RRB JE Complete Book PDF

CL MEDIA (P) LTD. Edition : 2019 © PU BLI SH ER Administrative and Production Offices No part of this book may be reproduced in a retrieval system Published by : CL Media (P) Ltd. or transmit ted, in any form or by any means, electr onics,...

CL MEDIA (P) LTD. Edition : 2019 © PU BLI SH ER Administrative and Production Offices No part of this book may be reproduced in a retrieval system Published by : CL Media (P) Ltd. or transmit ted, in any form or by any means, electr onics, A-45, Mohan Cooperative Industrial Area, mechani cal, phot ocopyi ng, r ecor di ng, scanni ng and or Near Mohan Estate Metro Station, without the wr itten permission of the publisher. New Delhi - 110044 M arketed by : G.K. Publications (P) Ltd. A-45, Mohan Cooperative Industrial Area, I SBN : 978-93-88426-79-4 Near Mohan Estate Metro Station, Typeset by : CL M edia DTP Unit New Delhi - 110044 I SBN -93-87444-84-3 For product information : Visit www.gkpublications.com or email to [email protected]  Preface Railway Recr uit ment Boar d (RRB) Junior Engineer Examinat ion 2019 is a combined t wo-st age examinat ion followed by Document Ver ificat ion conduct ed by t he r espect ive RRBs for r ecr uit ment of Junior Engineer in I ndian Railways (I R). I n ever y t wo year s, a lar ge number of candidat es appear for t his exam, compet i ng for a limit ed number of post s. Thus RRB(JE) is consider ed one of t he most sought exams in I ndia due t o it s low select ion r at io and t echnical nat ur e. Unlike befor e, t he RRB(JE) 2019 exam pat t er n and syllabus has been complet ely changed. The old pat t er n consist ed of single st age examinat ion wher ein t he candidat es wer e allot t ed depar t ment s in I ndian Rai lways aft er clear ing t he exam. But t he r evised pat t er n includes t wo st ages – CBT-I and CBT-I I followed by document ver ificat ion, t he candidat e is r equir ed t o qualify each st age in or der t o move on t o t he next st age. The pr elims stage includes Gener al I nt elligence and Reasoning, Quantit at ive Apt it ude, Gener al Science and Gener al Awar e- ness. H er e t he CBT-I is common for all t he br anches. The second st age, CBT-I I is of t he object ive t ype t o t est t he t echnical abilit y of t he r espect ive engineer ing discipline. GK Publicat ions has been t he ‘‘publisher of choice’’ t o student s pr epar ing for GATE, ESE and ot her technical t est pr ep examinat ions in t he count r y. GK P's RRB(JE) 2019 ser ies pr ovides a wide r ange of st udy mat er ial which is classified into guides and object ive solved paper books t o simplify t he ent ir e pr epar at ion. These books have been t hor oughly updat ed as per t he lat est pat t er n and syllabus t o pr ovide ever yt hing you need t o per fect your scor e. GK P has also launched an andr oid app t o pr ovide you wit h an updat e on all upcoming vacancies in t he technical segment and it also has a lot of added cont ent t o aid your pr epar at ion. We hope t his lit t le effor t of our s will be helpful in achieving your dr eams. I f you have any suggest ions for impr ovement of t his book, you can wr it e t o us at gkp@gkpublicat ions.com. All the Best! Team GKP  Contents  SYL L ABU S CBT – I MATHEMATICS 1. N umber Syst em 1.1 - 1.9 2. Per cent age and Rat i o & Pr opor t i on 2.1 - 2.16 3. Pr obl ems On Age 3.1 - 3.4 4. Al l i gat i ons & M i xt ur es 4.1 - 4.5 5. Ti me and Wor k 5.1 - 5.11 6. Ti me and Di st ance 6.1 - 6.10 7. Boat s and St r eams 7.1 - 7.3 8. Si mpl e I nt er est and Compound I nt er est 8.1 - 8.10 9. Pr ofi t , L oss and Di scount 9.1 - 9.9 10. Aver age 10.1 - 10.12 11. Algebr a 11.1 - 11.10 12. Tr i gonomet r i c Rat i os and H ei ght & Di st ance 12.1 - 12.13 13. Cl ock and Cal endar s 13.1 - 13.10 14. Geomet r y 14.1 - 14.18 15. M ensur at i on 15.1 - 15.12 GENERAL INTELLIGENCE & REASONING 1. Al phabet i cal and N umber Ser i es 1.1 - 1.7 2. Anal ogy 2.1 - 2.10 3. Odd One Out 3.1 - 3.4 4. Codi ng Decodi ng 4.1 - 4.10 5. Bl ood Rel at i ons 5.1 - 5.7 6. Di r ect i on Sense Test 6.1 - 6.4 7. Venn Di agr ams 7.1 - 7.7 8. Syl l ogi sm 8.1 - 8.10 9. St at ement Concl usi on 9.1 - 9.7 10. St at ement Assumpt i on 10.1 - 10.6 11. M at hemat i cal Puzzl es 11.1 - 11.6 12. Cubes and Di ce 12.1 - 12.4 13. Anal yt i cal Reasoni ng 13.1 - 13.6 14. Dat a Suffi ci ency 14.1 - 14.9  GENERAL SCIENCE* 1. Physi cs 1.1 - 1.32 2. Chemi st r y 2.1 - 2.22 3. Bi ol ogy 3.1 - 3.43 GENERAL AWARENESS* 1. Science and Technology 1.1 - 1.7 2. Spor ts 2.1 - 2.7 3. I ndian Hist or y 3.1 - 3.8 4. Geogr aphy 4.1 - 4.14 5. I ndian Polit y 5.1 - 5.6 6. Cur r ent Affair s 6.1 - 6.21 CBT – II BASICS OF ENVIRONMENT AND POLLUTION CONTROL Basics of Envir onment and Pollut ion Cont r ol 1 - 16 BASICS OF COMPUTERS AND APPLICATIONS Basics of Computer s and Applicat ions 1 - 15 MECHANICAL 1. Engi neer i ng M echani cs 1.1 - 1.17 2. Engi neer i ng M at er i al s 2.1 - 2.36 3. St r engt h of M at er i al s 3.1 - 3.30 4. M anufact ur i ng Engi neer i ng (M achi ni ng, Wel di ng and Fi ni shi ng Pr ocess) 4.1 - 4.51 5. M et r ol ogy and I nspect i on 5.1 - 5.41 6. Fl ui d M echani cs and H ydr aul i c M achi ner y 6.1 - 6.50 7. I ndust r i al Engi neer i ng 7.1 - 7.34 8. Ther mal Engi neer i ng 8.1 - 8.64 PRACT I CE PAPE R CBT-I 1 - 12 CBT-II 1 - 10 * Common for CBT-I and CBT-I I  SYL L ABU S RECRUITMENT PROCESS Only single online applicat ion {common to all the notified posts in opted RRB - Junior Engineer (JE), Junior Engineer (I nformat ion Technology) [JE(I T )], Depot M aterial Superintendant (DM S) and Chemical & M etallurgical Assistant (CM A)} has t o be submit t ed by t he candidat e t hr ough t he link pr ovided on t he official websit e of RRBs. The ent ir e r ecr uit ment pr ocess shall involve, 1st st age Comput er Based Test (CBT), 2nd st age CBT, and Document Ver ificat ion/M edical Examinat ion as applicable. Select i on is made st r i ct ly as per mer it , on t he basis of CBTs. The dat e, t ime and venue for al l t he act ivi t i es vi z CBTs and DV or any ot her addit ional act ivi t y as applicable shall be fixed by the RRB and shall be intimated to the eligible candidates in due cour se. Request for postponement of any of t he above act ivity or for change of venue, date and shift will not be enter tained under any cir cumstances. 1ST STAGE CBT (COMMON FOR ALL NOTIFIED POSTS OF THIS CEN) Duration : 90 minutes (120 Minut es for eligible PwBD candidat es accompanied wit h Scr ibe) No. of Quest ions : 100 The 1st st age CBT is of scr eening nat ur e and t he st andar d of quest ions for t he CBT wi ll be gener all y in confor mit y wit h t he educat ional st andar ds and/or minimum t echnical quali fi cat ions pr escr ibed for t he post s. The nor mali zed scor e of 1st st age exam shall be used only for shor t l ist ing of candi dat es for 2nd st age exam as per t heir mer i t. Candidat es who ar e shor t li st ed for 2nd st age CBT avail ing t he r eser vat i on benefi t s of a communi t y, PwBD and ExSM shall cont i nue t o be consider ed onl y against t hat communi t y for all subsequent st ages of r ecr ui t ment pr ocess. The Quest ions wi ll be of object ive t ype wit h mul t i pl e choices and ar e li kely t o i ncl ude quest ions per t aining t o: a. Mathematics : N umber syst ems, BODM AS, Deci mal s, Fr act i ons, L CM and H CF, Rat i o and Pr opor t i on, Per cent ages, M ensur at i on, Time and Wor k , Time and Dist ance, Simpl e and Compound I nt er est , Pr ofit and L oss, Algebr a, Geomet r y, Tr igonomet r y, El ement ar y St at ist i cs, Squar e Root , Age Cal cul at ions, Calendar & Clock , Pipes & Cister n. b. General Intelligence and Reasoning : Anal ogi es, Alphabet ical and Number Ser ies, Coding and Decodi ng, M at hemat ical oper at i ons, Rel at ionshi ps, Syll ogism, Jumbl ing, Venn Di agr am, Dat a I nt er pr et at i on and Suffi ci ency, Conclusions and Decision M ak ing, Si mi lar it ies and Di ffer ences, Analyt ical r easoni ng, Classifi cat ion, Dir ect ions, St at ement – Ar gument s and Assumpt ions et c. c. General Awareness : K nowledge of Cur r ent affair s, I ndian geogr aphy, cult ur e and histor y of I ndia including fr eedom st r uggle, I ndian Polit y and const itution, I ndian Economy, Envir onmental issues concer ning I ndia and t he Wor ld, Spor ts, Gener al scient ifi c and t echnologi cal development s et c. d. General Science : Physics, Chemi st r y and L ife Sci ences (up t o 10t h St andar d CBSE syll abus). The sect i on wi se Number of quest i ons and mar k s ar e as bel ow : Subjects N o. of M arks for each Questions Section Stage-I Stage-I M athematics 30 30 Gener al I nt ell igence & Reasoni ng 25 25 Gener al Awar eness 15 15 Gener al Sci ence 30 30 Tot al 100 100 Ti me i n M inut es 90 The sect ion wise dist r ibut ion given in t he above t able is only indicat ive and t her e may be some var i at ions in t he act ual quest ion paper s. M inimum per cent age of mar ks for eligibilit y in var ious cat egor ies: U R -40%, OBC-30%, SC-30%, ST -25%. These per cent age of mar ks for eli gibil it y may be r elaxed by 2% for PwBD candi dat es i n case of shor t age of PwBD candidat es against vacanci es r eser ved for t hem. 2nd Stage CBT Shor t l ist i ng of Candidat es for t he 2nd St age CBT exam shall be based on t he nor mali zed mar ks obt ained by t hem in t he 1st St age CBT Exam. Tot al number of candi dat es t o be shor t li st ed for 2nd St age shall be 15 t imes t he communi t y wi se t ot al vacancy of Post s not ifi ed agai nst t he RRB as per t heir mer i t i n 1st St age CBT. H owever, Railways r eser ve t he r ight t o i ncr ease/decr ease t his li mi t i n t ot al or for any speci fi c cat egor y(s) as r equir ed t o ensur e avai labil it y of adequat e candi dat es for al l t he not i fi ed post s. Dur ation : 120 minut es (160 M inut es for eligible PwBD candidat es accompanied wit h Scr ibe) No of Quest ions : 150 Syllabus The Quest ions wi ll be of object ive t ype wi t h mult i pl e choices and ar e l ik ely t o i nclude quest ions per t aining t o Gener al Awar eness, Physics and Chemist r y, Basics of Comput er s and Applicat ions, Basics of Envir onment and Pollution Contr ol and Technical abilities for the post. The syllabus for Gener al Awar eness, Physics and Chemistr y, Basi cs of Comput er s and Appl icat i ons, Basics of Envir onment and Pol lut ion Cont r ol is common for all not i fi ed post s under t his CEN as det ai led bel ow:- a) General Awareness K nowledge of Cur r ent affair s, I ndian geogr aphy, cult ur e and histor y of I ndia including fr eedom st r uggle, I ndian Polit y and const itution, I ndian Economy, Envir onmental issues concer ning I ndia and t he Wor ld, Spor ts, Gener al scient ifi c and t echnologi cal development s et c. b) Physics and Chemistry Up t o 10t h st andar d CBSE syllabus. c) Basics of Computers and Applications Ar chit ect ur e of Comput er s; i nput and Out put devi ces; St or age devi ces, Net wor k ing, Oper at i ng Syst em l i ke Windows, Unix, L inux; M S Office; Var ious dat a r epr esent at ion; I nt er net and Email; Websit es & Web Br owser s; Comput er Vir us. d) Basics of Environment and Pollution Control: Basi cs of Envi r onment ; Adver se effect of envi r onment al poll ut i on and cont r ol st r at egies; Ai r, wat er and N oi se poll ut i on, t heir effect and cont r ol; Wast e M anagement , Gl obal war mi ng; Aci d r ai n; Ozone deplet ion. e) Technical Abilities: The educat i onal quali fi cat ions ment ioned agai nst each post shown in Annexur e-A, have been gr ouped int o di ffer ent exam gr oups as bel ow. Quest i ons on t he Technical abili t i es wi ll be fr amed i n t he syll abus defined for var ious Exam Gr oups gi ven at Annexur e-VI I -A, B, C, D, E, F & G. The sect i on wi se Number of quest i ons and mar k s ar e as bel ow : Subjects N o. of M arks Questions for each Section Stage-I I Stage-I I Gener al Awar eness 15 15 Physics & Chemi st r y 15 15 Basi cs of Comput er s and Applicat ions 10 10 Basi cs of Envi r onment and Pollut i on Cont r ol 10 10 Technical Abilit ies 100 100 Tot al 150 150 Ti me i n M inut es 120 The sect ion wise dist r ibut ion given in t he above t able is only indicat ive and t her e may be some var i at ions in t he act ual quest ion paper s. M i nimum per cent age of mar k s for eligibi lit y in var ious cat egor i es: UR -40%, OBC-30%, SC-30%, ST -25%. This per cent age of mar ks for el igibi li t y may be r elaxed by 2% for PwBD candidat es, i n case of shor t age of PwBD candidat es against vacanci es r eser ved for t hem. Vi r t ual cal cul at or wi ll be made availabl e on t he Comput er M onit or dur ing 2nd St age CBT. D iscipline M apping Tables:- (I ) Sl. N o. T hree years Diploma in Engineering E xam Gr oup or Bachelor ’s D egr ee in Engineering/Technology 1. M echanical Engineer ing Pr oduct ion Engi neer ing Aut omobi le Engineer ing M anufact ur ing Engineer ing M echanical and Allied M echat r onics Engineer ing Engineer ing I ndust r ial Engineer ing M achining Engineer ing Tool s and M achini ng Engi neer i ng Tool s and Die M aki ng Engineer i ng Combinat i on of any sub st r eam of basic st r eams of above di scipl ines 2. Elect r ical Engineer ing Electr ical and Allied Combinat i on of any sub st r eam of basic st r eams of Elect r ical Engi neer i ng Engineer ing 3. Elect r oni cs Engineer ing I nst r ument at i on and Cont r ol Engineer i ng  Communicat i on Engineer i ng Comput er Science and Engineer i ng Elect r onics and Allied Comput er Engineer ing Engineer ing Comput er Sci ence I nfor mat i on Technology Combinat i on of sub st r eams of basic st r eams of above discipli nes. 4. Civil Engineer ing Combinat i on of any sub st r eam of basic st r eams of Civil Engineer ing Civi l and Allied Engineer ing B.Sc., in Civi l Engineer i ng of 3year s dur at i on 5. Pr int ing Technology/Engi neer i ng Pr int ing Technology (I I ) Sl. N o. E ducat ional Qual ificat ions E xam Gr oup 1. B.Sc., Chemi st r y and Physi cs CMA (I I I ) Sl. N o. E ducat ional Qual ificat ions E xam Gr oup 1. BE/B.Tech., (Comput er Science) BE/B.Tech., (I nfor mat ion Technol ogy) PGDCA Comput er Science and B.Sc. Comput er Sci ence I nfor mat ion Technology BCA DOEACC “ B” L evel Cour se of 3 year s dur at i on or equivalent Al l t he candidat es wi t h t he above qual ificat i on shall be t est ed i n t he Exam Gr oup mapped as per t he above char t. H owever, candi dat es wit h educat ional qualificat i on of BE/B.Tech (Comput er Science) or BE/B.Tech (I nfor mat ion Technology), appl ying for bot h t he post s of JE(S& T Depar t ment ) and JE (I T), have t o opt for ei t her El ect r oni cs and All ied Engineer ing Exam Gr oup or Comput er Science and I nfor mat ion Technology Exam Gr oup. The educat i onal quali fi cat ion for t he post of DM S (Depot M at er ial Super int endent ) i s Thr ee Year s Diploma in Engi neer i ng i.e a candidat e wit h Thr ee Year s Diploma in any of Engi neer i ng di sci pl ines, can appl y for t hese post s as appli cable. Candi dat es wi t h educat ional qual ificat i ons not figur i ng in t he above char t and el igi bl e for DM S post s have t o choose any one of t he above l ist ed Exam Gr oups ot her t han CM A Exam Gr oup, dur ing t he r egist r at ion for onl ine appl icat i ons of t his CEN. A candidat e possessing mor e t han one minimum educat i onal quali fi cat ion, mapped t o differ ent Exam Gr oups, can choose any one Exam Gr oup. These candi dat es woul d be el igibl e for all t he post s for which t hey possess minimum educat ional qualificat ions. SH ORTLI ST I N G OF CAN DI DAT ES FOR 2 nd STAGE CBT, DV AN D EM PAN ELM EN T: Shor t l ist i ng of Candidat es for t he DV shal l be based on t he nor mal ized mar ks obt ained by t hem i n 2nd St age CBT. The nor mali zat ion scheme t o be adopt ed for shor t l ist ing t he candidat es fr om 1st St age CBT t o 2nd St age CBT and for DV on t he basis of per for mance i n 2nd St age CBT is det ail ed below: N ORM ALI SAT I ON OF TH E M ARKS: Whenever CBT is conducted in multiple sessions for t he same syllabus, the r aw mar ks obtained by the candidat es in differ ent sessions wi ll be conver t ed t o nor malized mar k s. The r aw mar ks for single session paper and nor malized mar ks for mult i session paper will be used for comput ing M er i t I ndex, whi ch is a common benchmar k for gener at ing mer it for candi dat es fr om differ ent Exam Gr oups. CALCU LATI ON OF N ORM ALI ZED M ARKS FOR M U LT I -SESSI ON PAPERS: I n 1st St age CBT and for some Exam Gr oups i n 2nd St age CBT, t he exami nat ion may be conduct ed in mult i- sessions. H ence, for t hese mult isession paper s, a sui t able nor mal izat i on is appl ied t o t ake int o account any var i at i on i n t he di fficult y l evel s of t he quest i on paper s acr oss differ ent sessi ons. The for mula for cal culat ing t he nor mal ized mar ks for t he mul t i-session paper s is det ail ed below: Nor mal izat i on mar k of j t h candidat e i n i t h sessi on M̂ ij is given by : M tg  M qg M̂ ij  M ti  M i q  M i j  M i q   M qg M ij : is t he act ual mar ks obt ained by t he j t h candidat e i n i t h session. M tg : is t he aver age mar k s of t he t op 0.1% of t he candidat es consider i ng al l sessi ons. M qg : is t he sum of mean and st andar d devi at i on mar k s of t he candidat es in t he paper consider i ng al l sessi ons. M ti : is t he aver age mar k s of t he t op 0.1% of t he candidat es in t he i t h session or mar ks of t opper i f sessi on st r engt h is less t han 1000. M iq : is t he sum of t he mean mar ks and st andar d deviat ion of t he i t h session. CALCU LATI ON OF M ERI T I N DEX FOR ALL PAPERS : I n or der t o gener at e a common mer it l ist compr i si ng of candidat es who gave exami nat ion fr om differ ent Exam Gr oups, but el igi bl e for a common post , mer it i ndex wil l be comput ed. For all paper s for which t her e is only one session, act ual mar ks obt ained by t he candi dat es will be used for calcul at ing M er i t I ndex, whi le for paper s in mult i-sessi ons, nor mali zed mar ks wi ll be calcul at ed cor r esponding t o t he r aw mar ks obt ai ned by a candi dat e and t he M er it I ndex will be cal culat ed based on t he nor mal ized mar k s. The M er it I ndex wil l be comput ed usi ng t he for mula gi ven bel ow : M  Mq M er i t I ndex = Sq + (St – Sq) Mt  Mq (M er it I ndex i s t he r el at i ve scor e of a candi dat e wi t hi n t he discipli ne.) M : M ar ks obt ained by t he candi dat e (act ual/r aw mar ks for single session exam and nor malized mar ks for mult i session exam. M q: The qualifyi ng mar k s for gener al cat egor y candidat e in t he paper (40). M t : The mean mar ks of t op 0.1% or t op 10 whi chever is lar ger of t he candi dat es who appear ed in t he paper (in case of mult i sessi on exam incl udi ng all sessi ons) Sq: 350 i s t he scor e assigned t o M q. St : 900 i s t he scor e assi gned t o M t. The quali fying mar ks (M q) for gener al cat egor y candi dat e is 40. The M er it I ndex wi ll be calculat ed for UR, OBC, SC, ST candi dat es whose act ual mar k s for singl e session exam and nor mali zed mar ks for mul t i session exam ar e equal or above t he communi t y qual ifying mar ks pr escr ibed in Par a 13.2. Based on t he M er it I ndex gener at ed, a combined mer it list of t he candidat es of differ ent disciplines/ Exam Gr oup wil l be pr epar ed in t he descendi ng or der of mer i t and t he al lot ment of t he pr efer ence wil l be done on t he basi s of t his mer it l ist. CBT – I   1 CHAPTER Number System N U M BER SYSTEM ( ii ) P r i m e an d Com posi t e n u m ber s : Pr i m e I t divided int o r eal number and imaginar y number. number s ar e t hose, which does not have any fact or The number which we can put on a number line ar e except 1 and it self like 2,3,5,7,11,13 et c. Composit e 1 1 1 number s ar e number s ot her t han pr ime. r eal number , e.g. , 2, , , 2 et c. 2 3 4 Note : (1) 1 is neit her pr ime nor composit e. And ot her number s ar e i magi nar y number s, e.g. Note :(2) 2 is t he only even pr ime. 4 , 6 et c. Note : (3) Ther e ar e 25 pr ime number s fr om 1 t o 100 and 15 pr ime number s fr om 1 t o 50. H er e 1 is denot ed by i. U N I T’S DI GI T I N TH E PRODU CT REAL N U M BERS Real number s ar e divided int o r at ional number s and E xampl e. F i n d t h e u n i t ’s di gi t i n t h e pr odu ct ir r at ional number s. (256 × 27 × 159 × 182). 1. Rat ional N umbers. Solution : Pr oduct of unit ’s digit s in given number s p = (6 × 7 × 9 × 2) = 756 Rat ional number s can be put in t he for m of wher e q is not equal t o 0. q H ence, unit digit in t he given pr oduct is 6. 2 3 E xampl e. F i n d t h e u n i t ’s di gi t i n t h e pr odu ct e.g. , et c. 3 5 (367 × 639 × 753). Again r ational number s ar e divided into terminating Solution : Cl ear ly, unit di git i n 34 i s 1 decimals and non-t er minat ing decimals.  Unit di gi t i n 364 i s 1. e.g. 1/2 = 0.5 i s a t er mi nat i ng deci mal whi l e 1/3, 1/6 ar e non-t er minat ing decimal. Ter minat ing  U nit digit in 367 is 7. decimals ar e being identified by t he fact t hat t her e (U ni t digit in 1 × 3 × 3 × 3 i s 7) i s n o pr i m e f act or ot h er t h an 2 or 5 i n t h e Cl ear ly, uni t di git i n ever y power of 6 i s 6. denominat or of t he lowest fr act ion while if t her e  U nit digit in 639 is 6. is any ot her pr ime fact or ot her t han 2 or 5, t han non-t er minat ing but r epeat ing. We can put a bar Cl ear ly, unit di git i n 74 i s 1. on it for defi ni ng t he same. For exampl e, 2. 3  Unit di gi t i n 752 i s 1. —  U nit digit in 753 is 7. (U ni t digit in 1 × 7 i s 7) means 2.333........, 4. 67 means 4.676767........ Unit di git i n given pr oduct 2. I rr at ional N umbers. = Unit digit in (7 × 6 × 7) = 4 The number s which ar e non-t er minat ing and non r epeat ing ar e ir r at ional number s. FACTORI AL N U M BERS e.g. 2,3 et c. The hi ghest power of a pr i me number ‘a’ which i s Classificat ion of Rat ional numbers. cont ained in n ! is Rat ional number s ar e fur t her divided int o decimals and i nt eger s. Ther e ar e posi t i ve i nt eger s, 0 and n  n   n   a    2    3  ............ negat ive integer s. Posit ive int eger s ar e called natur al   a  a  number s. whi le posi t i ve i nt eger s wit h 0 ar e cal led whole numbers. wher e [ x ] r epr esent s t he gr eat est int eger less t han or equal t o x. Nat ur al number s fur t her classified int o t wo differ ent categor ies : Example. What is t he highest power of 2 cont ained ( i ) Even and Odd numbers : This concept of even in 70! ? and odd looks t o be ver y simple but certain t hings Sol ut i on. need t o be under st ood  70   70   70   70   70   70   70  i.e. E + E = E, E + O = O, O + O = E, E × E = E,  2    2    3    4    5    6    7 ........... O × E = E, O × O = O   2  2  2  2  2  2  0 (zer o) is neit her posit ive nor negat ive. = 35 + 17 + 8 + 4 + 2 + 1 + 0... = 67  1.2 Number System I DEN TI T I ES (General Formulae) e.g. 1  [1 + 1  {1 + 1  (1 + 1  2) } ]  (a + b)2 = a2 + 2ab + b2 1 = 1  [1 + 1  {1 + 1  (1 + )}]  (a – b)2 = a2 – 2ab + b2 2  (a + b) (a – b) = a2 – b2 3 = 1  [1 + 1  {1 + 1  }]  (a + b)2 + (a – b)2 = 2(a2 + b2) 2  (a + b)2 – (a – b)2 = 4ab 2 = 1  [1 + 1  {1 + }]  (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 3  (a + b)3 = a3 + b3 + 3ab (a + b) = a3 + b3 + 3a2b + 3ab2 5 = 1  [1 + 1  ]  (a – b)3 = a3 – b3 – 3ab (a – b) = a3 – b3 – 3a2b + 3ab2 3  a3 + b3 = (a + b) (a2 – ab + b2) = (a + b)3 – 3ab(a + b) 3 = 1  [1 + ]  a3 – b3 = (a – b) (a2 + ab + b2) = (a – b)3 + 3ab(a – b) 5 8  a2 + b2 = (a + b)2 – 2ab or (a – b)2 + 2ab =1 5  a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) 5  I f a + b + c = 0, t hen a3 + b3 + c3 = 3abc = 8  (x + a) (x + b) = x 2 + x (a + b) + ab RATI ON AL I SAT I ON  (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 I f pr odcut of t wo sur ds is r at ional, t hen one of t hem is  a4 + a2b2 + b4 = (a2 + ab + b2) (a2 – ab + b2) called r at ionalising fact or (R.F) of t he ot her.  a4 – b4 = (a + b) (a – b) (a2 + b2) I RRATI ON AL N U M BERS  (a + b)2 = (a – b)2 + 4ab TI PS : When doing t hese sort of problems, remember :  (a – b)2 = (a + b)2 – 4ab (1) When you conver t a fr act ion int o a decimal, you  a3 + b3 + c3 = (a + b + c)3 – 3(a + b) (b + c) (c + a) end up wi t h eit her a t er mi nat i ng decimal or a  (a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) – abc r ecu r r i n g deci m al. N u m ber s w h i ch can be expr essed as a fr action ar e called rational numbers.  a2 + b2+ c2 – ab – bc – ca 1/4 = 0.25 (Ter minat ing decimal) 1 1/3 = 0.33333..... (Recur r ing decimal) = [( a – b) 2  (b – c) 2  ( c – a) 2 ] 2 (2) Number s which cannot be expr essed as a fr act ion  I f a2 + b2 + c2 – ab – bc – ca = 0, t hen a=b=c ar e called ir r at ional number s. These number s go on for ever wit hout any appar ent pat t er n FG a  1 IJ 2 1 2a e.g : , 2 , 7 et c.  H bK = a2 + b 2 + b   has been calculat ed by comput er s t o millions of decimal places and st ill no pat t er n has been FG a – 1 IJ 2 1 2a found!  H bK = a2 + b2 – b  Squar e r oot s of all non-squar es ar e ir r at ional.  Cube r oot s of all non-cubes ar e ir r at ional. BOD M AS (3) The pr oduct of 2 ir r ational number s can sometimes To do t he simplificat ion, we should always car r y out be a r at i on al num ber. Remem ber on e of t he t he oper at ions in t he or der of each let t er of t he wor d examples below: ‘BODM AS’, wher e e.g., 3  12  36  6 B br acket s [{( )}] O of or 5  5  25  5 D division  (4) Expr essi ons which i nclude at l east 1 ir r at ional M multiplication  number ar e called sur ds. A addition + e.g., 2, 5  7 et c. S subtr action – I N DI CES e.g. 12 – 4  5 + 36  9 + 8 = 12 – 4  5 + 4 + 8 an is t he pr oduct of n fact or s each of which is ‘a’ called = 12 – 20 + 4 + 8 base of t he power and ‘n’ is any nat ur al number called index or exponent of t he power. = 24 – 20 e.g. I f a  a  a  a is wr it t en as a4, t hen a4 is called = 4. indices of base.  Number System 1.3 LAW OF I N DI CES M ethods to find H CF.  ao = 1 1. By met hod of fact orizat ion :  a1 = a 36 = 2 × 2 × 3 × 3  am  an = am+n 64 = 2 × 2 × 2 × 2 × 2 × 2  am  an  ap ... = am + n + p +...  HCF = 2 × 2 = 4 (Pr oduct of common fact or s)  (am )n = amn 2. By division method : Suppose t wo number s ar e  (ab)m = am bm given. Divide t he gr eat er number by t he lesser ; t he l esser by t he r emai nder ; di vi de t he fi r st p m n r emainder by t he new r emainder and so on t ill   a   = am  n  p ... t her e is no r emainder. The last divisor is t he H CF  am  an = am – n r equir ed.  I f am = an, t hen m = n Example. Find t he H CF of 64 and 36.  I f am = bm , t hen a = b Solution : 36)64(1  I f am = 1, t hen m = 0 36 32)36(1 1  a– m = 32 am 4)32(8 FG a IJ m am 32  H bK = bm 0 am The H CF of 64 and 36 is 4.  = am– n if m > n bn Example. Find t he H.C.F. of 126, 396 and 1080. Solution : Expr essing the number s in pr ime fact or s. am 1 = if n > m 126 = 2 × 32 × 7 an a n– m 396 = 22 × 32 × 11 n  a.n b  n ab 1080 = 23 × 33 × 5 The highest power of 2, which will divide 2, 22, and  m a.n a  mn am  n 23, is 2. The highest power of 3, which will divide 32 and mn  a  mn a 33, is 32, and t her e ar e no ot her common fact or s. Thus H.C.F. is 2 × 32, or 18. n a mn Example. Find t he H.C.F. of 440, 1800, 2800.  m  am  n a Solution : 440 = 10 × 11 × 4, and of t hese t hr ee fact or s 10 and 4 divide all t hr ee number s 11 does  p n a. q n b  pq n ab not.  H.C.F. = 10 × 4 = 40  I f x n = y, t hen x = y 1/n  x = n y 2 Some I mport ant Result s  e aj =a  The pr oduct of t wo number is equal t o pr oduct of t heir H CM and L CM. n  a p = ap/n  I f t wo number s divided by a t hir d number gives t he same r emainder, t heir differ ence is exact ly p m = divisible by t hat number.  n m a  mn a pm  I f ther e ar e mor e than 2 number s, say 4 number s. Find t he H CF of any 2 number s and t he H CF of n  a n b  an b the other 2 number s. The HCF of their HCFs gives t he H CF of all t he 4 number s. H I GH EST COM M ON FACTOR (H CF) LEAST COM M ON M U LTI PLE (LCM ) I t is t he gr eat est fact or common t o t wo or mor e given L east common mult iple (L CM ) of t wo or mor e given number s. I t is also called GCM (Gr eat est Common number is the least number which is exactly divisible Measur e). by each of t hem e.g. 4 is t he GCM of 12 and 16. e.g. 30 is t he L CM of 2, 3, 5, 6.  1.4 Number System M ethod to find LCM. 2. U si n g t h e f or m u l a : Pr odu ct of n u m ber s = 1. By met hod of fact orizations: Resol ve each one H CF × L CM of t he given number s int o pr ime fact or s, t hen t heir I f t wo number s ar e given, t heir L CM is given by L CM is t he pr oduct of highest power s of all fact or s, t hat occur in t hese number s. Pr oduct of t wo number LCM = Example. Find t he L CM of 36 and 64. H CF Solut ion : 36 = 2 × 2 × 3 × 3 For t he above example 64 = 2 × 2 × 2 × 2 × 2 × 2 36  64 LCM = = 576  L CM = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 = 576 4 DI VI SI BI LI TY RU LES. N umber Conditions divisible by 2 Last digi t is 0 or an even number. 3 Sum of all t he digit s of t he number is divisible by 3. 4 Last t wo digit s of t he number is divisible by 4 or 00. 5 Last digi t of t he number is 0 or 5. 6 Last digit i s 0 or an even number , and sum of all t he digit s of t he number is divisible by 3. 7 Differ ence bet ween digit /digit s in fr ont and doubled value of t he last digit is 0 (or ) is divisi ble by 7. 8 Last t hr ee digit s of t he number is divi sible by 8. 9 Sum of all t he digit s of t he number is divisible by 9. 10 Last digi t is 0. 11 I f ever y second digit is added and t hen subt r act ed sum of all ot her digit s, t he answer is 0, or divisible by 11. 12 Number is divisi ble by bot h 3 and 4. PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 4. I f n umer at or an d den omi nat or of a pr oper 1. I f t he following gr oups of fr act ions is ar r anged in fractions ar e incr eased by the same quantity, then ascending or der ? t he r esult ing fr act ion is 5 7 6 7 6 5 (a) always gr eat er t han t he or iginal fr act ion (a) , , (b), , (b) always less t han t he or iginal fr act ion 16 18 17 18 17 16 5 6 7 6 7 5 (c) always equal t o t he or iginal fr act ion (c) , , (d) , , 16 17 18 17 18 16 (d) none of t hese 9 2 8 5 5. I f x + y > 5 and x – y > 3, t hen which of t he 2. If f r act i on s , , , ar e ar r an ged i n following gives all possible values of x ? 13 3 11 7 ascendi ng or der, t hen t he cor r ect sequence (a) x > 3 (b) x > 4 (c) x > 5 (d) x < 5 is 6. I f x an d y ar e n egat i ve, t hen whi ch of t he 9 2 8 5 2 9 5 8 (a) , , , (b) , , , following st at ement s is/ar e always t r ue ? 13 3 11 7 3 13 7 11 I. x + y is posit ive 2 8 5 9 5 8 2 9 (c) , , , (d) , , , I I. xy is posit ive 3 11 7 13 7 11 3 13 3. Which one of t he following is t he lar gest ? I I I. x – y is posit ive (a) I only (b) I I only 2 5 , 6 3 , 3 7 and 8 2 (c) I I I only (d) I and I I I only (a) 8 2 (b) 2 5 (c) 6 3 (d) 3 7  Number System 1.5 7. The value of 3 0.000064 is 3. 10 = 3.1623(appr ox.). What is the appr ox, value 1 (a) 0.02 (b) 0.2 (c) 2.0 (d) N one of ? 8. I f 11, 109, 999 is divided by 1111, then what is t he 10 r emainder ? (a) 0.333 (b) 0.3162 (a) 1098 (b) 11888 (c) 1010 (d) 1110 (c) 0.3221 (d) 0.3437 [RRB JE 2014 GREEN SH I FT ] 1 1 1  of 4. Find the value of (2744)1/3 : 2 2 2 9. The value of is (a) 24 (b) 14 1 1 1  of 2 2 2 (c) 34 (d) 16 [RRB JE 2014 RED SH I FT ] 2 1 5. Find the L.C.M. of 148 and 185. (a) 2 (b) 1 (c) 1 (d) 3 3 3 (a) 680 (b) 740 10. Taking 2 = 1.414, 3 = 1.732, 5 = 2.236 and (c) 2960 (d) 3700 [RRB JE 2014 RED SH I FT ] 6 = 2.449, t hen t he value of 1 6. If 22n 1  , then the value of 'n' is : 9 2 9– 2 8n  3 + to three places of decimals is 5 3 5– 3 (a) 3 (b) 2 (c) 0 (d) –2 (a) 9.2321 (b) 13.716 (c) 10.723 (d) 15.892 [RRB JE 2014 RED SH I FT ] 11. The cube r oot t o 1.061208 7. What is the largest possible length of a scale that (a) 1.022 (b) 10.22 (c) 0.102 (d) 1.02 can be used to measure exactly the lengths 3 m, 12. The least number having four digit s which is a 5 m 10 cm and 12 m 90 cm ? per fect squar e is (a) 10 cm (b) 20 cm (a) 1004 (b) 1016 (c) 1036 (d) None of t hese (c) 25 cm (d) 30 cm [RRB JE 2014 YEL L OW SH I FT ] 13. The missing number in t he ser ies 8, 24, 12, 36, 18, 54,  is 8. After measuring 120 metres of a rope, it was (a) 27 (b) 108 discovered that the metre rod was 3 cm longer. The true length of the rope measured is : (c) 68 (d) 72 14. What is t he eight h t er m of t he sequence 1, 4, 9, (a) 116 m 40 cm (b) 121m 20 cm 16, 25,........ ? (c) 123 m (d) 123 m 60 cm (a) 8 (b) 64 [RRB JE 2014 YEL L OW SH I FT ] (c) 128 (d) 200 9. Solve 3 0.000064 = ? 15. Which of t he following is t he best appr oximat ion for t he following expr ession, (a) 0.4 (b) 0.04 (c) 0.004 (d) 0.0004 (7.9986 / 0.115)  19.97 ? [RRB JE 2014 YEL L OW SH I FT ] (a) 15 (b) 10 (c) 1.0 (d) 1.3 10. The HCF of two numbers is 6 and their LCM is LEVEL-1 72. If one number is 24, the other number is 1. The val ue of (1 +0.1 +0.11 +0.111) is (a) 12 (b) 18 (a) 1.321 (b) 1.211 (c) 36 (d) 72 (c) 1.111 (d) 1.331 [RRB JE 2015 26 th AU G 1 st SH I FT ] [RRB JE 2014 GREEN SH I FT ] 11. The largest number which divides by 72 and 125, 2. When a number is divided by 5, it gives r emainder leaving remainders 7 and 8 respectively is 3. What i s t he r emainder when squar e of t hat (a) 13 (b) 56 number is di vi ded by 5? (c) 65 (d) 900 (a) 9 (b) 3 [RRB JE 2015 26 th AU G 1 st SH I FT ] (c) 4 (d) 1 [RRB JE 2014 GREEN SH I FT ] 1.6 Number System 12. The HCF of two numbers is 12 and their LCM 5. Value of  (appr ox. value 3.14) is : is 72. If one number is 36, the other number is (a) Ter minat ing decimal (a) 12 (b) 24 (b) Recur r i ng decimal (c) 36 (d) 48 (c) Non-t er minat ing non-r epeat ing decimal [RRB JE 2015 26 th AU G 2 nd SH I FT ] (d) I ndet er minat e 13. The largest number which divides 81 and 108, [RRB SSE 2014 GREEN SH I FT ] leaving remainders 6 and 3 respectively is 6. Which one of the following is not a pr ime number ? (a) 9 (b) 15 (a) 71 (b) 91 (c) 18 (d) 515 (c) 61 (d) 31 [RRB JE 2015 26 th AU G 2 nd SH I FT ] [RRB SSE 2014 RED SH I FT ] 14. The HCF of two numbers is 15 and their LCM 7. Fi nd t he val ue of : is 270. If one number is 45, the other number is (a) 18 (b) 90  489  3752   489  375 2 (c) 81 (d) 675  489  375 [RRB JE 2015 26 th AU G 3 rd SH I FT ] (a) 144 (b) 864 15. The largest number which divides 247 and 319, (c) 2 (d) 4 leaving remainders 7 and 4 respectively is [RRB SSE 2014 RED SH I FT ] (a) 15 (b) 30 1 5 2 4 8. Fi nd t he L.C.M. of , , , : (c) 45 (d) 56 3 6 9 27 [RRB JE 2015 26 th AU G 3 rd SH I FT ] 1 10 (a) (b) LEVEL-2 54 27 3 3 20 1. 22  22   is equal t o: (c) 3 (d) None of t hese (a) 22 (b) 21 [RRB SSE 2014 RED SH I FT ] (c) 2– 2 (d) 2– 1 5 7 13 16 3 9. Ar r ange t he fr act ions , , , and in [RRB SSE 2014 GREEN SH I FT ] 8 12 16 29 4 2. Ar r ange t he fol l owi ng fr act i ons i n ascendi ng ascendi ng or der of magnit ude : or der. 16 7 5 3 13 7 3 4 (a)     , , 29 12 8 4 16 10 8 5 16 5 7 13 3 3 7 4 3 4 7 (b)     (a) , , (b) , , 29 8 12 16 4 8 10 5 8 5 10 4 3 7 7 3 4 3 13 7 5 16 (c) , , (d) , , (c)     5 8 10 10 8 5 4 16 12 8 29 [RRB SSE 2014 GREEN SH I FT ] 3 5 7 13 16 3. By what l east number shoul d 192,000 be divided (d)     4 8 12 16 29 so as t o become a per fect cube? [RRB SSE 2014 RED SH I FT ] (a) 2 (b) 5 x 1 x 3 (c) 3 (d) 7  a  b 10. I f     , t hen t he val ue of 'x ' i s : [RRB SSE 2014 GREEN SH I FT ]  b  a 4. Fi nd t he value of: 3 + 0.03 + 0.003 + 0.0003 1 (a) (b) 1 2 (a) 12 (b) 3.0333 (c) 3.3333 (d) 6.0333 (c) 2 (d) – 1 [RRB SSE 2014 GREEN SH I FT ] [RRB SSE 2014 RED SH I FT ]  Number System 1.7 11. Fi nd t he val ue of (0.000216)3 : 14. H CF of t wo number s, each consi st i ng of four (a) 0.6 (b) 0.06 di gi t s, i s 103 and t h ei r L CM i s 19261. The di ffer ence of t he number s is (c) 0.006 (d) 0.0006 [RRB SSE 2014 RED SH I FT ] (a) 6/8 (b) 5/9 12. I f a = 2129 × 381 × 5128 , b = 2127 × 381 × 5128 (c) 5/8 (d) 4/8 [RRB SSE 2015 1 st SEP 2 nd SH I FT ] c = 2126 × 382 × 5128 and d 2125 × 382 × 5129 t hen H CF of a, b ,c and d is 15. L et a  3129  5128  7 22 ,b  3128  5129  722 (a) 2125 × 381 × 5129 (b) 2125 × 381 × 5128 c  3128  5128  722 andd  3129  5128  723 (c) 2125 ×3 ×5 82 128 (d) 2129 × 382 × 5129 H CF of a, b, c and d is [RRB SSE 2015 1 st SEP 1 st SH I FT ] 128 13. L et x be a l east number which when divided by (a) 3129  5128  722 (b) 15   7 22 21,33,35 and 55 leaves i n each case a r emai nder 129 3, but is exact ly divisible by 67. The sum of digit s (c) 3128  5128  723 (d) 15  723 of x is [RRB SSE 2015 1 st SEP 2 nd SH I FT ] (a) 8 (b) 10 (c) 12 (d) 15 [RRB SSE 2015 1 st SEP 1 st SH I FT ] AN SWERS OBJECTI VE TYPE QU ESTI ON S 1. (c) 2. (b) 3. (a) 4. (a) 5. (b) 6. (b) 7. (b) 8. (d) 9. (a) 10. (c) 11. (d) 12. (d) 13. (a) 14. (b) 15. (b) LEVEL-1 1. (a) 2. (c) 3. (a) 4. (b) 5. (b) 6. (b) 7. (a) 8. (d) 9. (b) 10. (b) 11. (a) 12. (b) 13. (b) 14. (b) 15. (a) LEVEL-2 1. (a) 2. (a) 3. (c) 4. (b) 5. (b) 6. (b) 7. (d) 8. (c) 9. (a) 10. (c) 11. (b) 12. (b) 13. (d) 14. (a) 15. (d) EXPLAN ATI ON S OBJECTI VE TYPE QU ESTI ON S 3. 2 5 = 2  2.236 = 4.472 5 6 3 = 6  1.732 = 10.392 1. = 0.312 16 3 7 = 3  2.646 = 7.938 6 = 0.352 8 2 = 8  1.414 = 11.312 17 7 Clear ly 8 2 is lar gest. = 0.388 18 2 4. L et pr oper fr act ion = 9 2 3 2. = 0.692, = 0.666 21 3 13 3  Result ing fr act ion = = 31 4 8 5 2 3 = 0.727, = 0.714 H ence < 11 7 3 4 2 9 5 8 1 2 H ence ascending or der is , , ,  < 3 13 7 11 2 3 3 31 < 5 51 1.8 Number System 3 4 1  < et c. 6. 22n 1  5 6 3n  9 5. Solving x + y > 5 and x – y > 3 we get ,  22n 1  23n  9  20 x > 4.  5n – 10 = 0 6. Pr oduct of – ve number s is also +ve. 10 7. Given expr ession = 3 0.008 = 0.2 n 2 5 1 1 1 1 4 7. Requi r ed scal e has t o be of l engt h 10 cm    2 2 2 because 10 cm is t he shor t est lengt h in i n 9. Given expr ession = 1 1 1 = 2 1 +  3 gi ven quest ion. 2 2 2 4 8. Act ual lengt h has t o be 4 8 2 120m + (120 × 3) cm = 120 m + 360 cm = 2 = =2 3 3 3  123 m 60 cm. 9 2 6– 2 10. + 3 4 4 4 5 3 5 3 9. 0.000064  3   =.04 100 100 100 1 = 9 5  9 3  10 – 6 10. We know that, x × y = LCM × HCF 53 (x, y are the two distinct numbers) 6 5  6 3  10  3  x × 24 = 72 × 6 1  x = 18. = 15 5 – 3 5  2 6 2 11. In such a questions it is better to check the options. In this case 13 satisfies the given 1 = [15  2.236 – 3  1.732 – 2  2.449] condition. Rest of the numbers are large 2 enough to be eliminated easily. 1 = [33.540 – 5.196 – 7.898] 12. set the number be x 2  x × 36 = 12 × 72 = 10.732  x = 24 11. H er e 1.061208 = (1.02)3 13. Checking the options we get the correct Requir ed cube r oot = 1.02 answer as 15. 12. Requir ed number =1024 = (32)2 14. HCF × LCM = Product of numbers 13. Second t er m is 3 t imes of t he fir st t er m and t hir d 15  270 t er m i s hal f of t he second t er m, r epeat t hi s  other number = = 90 pr ocess t he missing t er m is half of 54, i.e. 27. 45 15. Numbers = 247, 319 14. Given sequence can be wr it t en as  remainders = 7 & 4, 12, 22, 32, 42, 52,... H ence it ’s eight h t er m = 82 = 64 247  7  240  319  4  315 are divisible  8 15. Rounding off, we get  20  100 HCF (240, 315) = 15 0.1 LEVEL-2 LEVEL-1 1. 1 + 0.1000 + 0.110 + 0.111 = 1.321 3 3 2. Let the number be 8. 1. 22  22   Thus, when 82 = 64 will be divided by 5, then = 28  26 = 22. remainder will be 4. 7 1 2.  0.7 3. will be less than 0.3333. 10 3.1623 4. 2744 i s a mul t i ple of 7. 3  0.375 H ence, t he answer has t o be 14. 8 5. 148 = 37 × 4 and 185 = 37 × 5 3 7 4 LCM = 37 × 4 × 5 = 740    8 10 5  Number System 1.9 4  a x 1  b x 3  0.8   5 10.    b  a 192000 3.  96,000  Not a perfect cube.  a x 1  a 3 x 2      b b 192000  64,000  A perfect cube 3   x  1  3  x  Ans = 3  2x = 4 4. 3 + 0.03 + 0.003 + 0.0003  x=2 = 3.0333 1 1 1 6. 91 is not a prime number 11.  0.000326  3   216  3  216 6  3 91 = 7 × 13 = 6 × 10–2 = 0.06  489  3752   489  3752 12. Highest power of 2 common in a1 b1 c1 d = 125 7.  489  375 Highest power of 3 common in a1 b1 c1 d = 81 Highest power of 5 common in a1 b1 c1 d = 128  a  b 2   a  b 2 i.e  HCF  2125  381  5128 ab 4ab x  21k1  3   4 x  33k 2  3  ab 13. A/Q x  35k 3  3   x x  55k 4  3 LCM 1,5,2,4  8. LCM  = LCM (211 331 351 55)k + 3 HCF  3,6,9,27  = 1155k + 3 20 A/Q x = 67 k5  3  67k 5  1155k  3 5 at k = 4 9.  0.625 8 Number = 4623 which 7 satisfies all conditions.  0.583  sum of digits = 4 + 6 + 2 + 3 = 15 12 14. Product of two Numbers = LCM × HCF 13  0.8125  x × y = 103 × 19261 16 = 103 × 103 × 11 × 17 = 1133 × 1751 16 6  0.551 29  Difference = (Ans) 8 3 15. Highest power of 3 common answer  0.75 4 a1 b1 c1 d = 129  option (a) is correct Highest power of 5 common answer a1 b1 c1 d = 129 Highest power of 7 common answer a1 b1 c1 d = 23  2 CHAPTER Percentage and Ratio & Proportion Percentage PERCEN T Per cent means out of hundr ed. For example if we say 50% Con ver si on t abl e of per cen t age i n t o 50 25 fract ion and vice-versa. (To learn) t h at m ean s , 25% m ean s. Or w e can say 100 100 1/2 50% 1/10 10% per cent age is a fr act ion in which denominat or is 100. 1 1 1/3 33 % 1/11 9 % Fir st lear n how to change per centage into fr action or fr action 3 11 1 int o per cent age. 1/4 25% 1/12 8 % 3 25 9 25% means or i n conver sion of % int o fr act i on 1/5 20% 1/13 7 % 13 100 2 1 divide t he number by 100. 1/6 16 % 1/14 7 % 3 7 I n conver sion of fr act ion int o per cent age mult iply by 2 2 100. 1/7 14 % 1/15 6 % 7 3 ¼ × 100 = 25% 1 (fr act ion) (per cent age) 1/8 12 ½% 1/16 6 % 4 1 1/9 11 % 9 Nor mally a st udent get s quest ions based on per cent ages given in per cent age t able. N ote : I f a fr act ion is given mor e and you have t o find for less per cent age, t hen decr ease t he fr act ion and a lower fr act ion is t he answer or vice-ver sa. U SE OF PERCEN TAGE TABLE TYPE I Example 1. I f t he salar y of A is 20% mor e t han B, t hen by what %, B's salar y is less t han A? 1 Solution : I f we see t he per cent age t able, 20% = 5 1 2 We have t o find a fr act ion less t han t his, i.e. , t he answer is 16 %. 6 3 Example 2. I f t he r at e has r educed by 10%, by what % t he consumpt ion has t o be incr eased, so t hat expendit ur e r emains t he same ? 1 1 1 Solution : I n percentage table, 10% =. We have to find increased fraction, i.e. or 11 % is the answer.. 10 9 9 Example 3. I f lengt h of a r ect angle is r educed by 40%, t hen by what % t he br eadt h has t o be incr eased so t hat ar ea r emains t he same ? FG x  100%IJ Solution : H 100  x K As 40% is not given in per cent age t able, we ar e asking for mor e, so denominat er must be less. FG 40  100IJ = 66 2 %. H (100 – 40) K 3 2.2 Percentage and Ratio & Proportion TYPE I I Ther e ar e t hr ee differ ent cases. We have t o solve in t he example : A + x % of A = B Case 1. Value of A and x are given and we have to find B. Example. Add 20% of a number t o a given number 600. 20 So, 600 + 20% of 600 = 600 + × 600 = 720. 100 Case 2. Value of A and B are given, we have to find the value of x. Example. What % has t o be added t o 200 t o get 250 ? H er e, A = 200, B = 250. Fir st of all , see t he change, i.e. 50 ( 250 – 200 ) I mportant  On which value it is changing on 200, or on 250. I n t his case it is 200. 50 1  = or 25% 200 4 Case 3. Value of x and B are given and we have to solve for A. Example. Adding 20% t o a number gives 480, what is t he number ? H er e, x = 20%, B = 480, 20FG  A = 480 IJ A+ 100 H K This is fr equent ly asked quest ion. Adding 20% or 1/5 t o a number gives B, means subt r act a lower fr act ion 1 i.e. 1/6 (lower t han 1/5) fr om t he number 480 (B) or 480 – × 480 = 400. 6 I f %age asked is not given in per cent age t able t hen. Example. A's salar y 30% mor e t han B, by what % B salar y less t han A ? Formulae : ( x /100 ± x ) × 100 % As we ar e asking for a lower per cent age, so denominat or must be mor e or (+) has t o be done. FG 30 IJ × 100 300 1 H 130 K = 13 = 23 13 %. TYPE I I I A×B=D I n t his case A and B ar e decr easing or incr easing hence changing t he value of C or find % change in C. I n quest ion of per cent age we can assume any t hing or ever yt hing t o be 100 inst ead of assuming x and y. Example. The lengt h and br eadt h of a r ect angle ar e incr eased by 30% and 20% r espect ively. What is %age change in Ar ea? Solution. I nst ead of assuming x and y as lengt h and br eadt h, assume 100 for lengt h and 100 for br eadt h. Ar ea = l × b L engt h incr eased by 30% means fr om 100 it became 130, Br eadt h incr eased by 20% means fr om 100 it became 120, so t he ar ea became 15600, which is 5600 mor e t han 10000, 5600 56 or = = 56%. 10000 100 TYPE I V We gener alize t he concept , A% of B = C  Percentage and Ratio & Proportion 2.3 I n t his concept : (i ) Eit her A and B ar e given and we have t o find t he value of C. (ii ) A and C ar e given and we have t o find t he value of B. (iii ) B and C ar e given and we have t o solve for A SOLVED EXAM PLES 1. (0.756 × ¾ ) is equivalent t o. FG0.756  3IJ × 100 = 56.7%. Solution. H 4K 2. Find (12 ½) % of 88. FG12 1 IJ % means 1 and 1 of 88 is 11. H ence t he answer is 11.1. Solution. H 2K 8 8 FG 2 IJ % of 75. 3. Find 6 H 3K Solution. FG 6 2 IJ % means 1 and 1 of 75 = 5. H 3K 15 15 4. F 1I Find GH 37 JK % of 48 2 FG12 1 IJ % is 1. Solution. H 2K 8 I f we mult iply by 3, 3 F 1I is GH 37 JK % ,so 3 × 48 = 18. 8 2 8 5. Find 43.5% of 20 Solution. 43.5% doesn't have any easy fr act ional r epr esent at ion x % of y = y % of x leads t o conclude 1 43.5% of 20= 20% of 43.5, i.e. × 43.5 = 8.7 5 FG 11 1 IJ % of which number is 12 ? 6. H 9K Solution. F 1 I 1 and C is 12 A = GH 11 JK % = 9 9 1 × B = 12 9 or B = 108. 7. 16% of ------- is 40. Solution. A = 16%, C = 40 16  × B= 40, or B = 250. 100 8. 24 is ------- % of 36. Solution. C = 24, B = 36. A t o find out. 2.4 Percentage and Ratio & Proportion A 2 × 36 = 24, or A = 66 %. 100 3 24 2 24 is of 36, i.e. of 36, 36 3 2 1 1 hence 66 as 33 % = mult iply by 2. 3 3 3 9. What is 3% of 5%? Solution. A = 3%, B = 5%, C t o find. 3 5  =C 100 100 or C =.0015 10. What % is 3% of 5% ? Solution. A t o be find out , B = 5%, C = 3%. A 5 3  = 100 100 100 or A = 60%. 11. 3% of what number is 5% ? Solution. A = 3%, B = t o be find out , C = 5%. 3 5 ×B = 100 100 5 or B = = 1.6 3 At t imes we have t o find t he fr act ion like 15%, 18% and so on. 12. Find 15% of 480 (concept for ment al calculat ion) Solut ion. Fi r st fi nd 10% of 480 = 48 and 5% sh oul d be hal f of t he val u e (48) whi ch i s 10%. So 5% = 24 add t he t wo, i.e. 72. 13. Find 18% of 450. 1 Solution. 10% of 450, i.e. of 450 = 45 mult iply by 2 it becomes 20% = 90 10 1 of 20% is 90 t han 2% is 9 as ( 2 is of 20) 10 I f one subt r act ion 2% fr om 20%, one get s 18%, so 90 – 9 = 81. PRACTI CE EXERCI SE OBJECTI VE TYPE QU ESTI ON S 3. What will be 160% of a num ber whose 200% is 140 ? 1. What per cent of 72 is 6 ? (a) 200 (b) 160 (a) 10% (b) 20% (c) 140 (d) 112 1 (c) 8 % (d) N one 1 3 4. I f A's in come is 33% mor e t han t hat of B , t hen 3 2. I f 30% of a num ber is 12.6, t hen t he num ber is how mu ch per cent is B 's in come less t han t hat of (a) 41 (b) 51 A? (c) 52 (d) 42 1 (a) 25% (b) 33 % 3 (c) 40% (d) N one  Percentage and Ratio & Proportion 2.5 4 15. 300 gms of sugar solut ion has 40% sugar in it. 5. 63% of 3 is H ow much sugar should be added t o make it 50% 7 in t he solut ion? (a) 2.25 (b) 2.40 (a) 10 gms. (b) 40 gms. (c) 2.50 (d) 2.75 (c) 60 gms. (d) 30 gms. 6. What per cent age is 3% of 5% ? LEVEL-1 (a) 60% (b) 50% 1. The salar ies of A and B t ogether amount to ` 2000. (c) 15% (d) 30% A spends 95% of his salar y and B 85% of his. I f now, t heir savings ar e t he same, t hen what is A's 1 7. A number i n cr eased by 37 % gi ves 33. The salar y ? 2 (a) ` 1500 (b) ` 1250 number is (c) ` 750 (d) ` 2600 (a) 27 (b) 25 2. I f t he r adius of a cir cle is decr eased by 50%, it s (c) 24 (d) 22 ar ea is r educed by 8. 40 quint als is what per cent of 2 m t ones ? (a) 25% (b) 50% (a) 20% (b) 2% (c) 75% (d) N one (c) 200% (d) 150% 3. The populat ion of a t own is 18000. I t incr eases by 9. I f 15% of 40 is gr eat er t han 25% of a number by 2, 10% during first year and by 20% during the second t hen number is year. The populat ion aft er 2 year s will be (a) 16 (b) 20 (a) 19800 (b) 21600 (c) 24 (d) 32 (c) 23760 (d) N one 1 10. 12.5 % of 192 = 50% of ? 4. By mixing wat er a man gains t he pr ofit of 11 %. 9 (a) 48 (b) 96 Find t he quant it y of wat er mixed by him. (c) 24 (d) N one 1 1 (a) lit er s (b) lit er s 11. x% of y is y% of ? 9 8 x (a) x (b) 1 1 100 (c) lit er s (d) lit er s 10 11 y (c) 100x (d) 5. M ohan’s expendit ur e and saving ar e in t he r at io 100 of 4:1. H is income incr eases by 20% if his saving 12. A st udent has t o secur e 40% mar ks t o pass. H e i ncr eases by 12%. By how much % shoul d hi s get s 178 m ar k s and f ai l s by 22 m ar k s. Th e expendit ur e incr eases ? maximum mar ks ar e (a) 21% (b) 20% (a) 200 (b) 500 (c) 42% (d) none (c) 800 (d) 1000 6. A candidat e has t o secur e 40% of t he t ot al mar ks 13. I n an exam i n at i on 1100 boys an d 900 gi r l s to pass. He gets 190 mar ks and fails by 190 mar ks. appear ed. 50% of t he boys and 40% of t he gir ls Find t he t ot al mar ks. passed t he exam. The per cent age of candidat es (a) 900 (b) 800 failed is (c) 700 (d) 950 (a) 45 (b) 45.5