Post on 02-Feb-2018
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MAT
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ICS 1 Life Mathematics
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
Chapter 1
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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.
Chapter 1
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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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ATICS
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
Chapter 1
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MAT
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ICS
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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ATICS
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?
Chapter 1
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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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ATICS
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?
Chapter 1
44
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ICS
` 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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ATICS
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.
Chapter 1
46
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ICS
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
Geometry
47
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ATICS
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
Chapter 2
48
MAT
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ICS
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
Geometry
49
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EM
ATICS
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
Chapter 2
50
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ICS
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
Geometry
51
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ATICS
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
Chapter 2
52
MAT
HE
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ICS
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?
Geometry
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ATICS
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?
Geometry
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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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ICS 4 PracticalGeometry
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
HE
MAT
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