasmean(A).ThisassumedvalueAisknownasassumedmean. Then we calculate the deviation , , , ,d d d d

, n1 2 3g of each of the variables

, , , ,x x x xn1 2 3

g from the assumed mean A. i.e. d x A1 1= - ,d x A2 2= - ,d x A3 3= - ,g ,d x An n= -

Now, multiply , , , ,d d d d, n1 2 3g respectively by , , , ,f f f f

n1 2 3g and add all these

valuestoget fdR .Now,

Arithmetic mean x =ffdAR

R+

x = fdA NR

+ (HereAisassumedmeanandN= fR )

Now, we can calculate the A.M. for the above problem (example 8.15) byassumed mean method.

TaketheassumedmeanA=15x f d = x – A f d5 4 –10 –4010 5 –5 –2515 7 0 020 4 5 2025 3 10 30

30 2 15 30

total n = 25 fdR = 15

ArithmeticMean=x =NfdA R

+

= 15+2515 =15+

53 =

575 3+ =

578

= 15.6.

3.5.2 Weighted arithmetic mean (W.a.m.)Sometimes the variables are associated with various weights and in

those cases the A.M. can be calculated, such an arithmetic mean is known as Weighted arithmetic mean (W.a.m.).

Forexample, letusassumethat thevariable x1 isassociatedwith theweight w1 ,x2 isassociatedwiththeweightw2 etc.andfinally,xn isassociatedwiththeweightwn then

W.A.M. =w w w w

w x w x w x w x

n

n n

1 2 3

1 1 2 2 3 3

g

g

+ + + +

+ + + + =wwxRR

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Example 3.16FindtheweightedA.Mofthepriceforthefollowingdata:

Food stuff Quantity(inkg)wi Priceperkg(in` ) xi

Rice 25 30Sugar 12 30Oil 8 70

SolutionHere the x-valuesarethepriceofthegivenfoodstuffandtheweightsassociated

arethequantities(inKg) Then,theW.A.M =

w w w w

w x w x w x w x

n

n n

1 2 3

1 1 2 2 3 3

g

g

+ + + +

+ + + +

=25 12 8

25 30 12 30 8 70# # #+ +

+ + =45

1670

= ` 37.11 .3.5.3 median

Anothermeasureofcentraltendencyisthemedian.3.5.3 (a) to find median for ungrouped data

Themedianiscalculatedasfollows: (i) Supposethereareanodd numberofobservations,writetheminascending

ordescendingorder.ThenthemiddletermistheMedian. Forexample:Considerthefiveobservations33,35,39,40,43.Themiddle

mostvalueoftheseobservationis39.ItistheMedianoftheseobservation.(ii) Suppose there are an even number of observations, write them in

ascendingordescendingorder.Thentheaverageofthetwomiddletermsis the Median.

Forexample,themedianof33,35,39,40,43,48is2

39 40+ =39.5.

note:TheMedianisthatvalueofthevariablewhichissuchthatthereareasmanyobservationsaboveandbelowit.

Example 3.17Findthemedianof17,15,9,13,21,7,32.SolutionArrangethevaluesintheascendingorderas7,9,13,15,17,21,32, Here,n = 7(oddnumber) Therefore,Median = Middlevalue = n

21 th+` j value=

27 1 th+` j value=4thvalue.

Hence,themedianis15.

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Example 3.18Acricketplayerhastakentheruns13,28,61,70,4,11,33,0,71,92.Findthe

median.SolutionArrangetherunsinascendingorderas0,4,11,13,28,33,61,70,71,92.Here n=10(evennumber).Therearetwomiddlevalues28and33.

\Median = Averageofthetwomiddlevalues

=2

28 33+ =261 =30.5.

3.5.3 (b) to find median for grouped data

cumulative frequency

cumulative frequencyofaclassisnothingbutthetotalfrequencyuptothatclass.

Example 3.19Findthemedianformarksof50students

Marks 20 27 34 43 58 65 89Numberofstudents 2 4 6 11 12 8 7

Solution

Marks(x) Numberofstudents(f) Cumulativefrequency

20 2 227 4 (2+4=)634 6 (6+6=)1243 11 (11+12=)2358 12 (23 + 12 = ) 3565 8 (35+8=)4389 7 (43+7=)50

Here,thetotalfrequency,N = fR =50 \

2N =

250 =25.

Themedianis 2N th

` j value = 25thvalue.Now, 25th value occurs in the cummulative frequency 35, whose

corresponding marks is 58.Hence,themedian=58.

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3.5.4 mode

modeisalsoameasureofcentraltendency.TheModecanbecalculatedasfollows:

3.5.4 (a) to find mode for ungrouped data (discrete data)

Ifasetofindividualobservationsaregiven,thentheModeisthevaluewhichoccurs most often.

Example 3.20Findthemodeof2,4,5,2,1,2,3,4,4,6,2.SolutionIntheaboveexamplethenumber2occursmaximumnumberoftimes.ie,4times.Hencemode=2.

Example 3.21Findthemodeof22,25,21,22,29,25,34,37,30,22,29,25.SolutionHere22occurs3timesand25alsooccurs3times ` Both22and25arethemodesforthisdata.Weobservethattherearetwo

modesforthegivendata.Example 3.22

Findthemodeof15,25,35,45,55,65,SolutionEachvalueoccursexactlyonetimeintheseries.Hencethereisnomodefor

this data.

3.5.4 (b) to find mode for grouped data (frequency distribution)

Ifthedataarearrangedintheformofafrequencytable,theclasscorrespondingtothemaximumfrequencyiscalledthemodalclass.Thevalueofthevariateofthemodalclassisthemode.

Example: 3.23

Findthemodeforthefollowingfrequencytable

Wages(`) 250 300 350 400 450 500Numberofworkers 10 15 16 12 11 13

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Solution

Wages(`) Numberofworkers250 10

300 15350 16

400 12

450 11500 13

Weobservefromtheabovetablethatthemaximumfrequencyis16. the value of the variate (wage) corresponding to the maximum frequency 16 is 350. Thisisthemodeofthegivendata.

exeRcISe 3.3 I. Problems on arithmetic mean 1. Findthemeanof2,4,6,8,10,12,14,16.

2. Iftheaverageofthevalues18,41,x,36,31,24,37,35,27,36,is31.Findthevalueofx.

3. Ifinaclassof20students,5studentshavescored76marks,7studentshavescored77marksand8studentshavescored78marks,thencomputethemeanoftheclass.

4. Theaverageheightof20studentsinaclasswascalculatedas160cm.Onverificationitwasfoundthatonereadingwaswronglyrecordedas132cminsteadof152cm.Findthecorrectmeanheight.

unimodal Bimodal trimodal multimodalIf there is only one mode in agivenseries,thenitiscalled unimodal.

If there are two modes in a givenseries,thenitiscalled Bimodal.

If there are three modes in a givenseries,thenitiscalled trimodal.

If there are more than three modes in the seriesitiscalled multimodal.

Example:10,15,20,25,15, 18,12,15.Here,Modeis15.

Example: 20,25,30,30, 15,10,25.Here25,30are Bimodal.

Example:60,40,85,30,85,45,80,80,55,50,60.Here60,80,85areTrimodal.

Example:1,2,3,8,5,4,5,4,2,3,1,3,5,2,7,4,1.Here1,2,3,4,5areMultimodal.

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5. CalculatetheArithmeticmeanofthefollowingdata:x 15 25 35 45 55 65 75 85f 12 20 15 14 16 11 7 8

6. Thefollowingdatagivethenumberofboysofaparticularageinaclassof40students.Calculatethemeanageofthestudents

age (in years) 13 14 15 16 17 18number of students 3 8 9 11 6 3

7. ObtaintheA.Mofthefollowingdata:marks 65 70 75 80 85 90 95 100number of students 6 11 3 5 4 7 10 4

8. Thefollowingtableshowstheweightsof12workersinafactoryWeight(inKg) 60 64 68 70 72Numberofworkers 3 4 2 2 1

Findthemeanweightoftheworkers.

9. Foramonth,afamilyrequiresthecommoditieslistedinthetablebelow.Theweightstoeachcommodityisgiven.FindtheWeightedArithmeticMean.

Commodity Weights(inkg) Priceperkg(in`)Rice 25 30

Wheat 5 20

Pulses 4 60

Vegetables 8 25

Oil 3 65

10. FindtheWeightedArithmeticMeanforthefollowingdata:Item NumberofItem Cost of Item

Powder 2 ` 45Soap 4 ` 12Pen 5 ` 15

Instrumentsbox 4 `25.50

II. Problems on median 1. Findthemedianofthefollowingsetofvalues:

(i)83,66,86,30,81.

(ii)45,49,46,44,38,37,55,51.

(iii)70,71,70,68,67,69,70.

(iv)51,55,46,47,53,55,51,46.

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2. Findthemedianforthefollowingdata:x 1 2 3 4 5 6 7 8

f 9 11 5 6 8 1 3 7

3. Theheight(incm)of50studentsinaparticularclassaregivenbelow:Height(incm) 156 155 154 153 152 151 150

Numberofstudents 8 4 6 10 12 3 7

Find the median.

4. Theheartsof60patientswereexaminedthroughX-rayandtheobservationsobtainedaregivenbelow:

Diameterofheart(inmm) 130 131 132 133 134 135Numberofpatients 7 9 15 12 6 11

Find the median.

5. Thesalaryof43employeesaregiveninthefollowingtable.Findthemedian.Salary(in`) 4000 5500 6000 8250 10,000 17,000 25,000

Numberofemployees 7 5 4 3 13 8 3

III. Problems on mode 1. Findthemodeofthefollowingdata:

i)74,81,62,58,77,74. iii) 43,36,27,25,36,66,20,25.ii)55,51,62,71,50,32. vi) 24,20,27,32,20,28,20.

2. Findthemodeforthefollowingfrequencytable:x 5 10 15 20 25 30f 14 25 37 16 8 5

3. Findthemodeforthefollowingtable:Temperaturein°c 29 32.4 34.6 36.9 38.7 40Numberofdays 7 2 6 4 8 3

4. Thedemandofdifferentshirtsizes,asobtainedbyasurvey,isgivenbelow.Size 38 39 40 41 42 43 44

Numberofpersons(wearingit) 27 40 51 16 14 8 6

CalculatetheMode.

IV. Problems on mean, median and mode 1. Findthemean,medianandmodeforthefollowingfrequencytable:

x 10 20 25 30 37 55f 5 12 14 15 10 4

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2. Theageoftheemployeesofacompanyaregivenbelow.Age(inyears) 19 21 23 25 27 29 31

Numberofpersons 13 15 20 18 16 17 13

Findthemean,medianandmode.

3. Thefollowingtableshowstheweightsof20students.Weight(inkg) 47 50 53 56 60

Numberofstudents 4 3 7 2 4

Calculatethemean,medianandmode.

Histogram and frequency polygon are the two types of graphicalrepresentationsofafrequencydistribution.

In thegraphical representationofhistogramandfrequencypolygontheclass-intervalsaretakenalongtheX-axisandthecorrespondingfrequenciesaretakenalongtheY-axis.

Inahistogramtherectanglesareplacedcontinuouslysidebysidewithnogapbetweentheadjacentrectangles.

Thepolygonobtainedbyjoiningthemidpointsofthetopoftheadjacentrectangles of a histogram and also the midpoints of the preceding and succeedingclassintervalsisknownasafrequencypolygon.

Thecentralangleofacomponent= 360Total value

Value of the component# c; E

Arthmeticmeanistheratioofthesumofalltheobservationstothetotalnumberofobservations.

FormulaforfindingA.M.

(i) x =nxR (ii) x =

ffxR

R

(iii) x = A+ffdR

R when A is the assumed mean and d=x–A.

TheweightedArithmeticmean(W.A.M.)=w

w x

i

i iR

R.

Themedian is thatvalueof thevariablewhich issuchthat thereareasmanyobservationsaboveandbelowit.

Modeisthatvaluewhichoccursmostfrequentlyinadistribution.

90

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4.1 Introduction

AncientEgyptiansdemonstratedpracticalknowledgeofgeometry

throughsurveyingandconstructionofprojects.AncientGreekspractised

experimentalgeometryintheirculture.Theyhaveperformedvarietyof

constructionsusingrulerandcompass.

Geometry is one of the earliest branches of Mathematics.

Geometry can be broadly classified into Theoretical Geometry and

PracticalGeometry.TheoreticalGeometrydealswiththeprinciplesof

geometrybyexplainingtheconstructionoffiguresusingroughsketches.

Practical Geometry deals with constructing of exact figures using

geometricalinstruments.

We have already learnt in the previous classes, the definition,

propertiesandformulaefortheareaofsomeplanegeometricalfigures.

Inthischapterletuslearntodrawconcentriccircles,

4.1 Introduction

4.2 ConcentricCirclesBrahmagupta[598-670A.D.]

Thegreat7thcentury Indian

mathematician of RajastanBrah-maguptawroteseveralbooksonMathematics and

astronomy. He becametheheadofastronomicalobservatoryatUjjain.Hewrotethefamousbook“Brahmasphuta-siddhanta”.

He found that 10-r and

the sum of the squaresofthe“n”naturalnumbers=( ) ( )n n n

61 2 1+ +

90

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4.2 concentric circlesInthissection,wearegoingtolearnaboutConcentricCircles.Alreadyweare

familiarwithCircles.

4.2.1. motivationWhenasmallstonewasdroppedinstillwater,youmighthaveseencircular

rippleswereformed.Whichisthecentreofthesecircles?Isitnottheplacewherethestonewasdropped?Yes.

Thecircleswithdifferentmeasuresofradiiandwiththesamecentrearecalledconcentriccircles.Thecentreisknownascommoncentre.

the concentric circles

circles drawn in a plane with a common centre and different radii are called concentric circles.SeeFig.4.1and4.2.

Lookatthefollowingtwofigures:

Fig.4.3representstwoconcentriccircles.InFig.4.4theareabetweenthetwoconcentriccirclesareshadedwithcolour.

Thecolouredareaisknownascircular ring.

Fig. 4.1 Fig. 4.2

Fig. 4.3 Fig. 4.4

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description : circular Ring

InFig.4.5,C1 and C2aretwocircleshavingthesamecentreOwithdifferentradiir1 and r2 respectively.

CirclesC1 and C2arecalledconcentriccircles.The area bounded between the two circles is

knownascircularring.Widthofthecircularring=OB–OA r r2 1= - (r2 >r1)

4.2.2. construction of concentric circles when the radii are given.

example 4.1Drawconcentriccircleswithradii3cmand5cmandshadethecircularring.

Finditswidth.SolutionGiven: Theradiiare3cmand5cm.

to construct concentric circles

Fig. 4.7

Fig. 4.5

Fig. 4.6

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Steps for constructionStep 1 : Drawaroughdiagramandmarkthegivenmeasurements.Step 2 : TakeanypointOandmarkitasthecentre.Step 3 : WithOascentreanddrawacircleofradiusOA=3cmStep 4 : WithOascentreanddrawacircleofradiusOB=5cm.ThustheconcentriccirclesC1 and C2 aredrawn. Widthofthecircularring =OB–OA =5–3 =2cm.

exeRcISe 4.1

1. draw concentric circles for the following measurements of radii. find out the width of each circular ring.

(i) 4 cm and 6 cm. (ii) 3.5 cm and 5.5 cm. (iii) 4.2 cm and 6.8 cm. (iv) 5 cm and 6.5 cm. (v) 6.2 cm and 8.1 cm. (vi) 5.3 cm and 7 cm.

Circles drawn in a planewith a common centre anddifferent radii arecalled concentric circles.

The area boundedbetween two concentric circles is known as circularring.

Width of the circular ring r r2 1= - ; (r2 >r1)

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anSWeRS

chapter 1

Exercise 1.1 1. i) D ii) C iii) B iv) B v) A

2. i) 200litres ii) 20,000km iii) ` 1,550

iv) 50minutes v) ` 50

3. ` 40,000 4. 5000,3750

5. i) 90% ii) 94% iii) 98% iv) 88% v) 95% vi) 93%

6. 5 7. ` 9,000 8.̀ 1,0209.180,1320 10.6kgs.

11. i) 26,100 ii) 5,220 12. 25%,` 8,600

13. Shescoredbetterinmathsby20%14. ` 6,250 15.20%

Exercise 1.2

1. i) ` 7490 ii) ` 500 iii) ` 9,000 iv)` 2,246 v)` 6,57,500

2. i) Profit` 64,Profit%=20% ii)Profit` 200,Profit%=8%

iii) Loss ` 19,Loss%=5% iv)S.P.=` 38,Loss%=5%

v) S.P.=` 5,500,Profit%=10%.

3. i) ` 787.50 ii) ` 1,260 iii) ` 2,835

4. ` 1,200 5. 3331 % 6. 25%

7. ` 22,80,000 8. ` 34,40,000

9. 11 91 % 10. Overallgain` 113.68

Exercise 1.3 1. i) A ii) D iii) B iv) B v) C2.` 360

3. ̀ 8,000 4. ` 49,220 5.` 18,433.40 6. ` 4,950

7. ` 13,000 8. 33% 9. ̀ 9,832.50 10.20%11.̀ 1,310.40pershirt

12. i) Amountofdiscount=` 460;S.P.=` 1,840

ii) Amountofdiscount=` 35;Rateofdiscount=25%

iii)M.P.=` 20,000;Amountofdiscount=` 4,000

iv) Rateofdiscount=5%;Amountofdiscount=` 725

v) Amountofdiscount=` 403;S.P.=` 2,821

Answers

95

MATH

EM

ATICS

Exercise 1.4

1. i) A=` 1,157.63,Interest=` 157.63

ii) A=` 4,840,Interest=` 840

iii) A=` 22,869,Interest=` 4,869

2. ` 2,125 3.i)̀ 88,200ii)̀ 4,4104.A=` 27,783,C.I.=` 3,783

5. ` 9,826 6.C.I.=` 1,951 7. ` 20,000 8. ` 36,659.70

9. i) ` 92,400 ii)` 92,610,Difference=` 210 10. ` 6

11. ` 25 12. ` 2,000 13. Sujapaysmoreinterestof` 924.10

14. P=` 1,25,000 15. 2years 16. 10%

Exercise 1.5

1. 2,205 2.` 2,55,150 3. ` 46,000 4.5,31,616.25

5. 5,415 6. ` 20,000 7. 10,000

Exercise 1.6

1. ` 27,000 2. ` 86,250 3. ` 10,800

4. ` 1,2505.E.M.I.=` 700,TotalAmount=` 19,404

6. E.M.I.=` 875,TotalAmount=` 8,750

Exercise 1.7

1. 24days 2. 10;1250 3. 10compositors 4. 15Workers

5. 24 days 6. ` 192

Exercise 1.8

1. 3days 2. 30days 3. 2days 4. 12minutes

5. A=` 360,B=` 2406.6 days 7. 1 hour

Answers

96

MAT

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ICS

chapter 2

Exercise 2.1

1.i)Dii)Ciii)Aiv)B

2.TheshortestsideisBC. 3. QR=26cm4.Itformsarightangledtriangle

5.QR=5cm 6. x=9m 7.Altitudex=5 3 cm

8. Yes 9. 2 51 ft

Exercise 2.2

1. i) D ii) D iii) C 2.radius=5cm.

chapter 3

Exercise 3.3

I. Problems on arithmetic mean

1. 9 2. x=25 3. 77.15 4. 161cm 5. 45

6. 15.45 7. 82.1 8. 65.33 9. ` 33 10. ` 21

II. Problems on median

1. i) 81 ii) 45.5 iii) 70 iv) 51

2. 3 3. 153 4. 132 5. ` 10,000

III. Problems on mode

1. i) 74 ii) Nomode iii) 25and36 iv) 20

2. 15 3. 38.7°C 4. 40

IV. 1.Mean28;Median25;Mode30

2.Mean25;Median25;Mode23

3.Mean53.05;Median53;Mode53

‘I can, I did’Student’s Activity Record

Subject:

Sl. No DateLesson

No.Topic of the Lesson Activities Remarks

97