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© VETRII IAS STUDY CIRCLEFirst Edition : 2015Second Edition : 2020

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My Dear Aspirants,

Greetings to all of you!

In all the TNPSC exams (Group I, II, IIA and IV), Aptitude & Logical Reasoning are of great significance. Questions are framed from the uniform system of school education (Samacheer Kalvi) book by the TNPSC.

This book is compiled by doing extensive research on standard textbooks for mathematics and also from the Previous year question paper of TNPSC. We have met the needs of what the TNPSC demands both from the old and new textbooks.

The answer have been provided for all the Questions in accordance with the recent trends in Aptitude and Logical reasoning.

I suggest all the aspirants to make the best use of this book for your success.

Our Journey towards your Success!!

M. SHUNMUGAM

Director,

Vetrii IAS Study Circle.

CONTENTS1. Number System and Number Series........................1

2. Simplification..........................................................33

3. Average and Problems on Ages...............................57

4. HCF and LCM.........................................................67

5. Ratio and Proportion...............................................83

6. Percentage..............................................................101

7. Simple and Compound Interest...................... ....... 121

8. Profit and Loss........................................................151

9. Probability...............................................................187

10. Time and Work.......................................................203

11. Area and Volume.....................................................215

12. Reasoning...............................................................327

13. Data Collections and Statistics..................................335

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Odd Numbers ▪ Anumberiscalledanoddnumberifthenumberwhichisnotdivisibleby'2'.

Eg., 1, 3, 5, 7, …▪ Alloddnumbersendwithanyoneoftheodddigits1,3,5,7or9.

Even Numbers▪ Anumberiscalledanevennumberifthenumberwhichisnotdivisibleby'2'.

Eg., 2, 4, 6, 8, ...▪ Allevennumbersendwithanyoneoftheevendigits0,2,4,6or8.

Natural Number▪Numberswhichweuseforcountingtheobjectsareknownasnaturalnumbers.Theyaredenotedby'n'. N = { 1, 2, 3, .... }

▪ Thefirstnaturalnumber1isoddandthefirstwholenumber0iseven.

Whole Number▪ Whenweinclude'Zero'inthenaturalnumbers,itisknownaswholenumbers.Theydenotedby'w'. w = { 0, 1, 2, 3 .... }

▪ Inwholenumbers,oddandevennumberscomealternatively.▪ 1isodd,itssuccessor2isevenandsoitspredecessor0isalsoeven.

Eg., 43 + 57 = 57 + 43; 12 × 15 = 15 × 12.

Associativity of Addition and MultiplicationWhen several numbers are added, the order in which the numbers are added does not matter. Thisiscalledassociativityofaddition.Similarly,whenseveralnumbersaretobemultiplied,theorderinwhichthenumbersaremultiplieddoesnotmatter.Thisiscalledassociativityofmultiplication.Eg., (43 + 57) + 25 = 43 + (57 + 25)

Chapter

1 Number System and

Number Series

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Vetrii IAS Study Circle Number System and Number Series

12 × (15 × 7) = (12 × 15) × 7

Note:SubtractionanddivisionareNOTassociative.

Factors

▪ Afactorisanumberthatdividesthegivennumberexactly(givesremainderzero).▪ Everynumberhastwofactorsthatis1andthenumberitself.▪ Everyfactorofanumberislessthanorequaltothatnumber.

Multiples

▪ Everymultipleofanumberisgreaterthanorequaltothatnumber. Multiplesof7are7,14,21,28,....Theyaregreaterthanorequalto7.▪Multiplesofanumberareendless.

Multiplesof5are5,10,15,20,.....Theyareendless.

Prime and Composite Numbers

▪ Anaturalnumbergreaterthan1,havingonlytwofactorsnamely1andthenumberitself,iscalledaprimenumber.

Eg.,2(1x2)isaprimenumberasis13(1x13).▪ All the prime numbers, except 2 and 3 can be expressed as 1more or 1 less to amultipleof6(Eg.,37=6x6+1).

▪ Explorethisstatementforotherprimes.▪ Anaturalnumberhavingmorethan2factorsiscalledacompositenumber.

Eg.,15isacompositenumber(15=1x3x5)asis70(1x2x5x7).

Anumber isaperfectnumber if thesumof its factorsother than thegivennumbergivesthesamenumber.Forexample,6isaperfectnumber,sinceaddingthefactorsof6(other than 6), namely 1, 2 and 3 gives the given number 6. i.e., 1 + 2 + 3 = 6 is the given number.

Twin Primes▪ Apairofprimenumberswhosedifferenceis2,iscalledtwinprimes.

Eg.,(5,7)isatwinprimepairasis(17,19).▪ Ifthreesuccessiveprimenumbersdifferby2,thentheprimenumbersformaprimetriplet.Theonlyprimetripletis(3,5,7).

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Vetrii IAS Study Circle Number System and Number Series

1.1 RULES FOR TEST OF DIVISIBILITY OF NUMBERSDivisibility by 2: Anumberisdivisibleby2,ifitsonesplaceisanyoneoftheevennumbers0, 2, 4, 6 or 8.Examples: 1.456368isdivisibleby2,sinceitsonesplaceiseven(8). 2.1234567isnotdivisibleby2,since,itsonesplaceisnoteven(7).

Divisibility by 3: Divisibilityofanumberby3isinteresting!wecanfindthat96isdivisibleby3.Here,notethatthesumofitsdigits9+6=15isalsodivisibleby3.Even1+5=6isalsodivisibleby3.Thisiscalledasiterativeorrepeatedaddition.So,Anumberisdivisibleby3ifthesumofitsdigitsisdivisibleby3.Examples: 1. 654321 is divisible by 3.

Here6+5+4+3+2+1=21and2+1=3isdivisibleby3.Hence,654321isdivisibleby3.

2.Thesumofanythreeconsecutivenumbersisdivisibleby3.(Eg., 33 + 34 + 35 = 102, is divisible by 3)

3. 107 is not divisible by 3 since, 1 + 0 + 7 = 8, is not divisible by 3.

Divisibility by 4: Anumberisdivisibleby4ifthelasttwodigitsofthegivennumberisdivisible by 4.

Note:Ifthelasttwodigitsofanumberarezeros,thenalsoitisdivisibleby4.

Examples:71628, 492, 2900 are divisible by 4, because 28 and 92 are divisible by 4 and 2900 is also divisible by 4 as it has two zeros.

Divisibility by 5: Anumberisdivisibleby5ifitsonesplaceiseither0or5.Examples: 5225 and 280 are divisible by 5.

Divisibility by 6: Anumberisdivisibleby6ifitisdivisiblebyboth2and3.Examples: 138, 3246, 6552 and 65784 are divisible by 6.

Divisibility by 8: Anumberisdivisibleby8ifthelastthreedigitsofthegivennumberisdivisible by 8.

Note:Ifthelastthreedigitsofanumberarezeros,thenalsoitisdivisibleby8.

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Vetrii IAS Study Circle Number System and Number Series

Examples: 2992 is divisible by 8 as 992 is divisible by 8 and 3000 is divisible by 8 as its last three digits are zero.

Divisibility by 9: Anumberisdivisibleby9ifthesumofitsdigitsisdivisibleby9.

Note:Thenumbersdivisibleby9aredivisibleby3.

Examples: 9567 is divisible by 9 as 9 + 5 + 6 + 7 = 27 is divisible by 9.

Divisibility by 10: Anumberisdivisibleby10ifitsonesplaceisonlyzero.Examples: 1. 2020 is divisible by 10 (2020 ÷ 10 = 202) where as 2021 is not divisible by 10. 2. 26011950 is divisible by 10 and hence, divisible by 5.Divisibility by 11: A number is divisible by 11 if the difference between the sum ofalternativedigitsofthenumberiseither0ordivisibleby11.Examples:▪ Consider thenumber 256795.Here, thedifferencebetween the sumof alternativedigits=(2+6+9)−(5+7+5)=17−17=0.

▪Hence,256795isdivisibleby11.

1.2 PRIME FACTORISATIONExpressingagivennumberasaproductoffactorsthatareallprimenumbersiscalledtheprimefactorisationofanumber.

Eg.,36canbewrittenasproductoffactorsas 36 = 1 × 36; 36 = 2 × 18; 36 = 3 × 12; 36 = 4 × 9; 36 = 6 × 6

Here,thefactorsof36canbefoundeasilyas1,2,3,4,6,9,12,18and36.Notethatnotallthefactorsof36areprimenumbers.Tofindtheprimefactorsof36,wedotheprimefactorisationbythefollowingmethods.

1. Division Method:

2 362 183 93 3

136 = 2 × 2 × 3 × 3

3 362 122 63 3

136 = 3 × 2 × 2 × 3

2. Factor Tree Method36

3 18

3 9

3 3

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Vetrii IAS Study Circle Number System and Number Series

1.3.1 Division and RemainderWhenagivennumber isnot exactlydivisibleby anynumber, then there is a remaindernumberattheendofsuchdivision.Supposewedivide25by7as,

7) 25 (321

4Then,divisor=7,dividend=25,Quotient=3andremainder=4So,wecanrepresentitas

Divisor) Dividend (Quotient _____________Remainder

Thus,divided=(divisor×quotient)+remainder.

1.3.2 Methods to find a Number Completely Divisible by AnotherConsider a given number x. When divided by d, it gives a quotient q and remainder ‘r’.Itimpliesthatthegivennumber‘x’isnotexactlydivisibleby‘d’.

d) x (q___r

Now,tofindanumberexactlydivisibleby‘d’,wecanuseeitherofthefollowingtwomethodstoreducetheremaindertozero.(Ifanumberisexactlydivisible,thenremainderiszero).

Method 1: Bysubtractingtheremainderfromthegivennumber(dividend).

∴Therequirednumberthatisexactlydivisibleby‘d’=divided – remainder

Hence,‘remainder’istheleastnumberthatcanbesubtractedfromdividedtomakeitexactlydivisible.

Method 2: Byaddingthe(divisor−remainder)tothegivennumber.

∴Therequirednumberthatisexactlydivisiblebyd=divided+(divisor–remainder)

Therefore,(divisor−remainder)istheleastnumberthatcanbeaddedtoanygivennumbertomakeitexactlydivisible.

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Example: Find the least number, that must be(a)subtractedfrom,(b)addedtoagivennumber5029,tomakeitexactlydivisibleby17.

Solution:Ondividing5029by17,wefindthat

17) 5029 (29534

162153998514

∴ remainder = 14.

(a) The least number to be subtracted to make it exactly divisible = remainder = 14. (By method 1).(b) The least number to be added to make it exactly divisible = divisor − remainder =17−14=3/(Bymethod2).

1.3.3 Greatest N-Digit and Least N-Digit Number Exactly Divisible by A Number

(a) Tofindoutthegreatestn-digitnumberexactlydivisiblebyadivisor‘d’,

∴Therequirednumber=Greatestn-digitnumber–remainder.

(b) To find out the least n-digit number exactly divisible by a divisor ‘d’,we useMethod 2 because ifwe usemethod 1, then subtracting any number from then-digitleastnumberwillreduceitto(n−1)digitnumber.

∴Therequirednumber=Greatestn-digitnumber+(divisor–remainder)

ExampleFind the

(a) greatest 3-digit number divisible by 13.(b) the least 3-digit number divisible by 13.

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Solution: (a) 13) 999 ( 11

∴ the required 3-digit greatest number =999−11=988.

(b) 13) 100 ( 9

∴the required 3-digit least number =100+(13−9) = 104.

1.3.4 Remainder RulesRule 1: Iftheremainderforthegreaterdivisor=r,andthesmallerdivisor=d,whenr > d, then the required remainder is ‘r’itself.

Example: Ifanumberisdividedby527,theremainderis42.Whatwillbetheremainderifitisdividedby 17?

Solution:Here,thesamenumberisdividedbytwodivisors;527and17.

Now, 52717

=31,so,527isamultipleof17,HenceRule1canbeapplied.

Remainderforthegreaterdivisor,(i.e.,for527)=42,Smallerdivisor=17. So, 17) 42 ( 34 8⇒requiredremainderforsmallerdivisor(i.e.,17)=8.

Rule 2: Iftwodifferentnumbersaandb,onbeingdividedbythesamedivisorleaveremaindersr1andr2respectively,thentheirsum(a+b)ifdividedbysamedivisorwillleave remainder R, given by

R=(r1+r2)−divisor

⇒ TherequiredremainderR=sumofremainders−divisor(Whensumisdivided).

Example:Two different numbers, when divided by the same divisor, leave remainders 15 and 39 respectively,andwhentheirsumwasdividedbythesamedivisor, theremainderwas7.What is the divisor?

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Vetrii IAS Study Circle Number System and Number Series

Solution:UsingtheRule27=(15+39)−divisor ⇒ Divisor = 47.

Example: Twodifferentnumbers,whendividedby47,leaveremainders13and23respectively.Iftheirsum is divided by the same number 47, what will be the remainder?

Solution:Therequiredremainder(13+23)−47=−11Since,theremainderis(−)ve,so,the actual remainder will be 23 + 13 = 36 (refertoNOTEunderRule2).

Rule 3: Whentwonumber,afterbeingdividedbythesamedivisorleavethesameremainder,thenthedifferenceofthosetwonumbersmustbeexactlydivisiblebythesamedivisor.

Example: Twonumbers147and225,afterbeingdividedbya2-digitnumber,leavethesameremainder,Find the divisor.

Solution:

Thedifferenceof225and147mustbeperfectlydivisiblebythedivisor. Thedifference=225−147=78 Now, 78 = 13 × 2 × 3 Thus,1-digitdivisors =2,3and2×3 2-digit divisors = 13 × 2, 13 × 3, 13, 13 × 2 × 3

∴ the possible divisors are 26, 39, 13, 78.

Rule 4: Trueremainder=(firstremainder)+(secondremainder×firstdivisor)+(thirdremainder×firstdivisor×seconddivisor)

Example: Anumber,beingsuccessivelydividedby5,7and11leaves3,1,2asremaindersrespectively.Findtheremainderifthesamenumberisdividedby385.

Solution:Here,thedivisoris385,whosefactorsare5,7and11.∴ByRule3,Trueremainder(i.e.,remainderwhendividedby385)

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= 3 + (1 × 5) + (2 × 5 × 7) = 3 + 5 + 70 = 78.

Rule 5: When (x + 1)n is divided by x, the remainder is always 1, where x and n are natural numbers.

Example: What will be the remainder when (17)21 is divided by 16?

Solution: (17)21 = (16 + 1)21

∴ When (16 + 1)21 is divided by 16, the remainder = 1.

Rule 6: When (x + 1)n is divided by x, then the remainder = 1, when n is an even natural number

Example:What will be the remainder when (29)75 is divided by 30?

Solution:(29)75=(30−1)75,hereindex=75(whichisodd)so,when(30−1)75 is divided by 30, the remainder will be x−1=30−1=29.

1.4 NUMBER SERIESInthenumberseries,somenumbersarearrangedinaparticularsequence.Allthenumbersformaseriesandchangeinacertainorder.Sometimes,oneormorenumbersarewronglyputinthenumberseries.Oneisrequiredtoobservethetrendinwhichthenumberschangeintheobservethetrendinwhichthenumberschangeintheseriesandfindoutwhichnumber/numbersmisfitintotheseries.Thatnumber/numbersistheODDNUMBERoftheseries.

1.4.1 Important Number SeriesFollowingaresomeoftheimportantrulesororderonwhichthenumberseriescanbemade.i). Pure Series: Inthistypeofnumberseries,thenumberitselfobeyscertainordersothat

thecharacteroftheseriescanbefoundout.The number it self may be:

▪Perfectsquare ▪Perfectcube▪Prime ▪Combinationofabove

ii). Different Series: Underthiscategory,thechangeinorderforthedifferencesbetweeneachconsecutivenumberoftheseriesisfound.

iii). Ratio Series: Under this category, the change in order for the ratios between eachconsecutivenumberoftheseriesisfound.