Work Time Problems: tathagat classes Problem 1: A takes 5 days to complete a piece of work and B takes 15 days to complete a piece of work. In how many days can A and B complete the work if they work together? Standard Solution: Let Solution: Let us consider Work to be 1 unit. o if W ! 1 "nit and A takes 5 days to complete complete the work then in 1 day A completes completes 1#5th of the work. imilarly B completes 1#15th of the work. If they work together$ in one day A and B can complete %1#5 & 1#15 ! '#15( of the work. o to complete 1 unit of work they will take 15#' days. New method: Let method: Let us assume W ! 15 units$ which is the L)* of 5 and 15. +i,en that total time taken for A to complete 15 units of work ! 5 days -- A/s 1 day work ! 15#5 ! 0 units +i,en that total time taken for B to complete 15 units of work ! 15 days -- B/s 1 day work ! 15#15 ! 1 unit --%A & B(/s 1 day work ! 0 & 1 ! ' units --15 units of work can be done in 15#' days. *any sol,e ime and Work problems by assuming work as 1 unit %first method( but I feel it is faster to sol,e the problems by assuming work to be of multiple units %second method(. his would be more e,ident when we sol,e problems which are little more comple2 than the abo,e one. Problem 2: 3 can do a work in 15 days. After working for 0 days he is 4oined by . If they complete the remaining work in 0 more days$ in how many days can alone complete the work? Solution: Assume Solution: Assume W ! 15 units. Note: ou can assume work to be any number of units but it is better to take %Note: ou the L)* of all the numbers in,ol,ed in the problem so that you can a,oid fractions( 3 can do 15 units of work in 15 days --3 can do 1 unit of work in 1 day %6ote7 If I had assumed work as 10 units for e2ample then 3/s 1 day work would be 10#15$ which is a fraction and hence I a,oided it by taking work as 15 units which is easily di,isible by 15 and 0( ince 3 worked for 8 days$ total work done by 3 ! 8 days ⋅ 1 unit#day ! 8 units. "nits of work remaining ! 15 9 8 ! : units. All the remaining units of work ha,e been completed by in 0 days −−>/s 1 day work ! :#0 ! 0 units. If can complete 0 units of work per day then it would take 5 days to complete 15 units of work. o takes 5 da ys to complete the work. Problem 3: A$ B and ) can do a piece of work in 15 days. After a ll the three worked for ; days$ A left. B and ) worked for 1< more days and B left. ) worked for another '< days and completed the work. In how many days can A alone complete the work if ) can complete it in =5 da ys?
Solution: Assume Solution: Assume the total work to be 8<< units. %L)* of a ll the numbers( hen )/s 1 day work ! > units. --%A & B & )(/s 1 day work ! '< units. A$ B$ ) work together in the first ; days --Work done in the first ; days ! '< ⋅ ; ! >< units ) alone works during the last '< days --Work done in the last '< days ! '< ⋅ > ! 0;< units emaining work ! 8<< 9 %0;< & ><( ! ;<< units his work is done by B and ) in 1< days. --%B & )(/s 1 day work ! ;< units --A/s 1 day work ! %A & B & )(/s 1 day work 9 %B & )(/s 1 day work ! '< units 9 ;< units ! ;< units --A can do the work of 8<< units in 0< days. Problem 4: +errard can dig a well in 5 hours. @e in,ites Lampard and ooney who can dig 0#'th as fast a s he can to 4oin him. @e also in,ites Walcott and abregas who can dig only 1#5th as fast as he can %Inefficient gunners you see ( to 4oin him. If the fi,e f i,e person team digs the same well and they start together$ how long will it take for them to f inish the 4ob? Solution: Let the work be be 1<< units. +errard/s 1 hour work ! 1<<#5 ! ;< units Lampard and ooney/s 1 day work ! 0#' ⋅ ;< ! 15 units. abregas and Walcott/s 1 day work ! 1#5 ⋅ ;< ! ' units. ⇒ In one day all fi,e of them can do ! ;< & 15 & 15 & ' & ' ! 5> units of work. @ence they can complete the work in 1<<#5> days. I hope you got the knack of it. Let us now see how to sol,e the second kind of problems in ime and Work 9 the MANA!S problems. MANA!S problems. In these kinds of problems we need to remember that the number of men multiplied by the number of days that they take to complete the work will gi,e the number of mandays reuired to complete the work. he number of mandays reuired to complete a piece of work will remain constant. We will try and understand this concept by applying it to the ne2t three problems. A Cery simple problem to start with7 Problem ": If 1< men take 15 days to complete a work. In how many days will ;5 men complete the work? Solution: +i,en Solution: +i,en that 1< men take 15 days to complete the work. o the number of mandays reuired to complete the work ! 1< ⋅ 15 mandays. o assume W ! 15< mandays. 6ow the work has to be done by ;5 men and since W ! 15< mandays$ the number of days to complete the work would be 15<#;5 ! 8 days. Problem #: A piece of work can be done by > boys in ' days working 8 hours a day. @ow many boys are needed to complete another work which is three times the first one in ;' days working > hours a day? Solution: Assume Solution: Assume the first piece of work to be > ⋅ ' ⋅ 8 ! 1:; boy-day-hours.
he second piece of work ! 0 %he first piece of work( ! 0 ⋅ 1:; ! 5=8 boy-dayhours. o W ! 5=8 boy-day-hours. If this work has to be completed in ;' days by working > hours a day the number of boys reuired would be 5=8#%;' ⋅ >( ! 0 boys. Problem $: 3 can do a piece of work in ;< days working = hours a day. he work is started by 3 and on the second day one man whose capacity to do the work is twice that of 3$ 4oined. Dn the third day another man whose capacity is thrice that of 3$ 4oined and the process continues till the work is completed. In how many days will the work be completed$ if e,eryone works for four hours a day? Solution: ince Solution: ince 3 takes ;< days working = hours a day to complete the work$ the number of day-hours reuired to complete this work would be 1'< day-hours. Like in the two problems abo,e$ this is going to be constant throughout. o$ W ! 1'< day-hours. Amount of work done in the 1st day by 3 ! 1day ⋅ ' hours ! ' day-hours ;nd day$ 3 does again ' day-hours of work. he second person is twice as efficient as 3 so he will do > day-hours of work. otal work done on second day ! > & ' ! 1; day-hours. Amount of work completed after two days ! 1; & ' ! 18 day-hours. 0rd day$ 3 does ' day-hours of work. econd Eerson does > day-hours of work. hird person who is thrice as efficient as 3 does 1; day-hours of work. otal work done on 0rd day ! ' & > & 1; ! ;' d ay-hours Amount of work completed after 0 days ! 18 & ;' ! '< day-hours imilarly on 'th day the amount of work done would be ' & > & 1; & 18 ! '< day-hours Work done on the 5th day ! ' & > & 1; & 18 & ;< ! 8< day-hours otal work done after 5 days ! ' & 1; & ;' & '< & 8< ! 1'< day-hours ! W. o it takes 5 days to complete the work. %emember that whene&er there is mone' in&ol&ed in a (roblem) the mone' earned should be shared b' (eo(le doing the work together in the ratio o* total work done b' each o* them+ Again , will e-(lain this with the hel( o* an e-am(le: Problem .: 3 can do a piece of work in ;< days and can do the same work in 0< days. hey finished the work with the help of F in > days. If they earned a total of s. 555<$ then what is the share of F? Solution: Let Solution: Let work W ! 1;< units. %L)* of ;<$ 0< and >( 3/s 1 day work ! 8 units /s 1 day work ! ' units %3 & & F(/s 1 da y work ! 15 units. o F/s 1 day work ! 15 9 %8 & '( ! 5 units In > days F would ha,e completed 5 un its#day ⋅ > days ! '< units of work ince F does '<#1;< ! 1#0rd of the work$ he will recei,e 1#0rd of the money$ which is 1#0 2 555< ! s. 1>5<. Pi(es and /isterns
Problem 0: here are three hoses$ A$ B and )$ attached to a reser,oir. A and B can fill the reser,oir alone in ;< and 0< mins$ respecti,ely whereas ) can empty the reser,oir alone in '5 mins. he three hoses are kept opened alone for one minute each in the the order A$ B and ). he same order is followed subseuently. In how many minutes will the reser,oir be full? Solution: hese kinds of problems can be sol,ed in the same way as we sol,e problems where one or more men are in,ol,ed. A$ B and ) are eui,alent to three people trying to complete a piece of work. he amount of work to be done would be the capacity of the reser,oir. Lets assume capacity of the reser,oir ! W ! 1>< %L)* of ;<$ 0<$ '5( litres. A can fill the reser,oir in ;< mins ⇒ In 1 min A can fill 1><#;< ! : L. B can fill 1><#8< ! 8 L in a minute. In one minute ) can empty 1><#'5 ! ' L from the reser,oir. 1st *inute ! A is opened ! fills : L ;nd *inute ! B is opened !fills another 8 L 0rd *inute ! ) is opened ! empties ' L @ence e,ery 0 minutes ! %: & 8 9 ' !( 11 litres are filled into the reser,oir. o in '5 minutes %11 ⋅ 15 !( 185 litres are filled. In the '8th minute A is opened and it fills : litres. In the '= th minute B is opened and it fills 8 litres. @ence the reser,oir will be full in '= minutes. Problem 1: here is an empty reser,oir whose capacity is 0< litres. here is an inlet pipe which fills at 5 L#min and there is an outlet pipe which empties at ' L#min. Both the pipes function alternately for 1 minute. Assuming that the inlet pipe is the first one to function$ how much time will it take for the reser,oir to be filled up to its capacity? Solution: he work to be done ! )apacity of reser,oir ! W ! 0< litres 1st *inute ! inlet pipe opened ! 5l filled ;nd minute ! inlet pipe closedG outlet pipe opened ! 'l emptied In ; minutes %5 litres -' litres !( 1l is filled into the reser,oir. It takes ; minutes to fill 1l ! it takes 5< minutes to fill ;5 litres into the tank. In the 51st minute inlet pipe is opened and the tank is filled.
Problem 11: ohan can work for three hours non-stop but then needs to rest for half an hour. @is wife can work for two hours but rests for 15 min after that$ while his son can work for 1 hour before resting for half an hour. If a work takes 5< man-hours to get completed$ then appro2imately how long will it take for the three to complete the same? Assume all of them all eually skilled in their work. %a( 15
%b( 1=
Solution: W ! 5< man-hours
%c( ;<
%d( ;'
ince all of them are eually skilledG in 1 hour they can do 0 man-hours of work if no one is resting. It will take them 5<#0 ! 18.8 hours to complete the work if they work continuously. But$ since they take breaks the actual amount of time would 1= hours. Dption %a( and %b( are ruled out. 6ow let us calculate the amount of work done in ;< hours. ohan does 0 man-hours in e,ery 0.5 hours %because he takes rest for half an hour on the ' th hour( In ;< hours %0.5 ⋅ 5 & ;.5( ohan completes ! 0 ⋅ 5 & ;.5 ! 1=.5 man-hours ---- %1 @is wife completes ; man-hours e,ery ;.;5 hours %because she rests on the 0rd hour( In ;< hours %;.;5 ⋅ > & ;( she completes ! ; ⋅ > & ; ! 1> man-hours. ---- %2 )hild completes 1 man-hours e,ery 1.5 hour. In ;< hours %1.5 ⋅ 10 & <.5( he completes 1 ⋅ 10 & <.5 ! 10.5 man-hours of work. ------ %3 Adding 1) 2 3 In appro2imately ;< hours ': man-hours will be completedG so the work can be completed in ;
nice article thank u so much problem = is bit problematic...' or = hours a day work is carried out not clear how parent eplye7 ime and Workby indhoor +randhi - ri$ 1< Hul<:$ 117;> A* hanks hi,am smile and read the last line of problem = .. it says e,eryone works for ' hours a day. how parent eplye7 imeand Workbyindhoor+randhi-ri$1< Hul<:$ 1170; A* Cishnu$ I am happy you liked it. hoping to see your article someday on + smile how parenteplye7imeandWorkbyindhoor+randhi - ri$ 1< Hul<:$ 117': A* an4eeb$I am short of good uestions in this topic too .. but will try and see if I can post more .. or may be you can post any that you encounter during your preparation. how parenteplye7ime and Workby oftware ngineer-ri$
hours. how parent eplye7 ime and Workby Ci,ek 6arain-ri$1< Hul<:$ <=7<' E* " rock dude. he article is ,ery helpful. Elease post in some more problems$ esp. on pipes n cisterns$ and those like the last problem where a person takes a break while working... hanks a ton.. how parent eplye7 ime and Workby kat goel - ri$1< Hul<:$ 1 <7<0 E* I agree with Hay . Kay in problem ' should be replaced by hour how parent eplye7 ime and Workby ayush roy-aturday$11Hul<:$ 1;70> A* grt article dude$feelin so confident nw.4ust replace 1><#8< wid 1><#0< in prb :$4ust a printin mistake.+ sir$e2pectin some challenging prbs frm u in this topic.
hanks$ ayush how parent eplye7 ime and Work by akash gupta-at$11Huly<:$ <17;> A* too good boss..,ery helpful how parent eply e7 ime and Workbyamit kheterpal-at$11 Hul<:$ 1<71< A* nice work bro..can u gi,e u s article on tsd$percentages$profit n loss..these article help us lot n sa,e our time.. sir#mam#se#sindhoor )an u make article on each n e,ery topic..n please post sum uestions too..i lo,e studying in this site..i ha,e learnt many things here more than my coaching institute.At the same time iam sad today i got backlog in my 8 thsem engineering e2ams my entire planning is now shatered.iam not able to understand wat i should do.. how parent eply e7 ime and Work by Euneet Aggarwal - at$ 11 Hul<:$ <=70< E* after this artice I am feeling lot more confident with time and work...thanks a lot how parent eply e7 ime and Work by uhin Baner4ee - at$ 11 Huly ;<<:$ 1<7;; E* @i indhoor$ +reat Article$ thanks a lot. Eersonally I like this chapter a lot$ could you please put some good uality uestions here specific to ime J Work? egards$ uhin how parenteplye7 ime and Work by Earag Earatkar-at$ 11 Hul<:$ 1170> E* @i indhoor$ hanks for the article..............smileMMMMM -Earag how parenteply e7ime and Work by 4oydeep sarkar-un$1; Hul<:$ 1<7'' A* here is a problem... A group of workers was put on a 4ob. rom the second day onwards$ one worker was withdrawn each day. he 4ob was finished when the last worker was withdrawn. @ad no worker been withdrawn at any stage$ the group would ha,e finished the 4ob in two-third of the time. @ow many workers were there in the group? how parenteplye7ime and Work by harshhal bandekar-un$1; Huly<:$ <5758 E* Kear indhoor 6ice article on time J work But I did not follow your third solution properly. I suppose whene,er we take the l)* of the indi,idualPs working days we do it to calcualte the total workdone in units for e2. in the first problem total workdone is 15units.
But in third problem you ha,e taken the L)* 15$1<$'< ;$=5.Hust want to know the logic behind it. Also Elease clarify why you ha,e taken the L)* of abo,e nos when the work done by A$B$) is A ! ; days B ! ; days with A$B & 1< days with ) ! 1; days A$B & 1
) ! ; days with
how parent eply e7 ime and Work by indhoor +randhi - *on$ 10 Huly<:$ 1;7'< E* @i harshhal$ orry for the late reply .. In problem 07 Whate,er you assume Work asG remember that you need to di,ide it with =5 days%to find )Ps 1 da y work( and 15 days% to find A&B&)Ps 1 day work(. o it is enough to take W as the L)* of =5 and 15 which is =5. ol,e the problem with W!=5 units and you will get the same answer. ou need to take a mental note of what you would be di,iding W with while you are sol,ing the problem. +ot it? ell me frankly if you still ha,e a doubt .. how parent eply e7 4oydeep by Oul,ir ingh - *on$ 10 Huly ;<<:$ 1;75; E* Is the answer three?? how parenteplye7ime and Work by indhoor+randhi-*on$10Hul<:$1;750 E* +uyQ$ ry this problem. Its a good one .. how parenteplye7ime and Workby +aura, *ittal-*on$10Huly<:$ <'7<< E* great work done sindhoor... hope to see some more articles like this from you in future... thanks a lot how parent eplye7 ime and Workby Oul,ir ingh-ues$1' Hul<:$ 1175= A* R indhoor.. 6ice article...Which problem you are talking about..?? how parenteplye7imeand WorkbyAmarOrKubedy-ue$1' Hul<:$ <07'> E* Ans is 8 ?? how parenteplye7imeandWork by 4oydeep sarkar-ue$1' Hul<:$ 1<7'0 E* the answer is 0... how parenteplye7imeandWorkbyCickramAsokan - Wed$15 Hul<:$ 1<751 A* +reat article...Oudos... Eost more challenging problems on the topic..hanks$CIckram.how parenteplyime and Work by hyam *anawat-Wed$ 15 Huly<:$<57<> E* ,ery nice artical. i learned a lot. i try to use it in each prob. than2 how parent eplye7 ime and Work by indhoor +randhi - Wed$15 Huly <:$ <57'8 E*
hanks Oul,ir$ I was telling about the problem posted by 4oydeep only and 0 is the right answer smile how parent eply e7 ime and Work by indhoor +randhi - Wednesday$ 15 Huly ;<<:$ <57'= E* hanks hyam how parent eply e7 imeandWorkby4oydeep sarkar-Wed$15 Hul<:$ <870' E* wo friends decide to work together and complete the construction of the four walls of a room which is in the shape of a suare.howe,er$as b fell sick$a started the work alone and completed the construction of one wall and took StS hours more than what they would ha,e taken if they had worked together.after this a left and b$working alone completed the construction of the second wall and took S%;5#18(tS hours more than what they would ha,e taken if they had worked together.inally$both of them worked together and completed the remaining two walls.the construction of the four walls was completed in 1;1 days. in how many days can b alone complete the construction of the four walls? if a total of 08<< is paid for the entire work$what is the share of a??? how parenteply e7ime and Work by Oul,iringh - hur$ 18 Hul<:$ 1;71; A* A can make ' walls in : days...how parent eplye7 ime and Work by 4yoti ,erma - riday$ 1= Huly ;<<:$ 1;71; A* hihanks for the post .6ice fundasMM how parenteply e7imeandWorkbyantriksh agarwal - ri$ 1= Hul<:$ <'7;= E* hi 4oydeep$kindly pro,ide a detailed solution of the abo,e problem %wall construction prob( as i am unable to sol,e it. thanks. Antriksh how parent eplye7imeandWorkbykumarswambhu-un$1: Hul<:$ 1<718 A* his article ha,e some really tricky uestions how parenteply e7 ime and Workby tarun bha,nani-un$1: Hul<:$ <17;1 E* ans is --- b can do it in '5 days.... sol.--say a can do wrk in l days and b can do it in m days... both can do it in n days....l-n!t and m-n ! %;5#18(t and l&m&n ! 1;1T;' %con,erting days to hrs( 0 euations n 0 unknowns...sol,e.. en4oy.. how parenteplye7ime and Workby @emant Ahire - un$ ;8 Hul<:$ <>710 E* 3 can do a piece of work in ;< days and can do the same work in 0< days. hey finished the work with the help of F in > days. If they earned a total of s. 555<$ then what is the share of F? Elease correct me if I m wrong.
1#;<&1#0<&1#F ! 1#> 1#F ! 1#> - %5<#8<<( 1#F ! 1#> - 1#1; 1#F ! 1#;' herefore F can do piece of work in ;' days. o the ration of 377Fwill be ;<70<7;'. herefore share of F will be 555<<. how parent eply e7 ime and Work by cc8>>8 cc-ue$ ;> Hul<:$ 1170: A* hi indhoor$ wont the ans be true for all multiples of 0 %regarding the no of workers problem(... i think 8$: would also be right...please correct if am wrong. how parenteplye7ime and Work by "mang *athur9Wed$;: Hul<:$ <'705 E* he solution to last problem seems problem specific. @ad the total man-hours reuired to complete the work were not 5< but some creepy number$ then directly multiplying by a number$ as in this case by ;<$ would not be easily possible$ rather$ we would ha,e then used$ hit and trial approach. Elease )orrect me if I am wrong. egards "mang how parent eply e7 ime and Workby diptadeep baner4ee Wednesday$ ;: Huly ;<<:$ 1<7;; E* hi @emant$ I think we need to consider the ratio of work done rather than the time taken in doing the work.. thatPs what i understand. )an anyone tell me how to sol,e 4oydeepPs first uestion please? how parenteplye7imeandWorkbyA4aykumarOamne-hur$0< Hul<:$ 1<7;1 A* hey tarun$
I had a doubt in ur solution for the uestion with ' walls.
1. as the room is a suare$ the amount of work red to do each of the wall would be eual. ;. and considering l$m$n to be the hours taken by a$b and a &b together to build one wall. the euations would be$ l-n ! t $ m-n ! ;5t#18 and l&m&;n ! 1;1T;' %not l&m&n ! 1;1T;' $ ;n because they build two walls together.( I ha,e a doubt in the third euation of your solution. )orrect me if am wrong. how parent eplye7 ime and Workby A4aykumar Oamne - hursday$ 0< Huly ;<<:$ 1<7': A* solution for 4oydeepPs first uestion$ let n be the no. of persons working on first day$ thn
no of persons working first day ! n no of persons working second day ! n-1 no of persons working third day ! n-; hence no of persons working on nth day ! 1 if all n persons kept working daily$ thn it took %;#0(n days. n & n-1 & n-; & ........ &1 ! %;#0(Tn using sum of arithmetic progressions and sol,ing this 1#;U;T%a(&%6-1(%d(V ! ;6#0 1#;U;T%1(&%n-1(%1(V ! ;n#0 1#;Un&1V ! ;n#0 n&1 ! 'n#0 hence n ! 0G and no$ n cannot be any multiple of 0 as can be seen here$G let 2 be 1$;$0..... anythingG then our euation would become$ 2n & 2n-1 & 2n-; &.....&1 ! %;#0(T2n and sol,ing similarly 1#;U;T%a(&%6-1(%d(V ! ;6#0$ here 6 ! 2n 1#;U;T%1(&%2n-1(%1(V ! ;%2n(#0 1#;U2n&1V ! ;2n#0 2n&1 ! '2n#0 2n ! 0. n ! 0#2 %not an integer( @ence n can only take ,alue 0G other ways of sol,ing this problem are welcome... smile how parenteplye7imeandWorkbybri4esh 4ada,-*on$0 Aug<:$1;71> E* +reat work and an eually great method... thnk u how parenteplye7 ime and Workby manish kakati - ue$ ' Aug<:$ <87;0 E* I getting the concepts...but some more problems will strengthen our basics and acumen...ELA EDCIK *D EDBs D6 I. how parent eplye7 ime and Workby harsh ana - ri$= Aug<:$ 1;71< A*
WowN.Awe-inspiring e2planation about ime and Work opic. As usual same like other co,ered topics on this site. I feel amaQed whene,er i read your topics here and you guys make me addicted of your mar,ellous site. his site really educate me for my *BA formulations. Hust$ a week before got to know about atha+at and now itPll become my diurnal habbit of surfing on this website. Infact$ i already send you a 18<<#- cheue for )A )B membership and number system or geometry e-book. @ope you will respond AAE. @ere$ i want to share some of my opinion about your site. irst of all$ if you can put a chat bo2 here so aspirants can easily interact with each other and discuss all there problem also. And$ i think you guys are fantabulous so donPt restrict or confine yourself with *anagement e2ams only. Ero,ide new courses also because our society e2tremely reuired your kinf off mentors. When i read your article about becoming teacher$ i was filled with complete emotions because in this fast-paced country e,eryone run like rat and want to win the race. hey all become so much self-interested and caught in this monetary-system. In my point of ,iew$ eachers and oldiers are two field which are completely neglected by our youth. I guess$ enough for today keep post like this which arousing or holding the attention of people like me. ou guyQ are so SDBA*A )heers how parenteplye7imeandWorkbyErimeDptimus-Wed$ 1; Aug<:$ 1;711 A* @i$ I had pre-concei,ed notion that ime and work is tedious.. but this article has made it simpler and lucid.. hanks -@ari how parent eply e7 ime and Workby punit badal - Wed$1;Aug<:$<0718 E* @i$could you put some probems which in,ol,e persons with different capabilities.. how parenteplye7ime and Workby 6ikhil Lingala-hur$10 Aug<:$ 1<7<< E* hat was awesomeMM 2plained beautifullyMM hanks so much for the post indhoor. mile 6ikhil how parent eply
e7 ime and Work by EACA6 ) - at$ 15 Aug<:$ <>71' E*
hank you ,ery much for this article. egard$ pa,an how parenteplye7imeandWorkby manish kakati-Wed$1:Aug<:$ 117;0 A*
Elease this for me7 here/s a lot of work in preparing a b irthday dinner. ,en after the turkey is in the o,en$ there/s still the potatoes and gra,y$ yams$ salad$ and cranberries$ not to mention setting the table. hree friends$ Asit$ Arnold$ and AfQal$ work together to get all of these chores done. he time it takes them to do the work together is si2 hours less than Asit would ha,e taken working alone$ one hour less than Arnold would ha,e taken alone$ and half the time AfQal would ha,e taken working alone. @ow long did it take them to do these chores working together? 1. ;< minutes ;. 0< minutes 0. '< minutes '. 5< minutes how parenteplye7imeandWorkbykanwar4itchadha-Wed$ 1: Aug<:$ <:71' E* Boss$ ans is '< min but this can be obtained by back calculation from gi,en an s only is there a better way to sol,e??? how parent eply e7 ime and Workbyabhishektripathi-ri$;1Aug<:$<;7;= A* thank u tg sir for dis article l oking forwards ' more of ur noble work.ha,e a great day ahead cheers abhisheksmile how parenteplye7imeandWorkbya,iBabu+udipudi-ri$;1Aug<:$<0718E* @i +$ hank you$ this artical is really good ... how parent eplye7 ime and Workby 6etra *ehta - at$;;Aug<:$ <;758 A* S n & n-1 & n-; & ........ &1 ! %;#0(Tn S I dint get this step in the solution to 4oydeepPs uestion..)an u e2plain me why u euated these two terms?? how parent eplye7 ime and Workby mitesh anand-*on$01Aug<:$ <17<= A* hi e,eryone$ the ans for 4oydeepPs ;nd prob wud be--b can complete in '5T'!1>< da ys%only I think u might ha, mistaken in writing hrs instead of dayQ at sum places( a-1#08$b-1#'5 so aPs share-%5#:(T08<
no. of prsons working on ;nd day!n-1--workdone-n-1 units no. of prsons working on nth day!1--workdone-1 unit euation becomes- n & n-1 & n-; & ..... &1!%;#0(Un&n&n...upto PnP timesV Un%n&1(V#;!%;#0(n; 'n;!0n;&0n n;!0n n!< or 0 discarding <$ans wud b 0 men.how parent eplye7 ime and Workby tti aturday$ 5 eptember ;<<:$ <17;< A* WonderfulMa ,ery good ,ariety..please post some more such problems.thanks a lotMM how parent eplye7 ime and WorkbyEriyeshungare-at$5 ept<:$<:7'' E* hiii...can anyone e2plain me the problems where two or more people are included and they work on alternate days??i am not able to sol,e this kind of problems.. pls help me... how parent eplye7 ime and Worky nitin bhat - at$ 5 ep<:$ 117<1 E* @i Eriyesh$ Eroblems that include ; or more people can be sol,ed in the similar fashion as you take the units of work. 2. A can finish a work in 1< days and B takes 15 days to complete the same work whereas ) takes ;5 days to do the same 4ob. ind the number of days it takes to finish the 4ob when7 a( A$B J ) all work on alternate days b(AJB work on the same day and ) works the ne2t day. oln.Let the total amount of work be 15< units%L)* of 1<$15$;5( then the respecti,e units of works of A-15$B-1< and )-8. a(Work completed 1st Kay-15%Dnly A works( ;nd day-1<%Dnly B works( 0rd day -8 %Dnly ) works( so at the end of 0 days the amount of work finished15&1<&8!01"nits.hence they would take Appro2.15 days%1'.51( to complete the work. b(When AJB work same day the work completed-;5 units on day 1 ne2t day since only ) works$ units of work completed on day ; is 8"nits. o at the end of ; days the unit of work completed is%;5&8(!01"nits. o to complete 15< units itPll take them Appro2. 1< Kays%:.81(.
I hope you ha,e understood by now....egards$ 6itin Bhat nitinr.bhatRgmail.com howparenteplye7imeandWorkbyEriyeshungare-at$5ept<:$ 1171: E* hey nitin...it may feel like stupidity...but can u e2plain one thing.. total units of work is 01 ... then how did u find that they wud take 15 days to complete the work??? how parenteplye7 ime and Workby nitin bhat-un$8 ept<:$ 1;7': A* Eriyesh$ 0Kays-01 "nits of work done$ ne2t 0 days 8; units done$ ne2t 0 days :0 units done. so on and so forth by the end of 15 days 155 units of work will be completed. our target is to complete 15< units of work. so itPll be little less than 15 days... egards$6itinhow parent eplye7 ime and Workby sindhu4a morampalli riday$ 11 eptember ;<<:$ <'7<5 E* really a great work$ thank u so much sir$ keep updating these kind of works how parent eplye7 ime and Workby 6eetendra *ishra - unday$ 10 eptember ;<<:$ <>751 E* gud work... how parent eplye7 ime and Workby ramkrishna roy - hursday$ 1= eptember ;<<:$ <075' E* for e2ample-0. $ how are we getting L)* to be 8< hours a day$ @ow many men working with 8 women and ; boys take to do a 4ob ' times larger$ if they work 5 hours a day for 1; day? DE @ first try it do by yourself. if u canPt solution is here. olution7 take lcm as '><. total work hours for 1st 4ob is 1;T>! :8 man hours. now e,ery group here takes :8 man hours$ so$ first group of men will do '><#:8 ! 5 units of work per hour. ame is true for group of women and boys.
ach man will do 5#;< unit work per hour. each Women do 5#;' unit work per hour$ each boy will do 5#'< unit per hour. now the new task is ' times bigger i.e '>< T ' it gi,es 2 ! 1;;$ which is the no of men reuired. en4oysmile how parent eplye7 ime and Workby 4aya dulani - hur$ 1 Dct<:$ <070; E* Awesome article smileMMMM "P,e made time J work probs much easier and interesting for me.. hanks a tonMMM Rarun- hey tarun$ i agree with A4ay..shudnPt that be ;n$ as both of them constructed two walls together. *oreo,er$ arenPt there ' unknowns and 0 euations? ElQ lead me to the correct solution. han2.. how parent eplye7 ime and Workby arpita bansal - at$ 0 Dct<:$ 1<7;= A* )an someone please help me in sol,ing this uestion7 a large tank of height 1m$ :.5m$ from the bottom. all the pipes are opened simultaneously$ with the tank being empty initially. in how much time :5Y of the tank can be filled? if the outlert pipe can empty half of the tank in 1 and half hours$ the outlet pipe in the middle can empty ; mins how parent eplye7 ime and Workby arnab das - ues$ 8 Dct<:$ <170; E* itPs ,ery helpful. hanks a lot.It will be ,ery good if you can kindly can upload an article based on time and distance%*ainly circle problem$ A$B$ ) running and how many times they meet$ kinna........(It will be ,ery helpful i f you do so..... how parent eplye7 ime and Workby supriya naidu - hur$>Dct<:$ 1;70' E* @i$ hanks for the information.I ha,e found one site snapwiQ.co.in there is lot of good stuff regarding this chapter in case if you are planning to prepare as a whole for )A i strongly recommend you to look snapwiQ .ake the tests there and compare your percentile score with respecti,e to each sub4ect. 6ow a days there is promotional offer going on.i registered through promotional code W1<<)A you can also try this how parent eplye7 ime and Workby Erudh,i + - hur$ > Dct<:$ <'70> E*
X1 by Ke,esh If ;< men or ;' women or '< boys can do a work in 1; days working for > hours a day$ @ow many men working with 8 women and ; boys take to do a 4ob ' times larger$ if they work 5 hours a day for 1 ; day? Alternate solution for this problem7Work ! 1; K 3 > @ ! :8 Kay-@ours As a group of ;< *en or ;' Women or '< Boys can do the work in same time ;< * ! ;' W ! '< B 5 * ! 8 W ! 1< B
%please do not blame me for gender bias -((
Work ! 1; 2 > 2 ;< man-day-hours 6ew work ! ' 2 pre,ious work ! ' 2 %1; 2 > 2 ;< ( *-K-@ Let the new work be done by *en alone. 6ew work done by7 8 W & ; B & a * ! 5 * & 1* & a * %relation btw *$W$B ( 6ew work ! ' 2 %1; 2 > 2 ;< ( ! %5&1&a( 2 5 21;G a ! 1;> - 8 ! 1;; men Z)heersZa4how parent eply e7 ime and Workby sushil krishna aturday$ 1< Dctober ;<<:$ <07<: A* than2 how parent eplye7 ime and Workby manish ma4umder - *onday$ 1; Dctober ;<<:$ <=7<= E* it was really a ,ery useful post for time and work preparation.how parenteply e7 ime and Workby C6" KK - Wednesday$ ;1 Dctober ;<<:$ 117;' A* ime and work articles is really good. hank you ,ery much for posting the article here how parent eplye7 imeandWorkby Eanka4 Oumar-hur$;: Dct<:$ <17<; E* great n summariQed piece on ime and Work$ + can you please help us on logarithms topic too.hanksMM how parent eplye7imeandWorkby smriti chhabra-Wed$ ' 6o,<:$1<750 A* hiM ds article s really helpfulM plQ g,e such articles n all important topics n )AM but i ,e sme doubts in problem : ,e understood upto dt le,et dt 11 litres are filled into d reser,ior after dt last ; steps plQ make me understand n problem 0 tooM how parent eplyime and Work e bookby prateek bansal - uesday$ 1< 6o,ember ;<<:$ <;715 A* tg sir
i ha,e sent u cheue for K and geometry e book bt ha,e stioll nt recei,ed .)A is , ery near.pls help prateek how parent eplye7imeandWorkebookby4atinarora- ue$1< 6o,<:$1;75< E* Kear ir $the problems can also be sol,ed using work efficiences also 1. A does work in 5 days so A effY ! 1<<#5 B does work in 15 days so B effY ! 1<<#15 ffY of A & ffY of B ! 1<<#5 & 1<<#15 ! '<<#15 Both can A and B can do work in ! 1<<#'<<#15 ! 15#' answer ;. A does work in 15 days so A effY ! 1<<#15!;<#0 A works for 8 days so work completed by A in 8 days ! %;<#0(T8 ! '
o B alone can do work in ! 1<<#;
answer
how parent eplye7imeandWorkbyamitOatiyar-*on$18 6o,<:$ <:70' E* @ello ir... *any many thanks to u............In Xuant$ime and Work was the only section where I ha,e sol,ed any uestion correctly..........But after going through ur sol,ing skills.........I can sol,e nW uestions............hanks a lot for this great tutorial.......... how parenteplye7imeandWorkbyAnkit+arg-ue$;' 6o,<:$ 1171; A* great article$ after reading i realiQed how simple it will be with taking L)* ... thanks a lot guys how parent eplye7 ime and Workby nidhi soni - un$ ;: 6o,<:$ <57;= E* thn2 sindhoor smile awesome artical smile e7 ime and Workby abhishek rai - uesday$ > Hune ;<1<$ <'701 A* I may be an idiot$ but please see the following.A group of men decided to do a 4ob in > days. But since 1< men dropped e,eryday$ the 4ob got completed in 1; days. @ow many men were at the beginning?I got the answer as 00<. Am I correct. I think the answer$ gi,en as 185 is wrong..@elpMMM how parent eplye7 ime and Workby +ul +ul - ue$ > Hune ;<1<$ <>75> A* Abhishek$he euation would be sth like this7
>2 ! 1;2 - 1<%1&;&...11(G
2! 185
how parent eplye7 ime and Workby abhishek rai - ue$ > Hun1<$ <17': E* hn2 +ulab...But accroindgly$ 88< men dropped in 1; days$ but only 185 men were at the beginingMMMMM houldnPt the no. of dropped people be 11<. hen$ >2!%2-11<(T1; or$ >2 ! 1;T11<-1;2 or$ '2!1;T11< or 2! 00<. 6ow plQ e2plainMMMalso$ initially$ the no. of mandays reuired 10;<$ if 185 is the answer and after after dropping of 11< people$ it becomes 5'<. omething fishyM how parent eplye7imeandWorkby miti chakraborty-ues$>Hun1<$ <;7;: E* a4ay$ i used simple trial nd error method ; sol,e dis ... here$ t1#t;! 1#%;#0( ! 0#;
therefore$ w1#w;! ;#0
nw capacity of each person is same. Let us assume capacity is 2 for all. nw using trial nd error.... w1! 02&;2&2!82 G w;! 02&02&02 ! :2G w1#w;!;#0 i knw$ it cant b d proper way ; sol,e ds ...but itPs time sa,ing... how parenteplye7imeandWorkby miti chakraborty - ues$> Hun1<$ <;758 E* abhisek$ the no. of men dropped during the process is 7- o for the 1st day 1< for the %;nd$ 0rd$ 'th$ 5th$ ........$ 1;th day( i.e. in total %1< 311(! 11< men were dropped according to the am i right$ gulab???? how parent eplye7 ime and Workby +ul +ul-ue$> Hun1<$ <>750 E* up *iti$ u r correct smile how parent eplye7imeandWorkby umang shastri-hur$ ; ept1<$ <071' E* prefectsmile how parent eplye7 ime and Workby *ohit harma-ue$0 Apr 1;$<:75> A*
R 4oydeep$ Let both the frnd can complete the wrk in 3 hrs$ then$ first frnd can complete wrk in! 3&t hrsG second frnd can complete wrk in! 3&;5t#18$ We also knw that$
1#3!1#%3&t(&1#%3&;5t#18(
from this we get$
'3!5t
6ow use$ %3&t(&%3&;5t#18(&;3! 1;1T;' hr Gand calculate the ,alue of 3 and t. how parenteply e7ime and Work by *ohit harma-hur$5 Apr1;$ 1<7;= E* @i Oamal ir$ Elease help me with the uestion. Xues7 *ini and Cinay are uiQ masters preparing for a u iQ. In 2 minutes$ *ini makes y uestions more than Cinay. If it were possible to reduce the time needed by each to make a uestion by two minutes$ then in 2 minutes *ini would make ;y uestions more than Cinay. @ow many uestions does *ini make in 2 minutes? a( 1#'U;%2&y(-srt%;2;&'y;(V b( 1#'U;%2-y(-srt%;2;&'y;(V c( ither a or b d( 1#'U;%2-y(-srt%;2;-'y;(V e( 6one of these how parenteplye7imeandWorkby*ohit harma-Wed$1>April1;$<875; E* hi Oamal ir$ Elease help me with the abo,e uestion. how parent eplye7 ime and Workby Oirti ahoo - Wed$ 8 Hun1;$ 1<701 A* @ello +$ I think therePs a bit of an anomaly in the solution for Eroblem '. In the solution itPs gi,en that Lapmpard and ooneyPs 1 day work ! 15 units but i think it should be 1 hours work and same with abregas and Walcott. o there combined 1 hours work should be 0: units and thus they can complete the work in 1<<#0: hours or ;.58 hrs As per the solution pro,ided here the answer 1<<#5> days is absurd because thatPs appro2. 1.=; which is greater than 1 day which canPt be the case as +errard himself takes 5 hours to complete the work so with the help of ' other people he should be able to complete the work in less time. Elease check the solution and correct me if iPm wrong. smile
hanks.how parenteplye7imeandWorkbykirti,ardhan-un$;:Hul1;$<'70: E* my first post on +7 here is another concept that can be used to sol,e time work problems and that is the A)ID6 "L. I will post a uestion here.i will gi,e you the easy method.;5 men and ;< days to construct a 1< mt wall.how many men are reuired to make an > mt wall if it is planned that the work be completed in 1< days.. A)ID6 "L7take the ,alue of the uantity that is to be calculated as reference..here ;5. then frame the fractions by the following7 if number of days decreases %;< to 1<( the number men has to inclrease so the fraction that has to be multiplied should be greater than 1 so ;<#1<. ne2t for constructin an > mt wall lesser number of men are reuired as compared to a 1 #1<$ o men reuired ! ;5T%;<#1<(T%>#1<(! '< men%answer( hope this helps u to crack uestions faster. how parent eplye7 ime and Workby Eratibha B - riday$ 0 August ; <1;$ 1171= A* indhoor$ I ha,e a confusion in problem 6o.=. he uestion says Se,eryone works for four hours a dayS. I donPt understand why$ then$ other workers apart from 3 work more than four hours a dayUas per the solutionV. Elease clarify. how parent eplye7 ime and Workby hythm +oyal-un$5Aug1;$ 1<75= E* @I @emant$ F alone Scan doS work in ;' days. But he worked only > days.Also payment does not depend on number of days worked for$ but on the amount of work.hus in > days he worked 1#;'T> part of work i.e. 1#0 of work.thus payment too 1#0rd only. how parenteplye7imeandWorkbysameersapre-un$5 Aug1;$ 1171= E* thank you sir...MMgr> and laudable work. thanks a lot....MM smile smile how parenteplye7imeandWorkbyAshwani Oushwaha-ri$;' Aug1;$<07;0 E* cud anyone out here pro,ide me the link to the atio and proportion forum. I canPt find it. It would be of great help. han2 how parent eplye7 ime and Workby anil mosali-*on$1<ept1;$<=7'0 A* u r genius boss. Actually speaking u r total n we are gadhas.. thats our site. help us impro,e still more with concepts of deeper comple2ity..smile how parent eplye7 imeandWorkby4yotsnachauhan-*on$;; Dct1;$<:7<< E*
hello indhoor$
Elease sol,e this uestion for me . hank you.
our pipes can fill a reser,oir in 15$ ;<$ 0< and 8< hours respecti,ely. he first one was opened at 8 A*$ second at = A*$ third at > A* and the fourth at : A*. When will the reser,oir be filled ? how parent eplye7 ime and Workby ,i,ek ;01 - ues$;0 Dct1;$ <'7;= E* hi $ let the capacity of tank be 8< %l.c.m of 15$;<$0<$8<( and the pipe A will fill at a rate ' litres per hour B at 0$c at ; A6K K A ;$ in the first hour only A is opened so 'litrs will be filled in the second A and B so = in the third hour a$b$and c : in the fourth a$b$cand K 1< now 0< litres will be filled at 1 ept10$ 1<71> E* hi$ can some one please sol,e this uestion7 he rates at which Alok and Bhaskar work are in the ratio 5 7 '. hey work on alternate days to complete a 4ob. Bhaskar started the 4ob and he worked on the last day. he 4ob was completed in 1 day more than the time that Alok alone would take to complete it. ind the time taken %in days( by them working together to complete the 4ob. how parent eplye7ime and Workby debaditya khan-*on$:ept10$<17'0 E* fficiency of A7 B! 57' let total work is 1<<2 total wok!%5&'(2!:2 now total work!:2T11!::2% in ;; days( in ;0 days the rest work i$e. 2 will be completed by B R%2#'2(
therefore total work will be completed in ;0 days. how parent eplye7 ime and Workby ankit A@A - *on$:ept10$ <=701 E* than2 for the response but ans is gi,en as '<#: days.
Man Hour Work Mathematics
Most of the questions asked in CAT from the chapter MAN HOUR WORK are complicated in nature and requires a good reasoning skill !irst tr" to understand the #asics and sol$e the questions gi$en here as e%ample #efore mo$ing to the questions in &uestion 'ank
CONCEPT
(et)s take an e%ample of #uilding a #ridge *ou ha$e #een told that +, men -orked for +, da"s and . hours dail" to complete the a#o$e task No- lets anal"/e fe- different cases #" changing some of the a#o$e $aria#les associated -ith the task Case1: *ou Case1: *ou are asked to #uild a similar #ridge at some other place #ut -ith less num#er of people 0sa" .1 then it2s quite o#$ious that the num#er of da"s required shall increase 0Note3 Here -e are assuming that each man does same amount of -ork in a gi$en time frame1 Case 2: *ou 2: *ou ha$e to #uild one more #ridge -ith +, men #ut the num#er of -orking hours reduced to 4 hr dail" 5n this case the num#er of da"s)ll also increase Case 3: (et2s 3: (et2s consider another case -here -e are in a hurr" and required to finish the -ork in less num#er of da"s 0sa" 61 Then -hat -e)ll ha$e to do7 We ha$e to increase either the num#er of -orkers or the num#er of -orking hours per da" or #oth Case 4: !inall" 4: !inall" let2s consider a situation -here -e are required to #uild more than one #ridge 0sa" 81 Here if -e -ant to #uild the #ridge in +, da"s then -e -ould #e required to increase either the num#er of -orkers or the num#er of -orking hours per da" or #oth 5f -e don2t increase the num#er of -orkers or -orking hour then the time taken to complete the #ridge -ill #e definitel" more !rom all the a#o$e cases it can #e concluded that each task9-ork can #e represented -ith : $aria#les M ; num#er of men < ; num#er of da"s H ; num#er of hours per da" W ; amount of -ork And the #asic #asic relationship relationship among the $aria#les $aria#les is
MDH/W = Constant 5n the a#o$e specified situation M ; +, = < ; +,= H ;. and W ; + No- tr" to ans-er the follo-ing questions + Ho- long -ill it take . men to complete the #ridge -orking 4 hours per da"7 8 5f the #ridge is required to #e completed in 6 da"s then ho- man" -orkers need to -ork 4 hour per da" to complete the >o# ? Ho- man" da"s it ll take for +, men -orking +, hours per da" to #uild ? similar #ridges7 Ans-er +3 M+;. = H+;4 and W+;+ Using the formula M
Problem Sol!n"
With the a#o$e mentioned concept "ou can sol$e an" kind of MAN HOUR WORK pro#lem #ut for fespecific pro#lems it is easier to attempt the pro#lem in a different -a" 5f fi$e men finish a -ork in 4 hours then ? men -ill finish the same -ork in BBBBBBBBBB man" hours7 Met#o$ 1: This 1: This pro#lem assumes that all men do the same amount of Work in an hour= and that the" -ould do= sa"= t-ice as much -ork in t-o hours The first phrase3 !i$e men finish the -ork in 4 hours is simpl" a tatement of ho- much -ork is to #e done We kno- ho- man" menD -e kno- ho- man" hoursD #ut -e are not told ho- much -ork That is a clue that the pro#lem demands us to figure out -hat sort of num#er= -ith -hat sort of units= can #e attached to the -ord -ork Each men -orks for 4 hours and there are 6 men o the total effort gi$en is ?, man hour No- that -e kno- ho- much -ork three men must do= it is not so hard to figure out ho- long it -ill take them to do it impl" di$ide ?, #" ? and the ans-er is +, hour *ou can use the Mo#
6 men complete the >o# in 4 hour ;@ One man can complete the >o# in ?, hours ;@ One man does +9?,th of the >o# each hour
;@ ? men )ll do ?9?,0;+9+,1th >o# in each hour ;@ The >o# can #e done in +, hours
0Read the a#o$e lines carefull" and tr" to understand it thoroughl" as it)ll help "ou in sol$ing man" MHW pro#lem1 (et2s tr" to sol$e one more pro#lem -ith method 8 &uestion3 Rama is a#le to do a >o# in +, da"s -orking alone and Anil is a#le to do the same >o# in 8, da"s= -orking alone 0i1 Ho- long -ill the" take to do the >o# if #oth of them -ork together 7 0ii15f the" -ork alternatel" 0pro$ided Rama starts17 ol0i1 Rama completes the >o# in +, da"s ;@ Rama does +9+,th of the >o# in one da" Anil completes completes the >o# in 8, 8, da"s ;@ Anil does does +98,th of of the >o# in one one da" o total >o# done in one da" 0-hen #oth -ork together1 ; +9+, F +98, ; ?98, Num#er of da"s taken #" them -ould #e 8,9? ol 0ii1 When the" -ork on alternate da"s= starting -ith Rama= it means that on first da"= Rama -orks On second da"= Anil -orks= on third da"= Rama -orks= and so on Therefore= in first t-o da"s= the fractional contri#ution to the amount of -ork is ?98, o in 4 t-o da"s the" -ould complete +.98,th of the -ork 0ie +8 -orking da"s1 o total >o# left after +8th da" ; + G +.98, ; +9+, And on +? th da" Rama -ould #e -orking -orking -ho completes +9+,th +9+,th of the >o# in a da" o the task -ould #e completed on +8 F + ; +? th da"
E%&MP'E
E%ample +3 (et there #e three men for a construction >o# A completes the >o# in ?, da"s= ' in 8, da"s and C demolishes the construction in 4, da"s= each person -orking alone When -ill the construction #e complete if 0a1 the" all -ork together 0#1 the" -ork on alternate da" starting -ith A= then ' and then C ol0a1 When the" all -ork together= fractional -ork done on each da" ; +9?, F +98, G +94, ; :94, ; +9+6 Therefore= the" -ill take +6 da"s to complete the >o# 0#1When the" -ork on alternate da"s= then -e o#ser$e that in the first ? da"s= the fractional -ork done is :94, 0from 0a11 The greatest multiple of this fraction such that the fraction is less than + is +: times this fraction= ie= 6494, Therefore= in +: % ? ; :8 da"s= the fractional -ork done is 6494, and the -ork left is :9:, or +9+6 On :?rd da"= A -ill do +9?,th of the -ork and -e -ill #e left -ith +9?,th -ork= -hich -ill #e done #" ' in 8,9?, da"s or 89?rd of the da" Therefore= total num#er of da"s taken -ould #e :? 89? da"s
E%ample 83 A tank is fitted -ith . pipes= some of them that fill the tank and others that are -aste pipe meant to empt" the tank Each of the pipes that fill the tank can fill it in . hours= -hile each of those that empt" the tank can empt" it in 4 hours 5f all the pipes are kept open -hen the tank is full= it -ill take e%actl" 4 hours for the tank to empt" Ho- man" of these are fill pipes7 0+18 081: 0?14 0:16 olution3 (et the num#er of fill pipes #e )n Therefore= there -ill #e .In= -aste pipes Each of the fill pipes can fill the tank in . hours Therefore= each of the fill pipes -ill fill +9.th of the tank in an hour Hence= n fill pipes -ill fill n9.th of the tank in an hour imilarl"= each of the -aste pipes -ill drain the full tank in 4 hours That is= each of the -aste pipes -ill drain +94th of the tank in an hour Therefore= 0.In1 -aste pipes -ill drain 00.In1941th of the tank in an hour 'et-een the fill pipes and the -aste pipes= the" drain the tank in 4 hours That is= -hen all . of them are opened= +94th of the tank gets drained in an hour 0Amount of -ater filled #" fill pipes in + hour I Amount of -ater drained #" -aste pipes + hour1 ; +94th capacit" of the tank drained in + hour E%ample ?3 A takes 8 da"s to finish a task= ' takes t-ice that amount of time= C takes t-ice of '= and < takes t-ice of C 5f the" do it in pairs= one pair takes t-ice the amount of time as the other== -hich is the pair that takes longer7 a A= ' # '= C c '= < d C= < AN3 # E%ample :3 ?, co-s gra/e a piece of grassland in 4, da"s= -hereas :, co-s gra/e in :, da"s 5n ho- man" da"s -ill 8, co-s gra/e the grass 0assuming the grass gro-s e$er" da"17 ol(et a #e the initial quantit" 0in rele$ant units1 of grass and # #e the amount of grass eaten per da" Then= 4, % ?, ; +.,, ; a F 4,# and :, % :, ; +4,, ; a F :,# ol$ing these equations= -e get a ; +8,, units and # ; +, units E%ample 63 5f A and ' -ork together= the" -ill complete a >o# in J6 da"s Ho-e$er= if A -orks alone and completes half the >o# and then ' takes o$er and completes the remaining half alone= the" -ill #e a#le to complete the >o# in 8, da"s Ho- long -ill ' alone take to do the >o# if A is more efficient than '7 0+18, da"s 081:, da"s 0?1?, da"s
0:18: da"s olution3 (et a #e the num#er of da"s in -hich A can do the >o# alone Therefore= -orking alone= A -ill complete +9a of the >o# in a da" imilarl"= let # the num#er of da"s in -hich ' can do the >o# alone Hence= ' -ill complete +9# of the >o# in a da" Working together= A and ' -ill complete 0+9a F +9#1 of the >o# in a da" The pro#lem states that -orking together= A and ' -ill complete the >o# in J6 or +698 da"s ie the" -ill complete 89+6th of the >o# in a da" Therefore= 0+9a F +9#1;89+6 IIIII0i1 !rom statement 8 of the question= -e kno- that if A completes half the >o# -orking alone and ' takes o$er and completes the ne%t half= the" -ill take 8, da"s As A can complete the >o# -orking alone in )a2 da"s= he -ill complete half the >o#= -orking alone= in a98 da"s imilarl"= ' -ill complete the remaining half of the >o# in #98 da"s Therefore= or a98 F #98 ; 8, ;@ a F # ; :, ;@a ; #I :, IIIIIII 081 !rom 0+1 and 081 -e get= ;@ 4,, ; .,# I 8#8 ;@ #8 I :,# F ?,, ; , ;@ 0# I ?,10# I +,1 ; , ;@ # ; ?, or # ; +, 5f # ; ?,= then a ; :, I ?, ; +, or 5f # ; +,= then a ; :, I +, ; ?, As A is more efficient then '= he -ill take lesser time to do the >o# alone Hence A -ill take onl" +, da"s and ' -ill take ?, da"s 5f )A2 takes +, da"s to do a >o#= he -ill do +9+,th of the >o# in a da" imilarl"= if 896ths of the >o# is done in a da"= the entire >o# -ill #e done in 698 da"s E%ample 43 Mark can dig a ditch in : hours reg can dig the same ditch in ? hours Ho- long -ould it take them to dig it together7 olution3 (et % ; num#er of hours to dig the ditch together 5f Mark takes : hours to dig the ditch= he can dig +9: of it in + hour reg can dig +9? of it in one hour Marks rate is +9: and regs rate is +9? 5f it takes them % hours to dig it together= the" can dig +9% part of it in + hour together The total of the fractional part each can dig or +9? F +9: ; the fractional part the" can dig together in + hour +9? F +9: ; +9% Multipl" #e +8% to clear the fractions :% F ?% ; +8 J% ; +8 % ; +89J or + 69J
S(MM&)*
+ To sol$e -ork pro#lems= "ou need to -ork -ith the same unit of measure -ithin each pro#lem !or e%ample= "ou cannot mi% hours and minutes in the same equation 8 *ou need to find the fractional part of the >o# that -ould #e done in one unit of time= such as + minute or + hour 5f a person can do a complete >o# in ? da"s= he can do +9? of it in + da" ? The fractional part of the >o# one person can do in + da" plus the fractional part another person can do in + da" equals the fractional part of the >o# the t-o can do together in + da" E%ample3 5f 'ill can #uild +9? of a dog house in + da" and ar" can #uild +96 of it in + da"= together the" can #uild +9? F +96 of the dog house in + da" : Rate of -ork % time ; -ork done 6 M
T!me + S,ee$ an$ D!stan-e Mathematics
We define speed as distance di$ided #" time= S,ee$ = $!stan-e /t!me .D=ST #ut once -e ha$e the equation= -e can use an" of its $ariations= speed ; distance 9 time = distance ; speed L time and time ; distance 9 speed to compute an" one of the quantities -hen -e happen to kno- the other t-o !or e%ample= suppose -e dri$e for 8 hours at ?, miles per hour= for a total of 4, miles 5f -e kno- the time and the speed= -e can find the distance3 8 hours L ?, miles9hour ; 4, miles 5f -e kno- the time and the distance= -e can find the speed3 4, miles 9 8 hours ; ?, miles9hour D!stan-e !s $!re-tl0 ,ro,ort!onal to elo-!t0 #en t!me !s -onstant + A car tra$els at ?,km9hr for the first 8 hrs then :,km9h for the ne%t 8hrs !ind the ratio of distance tra$eled +98;+98;?9: 8 T-o cars lea$e simultaneousl" from points A ' 0+,,km apart1 the" meet at a point :, km from A What is a9#7 T is constant so +98;+98;:,94,;:94 ? A train meets -ith an accidient and mo$es at ?9:th its original speed ourne" #e"ond the point of accident7 Method+ 3 Think a#out 8 diff ituations = +st -ith accident and another -9o accident = then is constant
o +98;T+9T8 ;@+9?9:1L8P;0T+F8,19T+
;@:9?;0T+F8,19T+ ;@T+;4,
Method 83 elocit" decreases #" 86Q so time -ill increase #" ???Q ???Q;8, mins ;@+,,Q;4, mins CONE)SON: +km9hr;+,,,m9h;+,,,9?4,,m9sec;69+.m9sec
)E'&TE SPEED
Caes1: T-o #odies are mo$ing in opposite directions at speed + 8 respecti$el" The relati$e speed is defined as r = 1 2 Case2: T-o #odies are mo$ing in same directions at speed + 8 respecti$el"The relati$e speed is defined as r = 51 6 25 Tra!n Problems The #asic equation in train pro#lem is the same ;T The follo-ing things need to #e kept in mind -hile sol$ing the train related pro#lems When the train is crossing a mo$ing o#>ect = the speed has to #e taken as the relati$e speed of the train -ith respect to the o#>ect The distance to #e co$ered -hen crossing an o#>ect= -hene$er trains crosses an o#>ect -ill #e equal to3 (ength of the train F (ength of the o#>ect 7oats 8 Streams (et U; elocit" of the #oat in still -ater ;elocit" of the starem While mo$ing in upstream= distance co$ered ; 0UI1 T 5ncase of do-n stream= distance co$ered ; 0UF1 T C!r-ular Mot!on The relati$e $elocit" of 8 #odies mo$ing around a circle in the same direction is taken as 0+ S 81 and -hile mo$ing in opposite direction is taken as 0+F81 !irst Meeting Three or more #odies start mo$ing simultaneousl" from the same point on the circumference of the circle The" -ill +st meet again in the (CM of the times that the fastest runner -ill takes in totall" o$erlapping each of the slo-er runners !irst meeting time ; Circumference 9 Relati$e $elocit" !irst Meeting at starting point The first meeting at the starting point -ill occur after a time that is o#tained #" the (CM of the times that each of the #odies takes the complete one full round
C'OC9 !or clock pro#lems consider the clock as a circular track of 4,kmMin hand mo$es at the speed of 4,km9hr 0think min hand as a point on the track1 and hour hand mo$es at 6km9hr and second hand at the speed of ?4,, km9hrRelati$e speed #et-een hr hand and mins hand ; 66 uest!onS +A train tra$eling at J8 kmph crosses a platform in ?, seconds and a man standing on the platform in +. seconds What is the length of the platform in meters7 The correct ans-er is 8:, meters 8 A train tra$eling at +,, kmph o$ertakes a motor#ike tra$eling at 4: kmph in :, seconds What is the length of the train in meters7 Correct ans-er is :,, meters ? im tra$els the first ? hours of his >ourne" at 4, mph speed and the remaining 6 hours at 8: mph speed -hat is the a$erage speed of imSVs tra$el in mph7 Correct ans-er is ?J6 mph : A runs 86Q faster than ' and is a#le to gi$e him a start of J meters to end a race in dead heat What is the length of the race7 The correct ans-er is ?6 meters 6 ane co$ered a distance of ?:, miles #et-een cit" A and cit" taking a total of 6 hours 5f part of the distance -as co$ered at 4, miles per hour speed and the #alance at ., miles per hour speed= ho- man" hours did she tra$el at 4, miles per hour7 The correct ans-er is ? hours 4 te$e tra$eled the first 8 hours of his >ourne" at :, mph and the last ? hours of his >ourne" at ., mph What is his a$erage speed of tra$el for the entire >ourne"7 The correct ans-er is 4: mph
1. A can do a work in 15 days and B in ;< days. If they work on it together for ' days$ then the fraction of the work that is left is 7 A+
/+
1 ' = 15
+
+
1 1< > 15
;. A can lay railway track between two gi,en stations in 18 days and B can do the same 4ob in 1; days. With help of )$ they did the 4ob in ' days only. hen$ ) alone can do the 4ob in7 A+
1 : days 5
+
; : days 5
/+
0 : days 5
+
1<
ie- Ans-er Workspace Report
0. A$ B and ) can do a piece of work in ;<$ 0< and 8< days respecti,ely. In how many days can A do the work if he is assisted by B and ) on e,ery third day?
A+
1; days
+
15 days
/+
18 days
+
1> days
ie- Ans-er Workspace Report
'. A is thrice as good as workman as B and therefore is able to finish a 4ob in 8< days less than B. Working together$ they can do it in7 A+
;< days
+
1 ;; days ;
/+
;5 days
+
0< days
ie- Ans-er Workspace Report
5. A alone can do a piece of work in 8 days and B alone in > days. A and B undertook to do it for s. 0;<<. With the help of )$ they completed the work in 0 days. @ow much is to be paid to )? A+
s. 0=5
+
s. '<<
/+
s. 8<<
+
s. ><<
8. If 8 men and > boys can do a piece of work in 1< days while ;8 men and '> boys can do the same in ; days$ the time taken by 15 men and ;< boys in doing the same type of work will be7 A+
' days
+
5 days
/+
8 days
+
= days
ie- Ans-er Workspace Report
=. A can do a piece of work in ' hoursG B and ) together can do it in 0 hours$ while A and ) together can do it in ; hours. @ow long will B alone take to do it? A+
> hours
+
1< hours
/+
1; hours
+
;' hours
ie- Ans-er Workspace Report
>. A can do a certain work in the same time in which B and ) together can do it. If A and B together could do it in 1< days and ) alone in 5< days$ then B alone could do it in7 A+
15 days
+
;< days
/+
;5 days
+
0< days
ie- Ans-er Workspace Report
:. A does >
;0 days
+
0= days
/+
0=
+
'< days
ie- Ans-er Workspace Report
1<. A machine E can print one lakh books in > hours$ machine X can print the same number of books in 1< hours while machine can print them in 1; hours. All the machines are started at : A.*. while machine E is closed at 11 A.*. and the remaining two machines complete work. Appro2imately at what time will the work %to print one lakh books( be finished ? A+
1170< A.*.
+
1; noon
/+
1;70< E.*.
+
17<< E.*.
11. A can finish a work in 1> days and B can do the same work in 15 days. B worked for 1< days and left the 4ob. In how many days$ A alone can finish the remaining work? A+
5
+
1 5 ;
/+
8
+
>
ie- Ans-er Workspace Report
1;. ' men and 8 women can complete a work in > days$ while 0 men and = women can complete it in 1< days. In how many days will 1< women complete it? A+
05
+
'<
/+
'5
+
5<
ie- Ans-er Workspace Report
10. A and B can together finish a work 0< days. hey worked together for ;< days and then B left. After another ;< days$ A finished the remaining work. In how many days A alone can finish the work? A+
'<
+
5<
/+
5'
+
8<
ie- Ans-er Workspace Report
1'. E can complete a work in 1; days working > hours a day. X can complete the same
work in > days working 1< hours a day. If both E and X work together$ working > hours a day$ in how many days can they complete the work? A+
5 5 11
/+
5 8 11
+
8 5 11
+
8 8 11
ie- Ans-er Workspace Report
15. 1< women can complete a work in = days and 1< children take 1' days to complete the work. @ow many days will 5 women and 1< children take to complete the wor k? A+
0
+
5
/+
=
+
)annot be determined
5+
6one of these
18. 3 and can do a piece of work in ;< days and 1; days respecti,ely. 3 started the work alone and then after ' days 4oined him till the completion of the work. @ow long did the work last? A+
8 days
+
1< days
/+
15 days
+
;< days
ie- Ans-er Workspace Report
1=. A is 0
11 days
/+
;<
0 1=
days
+
10 days
+
6one of these
ie- Ans-er Workspace Report
1>. a,i and Oumar are working on an assignment. a,i takes 8 hours to type 0; pages on a computer$ while Oumar takes 5 hours to type '< pages. @ow much time will they take$ working together on two different computers to type an assignment of 11< pages? A+
= hours 0< minutes
+
> hours
/+
> hours 15 minutes
+
> hours ;5 minutes
ie- Ans-er Workspace Report
1:. A$ B and ) can complete a piece of work in ;'$ 8 and 1; days respecti,ely. Working together$ they will complete the same work in7 1
day
A+
;'
/+
0 0 days =
=
day
+
;'
+
' days
ie- Ans-er Workspace Report
;<. akshi can do a piece of work in ;< days. anya is ;5Y more efficient than akshi. he number of days taken by anya to do the same piece of work is7 A+
15
+
18
/+
1>
+
;5
;1. A takes twice as much time as B or thrice as much time as ) to finish a piece of work. Working together$ they can finish the work in ; days. B can do the work alone in7 A+
' days
+
8 days
/+
> days
+
1; days
ie- Ans-er Workspace Report
;;. A and B can complete a work in 15 days and 1< days respecti,ely. hey started doing the work together but after ; days B had to lea,e and A alone completed the remaining work. he whole work was completed in 7 A+
> days
+
1< days
/+
1; days
+
15 days
ie- Ans-er Workspace Report
;0. A and B can do a piece of work in 0< days$ while B and ) can do the same work in ;' days and ) and A in ;< days. hey all work together for 1< days when B and ) lea,e. @ow many days more will A take to finish the work? A+
1> days
+
;' days
/+
0< days
+
08 days
ie- Ans-er Workspace Report
;'. A works twice as fast as B. If B can complete a work in 1; days independently$ the number of days in which A and B can together finish the work in 7
A+
' days
+
8 days
/+
> days
+
1> days
ie- Ans-er Workspace Report
;5. wenty women can do a work in si2teen days. i2teen men can complete the s ame work in fifteen days. What is the ratio between the capacity of a man and a woman? A+
07'
+
'70
/+
570
+
Kata inadeuate
;8. A and B can do a work in > days$ B and ) can do the same work in 1; days. A$ B and ) together can finish it in 8 days. A and ) together will do it in 7 A+
' days
+
8 days
/+
> days
+
1; days
ie- Ans-er Workspace Report
;=. A can finish a work in ;' days$ B in : days and ) in 1; days. B and ) start the work but are forced to lea,e after 0 days. he remaining work was done by A in7 A+
/+
5 days
1< days
+
8 days
+
1 1< days ;
ie- Ans-er Workspace Report
;>. 3 can do a piece of work in '< days. @e works at it for > days and then finished it in 18 days. @ow long will they together take to complete the work? 1
A+
10
/+
;< days
0
days
+
15 days
+
;8 days
ie- Ans-er Workspace Report
;:. A and B can do a 4ob together in = days. A is 1 can be done by A alone in 7 A+
1 : days 0
/+
1;
1 '
days
times as efficient as B. he same 4ob
+
11 days
+
1 18 days 0
ie- Ans-er Workspace Report
0<. A and B together can do a piece of work in 0< days. A ha,ing worked for 18 days$ B finishes the remaining work alone in '' days. In how many days shall B finish the whole work alone? A+
0< days
+
'< days
/+
8< days
+
=< days
1+ Work *rom a's: If A can do a piece of work in
days$ then APs 1 dayPs work !
n
1 . n
2+ a's *rom Work: If APs 1 dayPs work !
1
$ then A can finish the work in
days.
n
n
3+ %atio: If A is thrice as good a workman as B$ then7 atio of work done by A and B ! 0 7 1. atio of times taken by A and B to finish a work ! 1 7 0 .
Each of the questions gi$en #elo- consists of a statement and 9 or a question and t-o statements num#ered 5 and 55 gi$en #elo- it *ou ha$e to decide -hether the data pro$ided in the statement0s1 is 9 are sufficient to ans-er the gi$en question Read the #oth statements and •
•
•
•
•
i$e ans-er 0A1 if the data in tatement 5 alone are sufficient to ans-er the question= -hile the data in tatement 55 alone are not sufficient to ans-er the question i$e ans-er 0'1 if the data in tatement 55 alone are sufficient to ans-er the question= -hile the data in tatement 5 alone are not sufficient to ans-er the question i$e ans-er 0C1 if the data either in tatement 5 or in tatement 55 alone are sufficient to ans-er the question i$e ans-er 0<1 if the data e$en in #oth tatements 5 and 55 together are not sufficient to ans-er the question i$e ans-er0E1 if the data in #oth tatements 5 and 55 together are necessar" to ans-er the question
1. A and B together can complete a task in = days. B alone can do it in ;< days. What part of the work was carried out by A?
I. A completed the 4ob alone after A and B worked together for 5 days.
II. Eart of the work done by A could ha,e bee n done by B and ) together in 8 days.
A+
I alone sufficient while II alone not sufficient to answer
+
II alone sufficient while I alone not sufficient to answer
/+
ither I or II alone sufficient to answer
+
Both I and II are not sufficient to answer
5+
Both I and II are necessary to answer
;. @ow long will *achine $ working alone$ take to produce
candles?
x
I. *achine 3 produces x candles in 5 minutes.
II. *achine 3 and *achine working at the same time produce
A+
I alone sufficient while II alone not sufficient to answer
+
II alone sufficient while I alone not sufficient to answer
/+
ither I or II alone sufficient to answer
+
Both I and II are not sufficient to answer
5+
Both I and II are necessary to answer
candles in ; minutes.
x
Each of the questions gi$en #elo- consists of a question follo-ed #" three statements *ou ha$e to stud" the question and the statements and decide -hich of the statement0s1 is9are necessar" to ans-er the question 1. In how many days can 1< women finish a work?
I. 1< men can complete the work in 8 days.
II.
1< men and 1< women together can complete the work in 0
0 =
days
III. If 1< men work for 0 days and thereafter 1< women replace them$ the remaining work in completed in ' days.
A+
Any two of the three
+
I and II only
/+
II and III only
+
I and III only
5+
6one of these
;. @ow many workers are reuired for completing the construction work in 1< days?
I. ; workers in > days.
II. ;< workers can complete the work in 18 days.
III. Dne-eighth of the work can be completed by > workers in 5 days.
A+
I only
+
II and III only
/+
III only
+
I and III only
5+
Any one of the three
D!re-t!ons to Sole Each of these questions is follo-ed #" three statements *ou ha$e to stud" the question and all the three statements gi$en to decide -hether an" information pro$ided in the statement0s1 is redundant and can #e dispensed -ith -hile ans-ering the gi$en question 1. > men and 1' women are working together in a field. After working for 0 days$ 5 men and > women lea,e the work. @ow many more days will be reuired to complete the work?
I. 1: men and 1; women together can complete the work in 1> days.
II. 18 men can complete two-third of the work in 18 days.
III. In 1 day$ the work done by three men in eual to the work done by four women.
A+
I only
+
II only
/+
III only
+
I or II or III
5+
II or III only
Pipes cisterns, Work time Sample Question The given question is a work time question and tests your ability to relate to the relation between the number of days it takes two people to do a task to their efficiency of completing the task.
Question 1 Ram completes 4,Q of a task in +6 da"s and then takes the help of Rahim and Rachel Rahim is 6,Q as efficient as Ram is and Rachel is 6,Q as efficient as Rahim is 5n homan" more da"s -ill the" complete the -ork7
+
8
?
: 6
None of these
Correct choice is .3 Correct Ans-er is Explanatory Answer Ram completes 4,Q of the task in +6 da"s ie= he completes :Q of the task in a da" Rahim is 6,Q as efficient as Ram is Therefore= Rahim -ill complete 8Q of the task in a da" Rachel is 6,Q as efficient as Rahim is Therefore= Rachel -ill complete +Q of the task in a da" Together= Ram= Rahim and Rachel -ill complete : F 8 F + ; JQ of the -ork in a da" The" ha$e another :,Q of the task to #e completed Therefore= the" -ill take
more da"s to complete the task
Question 4 the day: April 8, 2002 A tank is fitted with 8 pipes, some of them that fill the tank and others that are waste pipe meant to empty the tank. Each of the pipes that fill the tank can fill it in 8 hours, while each of those that empty the tank can empty it in 6 hours. f all the pipes are kept open when the tank is full, it will take e!actly 6 hours for the tank to empty. "ow many of these are fill pipes#
$% &
'
$' &
(
$) &
6
$( &
*
+orrect Answer (2) Solution:-et the number of fill pipes be n/. Therefore, there will be 8n, waste pipes. Each of the fill pipes can fill the tank in 8 hours. Therefore, each of the fill pipes will fill %08th of the tank in an hour. "ence, n fill pipes will fill n08th of the tank in an hour. 1imilarly, each of the waste pipes will drain the full tank in 6 hours. That is, each of the waste pipes will drain %06th of the tank in an hour. Therefore, $8n& waste pipes will drain $$8n&06& th of the tank in an hour.
2etween the fill pipes and the waste pipes, they drain the tank in 6 hours. That is, when all 8 of them are opened, %06 th of the tank gets drained in an hour. $Amount of water filled by fill pipes in % hour Amount of water drained by waste pipes % hour& 3 %06th capacity of the tank drained in % hour.
4ote5 n problems pertaining to ipes and +isterns, as a general rule find out the amount of the tank that gets filled or drained by each of the pipes in unit time $say in % minute or % hour&.
Question 4 the day: April 25, 2002 %. 7orking together, A and 2 can do a ob in 6 days. 2 and + can do the same ob in %9 days, while + and A can do it in :.* days. "ow long will it take if all A, 2 and + work together to complete the ob#
$% &
8 days
$' &
* days
$) &
) days
$( &
: days
'. ). "ow long will it take for A alone to complete the ob# $% &
8 days
$' &
6 days
$) &
%9 days
$( &
'9 days
+orrect Answers % ' $'& $)& . . Solution:
Even before you start working on the problem, check out if you can eliminate some answer choices as impossible.
n question $%&, we know that if A and 2 alone work, they can complete the ob in 6 days. Therefore, if all three of them A, 2 and + work together the number of days it will take to complete the ob will surely be less than 6 days. "ence, we can eliminate answer choices $%& and $(& right away. 1imilarly in question $'&, we know that A and 2 together take 6 days to complete the ob. Therefore, A alone will take more than 6 days to complete the ob. Therefore, we can eliminate answer choice $'&. n any question, as a rule spend about * seconds to see if the answer choices provide any clue to solve the question or help in eliminating one or more obviously absurd choices. This will help you $%& in reducing the time it will take to do the problem and $'& in increasing your probability of success should you choose to take a guess without actually solving the problem.
uest!on 1 -et A be the number of days that A will take to complete the ob alone, 2 days for 2 to complete the ob alone and + days for + to complete the ob alone.
A and 2 can do a ob in 6 days. They complete %06 th of the ob in a day. i.e. 1imilarly, 2 and + will complete %0%9th of the ob in a day. i.e
$%&
$'&
And + and A will complete %0:.* or '0%*th of the ob in a day i.e
$)&.
Adding $%&, $'& and $)& we get 33; or . i.e working together, A, 2 and + complete %0*th of the ob in a day. Therefore, they will complete the ob in * days.
uest!on 2 1ubtracting eqn $'& from eqn $%& we get
$(&
Adding eqn $(& and eqn $)& we get, or . i.e. A does %0%9 of the ob in a day and therefore, will take %9 days to complete the ob working alone.
6uestion 4 the da': 7une #) 22 he uestion for the day is from the topic work and time. our men and three women can do a 4ob in 8 days. When fi,e men and si2 women work on the same 4ob$ the work gets completed in ' days. @ow long will a woman take to do the 4ob$ if she works alone on it? $% &
%8 days
$' &
)orrect Answer - 83 Solution:
)6 days
$) &
*( days
$( 4one of these &
Let the amount of work done by a man in a day be [m/ and the amount of work done by a woman in a day be [w/. herefore$ ' men and 0 women will do 'm & 0w amount of work in a day. If ' men and 0 women complete the entire work in 8 days$ they will complete 1#8th of the work in a day. @ence en %1( will be 'm & 0w ! and from statement %;($ en %;( will be 5m & 8w ! ol,ing en %1( and en %;($ we get 0m !
or m ! 1#08. i.e. a man does
1#08th of the work in a day. @ence he will take 08 days to do the work. ubstituting the ,alue of m in en %1($ we get ! 0w !
or w ! 1#5'. i.e. a woman does 1#5'th of the work in
a day. @ence she will take 5' days to do the entire work 6uestion 4 the da': 7une 10) 22 he uestion for the day is from the topic - Work and ime. hyam can do a 4ob in ;< days$ am in 0< days and inghal in 8< days. If hyam is helped by am and inghal e,ery 0 rd day$ how long will it take for them to complete the 4ob? $% &
%' days
$' &
$) &
%6 days
$( &
%* days
%9 days
)orrect Answer - 83 Solution: As hyam is helped by am and inghal e,ery third day$ hyam works for 0 days while am and inghal work for 1 day in e,ery 0 days. herefore$ the amount of work done in 0 days by hyam$ am and inghal !
th of the 4ob. @ence$ it will take them 5 times the
amount of time ! 0T5 ! 15 days. The question for the day is a pipes and cisterns problem. ipe A usually fills a tank in ' hours.
' hours )9 minutes
+orrect Answer (4)
$' &
* hours
$) &
( hours
$( &
%9 hours
Solution:
ipe A fills the tank normally in ' hours. Therefore, it will fill = of the tank in an hour. -et the leak take ! hours to empty a full tank when pipe A is shut. Therefore, the leak will empty
of the tank in an hour.
The net amount of water that gets filled in the tank in an hour when pipe A is open and when there is a leak 3 of the tank. > $%& 7hen there is a leak, the problem states that ipe A takes two and a half hours to fill the tank. i.e.
hours. Therefore, in an hour,
Equating $%& and $'&, we get
of the tank gets filled. ? $'& 3;
3; ! 3 %9 hours.
The problem can also be mentally done as follows. ipe A takes ' hours to fill the tank. Therefore, it fills half the tank in an hour or *9@ of the tank in an hour. 7hen there is a leak it takes ' hours )9 minutes for the tank to fill. i.e or
hours to fill the tank
or (9@ of the tank gets filled.
Question 4 the day: April 15, 2003
The question for the day is from the topic of 7ork and Time. A, 2 and + can do a work in * days, %9 days and %* days respectively. They started together to do the work but after ' days A and 2 left. + did the remaining work $in days& $% &
%
$' &
)
$) &
*
$( &
(
+orrect Answer (4) Solution:
f A, 2 and + work together for a day then they will finish $%0* %0%9 %0%*& th work 3 %%0)9th work. Therefore working together for two days they will finish ' B %%0)9 3 %%0%* th work. + alone does remain $%%0%* ? %& (0%* th work. 2ut + finishes %0%* th work in one day. Therefore + will finish (0%* th work in ( days.
Question 4 the day: June 13, 2003
The question for the day is from the topic of Time and 7ork. C alone can do a piece of work in %* days and D alone can do it in %9 days. C and D undertook to do it for s. :'9. 7ith the help of F they finished it in * days. "ow much is paid to F# $% &
s. )69
$' &
s. %'9
$) &
s. '(9
$( &
s. )99
+orrect Answer (2) Solution:
n one day C can finish %0%* th of the work. n one day D can finish %0%9 th of the work. -et us say that in one day F can finish %0F th of the work. 7hen all the three work together in one day they can finish %0%* %0%9 %0F 3 %0* th of the work. Therefore, %0F 3 %0)9. atio of their efficiencies 3 %0%*5 %0%95 %0)9 3 '5 )5 %.Therefore F receives %06 th of the total money. According to their efficiencies money is divided as '(95 )695 %'9. "ence, the share of F 3 s. %'9.
Question 4 the day: Oto!er 23, 2003
The question for the day is from the topic of Time and 7ork. A and 2 can do a piece of work in '% and '( days respectively. They started the work together and after some days A leaves the work and 2 completes the remaining work in G days. After how many days did A leave# $% &
*
$' &
:
$) &
8
$( &
6
+orrect Answer (2) Solution:
n G days 2 completes G0'( i.e., )08 th of the work. "ence % )08 3 *08 th of the work is completed by A and 2 together. i.e., ! $%0'% %0'(& 3 *08 3; ! 3 : days.
CAT Sample Questions - Question 4 the day : July 07, 2004
he uestion for the day is from the topic Work and ime.. A can complete a pro4ect in ;< days and B can complete the same pro4ect in 0< days. If A and B start working on the pro4ect together and A uits 1< days before the pro4ect is completed$ in how many days will the pro4ect be completed? $%&
%8 days
$'&
': days
$)&
'6.6: days
$(&
%6 days
/orrect choice 9 81 /orrect Answer 981. da's Explanatory Anser
If A can do complete a pro4ect in ;< days$ then A will complete in a day.
th of the pro4ect
imilarly$ B will complete th of the pro4ect in a day. Let the total number of days taken to complete the pro4ect be 2 days. hen$ B would ha,e worked for all 2 days$ while A would ha,e worked for %2 9 1<( days.
herefore$ A would ha,e completed complete
th of the pro4ect and B would ha,e
th of the pro4ect.
i.e.$ ol,ing for 2$ we get 2 ! 1>. TANCET 2008 Quant Question 4 : Work Time Question : A father can do a certain job in x hours. His son takes twice as long to do the job.
Working together, they can do the job in 6 hours. How many hours does the father take to do the_job? + 8 ? :
+. +8 8,
6 +4 Corre-t &nser : hours Choice 0+1
Explanator Ans!ers The father completes the >o# in % hours o= the son -ill take 8% hours to complete the same >o# 5n an hour= the father -ill complete 5n an hour= the son -ill complete
of the total >o# of the total >o#
o= if the father and son -ork together= in an hour the" -ill complete ie= in an hour the" -ill complete
F
of the >o#
of the >o#
The question states that the" complete the entire task in 4 hours if the" -ork together ie= the" complete
of the task in an hour
Equating the t-o information= -e get
;
Cross multipl"ing and sol$ing for % -e get 8% ; +. or % ; The father takes hours to complete the >o# Correct Ans-er 0+1
Spee", Time, #istance : CAT 20$2 %nline Preparation uest!on 4 t#e $a0 : Ma0 4+ 2;;4 The question for the da" is from the topic speed= time and distance and is a#out finding the distance co$ered #" a train -hile crossing a stationar" o#>ect -hose length is compara#le to the
length of the train uest!on: A train tra$eling at J8 kmph crosses a platform in ?, seconds and a man standing on the platform in +. seconds What is the length of t he platform in meters7 + 8:, meters 8 ?4, meters ? :8, meters : 4,, meters Correct Ans-er is 24; metres Choice .1 is right E<,lanator0 &nser When the train crosses a man standing on a platform= the distance co$ered #" the train is equal to the length of the train Ho-e$er= -hen the same train crosses a platform= the distance co$ered #" the train is equal to the length of the train plus the length of the platform The e%tra time that the train takes -hen crossing the platform is on account of the e%tra distance that it has to co$er ; length of the platform Therefore= length of the platform ; speed of train L e%tra time taken to cross the platform (ength of platform ; J8 kmph L +8 seconds Con$erting J8 kmph into m9sec= -e get J8 kmph ;
; 8, m9sec
Therefore= length of the platform ; 8, L +8 ; 8:, meters
Pa"al"u0-om
"i#e, speed, distane $ #otion o% t&o !odies in a strai'ht line by avi "anda in A *ntrane +as H '8 Ianuary /%)
Photo Credit: Urban Hafner There are some topics in Juantitative Aptitude, like permutation K combination, where you can easily find out the answer, $it is in the options& but it turns out to be wrong. And then there are some topics in which you read the question, understand it but cannot even begin solving it. Dou get stuck at the first step and you have no idea about how to even approach the question. The irritating fact is that you understood the question properly. t happens very frequently with questions on Time 1peed K Listance $T1L&. have always been a big advocate of skipping questions which you cannot solve. More often than not, T1L questions should be skipped if you cannot figure out how to start within the first minute. Typically the questions on T1L are based upon certain ideas, which if you are not aware of can make solving the question e!tremely difficult and time consuming. do not think that am even aware of all ideas 0 types of questions in T1L but there are a few popular ones which have been doing the rounds in the past few years. am going to cover some of them in this post and probably revisit some more in the months to come. am also going to discuss the reasoning behind those ideas. t is very important that you understand the logic behind the formulae before you actually start using them. f you donNt, there is a very high probability that you will make a mistake. To begin with, some of the very basic ideas that you should be aware of are5 •
1peed 3 Listance 0 Time
•
f Listance is constant, then 1peed and Time are inversely proportional to each other
•
Two bodies moving in the same direction would have the relative speed of 1$%& ? 1$'& and two bodies moving in the opposite direction would have the relative speed of 1$%& 1$'& n this particular post, am going to talk about the motion of two bodies in a straight line starting from opposite ends. -ase 15
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2) The! ta"e times of T(1) & T(2) to reach their destinations n such a case, the relationship between the times taken and the speeds will be S(1) . S(2) / "(2) . "(1)
The distances that both bodies have travelled are J and J respectively. J 3 1$%& T$%& J 3 1$'& T$'& Also, J 3 J so the above equations can be equated to get the desired result.
-et us say that they meet at a point in between, then &ould diide the distane Q in the ratio o% S(1):S(2)
They started at the same time and they are meeting at point , so the time taken by both the bodies will be the same. f we assume that time as T 3 1$%& T J 3 1$'& T Lividing the above two equations would give us the desired result. -ase 25
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2) After meeting each other # the! ta"e times of T(1) & T(2) to reach their destinations n such a case, the relationship between the times taken and the speeds will be
Also, the time taken for the two bodies to to meet will be -et us assume that the two bodies meet at a point , after time T. Oor the first body, 3 1$%&T and J 3 1$%&T$%& Oor the second body, J 3 1$'&T and 3 1$'&T$'& 7e know that 3 3; 1$%&T 3 1$'&T$'& 7e know the J 3 J 3; 1$'&T 3 1$%&T$%& Lividing the above two equations, we will get
Psing the above result, we can obtain the value of T
-ase 35
Two bodies start from opposite ends P & Q to reach their destinations at the same time and move towards each other with speeds S(1) & S(2) $efore meetin%# the! ta"e times of T(1) & T(2) to reach the meetin% point n such a case, the relationship between the times taken and the speeds will be
Also, the time taken for the two bodies to reach their destinations after meeting will be The logic for this would be e!actly the same as +ase '. Try working this out on your own. -ase 4:
Two bodies start from opposite ends P & Q at the same time and move towards each other with speeds S(1) & S(2) The! reach the opposite ends and reverse directions The! contine this to and fro motion f the distance between the two bodies in the beginning is L, then
Time taken by them to meet for the first time, T 3 L 0 1$%& 1$'& f is the first meeting point, 0 J 3 1$%& 0 1$'& The idea that am going to discuss now is only valid in the case when the 'reater speed is less than t&ie o% the lesser speed . 'f S(1) S(2)# then S(1) * 2S(2)+
After the first meeting, if the bodies continue to move they will reach their respective ends and start the return ourney. They will meet again on the return ourney and then proceed further. After the first meeting, they would have covered 2 distane. 1ince the distance has doubled, time taken will also double. 1o, the total distance covered by the bodies for their first, second, third Q nth meetings will be5 L, )L, *L Q and $'n%&L 1o, the total times taken by the bodies for their first, second, third Q nth meetings will be5 L 0 1$%& 1$'&, ) L 0 1$%& 1$'&, * L 0 1$%& 1$'& Q and $'n%& L 0 1$%& 1$'& n case you need to figure out the point at which the bodies meet for the nth time, consider only one of the bodies, say %. Listance covered by % till the nth meeting 3 R1$%&0$1$%&1$'&&SB$'n%&L The remainder of the above when divided by 'L will give you the e!act location of the nth meeting point. Oor e!ample, if the distance covered by % till the nth meeting comes out as :99 meters and L 3 %*9 meters, the nth meeting point can be obtained by the remainder of :990)99 which is %99. 1o, the nth meeting point will be %99 m from . know you would hate me for saying this, but this is not the end of T1L. This is not even the end of motion of two bodies in a straight line. The destination is still far away and hopefully we will reach there in time.
C&T uant: Problems on Es-alators >rom T!me+ S,ee$ an$ D!stan-e
Quant problems on Time, Speed and Distance make frequent appearances in management entrance exams. To my collection of posts on PaGaLGu ! detailed list", # $ould like to add a fe$ t%oug%ts about sol&ing escalator'related problems. (t t%e basic le&el, escalator related problems aren)t too different from boats and streams problem. T%ink of t%e escalator as *ust a replacement for a ri&er, t%e only difference being t%at escalators mo&e in bot% directions $%ereas ri&ers only flo$ do$nstream. # categorise escalator problem into t$o categories,
+. %en one person is mo&ing -. %en t$o people are mo&ing T%e problems are related to eac% ot%er and some of t%em use data from t%e pre&ious question.
Type 1: When one person is moving 1. Ravi takes 40 seconds to walk up on an escalator which is moving upwards but he takes 60 seconds to walk up on an escalator which is moving downwards. ow much time will he take to walk up i! the escalator is not moving" This is e,act-! -i"e a boats and streams prob-em# and we are %iven the times ta"en b! a person rowin% downstream and pstream -et us assume that the speed of avi is rN and the speed of the escalator is !N.
#n t%e first case, $%en t%e escalator is mo&ing up, a&i/s effecti&e speed is r 0 x as t%e motion of t%e escalator is assisting %is mo&ement. #n t%e second case, $%en t%e escalator is mo&ing do$n, a&i/s effecti&e speed is 1r 2 x/ as t%e motion of t%e escalator is a %indrance to %is mo&ement. %en t%e escalator is not mo&ing, %is speed $ould be 1r/. T%e distance co&ered in eac% case is constant. T%e t%ree speeds 1r 0 x/, 1r/ and 1r 2 x/ are in an (rit%metic Progression. 34 T%e times taken $ill be in a 5armonic Progression. 34 Time taken $%en t%e escalator is not mo&ing $ill be t%e 5armonic mean of t%e ot%er t$o times gi&en. 34 Time taken $%en escalator is not mo&ing 3 -678698:!78098" 3 4# seconds.
$. Ravi takes 60 seconds on an escalator which is moving down when he walks down but takes 40 seconds when he runs down. e takes $0 steps
when he walking whereas he takes %0 steps when he is running. What is the total number o! steps in the escalator"
Let us say t%at t%e speed of t%e escalator is 1x/ steps per second. Distance co&ered by a&i is t%e same $%et%er %e is $alking or running. Distance $%en %e is $alking 3 -8 0 98x !98x is co&ered by escalator" Distance $%en %e is running 3 ;8 0 78x !78x is co&ered by escalator" 34 -8 0 98x 3 ;8 0 78x 34 x 3 8.< Total number of steps 3 -8 0 98!8.<" 3 &0 steps
Type $: When two people are moving 1. Ravi and Rakesh are climbing on a moving escalator that is going up. Ravi takes 10 seconds to reach the top but Rakesh takes # seconds to reach the top. This happens because Rakesh is !aster than Ravi. Rakesh takes 4 steps whereas Ravi can take only % steps in one second. What is the total number o! steps in the escalator" This .estion can be interpreted as two peop-e rowin% a boat downstream -et us assume that the escalator moves at the rate of ! steps per second.
Distance co&ered by bot% of t%em is same 34 +8!; 0 x" 3 =!7 0 x" 34 +8x 2 =x 3 ;- 2 ;8 34 x 3 + step per second >umber of steps in t%e escalator 3 +8!; 0 +" 3 = !7 0 +" 3 40 steps Another variation of this problem could be if we were not given the times taken by Rakesh and Ravi but the number of steps that they took to reach the top. Let us look at that problem.
$. Ravi and Rakesh are climbing on a moving escalator that is going up. Ravi takes %0 steps to reach the top whereas Rakesh takes %$ steps !or the same. This happens because Rakesh is !aster than Ravi. Rakesh takes 4 steps whereas Ravi can take only % steps in one second. What is the total number o! steps in the escalator" Before moving ahead, please notice the difference between previous question and this one.
T%e first t%ing t%at $e $ill do is to figure out t%e time taken by akes% and a&i to reac% t%e top. akes% takes ;-:7 3 = seconds $%ereas a&i takes ;8:; 3 +8 seconds. ?rom no$ on, e&eryt%ing is same as t%e pre&ious problem. Let us assume t%at t%e escalator is mo&ing $it% a speed of x steps per second. Distance co&ered by bot% of t%em is same 34 +8!; 0 x" 3 =!7 0 x" 34 +8x 2 =x 3 ;- 2 ;8 34 x 3 + step per second >umber of steps in t%e escalator 3 +8!; 0 +" 3 = !7 0 +" 3 40 steps A slightly more difficult version would be when we do not know the speeds of Rakesh and Ravi but only the ratio. Let us look at that in the next problem.
%. Ravi and Rakesh are climbing on a moving escalator that is going up. Ravi takes %0 steps to reach the top whereas Rakesh takes %$ steps !or the same. This happens because Rakesh is !aster than Ravi. 'or every 4 steps that Rakesh takes( Ravi takes only % steps. What is the total number o! steps in the escalator" f you notice the difference between the previous question and this one, the problem is that now we don!t have a definite amount of time taken so we have to solve this by ratios.
?rom t%e gi&en data, $e kno$ t%at t%e ratio of speeds of akes% and a&i is 7 @ ;
To co&er distance $%ic% is in t%e ratio ;- @ ;8 or +9@+<, t%en $ill take times in t%e ratio 3 +9:7 @ +<:; 3 7 @ < #f t%e total number of steps is n, in case of akes% t%e escalator co&ers 1n 2 ;-/ and in case of a&i t%e escalator co&ers 1n 2 ;8/ T%ese $ill be in t%e same ratio as t%e times taken by akes% and a&i 34 !n 2 ;-":!n 2 ;8" 3 7:< 34
4. Ravi is climbing on a moving escalator that is going up and takes %0 steps to reach the top. Rakesh on the other hand is coming down on the same escalator. 'or every & steps that Rakesh takes( Ravi takes only % steps. *oth o! them take the same amount o! time to reach the other end. a+ What is the total number o! steps in the escalator" b+ What is the di!!erence in the number o! steps that both o! them had taken when they crossed each other"
T%e extra information in t%is question is "Both of them take the same amount of time to reach the other end# and t%at is really t%e key to sol&ing t%is question. Let us assume t%eir speeds are
34 %en a&i takes ;8 steps, t%e escalator takes +8 steps. 34 Total number of steps 3 ;8 0 +8 3 40 steps.
2oth of them would have covered '9 steps when they crossed each other.
a&i going up $ould %a&e taken +< steps, $%ereas t%e escalator $ould %a&e taken < steps for %im. akes% coming do$n $ould %a&e taken -< steps, out of $%ic% t%e escalator $ould %a&e nullified t%e mo&ement of < steps for %im. Difference in t%e number of steps 3 -< 2 +< 3 10 steps. /nother wa! of so-vin% this .estion is#
a&i %as taken ;8 steps in t%e full *ourney, $%ereas akes% %as taken ;86!<:;" 3 <8 steps. T%e difference in t%e number of steps in full *ourney 3 <8 2 ;8 3 -8 steps. T%e difference in t%e number of steps $%en t%ey cross eac% ot%er, $%ic% is exactly %alf of t%e *ourney 3 -8 : - 3 10 steps. # %ope in t%is post # %a&e co&ered most of t%e type of escalator problems $%ic% %a&e been asked and using t%e abo&e ideas and concepts you $ill be able to sol&e suc% problems easily in future. Sol!n" T!me6S,ee$6D!stan-e ,roblems !t#out us!n" e?uat!ons
(Photo credit: 0ichae- a--acher ) guess my first fascination with problems of Time, 1peed and Listance began when first watch a film called "enna . An important portion of the plot, if you can call it that, had ishi apoor floating from ndia to akistan in a river without drowning. remember arguing with my friends
that if he could float that long he could swim back to ndia as well. My friends nullified the argument by saying, 1peed iver ; 1peed ishi apoor
# kno$ t%at t%e reference is a little dated for most readers, but Aeba B%aktiyar made me look beyond reason. #n t%is post $e $ill discuss some of t%e ideas t%at %a&e %elped me sol&e TSD problems $it%out forming too many equations . unda 1@ Aera'e Speed
e kno$ t%at t%e a&erage speed during a *ourney is gi&en by !Total Distance Co&ered" : !Total Time Taken" but t%ere are a fe$ special cases $%ic% mig%t %elp in sol&ing questions, f the distance covered is constant (d 1 3 d 2 3 d 4 3 d n ) in each part of the 5orne!# then the avera%e speed is the Harmonic 0ean of the va-es Speed /v% 3 n 6 (16s 1 7 16s2 7 16s4 7 16s n ) f the time taken is constant (t 1 3 t 2 3 t 4 3 t n ) in each part of the 5orne! then the avera%e speed is /rithmetic 0ean of the va-es Speed /v% 3 (s1 7 s2 7 s4 7sn )6n unda 2: sin' ro'ressions (Arith#eti 6 +ar#oni) n many questions, you will come across a situation when a person is going from point A to point 2 at various speeds and taking various times. 7e know that if distane is onstant, speed and time are inversely proportional to each other. 2ut this information can also be used to deduce the following two facts,
f the varios speeds which are mentioned are in /P# then the correspondin% times ta"en wi-- be in HP f the varios speeds which are mentioned are in HP# then the correspondin% times ta"en wi-- be in /P
Let us use t%ese ideas to sol&e couple of quant questions. *7a#ple 1 Arun, 2arun and iranmala start from the same place and travel in the same direction at speeds of )9, (9 and 69 km per hour respectively. 2arun starts two hours after Arun. f 2arun and iranmala overtake Arun at the same instant, how many hours after Arun did iranmala start#
(Some se-ess information: /rn $arn 8iranma-a is a 19; $an%-adeshi fi-m
5ence, Eiranbala started 7 %ours after (run. *7a#ple 2
is%i Eapoor can s$im a certain course against t%e ri&er flo$ in =7 minutes %e can s$im t%e same course $it% t%e ri&er flo$ in F minutes less t%an %e can s$im in still $ater. 5o$ long $ould %e take to s$im t%e course $it% t%e ri&er flo$ Solution
Let us say Speed of is%i Eapoor in still $ater is E and Speed of t%e ri&er is . 5ence, is%i Eapoors speeds against t%e ri&er flo$, in still $ater and $it% t%e ri&er flo$ are, , and . As you can see, they are in A."ence, the corresponding times taken will be in ". -et us say that the time taken to row down with the stream is t, then 8(, tG a nd t are in ". 1o,
t 7 9 3 (2 = ;> = t) 6 (;> 7 t) ? t 2 7 94t 7 @A 3 1;t ? t 2 @At 7 @A 3 B ? t 3 4 or 12 unda 3: Speial -ase
Let us say t%at t$o bodies a H b start at t%e same time from t$o points P H Q to$ards eac% ot%er and meet at a point in bet$een. (fter meeting at , a takes ta time to reac% its destination !Q" and b takes t b time to reac% its destination !P". T%en, S a 6 S b 3 ?(t b 6 t a) Also, the time taken by a K b to meet $i.e. to reach point from K J respectively& is given by,
t 3 ?$ta B t b&
/thor Ravi Handa has ta%ht Qantitative /ptitde at 0S for > !ears /n a-mns of T 8hara%pr where he stdied a da-de%ree in compter science# he has a-so written a boo" on bsiness awareness T!me s,ee$ $!stan-e + Trains are mo$ing from A to ' and ' to A at regular inter$al of +hr The" complete their >ourne" in 6 hr Ho- man" trains comming from station ' -ill cross the train comming from station A that started at +,am77Assume the trains starts from #oth the stations at the same time olution3++ trainsThe train -hich started at +, am from station A -ill reach station ' at ? pm ourne"= it -ill meet all the trains from ' starting at 6 am 0Train A meets it at +, am at station A itself1 to ?pm at an inter$al of + hr cheers
&ll!"at!ons an$ M!
after a particular time 8nd -orker came thus the road -as completed in +,, da"s lets assume that %Q of -ork -as completed #" single -orker alone num#er of da"s single -orker -orked ; %90:9J1 num#er of da"s 8 -orkers -orked ; 0+,,I%190:9J1 thus 0%90:9J1 F 00+,,I%1908L0:9J111 ;+,, on sol$ing this the ans-er is +,,9JQ
Xipes A and ' can fill a cistern in 8, and ?, minutes and C can empt" it in +6 minutes 5f three are opened and closed one after the other successi$el" for + min each in that order ho- soon the cistern -ill #e filled7 +1 +6, minutes 81 +48 minutes ?1 +4J minutes :1 +J6 minutes A and ' are running a circular tracks in opposite directionthe" meet at a point :6, m from starting point and continued running the" no- meet at a point ?,, m from the starting point #ut in the opposite direction as #efore -hats the length of the track +1 +,,, m 81 +8,, m ?1 +?6, m :1 +6,, m 5n a la#orator"= three containers= A= ' and C= ha$e equal $olumes of different mi%tures of liquid o%"gen and liquid nitrogen The concentration 0#" $olume1 of liquid o%"gen in the three mi%tures is 8,Q= :,Q and .,Q respecti$el" !irst= oneIfifth of the contents of A are poured into ' and then t-oIfifths of the contents of ' are poured into C The final concentration 0#" $olume1 of liquid o%"gen in C is appro%imatel" +1 48:Q 81 4+JQ ?1 4.4Q :1 46Q A>a"= 'hanu and Chandu plan to tra$el from X and reach & at :3,, pm 'hanu starts e%actl" 8: minutes after A>a" started from X Had Chandu started +8 minutes after 'hanu= all of them -ould ha$e reached & at :3,, pm 'ut #efore Chandu left X= he recei$ed a call from A>a" and -as told to start immediatel" to-ards & and at e%actl" the same time A>a" re$ersed his direction and tra$elled to-ards X All the three of them meet at ?38: pm at R 0some-here #et-een X and &1 Each of them tra$elled -ith his o-n uniform speed +1 83+8 81 83?4 ?1 83:.
:1 cant #e determined Ar$il=anil and rihant can completel" sol$e a pro#lem together in : hoursAr$il and rihanth take +6 hours less than anil -orking aloneAnil -orks on t he pro#lem for first 8 hours and then ar$il and rihanth >oin himAfter another t-o hours=anil quits5n ho- man" hours is the pro#lem actuall" sol$ed7 +1 81 : ?1 6 :1 4 61 . The diluted -ine contains onl" . litres of -ine and the rest is -ater A ne- mi%ture -hose concentration is ?,Q=is to #e formed #" replacing -ineHo- man" litres of mi%ture shall #e replaced -ith pure -ine if there -as initiall" ?8 litres of -ater in the mi%ture70use #ucket method for sol$ing these t"pe of mi%tures 1 +1 : 81 6 ?1 . :1 none of these The ratio of e%penditure and sa$ings is ?385f the income increases #" +6Q and the sa$ings increases #" 4Q=then #" ho- much percentage should his e%penditure increases7 +1 86 81 8+ ?1 88 :1 8: A rail-a" track runs parallel to a road and a c"clist -hose speed is +8kmph meets a trains at the crossing=same time e$er"da" One da" c"clist started 86min late and met the train 4km ahead of the crossing What is the speed of the train7
pl/ e%plain
+1 J8kmph 81 4,kmph ?1 4kmph A and ' are participating in a race around a circular track of +,,,m The race is of total +,kms A completes one round in 8,,sec=' completes one round in :,,sec After ho- much time -ill A meet ' for the last time7 +1 8,,,sec 81 +,,,sec
?1 .,,sec :1 ?,,,sec A car co$ers the first half of the distance #et-een t-o places at :, km9hr and the second half of the distance at 4, km9hr so -hat is the a$erage speed of the car7 a :6 km9hr #:. km9hr c6, km9hr d4, km9hr 5n a kilometer race= if A gi$es ' a :, m start= A -ins #" + s 'ut if A gi$es ' a ?, s start= ' -ins #" :, m !ind the time taken #" ' to run 6,,, m 7 +1 +6, s 81 :6, s ?1 J6, s :1 .86 s 61 46, s the ratio of quantit" of fuel consumed #" an aeroplane0-hen tra$elling the same distance1 at ,, km9hr= J6, km9hr and 6,, km9hr is ?383+ -hat -ill #e the distance of the aeroplane tra$elled on ? litres of fuel at ,,km9hr= : lit of fuel at J6,km9hr and 4 lit of fuel at 6,, km9hr77 pl/ gi$e e%planation +1 +383? 81 +3?36 ?1 +3?34 :1 +3834 p adn & can do a certain -ork together in +8 da"s= & and R in 8, da"s= and p nad R in +6da"s X=& nad R start -orking together The" -ork for 8 da"s after -hich & lea$es After +, more da"s= & re>oins and X lea$es & -orks for 8 da"s= along -ith R and then he lea$es the remaining -ork is completed #" R 5n ho- man" da"s -ll the -ork #e completed7 +1 +8 da"s 81 +4da"s ?1 8,da"s :1 8:da"s A starts from home for his office He tra$els do-nhill= then on flat ground and then uphill to reach his office
C&T uant: Sol!n" C!r-ular Mot!on Problems
$%hoto& $Dr"r Si%rb5Drnsson )
( common question template in t%e Time, Speed and Distance topic of t%e quantitati&e ability section in t%e Common (dmission Test !C(T" is based on t$o or t%ree ob*ects mo&ing around a circular track. Let us assume t%at t%e ob*ects %a&e speeds 1a/, 1b/ and 1c/ and are mo&ing on a track of lengt% 1L/. Gi&en belo$ is a set of formulae t%at can be used for sol&ing suc% questions. # %a&e assumed t%at a 4 b 4 c.
These formulae use the concepts of relative speed. %. +ase % K ) ? 7hen two bodies are moving in the same direction, their relative speed is $ab&. '. +ase ' ? 7hen two bodies are moving in opposite direction or towards each other, their relative speed is $a b&. ). Meeting at the starting point depends on when a particular body comes back at the starting point and not on the direction.
(. Oor comple! cases, like two bodies are moving clockwise and one is anticlockwise, ust use the concept of relative distance 0 relative speed.
,umber o! distinct points at $%ic% t%ey !t%e bodies in circular motion"
$ill meet can be determined by finding out t%e reduced ratios o! their speeds -a:b+ . %. f they are moving in the same direction, then they will meet at a ! distinct points on the track. '. f they are moving in opposite directions, they will meet at a 9 ! distinct points on the track. ). All these points will be equidistant from each other and will include the starting point.
?or example, if t$o bodies are mo&ing at speeds of F m:sec and +< m:sec, t%e ratio of t%e speeds is ;@<. T%ey $ill meet at < 2 ; 3 - distinct points $%en mo&ing in same direction and < 0 ; 3 = distinct points in opposite direction. #f t%ere are t%ree bodies in motion t%en to find out t%e number of distinct points on t%e track at $%ic% t%ey $ill meet, you $ill first need to find out pair$ise distinct meeting points. T%e o&erall ans$er $ill be t%e 5ig%est Common ?actor of t%e pair$ise &alues. Ixample@ suppose (, B and C are running $it% speeds of <, J and ++ on a circular track. ( and B are running clock$ise, $%ereas C is running anti'clock$ise. ( and B $ill %a&e J 2 < 3 - distinct meeting points. ( and C $ill %a&e < 0 ++ 3 +9 distinct meeting points. B and C $ill %a&e J 0 ++ 3 += distinct meeting points. (, B and C $ill %a&e 5C? !-, +9, +=" 3 - distinct meeting points. hope that in this post have covered most of the type of circular motion problems which get asked and using the above ideas and concepts you will be able to solve such problems easily in future.
Sol!n" @r!$ C#essboar$ relate$ ,roblems >or C&T
Problems based on grids:c%essboards %a&e been asked in t%e C(T and ot%er KB( entrance exams o&er t%e years. T%ey seem really difficult $%en you encounter t%em for t%e first time but once you get t%e %ang of t%ings t%ey
become really simple. T%e key lies in understanding t%e basic concepts in&ol&ed. T%e most common grid structure t%at $e are all familiar $it% is t%e c%essboard. Let us look at some of t%e common questions based upon grid. Q1 %at is t%e number of squares on a c%essboard
Squares of sie +x+ 3 =M- 3 97 Squares of sie -x- 3 JM- 3 7F . .. Squares of sie =x= 3 +M- 3 + Total number of squares 3 1 9 4 9 ; 4; 9 <4 / 204 Eo co-d have a-so sed the form-a 1F2 7 2F2 7 4F2 G nF2 3 n(n71)(2n71)6 n this case# !o wo-d have %ot 3 ;=9=1@6 3 12=1@ 3 2B> Q2 %at is t%e total number of rectangles on a c%essboard
To form a = x = c%essboard, $e need F %oriontal and F &ertical lines. #f $e select any - lines from t%e F %oriontal lines and any - lines from t%e F &ertical lines, $e $ill get a rectangle Total number of rectangles 3 ;-2 = ;-2 / 3<=3< / 12;< Q3 #n %o$ many $ays can you place - rooks on a c%essboard suc% t%at t%ey are not in attacking positions T%e first rook can be placed in 97 $ays T%e second rook cannot be placed in t%e same ro$ or t%e same column. So, it %as J ro$s and J columns left for it. #t can be placed in 7F $ays. But t%e order in $%ic% t%e rooks are placed is not important. So, it $ill be di&ided by -N Total $ays 3 <4=4;.2 / 15<8 Q4 #n %o$ many $ays can you place = rooks on a c%essboard suc% t%at t%ey are not in attacking positions
T%e first rook can be placed in 97 $ays. T%e second rook can be placed in 7F $ays. . . T%e eigt% rook can be placed in + $ay But t%e order in $%ic% t%e rooks are placed is not important. So, it $ill be di&ided by =N Total $ays 3 <4=4;=3<>4=1.8? / 8? / 40320 Q5 #n %o$ many $ays is it possible to c%oose a $%ite square and a black square on a c%ess board so t%at t%e squares must not lie in t%e same ro$ or column @-A" 2002 T%e $%ite square can be c%osen in ;- $ays. #f $e remo&e t%e ro$ and t%e column $%ic% contains t%e c%osen $%ite square, $e $ill be left $it% J o$s H J Columns containing a total of 7F squares !-7 black and -< $%ite". e $ould %a&e remo&ed +< squares !J $%ite and = black" equired $ays 3 32=24 / B<8 Q< #n %o$ many $ays can you go from ( to B !S%ortest pat%"
To go from ( to B, you need to make 7 rig%t mo&es !" and 9 up mo&es!O". ne of t%e possible pat%s could be OOOOOO (ny rearrangement of t%e abo&e $ould gi&e you a different s%ortest pat% from ( to B. OOOOOO can be rearranged in 10?.(4?
To go from ( to B, you $ill %a&e to follo$ t%e pat% (PQB ( to P can be done in 7C- 3 9 $ays !Select - rig%t mo&es from a total of 7 mo&es" P to Q is only one $ay !T%e pat% t%roug% t%e red rectangle"
Q to B can be done in ;C+ 3 ; $ays !Select + rig%t mo&e from a total of ; mo&es" Total $ays 3 < 7 1 7 3 / 18 &ays Q8 #n %o$ many $ays can you go from ( to B !S%ortest pat%"
Total $ays !if t%e red rectangle $as not t%ere" 3 +8C7 3 -+8 PQ is t%e road $%ic% %as been remo&ed. #t $ould make all t%e routes $%ic% included t%e road PQ as in&alid. So, all routes consisting of (PQB $ill be in&alid ( to P 3
Q; >eelam rides %er bicycle from %er %ouse at ( to
%er office at B, taking t%e s%ortest pat%. T%en t%e number of possible s%ortest pat%s t%at s%e can c%oose is
Let us call t%e road t%roug% park as K>. T%e pat% >eelam $ill take is (K>B. >eelam can go from ( to K in 7C- 3 9 $ays. K to > in + $ay > to B in 9C- 3 +< $ays. >umber of possible s%ortest pat%s 3 <=15 / ;0 Q10. >eelam rides %er bicycle from %er %ouse at ( to %er club at C, &ia B taking t%e s%ortest pat%. T%en t%e number of possible s%ortest pat%s t%at s%e can c%oose is
>eelam can go from ( to B in F8 $ays. ?rom B to C &ia > 3 9 !and not &ia K" ?rom B to C &ia K 3 J $ays #n all ;0=(<9B) / 11B0 &ays # %ope you en*oyed t%is post and if you get a grid based question in C(T, you $ill be able to crack it. Ae "ot t#e Poer: Work!n" !t# Ba-tor!als >or C&T 2;11 ?uant
e all kno$ $%at factorials !nN" are. T%ey look friendly and %elpful but looks can be decei&ing, as many quant problems %a&e taug%t us. #t is probably because ?actorials are simple looking creatures, t%at most students prefer attempting questions based on t%em rat%er t%an on Permutation H Combination or Probability. # $ill co&er PHC and Probability in a later article but in todays post # $ould like to discuss some fundas related to factorials, $%ic% as a matter of fact form t%e basis of a large number of PHC and Probability problems.
1ome of the factorials that might speed up your calculation are5
B 3 1I 1 3 1I 2 3 2I 4 3 I > 3 2>I A 3 12BI 3 @2BI @ 3 AB>B unda 1: i'ht#ost non$Cero di'it o% n? or (n?)
(n) 3 Jast Ki%it of L 2 a , (a) , (b) M where n 3 Aa 7 b
Ixample@ %at is t%e rig%tmost non'ero digit of ;JN !;JN" 3 Last Digit of - J x !JN" x !-N" R !;JN" 3 Last Digit of = x 7 x - R 3 7
E!ample5 7hat is the rightmost nonUero digit of %)(V # $%)(V& 3 -ast Ligit of R ' '6 ! $'6V& ! $(V& S
!+;7N" 3 Last Digit of 7 x !-9N" x 7 R
e need to find out !-9N" 3 Last Digit of -< x !
unda 2: o&er o% a pri#e p in a %atorial (n?)
The bi%%est power of a prime p that divides n (or in other words# the power of prime p in n) is %iven b! the sm of .otients obtained b! sccessive division of n b! p
Ixample@ %at is t%e %ig%est po$er of J t%at di&ides +;7-N R%)(' 0 :S 3 %G% R%G% 0 :S 3 ': R': 0 :S 3 ) ower of : 3 %G% ': ) 3 ''%
Ixample@ %at is t%e %ig%est po$er of 9 t%at di&ides +;7N As 6 is not a prime number, we will divide it into its prime factors. ) is the bigger prime, so its power will be the limiting factor. "ence, we need to find out the power of ) in %)(V
+;7:;R 3 77 R((0)S 3 %(
+7:;R 3 7 R(0)S 3 %
Po$er of ; in +;7N 3 77 0 +7 0 7 0 + 3 9; E!ample5 7hat is the highest power of G that divides %)(V #
(s F is not a prime number, $e $ill di&ide it into its prime factors. F is actually ;-. T%e number of ;s a&ailable is 9;, so t%e number of Fs a&ailable $ill be 9;:-R 3 ;+. 5ig%est po$er of F t%at di&ides +;7N is ;+. 5ig%est po$er of += and ;9 $ill also be ;+. 5ig%est po$er of -J $ill be 9;:;R 3 -+.
<mber of Neroes is %iven b! the sm of the .otients obtained b! sccessive division of n b! A
T%is is actually an extension of ?unda +. >umber of ending eroes is not%ing else but t%e number of times nN is di&isible by +8 or in ot%er $ords, t%e %ig%est po$er of +8 t%at di&ides nN. +8 is not a prime number and its prime factors are - and <. < becomes t%e limiting factor and leads to t%e abo&e'mentioned idea. Ixample@ %at is t%e number of ending eroes in +;7N R%)(0*S 3 '6 R'60*S 3 * R*0*S 3 %
>umber of ending eroes 3 -9 0 < 0 + 3 ;# %ope t%at t%is gets you started $it% factorials and you mig%t start singing this song.
E ail(Dail)ed by Mavericks in Odds and *nds H %) Ianuary /%(
+AT results are ust few hours away from now and many of us would be going through different sets of emotions, which is understandable. Many of us would have put in a lot of effort and their high level of e!pectation is very much anticipated and obvious.
who dont even make it to 'nd round# canNt say about others but this is what i will go through, if i canNt make it to ne!t level. will feel bad, in fact very very bad. will even cry and there is no harm in crying. t will take weeks, even a month or so to get back to normal, but thats ok. My personal and professional life both will be impacted. donNt know if i did ustice with my preparation or not but failure is something that no one likes to have. will even blame the normaliUation and diversity factor. will blame my state board for being so harsh on me. will curse myself for wasting time on unnecessary things in past and not putting that into study. will start doubting my abilities, my preparation strategy if had any, my dedication, commitment everything. 1o, what can do fr om here# And 7hat need to do from here# know canNt do anything about my average academic record, normaliUation and diversity factor. 2ringing my so called failure to my personal and professional life is also not a good thing to do and it will further impact me. 1o will try to move on as quickly as possible from my failure in +AT although it will take some time, but will have to do that. will then start looking at my loop holes, weak points, where missed, what went wrong, how can improve myself, my strengths, weaknesses and many other things. will not let my dream to die. There are two factors, one those are controllable like my hard work, dedication and commitment and then there are things on which donNt have any control like academics, normaliUation factor and diversity factor etc. 1o will not waste my energy and time on things which cannot control rather will focus only on things which can control. To be very honest can only try and hope for the best and thatNs all anyone can do. n terms of ". 1tanley Iudd ? WLon/t be afraid to fail. Lon/t waste energy trying to cover up failure. -earn from your failures and go on to the ne!t challenge. t/s < to fail. f you/re not failing, you/re not growing. X
T)C9
Dou need to understand one simple concept f A can do a ob in %9 day then in one day A can do %0%9th of ob
SHO)TC(T 2est trick that use in e!ams myself is by finding the efficiency of workers in percent. f A can do a ob in ' days then he can do *9@ in a day.
n
+9n
+,,9n
n 1/n 1;;/n 1 1/1 1;; 2 1/2 ; 3 1/3 3333 4 1/4 2 1/ 2; 1/ 1 F 1/F 142G G 1/G 12 1/ 1111 1; 1/1; 1; 11 1/11 ;
NOW 'ETAS SO'E (ESTONS WTH THS T)C9 uest!on 6 & take $a0s to -om,lete a Iob an$ 7 takes 1; $a0s to -om,lete t#e same Iob n #o mu-# t!me t#e0 !ll -om,lete t#e Iob to"et#er J Solut!on 6 &As e>>!-!en-0 = 2;+ 7As e>>!-!en-0 = 1; > t#e0 ork to"et#er t#e0 -an $o 3; o> t#e Iob !n a $a0 To -om,lete t#e Iob t#e0 nee$ 333 $a0s uest!on 6 & !s t!-e as e>>!-!ent as 7 an$ -an -om,lete a Iob 3; $a0s be>ore 7 n #o mu-# t#e0 -an -om,lete t#e Iob to"et#er J Solut!on 6 'et e>>!-!en-0 ,er-enta"e as < &As e>>!-!en-0 = 2< an$ 7As e>>!-!en-0 = < & !s t!-e e>>!-!ent an$ -an -om,lete t#e Iob 3; $a0s be>ore 7 So+ & -an -om,lete t#e Iob !n 3; $a0s an$ 7 -an -om,lete t#e Iob !n ; $a0s &As e>>!-!en-0 = 1/3; = 333 7As e>>!-!en-0 = 1/; = 1 7ot# -an $o . 333 1 o> t#e Iob !n 1 $a0 So t#e -an -om,lete t#e #ole Iob !n 2; $a0s .1;;/
uest!on 6 & tank -an be >!lle$ !n 2; m!nutes T#ere !s a leaka"e #!-# -an
em,t0 !t !n ; m!nutes n #o man0 m!nutes tank -an be >!lle$J Solut!on 6 Met#o$ 1 E>>!-!en-0 o> >!ll!n" ,!,e = 2; m!nutes = 1/3 #our = 3;; E>>!-!en-0 o> leaka"e = ; m!nutes = 1;; We nee$ to $e$u-t e>>!-!en-0 o> leaka"e so >!nal e>>!-!en-0 !s 2;; We are tak!n" 1;; = 1 Hour as base so anser !s 3; m!nutes (,$ate 6 ;6;62;13 . &s S#ob#na an$ &s!n are >a-!n" ,roblem !n sol!n" t#!s ?uest!on+ am sol!n" t#!s ?uest!on !t# se-on$ met#o$ #!-# !s also er0 eas0+ #o,e t#!s !ll make t#e solut!on lot eas!er Met#o$ 2 E>>!-!en-0 o> >!ll!n" ,!,e = 1;;/2; = E>>!-!en-0 o> leaka"e ,!,e = 1;;/; = 1 Net >!ll!n" e>>!-!en-0 = 333 So tank -an be >!lle$ !n = 1;;/333 = 3; m!nutes *ou -an -#an"e t#e base to m!nutes or een se-on$s *ou -an sole eer0 t!me an$ ork ?uest!on !t# t#!s tr!-k n aboe e t!me
*ou -an >!n$ more tr!-ks l!ke t#!s !n ?uant!tat!e a,t!tu$e se-t!on Comment belo !n -ase o> an0 ?uer0+ ,rom!se to re,l0 !t#!n 24 #ours (,$ate ; O-tober 2;13 6 uest!on re?ueste$ b0 C#!tra Sal!n uest!on 6 4 men an$ omen ork!n" to"et#er -an -om,lete t#e ork !t#!n 1; $a0s 3 men an$ F omen ork!n" to"et#er !ll -om,lete t#e same ork !t#!n G $a0s n #o man0 $a0s 1; omen !ll -om,lete t#!s ork J Solut!on 6 'et number o> men =<+ number o> omen = 0 E>>!-!en-0 o> 4 men an$ omen = 1;;/1; = 1; so+ 4<0 = 1; &boe e?uat!on means 4 men an$ omen -an $o 1; o> a t#e Iob !n one $a0 E>>!-!en-0 o> 3 men an$ F omen = 1;;/G = 12 so+ 3<F0 = 12 70 sol!n" bot# e?uat!ons e "et+ < = 6; an$ 0 = 2 E>>!-!en-0 o> 1 oman.0 = 2 ,er $a0 E>>!-!en-0 o> 1; omen ,er $a0 = 2; So 1; omen -an -om,lete t#e Iob !n 1;;/2; = $a0s (,$ate 1161162;13 6 uest!on re?ueste$ b0 Pra!s0 uest!on 6 & an$ 7 to"et#er -an -om,lete a task !n 2; $a0s 7 an$ C
to"et#er -an -om,lete t#e same task !n 3; $a0s & an$ C to"et#er -an -om,lete t#e same task !n 3; $a0s W#at !s t#e res,e-t!e rat!o o> t#e number o> $a0s taken b0 & #en -om,let!n" t#e same task alone to t#e number o> $a0s taken b0 C #en -om,let!n" t#e same task aloneJ Solut!on 6 E>>!-!en-0 o> & an$ 7 = 1/2; ,er $a0 = ,er $a0 KKKKKKKKKKKKKKKK1 E>>!-!en-0 o> 7 an$ C = 1/3; ,er $a0 = 333 ,er $a0KKKKKKKKKKKKKK2 E>>!-!en-0 o> C an$ & = 1/3; ,er $a0 = 333 ,er $a0KKKKKKKKKKKKKK3 Tak!n" e?uat!on 2 an$ 3 to"et#er 7 C = 333 an$ C & = 333 C an$ 333 !ll be remoe$ Hen-e & = 7 E>>!-!en-0 o> & = 7 = /2 = 2 = 1/4; E>>!-!en-0 o> C = 333 6 2 = ;G33 = 1/12; & -an $o t#e Iob !n 4; $a0s an$ C -an $o t#e Iob !n 12; $a0s #e t#e0 ork G/2;14 T!me an$Work6 S#ort-uts an$Tr!-ks 5 7ankE
)at!o o> number o> $a0s !n #!-# & an$ C -an -om,lete t#e Iob 1:3
e$!- Mat#emat-!s61 Ti(s To %emember Some alues
he ,alues of reciprocal percentages %E( for 8 is e2actly half that for 0 %half of 00.00 ! 18.88( he E for > is e2actly half of ' %half of ;5 ! 1;.5( e,en is easy to remember 4ust = into ; %1'($ followed by 1' into ; %;>( which makes it 1'.;> : is one-third of 0 %00.00 di,ided by 0 ! 11.11( Elease start with the ne2t ten only after becoming absolutely comfortable with the first ten Number
%eci(rocal
ecimal
Percentage
11
1#11
<.<:<:<
0+0
12
1#1;
<.<>00
.+33
13
1#10
<.<==
$+$
14
1#1'
<.<=1'
$+14
1"
1#15
<.<888
#+##
1#
1#18
<.<8;5
#+2"
1$
1#1=
<.<5>>
"+..
1.
1#1>
<.<555
"+""
10
1#1:
<.<5;8
"+2#
2
1;2
+"
"
By now you would ha,e figured out that the difficult ones are the prime numbers. We ha,e already dealt with =. 6ow we need to work out 11$ 10$ 1=$ 1:$ ;0 and ;:. : and 11 are interrelated as 1#: is 11.11 and 1#11 is :.<:
10 is considered unlucky. he way you remember it is through the year 1:==$which pro,ed unlucky for Indira +andhi and )oca )ola. All e,en numbers can be worked out by di,iding the E for the number that was their half or uarter by two or four respecti,ely. or e2ample$ 1; is half of 8 %half of 18.88 ! >.00( Workout the rest of the primes and your own uniue way to remember them. We cannot emphasiQe the importance of ha,ing the percentages of the 1st 0< reciprocals on your fingertips.
Number
%eci(rocal
ecimal
Percentage
21
1#;1
<.<'=8
4+$#
22
1#;;
<.<'5'5
4+"4
23
1#;0
<.<'0'
4+34
24
1#;'
<.<'18
4+1#
2"
1#;5
<.<'
4
2#
1#;8
<.<0>'
3+.4
2$
1#;=
<.<0=<
3+$
2.
1#;>
<.<05=
3+"$
20
1#;:
<.<0''
3+44
3
1#0<
<.<000
3
e$!- Mat#emat!-s /om(osite Table
Sl+No
%
< %
S=uare
/ube
>ourth
S=uare
/ube
term
%oot
%oot
1
1.<<
1<<
1
1
1
1.<<
1
2
<.5<
5<.<<
'
>
18
1.'1
1+2"
3
<.00
00.00
:
;=
>1
1.=0
1+44
4
<.;5
;5.<<
18
8'
;58
;.<<
1+".
"
<.;<
;<.<<
;5
1;5
8;5
;.;'
1+$
#
<.1=
18.8=
08
;18
1;:8
;.'5
1+.1
$
<.1'
1'.;:
':
0'0
;'<1
;.85
1+01
.
<.10
1;.5<
8'
51;
'<:8
;.>0
2
0
<.11
11.11
>1
=;:
8581
0.<<
2+.
1
<.1<
1<.<<
1<<
1<<<
1<<<<
0.18
2+1"
e$!- Mat#emat!-s
e$!- Mat#emat!-s ase Method or *ultiplicationhis is ,ery suitable when numbers are close to a base like 1<$ 1<<$ 1<<< or so on. LetPs take an e2ample7 1<8 2 1<> @ere the base is 1<< and the PsurplusP is 8 and > for the two numbers. he answer will be found in two parts$ the right-hand should ha,e only two digits %because base is 1<<( and will be the product of the surpluses. hus$ the right-hand part will be 8 % >$ i.e. '>. he lefthand part will be one multiplicand plus the surplus of the other multiplicand. he left part of the answer in this case will be 1<8 & > or for that matter 1<> & 8 i.e. 11'. he answer is 11''>. 1; 3 1'. 1< would the most suitable base. In the current e2ample$ the surplus numbers are &; and &'. If >2= were to be performed and base of 1< were chosen$ then -; and -0 would ha,e been the deficit numbers. ry the following numbers %a( 10 3 18 %b( 18 3 1> %c( 1> 3 1: %d( ;; 3 ;' Dnce you get comfortable$ do not use any paper or pen. ?S,N@ TB5% AS5S
In '8 3 '>$ the base chosen is 5< and multiplication of '' by 5< is better done like this7 take the half of '' and put two Qeros at the end$ bec ause 5< is same as 1<<#;. herefore$ product will be ;;<<. It would be lengthy to multiply '' by 5 and put a Qero at the end. In general$ whene,er we want to multiply anything by 5$ simply hal,e it and put a Qero. *ultiply 0; by ;5. *ost of the students would take 0< as the base. he method is correct but nonetheless lengthier. Better techniue is to understand that ;5 is same as one-fourth. herefore$ one-fourth of 0; is > and hence the answer is ><<. An application of Base *ethod to learn multiplications of the type 0;0>$ where unitPs digit summation is 1< and digits other than unitPs digit are same in both the numbers. In the abo,e e2ample$ ; & > ! 1< and 0 in 0; is same as 0 in 0>. herefore method can be applied. he method is simple to apply. he group of digits other than unitPs digit$ in this case 0$ is multiplied by the number ne2t to itself. herefore$ 0 is multiplied by ' to obtain 1;$ which will form the left part of the answer. he unitPs digits are multiplied to obtain 18 %in this case($ which will form the right part of the answer. herefore$ the answer is 1;18. ry these now 50 3 5= :1 3 :: 1<8 3 1<' 1;0 3 1;= he rule for suares of numbers ending with 5 . e.g.$ 85;. his is same as 85 3 85 and since this multiplication satisfies the criteria that unitPs digit summation is 1< and rest of the numbers are same$ we can apply the method. herefore$ the answer is '; # ;5 ! ';;5. ry these7 05; :5; 1;5; ;<5; /?,N@ inding the cubes of numbers close to the powers of 1<. e.g.$ cubes of ::>$ 1<<'$ 1<<<1;$ 1<<<=$ ::8$ ::>>$ etc. ome of the numbers are in surplus and others are in deficit. 2plain the method as gi,en below. ind %1<<<'(0 tep %I( 7 Base is 1<<<<. Ero,ide three spaces in the answer. he base contains ' Qeros. @ence$ the second and third space must contain e2actly ' digits. 1 < < < ' ! ?# ?# ? tep %II( 7 he surplus is %&'(. If surplus is written as PaP$ perform the operation P0aP and add to the base 1<<<< to get 1<<1;. Eut this in the 1st space. 1 < < < ' ! 1 < < 1 ; #?#? tep %III( 7 he new surplus is %&1;(. *ultiply the new surplus by the old surplus$ i.e. %&'( %&1;( ! %&'>(. According to the rule written in the step %I($ '> is written as <<'>. 1 < < < ' ! 1 < < 1 ; # < < ' > #? tep %IC( 7 he last space will be filled by the c ube of the old surplus %&'(. herefore$ '0 ! 8'$ which is written as <<8'. 1<<<'!1 <<1;#<<' >#<<8' herefore$ the answer is 1<<1;<<'><<8'. ind %::>(0 tep %I( 7 Base ! 1<<<. @ence$ e2actly 0 digits must be there in the ;nd and 0rd space.he deficit ! %&;( : : > ! ?#?#? tep %II( 7 *ultiply the deficit by 0 and subtract %because this is the case of deficit( from the base. : : > ! : : ' #?#? tep %III( 7 %old deficit( 2 %new deficit( ! ; 2 8 ! 1;
: : > ! : : ' # < 1 ; #? tep %IC( 7 he cube of the old deficit ! >. ince it is the case of deficit$ -> s hould be written. All that you need to do to write the negati,e number in the third space is to find the complement of the number$ in this case >. But since the third space must ha,e e2actly 0 digits$ the complement of <<> must be calculated. he complement of <<> is ::;. KonPt forget to reduce the last digit of the seco nd space number by 1 ::>!::' #<1;#:: ; -1 ::'#<11#::; herefore$ the answer is ::'<11::; As an e2ercise$ try the following 7 ::::'0 ! : : : > ; # < < 1 < > # < < ; 1 8 ! :::>;#<<1<=#::=>' 1<<<50 ! 1 < < 1 5 # < < = 5 # < 1 ; 5 ! 1<<15#<<=5#<1;5 1<<15#<<=5# <1;5 1<<<;50 ! 1 < < < = 5 # < 1 > = 5 # 1 5 8 ; 5 ! 1<<<=5#<1>=5#158;5 :::::>>0 ! : : : : : 8 ' # < < < < ' 0 ; # < < < 1 = ; > ! :::::8'#<<<<'01#:::>;=; Multi(l'ing numbers Cust o&er 1+ 1. 1<0 2 1<' ! 1<=1; he answer is in two parts7 1<= and 1;$ 1<= is 4ust 1<0 & ' %or 1<' & 0($ and 1; is 4ust 0 2 '. ;. imilarly 1<= 2 1<8 ! 110'; 1<= & 8 ! 110 and = 2 8 ! '; >ew more ,deas Again$ 4ust for mental arithmetic $uick way to suare numbers that end in 5 using the formula B D6 *D @A6 @ D6 BD. 1. =5; ! 58;5 =5; means =5 2 =5. he answer is in two parts7 58 and ;5 . he last part is always ;5. he first part is the first number$ =$ multiplied by the number Sone moreS$ which is >7 so = 2 > ! 58 ;. 0; 2 0> ! 1;18 Both numbers here start with 0 and the last figures %; and >( add up to 1<. o we 4ust multiply 0 by ' %the ne2t number up( to get 1; for the first part of the answer. And we multiply the last figures7 ; 2 > ! 18 to get the last part of the answer. Method *or di&ing b' 0+ 1. ;0 # : ! ; remainder 5 he first figure of ;0 is ;$ and this is the answer. he remainder is 4ust ; and 0 added upM ;. '0 # : ! ' remainder = he first figure ' is the answer and ' & 0 ! = is the remainder r emainder - could it be easier? 0. 10' # : ! 1' remainder > he answer consists of 1$' and >. 1 is 4ust the first figure of 10'. ' is the total of the first two figures 1& 0 ! '$ and > is the total of all three figures 1& 0 & ' ! >
ill now$ you were multiplying like this7 Xuestion7 *ultiply '0; by 81=. Answer7 '0; 2 81= 0<;' '0; ;5:; ;885'' *ore the number of digits in the numbers$ more lines and time you consume. 6o moreM "sing the utra SCertically and )rosswiseS$ you ha,e tep 1 %mentally$ donPt write on notebook( 7 ,ertically %last digits( 7 ;2=!1'G write ' carry 1 tep ; %mentally( 7 crosswise %last two digits( 7 02= &;21 ! ;0 &carry 1 ! ;'G write ' carry ; tep 0 7 ,ertically and crosswise %three digits( 7 '2= & 021 &;28 ! '0 &carry ; ! '5G write 5 carry ' tep ' 7 %mo,e leftG first two digits( 7 '21 &028 ! ;; &carry ' ! ;8G write 8 carry ; tep 5 7 %mo,e leftG first digit of each number( 7 '28 ! ;' &carry ; ! ;8. nd. Write answer 7 ;885'' his is how it appears on notebook 7 '0; 2 81= ;885'' *ultiply ;0=8 by '<8< ;0=8 2 '<8< 82< ! &carry 1 ! :G write :. nd. Answer is :8'858< Submitted b' !itesh" b' !itesh"
roblems on Time and Work: /T uantitative ptitude Publ!s#e$ on: 'GMAD'9%' This article takes "ou through the pro#lems of Time= Work Wages This is i s an important topic and CAT has gi$en questions in the past e%ams= though as is the practice in CAT= the questions ha$e not #een directl" #ased on the formulae (et us go through the follo-ing pro#lems and solutionsY E
5n this case= A alone -orks for 8 da"s #efore ' >oins him= -hich means if A2s efficienc" is taken for first 8 da"s= -e get
o= the remaining
therefore Time taken together
this -ill #e less tha 4 da"s1 Thus= the total required time E
+P 6 8P +, ?P +6 :P 8, Solut!on: This is a trick"= #ut a $er" eas" pro#lem Had all the -orkers -orked and no -orker -as -ithdra-n= then the -ork -ould ha$e finished in 66Q of the time it actuall" took to finishZ this means3 5f the -ork -as to #e finished -ithout -ithdra-ing an" man it -ould take 66 ManI
+,, ManI
0+, ManI
On da" 8 there are men -orking
0 ManI
On da" ? there are . men -orking
0. ManI
3
3
3
3
3
3
On da" + there is + man -orking
0+ ManI
This means the contractor had a total of +, men in his groupTherefore= the correct option is 8P
2earn and understand problems on Relative 3otion: /T uantitative ptitude This article focuses on the concept of )Relati$e Motion 'et-een T-o 'odies2 uch questions ha$e #een part of the CAT paper and should not #e neglected as the" are reall" eas" to score on 7as!- Con-e,ts .T!me+ S,ee$ an$ D!stan-e 5f the speed of a #od" is changed in the ratio %3"= then the time taken #" the #od" is changed in the ratio "3%= assuming the distance is constant 5f the speed remains constant and the time for -hich a #od" is in motion is changed in a certain ratio %3" then the distance is changed in the same ratio %3" 5n a fi%ed time and gi$en a change in distance in the ratio %3"= the speed is also changed in the ratio %3"
Change in peed3At times during a >ourne"= the speed of a #od" -ill $ar" 0increase or decrease1 5n this case= the >ourne" should #e #roken do-n into sections Each section of the >ourne" should #e such that the speed of the #od" in that section is a constant
Average 1peed5f the speed changes several times during a ourney, then we introduce the idea of average speed. The formula for average speed, which we will be denote by 1peed avg or 1avg is,
)elat!e S,ee$ The idea of relati$e speed arises -hen t-o #odies are in motion The relati$e speed= -hich is denoted #" r = is defined as follo-s3
(et A and ' #e the t-o #odies (et their speeds #e A and ' respecti$el" Then= 0i1
r ; A F '= if the" are tra$elling in opposite directions0in the opposite direction the
relati$e speed or effecti$e speed #et-een t-o #odies is the sum of their speeds1 0ii1
r ; A I '= if the" are tra$elling in the same direction0in the same direction the relati$e
speed or effecti$e speed #et-een t-o #odies is the difference of their speeds1
Note:The concept of )to and fro& motion in a straight line= is also an e%tension of the concept of relati$e motion #et-een t-o #odies E
Solut!on: The cars tra$el in the same directionThe relati$e speed is +8km per hour Thus= the speed of the slo-er car is +4: I +8 ; +68 km per hour
T!me an$ Work: %ule : If A can do a piece of work in K days$ then APs 1 dayPs work ! 1#K %ule : If APs work efficiency is ; times as compared to B then atio of work done by A and B is ; 71 atio of times taken by A and B to finish a work ! 1 7 ; %in,erse ratio( 5-+ A does a work in 1< days and B does the same work in 15 days. In how many days they together will do the same work ?
o both will finish the work in 8 days 5-+ If A and B together can do a 4ob in 2 days and A alone can do the same 4ob in y days$ then how many days it taken by B to complete the same 4ob ?
%ule : Let *1 person can do W1 work in K1 days with time 1 and *; person can do W; work in K; days with time ;$ then the relationship can be written as 7
5-+ If '< people can make 8< toys in > hrs$ if > people lea,e the work$ how many toys can make in 1; hrs ? olution 7 here *1! '<$ K1! > hrs$ W1! 8<$ *;! 0;$ K; !1; hrs$ W;! ?
%ule : If A$ B and ) can do a work in 2$ y and Q days respecti,ely then all of them working together can finish the work in 7
5-+ A and B can do a piece of work in 1; days $ B and ) in 15 days and ) and A can do in ;< days. @ow long would each take seperately to do the same work ?
A&B&) can do the work in 5T;! 1< days
5-+ If 0 men or ' women can do a work in '0 days$ how long will = men and 5 women take to comlete the work ? Solution: 0 men can complete 1 # '0 of work in a day 1 man can complete 1 #%'0T0( work in a day ' women can complete 1# '0 of work in a day 1 woman can complete 1 # %'0T'( of the work in a day
o$ = men and 5 women will complete the work in 1; days 5-+ e,en men can complete a work in 1; days. hey started the work and after 5 days$ two men left. In how many days will the work be completed by the re maning men ? Soution : = men 1 dayPs work will be 1 # 1; 1 man 1 day work will be 1 # %1;T=( ! 1 # >' = men 5 days work will be 5 # 1;$ so remaning work will be %1- 5# 1;( ! = # 1; 5 mens 1 day work will be 5 # >'
5-+ A man$ a women and a boy can complete a 4ob in 0$ ' and 1; days respecti,ely. @ow many boys must assist 1 man and 1 woman to complete the 4ob in 1#' of a day ? Solution : 1 manPs & 1 womanPs one day work ! 1 # 0 & 1 # ' ! = # 1; work done by 1 man & 1 woman in 1 # ' day ! %= # 1; (T% 1 # ' (! = # '> o the work remains!%1-=#'>(! '1 # '> Also work done by one boy in 1 # ' day ! %1 # 1; (T%1 # ' ( ! 1 # '> ince 1 # '> work is done by 1 boy in 1 # ' days o '1 # '> work can be done by % '1 # '> ( T %'> # 1( ! '1 5-+ 1< men and 15 women together can c omplete a work in 8 days. It takes 1<< days for one man alone to complete the same work. @ow many days will be reuired for o ne woman alone to complete the same work ? Solution : 1 manPs one day work ! 1 # 1<< %1< menPs & 15 womenPs ( 1 day work! 1 # 8 15 womenPs one days work ! %1 # 8 (- %1< # 1<< ( ! 1 # 15 1 woman 1 day work ! % 1 # 15T 15 ( ! 1 # ;;5 o one woman will complete the work in ;;5 days 5-+ If 8 men and > boys can do a piece of work in 1< days while ;8 men and '> boys can do the same in ; days$ the time taken by 15 men and ;< boys in doing the same type of work will be 7 Solution : %8 men and > boys ( 1 day work ! 1 # 1< hie can be written as eueation form$ let 1 manPs one day work is * and one boyPs one day work is B 8 * & > B ! 1 # 1< imilarly 15 * & ;< B ! 1 # ; on sol,ing * ! 1 # 1<<$ B ! 1 # ;<< 6ow % 15 men & ;< boy ( 1 dayPs work ! % 15 # 1<< (& % ;< # 1<< ( ! 1 # '
o 15 men and ;< boys can do the work in ' days.
imple 5nterest and Compound 5nterest3 5nterest is mone" paid to the lender #" the borrower for using his mone" for a specified period of time This article gi$es the #asic fundamentals of 5nterest= specificall" the imple 5nterest and the Compound 5nterest The terms used in this topic and their general representation are3 5nterest '()= Xrincipal '*)= Time 'n+ n is e%pressed in num#er of periods= -hich i s normall" one "ear1= Rate of 5nterest 'r+ r is taken as the rate of interest on Re + for one "ear= instead of the rate on Rs +,, 1= Amount 'A) Mo&ing Trains :
%ule : When two trains are mo,ing in opposite diections$ then relat,e speed will be the addition of their indi,idiual speeds. %ule : When two trains are mo,ing in same diection$ then relat,e speed will be the subtrction of their indi,idiual speeds. %ule : Dn passing a platform by a certain train the net distance tra,elled is the sum of length of train and the length of platform both. %ule : When a train passes through a pole or person standing$ net distance tra,elled to pass is the length of the train 5-+ A train 1;< m long is running at the speed of 5' km #hr. ind the time taken it to pass a man standing near the railway track. olution 7 speed of train ! U5' T % 5 # 1> ( V ! 15 m # sec length of train ! 1;< m $ o reuired time 7
5-+ A train is mo,ing at a speed of 5' km # hr. If the length of the train is 1<< meters$ how long will it take to cross a railway platform 11< meters long ? olution 7 speed of train ! U5' T % 5 # 1> ( V ! 15 m # sec Kistance co,ered in passing the platform ! 1<< & 11< ! ;1< m
5-+ wo trains 1;5 m and 1<< m in length respecti,ely are running in opposite directions$ one at the rate of 5< km # hr and the other at the rate of '< km #hr. At what time they will clear each other from the moment they meet ? Solution : %elati&e s(eed of trains ! %5< & '<( km # hr ! U:< T % 5 # 1> ( V ! ;5 m # sec otal length to be tra,elled ! 1;5 & 1<< ! ;;5 m
5-+ wo trains 11< m and 1<< m in length respecti,ely are running in same directions$ one at the rate of 1<< km # hr and the other at the rate of 8' km # hr. At what time faster train will clear other train from the moment they meet ? Solution : ince trains are running in same direction$ so relati&e s(eed ! 1<<-8' ! 08 km # hr ! U 08 T % 5 # 1> (V ! 1< m # sec
otal length to be tra,elled ! 11< & 1<< ! ;1< m
5-+ A train 1;5 m in length$ mo,es at a speed of >; km # hr $ In what time the train will cross a boy who is walking at > km # hr in opposite direction ? Solution : %elati&e s(eed ! >;&> ! :< km # hr ! U :< T % 5 # 1> (V ! ;5 m # sec
5-+ A train passes a standing pole on the platform in 5 seconds and passes the platform completely in ;< seconds. If the length of the platform is ;;5 meters. hen find the length of the train ? Solution : Let the length of the train is 2 meter o speed of train !% 2 # 5 ( m # sec Also speed of train ! % ;;5 & 2 ( # ;< m#sec
5-+ wo trains of length 115 m and 11< m res pecti,ely run on parallel rails. When running in the same direction the faster train passes the slower one in ;5 seconds$ but when they are running in opposite directions with the same spee ds as earlier$ they pass each other in 5 seconds. ind the speed of each train ? Solution : Let the speed of trains be 2 m#sec and y m#sec especti,ely. When they mo,e in same direction their relati,e speed is 7 2 - y When they mo,es in opposite direction their relati,e speed is 7 2 & y
Dn sol,ing two euations 2!;= m#s and y!1> m#sec
Tabulation his page contains abulation utorial$ uestions based on tabulation$ freuently asked ables and +raphs uestions. ips for sol,ing tables and graphs problems %1(! ead and ,iew tables and diagram properly. %;(! Eut proper attention$ what sum rows and sum of columns represents. %0(! ake care regarding of units. %'(! ry to understand the uestion$ sometimes you can sol,e in your mind by 4ust looking at data. %5(! *ost of the uestions can be sol,ed by appro2imation$ thus you can sa,e time by
a,oiding calculation. 5-+ tudy the table and answers the uestions. inancial tatement of A company D,er the years %upees in Lakhs ( !ear
Pro*it be*ore @ross interesrt and Turno&er de(reciation
Net ,nterest e(rciation (ro*it %s+ %s+ %s+
;<<<-<1 108<.<<
0>1.:<
0<<.<<
=<.<<
1<.88
;<<1-<; 1'<;.<<
'<0.:;
015.05
=1.1<
1>.'5
;<<;-<0 150>.'<
5;<.<5
0:<.><
><.<1
':.15
;<<0-<' ;118.0'
8<<.<;
''<.>:
>:.<
88.<<
;<<5-<8 ;5;1.<<
>1<.<<
5<<.:<
:1.:;
;1;.><
;<<8-<= ;=5>.::
:;<.<<
8<<.<<
::.<<
;;<.><
6 81 Kuring which year did the P6et Erofit P e2ceed s. 1 crore for the first time ? %a( ;<<8-<= %d( ;<<;-<0
%b(;<<5-<8
%c( ;<<0-<'
%e( 6one of these
Solution :By looking at the table we find option %b( as correct 682 Kuring which year was the S +ross urno,er S closet to thrice the PErofit before Interest and KepreciationP ? %a( ;<<8-<=
%b(;<<5-<8
%d( ;<<;-<0
%e( ;<<1-<;
%c( ;<<0-<'
Solution : @ere we ha,e to find out the ratio of S+ross turno,er S to the SErofit before Interest and Kepreciation S for ;<<8-<= ratio ! ;=5>.:: # :; < ! 0.< $ since we get answer by hit and trial of first option only then we need not ha,e to find other option. o$ correct option is %a( ;<<8-<= 683 Kuring which of the gi,en year did the P6et profit P form the highest proportion of the P Erofit before Interest and Kepreciation ? %a( ;<<5-<8 %d( ;<<1-<;
%b( ;<<0-<'
%c( ;<<;-<0
%e( ;<<<-<1
Solution : @ere we will find the re,erse ths ratio between % Erofit before Interest and Kepreciation # 6et Erofit ( and try to find lowest ratio also we use appro2imation for sa,ing time or ;<<5-<8$ >1 # ;1 ! ' appro2. or ;<<0-<'$ 8< # 8.8 ! : appro2 or ;<<;-<0$ 5;< # 5< ! 1< appro2 or ;<<1-<;$ '< # 1.> ! ;; appro2
or ;<<<-<1$ 0>< # 1< ! 0> appro2 o$ lowest ratio is for ;<<5-<8 it means re,erse ratio between %6et Erofit # Erofit before Interest and Kepreciation ( will be highest$ so option %a( is correct 684 Which of the following registered the lowest increase in terms of rupees from the year ;<<5-<8 to the year ;<<8-<= ? %a( +ross turno,er %d( Interest
%b( Erofit before interest and depreciation
%c( Kepreciation
%e( 6et Erofit
Solution : We calculation with appro2imation in mind %a( ;=5> - ;5;1 !11! 11< appro2 %c( :: - :1.:; ! <>.<: appro2 %d( 8<< - 5<5 ! :5 appro2 %e( ;;< - ;1;! <> appro2 o correct option is %e( 6et profit 68" he gross urno,er for ;<<;-<0 is of what percentage of the P+ross urno,er P for ;<<5-<8 ? %a( 81
%b( 180
%d( 0:
%e( <.<<8
%c( <.811
Solution : ;<<;-<0 +ross urno,er 150>.'< or ;<<5-<8 +ross urno,er ;5;1.<<
o option %a( is correct 5-+ Marks otained b' di**erent students in di**erent subCects SubCect 8Ma-imum Marks Students @indi nglish *aths ocial std. %1<<( %1<<( %1<<( %1<<(
Ehy. cience anskrit du %=5( %5<( %=5(
Anupam
>5
:5
>=
>=
85
05
=1
Bimal
=;
:=
55
==
8;
'1
8'
)haman
8'
=>
='
80
55
;5
50
Ke,endar
85
8;
8:
>1
=<
'<
5<
+irish
:;
>;
>1
=:
':
0<
81
Ci,ek
55
=<
85
8:
''
;>
0<
681 @ow many students ha,e scored the lowest marks in two or more sub4ects ? %a( ; %b( 0 %c( 1 %d( < %e(' Solution :@ere option %a( is correct by looking at table we find )haman and Ci,ek has scored lowest marks in two or more sub4ects Ci,ek in @indi$ cience$ Ehy. du )haman in ocial std$ anskrit 682 Who has scored the highest marks in all the sub4ects together ? %a( Ke,endar %b( )haman %c( Anupam %d( +irish %e(Bimal Solution : or this calculate total toatal of each students in options Ke,endar ! 85 & 8; & 8: & >1 & =< & '< & 5< ! '0= )haman ! 8' & => & =' & 80 & 55 & ;5 & 50 ! '1; Anupam ! >5 & :5 & >= & >= & 85 & 05 & =1 ! 5;5 +irish ! :; & >; & >1 & =: & ': & 0< & 81! '=' Bimal ! =; & := & 55 & == & 8; & '1 & 8' ! '8> o correct option is %0( Anupam 6 83 What is the percentage of Ke,endarPs marks %upto two digits after decimal( in all the sub4ects together ? %a( >>.80 %b( ==.0> %c( 8=.>0 %d( 8;.>0 %e(=;.>0 Solution : Ke,endarPs total marks ! '0=
6 84 *arks obtained by )haman in @indi are what percentage of marks % upto two digits after decimal ( obtained by Anupam in the same sub4ect ? %a( =5.:; %b( =>.0> %c( ==.;: %d( =5.;: %e(=;.>0 Solution: *arks obtained in @indi by )haman ! 8'$ by Anupam! >5
6 8" What are the a,erage marks obtained by all the students together in cience ? %a( 55.=5
%b( 5=.5<
%d( 5:.5<
%e(5>.<<
%c( 8<.<<
Solution : A,erage maks obtained in cience !
o$ option %b( true enn iagram enn iagram : Eictorial representation of sets by means of diagrams is termed as Cenn Kiagrams. 5lements o* Sets : he ob4ects in a set are termed as elements or members of sets. Let A and B are two sets$ such that A! \ 1$ ;$ 0$ '$ 5$ 8$ =$ >$ :$ 1< ] B! \ ;$ '$ 8$ >$ 1<$ 1;$ 1'$ 18$ 1>$ ;< ] or this Cenn Kiagram representation will be 7
Where$ A - B ! his set has elments which are only in A B - A ! his set has elments which are only in B A [ B is set which has comman elements both from A and B Also number of elements in A ∪ B is same as number of elements in B ∪ A o$ n%A ∪ B( ! n%B ∪ A( Also$ n%A [ B( ! n%B [ A( rom Cenn Kiagram we can see that
n%A( ! n%A-B( & n%A [B( ...........%a(
imilarly$ n%B( ! n%B-A( & n%B [A( ............%b( Also from Kiagram we can write$ n%A∪B( ! n%A-B( & n%A[B( & n%B-A( .............%c( Dn adding %a( and %b( n%A( & n%B( ! n%A-B( & n%B-A( & n%A [B( & n%A [B( or n%A( & n%B( - n%A[B( ! n%A-B( & n%B-A( & n%A [B( ..................%d( rom euation %c( and %d( we can write
n8A D n8A E n8 9 n8A L 5-+ Among 58 people collected in a dinner party$ ;' eats non ,eg food but not ,eg food and ;> eats non-,eg food. 681D ind out how how many eat ,eg and non ,eg both ? Solution : here n%6∪C( ! 58 $ n%6-C( ! ;' and n%6( ! ;> 6ow n%6( ! n%6-C(& n%6 [C( ;>!;'& n%6 [C( o$ n%6 [C( !' $ @ence ' people eat ,eg and non ,eg both 682 ind out how many of them eat Ceg but not non ,eg ? Solution : We can write n%6 ∪C( ! n%6( & n%C( - n%6 [C( 58 ! ;> & n%C( - ' n%C( ! 0; Also$ n%C( ! n%C-6( & n%C [6( 0; ! n%C-6( & ' n%C-6( ! ;>$ @ence ;> people eats Ceg but not non ,eg 5-+ In a club of '> people$ ;' plays cr icket and 18 plays cricket but not hockey. ind the number of people in club who plays hockey but not cricket ? Solution :Let ) denotes the cricket and @ denotes hockey$ according to uestion$ n%)∪@(!'>$ n%)(!;' n%)-@(!18 6ow n%)(! n%) - @( & n%) [@( ;' ! 18 & n%) [@( n%)[@(! > 6ow$ n%) ∪@(!n%)(&n%@(-n%)[@( '> ! ;' & n%@( - > n%@(!0; n%@(!n%@-)(& n %@ [)( 0; ! n%@-)( & > n%@-)(! 0;-> ! ;' o$ people in club who plays hockey but not cricket are ;' 5-+ In a society of >< people$ '; read imes Df India and 05 read he @indu$ while > people don not read any of the two news papers. 681 ind the number of people $ who read at least one of the two news papers . Solution : @ere total number of people are >< out of which > do not read any news paper$ so >< - > ! =; people read remaning two news papers o$ n%∪@(!=;$ n%(!';$ n%@(!05 o$ the number of people $ who read at least one of the two news papers ! n% ∪@(!=; 682 ind the number of people in society $ who read both news papers . Solution :n%∪@( ! n%( & n%@( - n%[@( =; ! '; & 05 - n% [@(
n%[@( ! == - =; ! 5 o$ the number of people in society $ who read both news papers ! 5 5-+ In a society 5< Y people read imes Df India$ ;5 Y read he @indu. ;< Y read both news papers. What Y of people read neither imes Df India nor he @indu ? Solution : n%(!5<$
n%@(!;5$
n%[@(!;<
n%∪@( ! n%( & n%@( - n% [@( n%∪@( ! 5< & ;5 - ;< ! 55 ince 55 Y people read either imes Df India or he @indu $ so remaning 1<< - 55 ! '5 Y o$ '5 Y of people read neither imes Df India nor he @indu ar @ra(hs 7 5-+ Eroduction of steel by si2 different companies in three cosecuti,e years 1::'-:5-:8% In Lakh onnes ( are being gi,en
681 What is the difference between a,erage production of the si2 companies in 1::5 and a,erage production of the same companies in 1::' ? Solution : um of production in 1::5! 55&55&'<&=<&=<&'5!005 um of production in 1::'!5<&'5&0<&5<&=<&05!;><
Kifference!:18888 tonnes.
682 What is the Y decline in production by company ) from 1::5 to 1::8 ? Solution :or company c production in 1::5! '< and in 1::8!05
683 Which of the following companies recorded the minimum Y growth from 1::' to 1::5 ? %a( A %b( B %c( ) %d( K %e( Solution :Dn seeing graph it is clear that company has re corded minimum growth from 1::' to 1::5 684 Eroduction of company P ) P in 1::5 and production of company P P in 1::' together is what Y of production of B in 1::8 ? Solution : Eroduction of ) in 1::5 ! '< $ Eroduction of in 1::' ! 05 um ! =5$ production of B in 1::8 ! 8<
68" In which of the following pairs of companies the difference between a,erage production for the three years is ma2imum ? %a( and %b( K and %c( and ) %d( A and %e( 6one of these Solution : rom the graph we can see that highest pr oduction a,erage is for company and Lowest production a,erage is for company )$ so difference between a,erage production for the three years is ma2imum for company and )$ o$ option %c( is correct
5-+ oatal sale of nglish and @indi 6ewspapers in i,e different Localities of a city are gi,en. % Note : Ko not get confused by a2is of data$ it is shown in cylendrical way$ so ,alue is same as shown in y a2is (
681 What is the difference between the total sale of english 6ewspapers and total sale of @indi newspapers in all the localities together ? %a( 8<<< %b( 85<< %c( =<<<< %'(=5<< %5( 6one of these Solution : otal um of all the nglish 6ews papers ! =5<< & :<<< & :5<< & =<<< & 85<< ! 0:5<< otal um of all the @indi 6ews papers ! 55<< & >5<< & '5<< & :5<< & 5<<< ! 00<<< Kifference ! 0:5<< - 00<<< ! 85<<$ so o(tion 8b is correct
682 he sale of nglish 6ews paper in locality A is appro2imatley what Y of the total sale of english newspapers in all the localities together ? %a( 5;= %b( ;5 %c( 111 %'(;08 %5( 1: Solution : otal sum of sale of english newspapers in all the localities together ! 0:5<< as calculated abo,e ale of nglish 6ews paper in locality A ! =5<< % from fig. abo,e (
o o(tion 8" is correct.
683 What is the respecti,e ratio of the sale of @indi 6ewspapers in locality A to the sale of @indi 6ewspapers in locality K %a( 11 7 1: %b( 8 7 5 %c( 5 7 8 %'(1: 7 11 %5( 6one of these Solution :sale of @indi 6ewspapers in locality A!55<< sale of @indi 6ewspapers in locality K!:5<<
o$ o(tion 8a is correct 684 he sale of nglish 6ewspaper in localities B and K together is appro2imately what Y of the sale nglish 6ewspaper in localities A$ ) and together ? %a( 18; %b( >' %c( 8> %'(1;1 %5( 1'= Solution : um of the sale of nglish 6ewspaper in localities B and K together ! :<<< & =<<< ! 18<<< um of the sale of nglish 6ewspaper in localities A$ ) and together ! =5<< & :5<< & 85<< ! ;05<<
o$ option 8c is correct 68" What is the a,erage sale of hindi news papers in all the localities together ? %a( 88<<
%b( >;5<
%c( 55<<
%'('=15
%5( 6one of these
Solution : um of the sale of @indi 6ewspapers in all localities ! 00<<< as calculated in X%1(
o$ o(tion 8a correct. Tutorial9Fine @ra(hs 7 5-+ 6umber of hotels in a state$ according to years are gi,en % tudy the gi,en chart carefully and then answer the uestions accordingly (
681 he appro2imate Y increase in hotels from year 1:>: to 1::' was %a( =5 %b( 1<< %c( 1;5 %d( 15< %e( 1=5 olution 7
Dption %b( is correct 682 If the number of newly made hotels in 1::1 was les s by 1< then what is the ratio of the number of hotels in 1::1 and that in 1::< ? %a( 1' 7 11 %b( 0 7 ' %c( ' 7 5 %d( 5 7 ' %e( 1 7 ' Solution : 6o of hotels in 1::1 will be 5=< - 1< ! 58<
Dption %a( is correct 683 If the Y increase in the number of hotels from 1::0 to 1::' continued up to 1::5 then what is the number of hotels built in 1::5 ? %a( *inimum =5 %b( *inimum =< %c( *inimum 5< %d( *inimum 10: %e( *inimum >< Solution : irst find Y increase in 1::' from 1::0
6ow 1> Y of >0> ! 15< o$ nearest option is %d( *inimum 10: hotels 684 In which of the gi,en years increase in hotels in comparison to the pre,ious year is the ma2imum ? %a( 1::< %b( 1::1 %c( 1::; %d( 1::0 %e( 1::' Solution :
o option %b( is correct 68" If increase in hotels from 1::1 to 1::; is E Y and increase in hotels from 1::; to 1::' is X Y $ then which of the following relations between E and X is true ? %a( Kata is inadeuate %b( E ^ X %c( E ! X %d( E X %e( 6one of these Solution : E Y ! ;'.58 Y from uestion %'( X Y ! 1>.<5 Y o E X$ option %d( correct
5-+ Y profit earned by two companies o,er the years is gi,en in graph. Also
5-+ If the e2penditure of )ompany B in ;<<< was s. ;<< crores$ what was its income ? %a( s. ;'< crores %b( s. ;;< crores %c( s. 18< crores %d( )an not be determined %e( 6one of these.
Solution : Let income be s 2 crores so$ we can use the abo,e formula as
'< ! 2 - ;<< 2! ;'< crores 5-+ If the income of company A in ;<<; was s. 8<< crores. What was its e2penditure ? %a( s. 08< crores %b( s. '>< crores %c( s. 0=5 crores %d( )an not be determined %e( 6one of these. Solution : or company A in ;<<; Y profit was 8< Y $ Let e2penditure be 2 crores $ so
5-+ If the income of a company B in 1::> was s. ;<< crores$ what was its profit in 1::: ? %a( s. ;1.5 crores %b( s. 150 crores %c( s. '8.15 crores %d( )an not be determined %e( 6one of these. Solution : Erofit can be calculated only when Income and e2penditure of the gi,en year should be known. o$ o(tion 8d is correct 5-+ If the Incomes of the two companies in 1::> were eual$ what was the ratio of their 2penditure ? %a( 1 7 ; %b( ;8 7 ;= %c( 1<< 7 8= %d( )an not be determined %e( 6one of these. Solution : Let$ the income of both companies be E$ e2penditure of A is 1 and e2penditure of B is ; 6ow we can write$
o$ option %b( is correct
5-+ What is the Y increase in Y profit for company B from year ;<<< to ;<<1 ? %a( =5
%b( 1=5
%c( ';.>8
%d( )an not be determined
%e( 6one of these.
Solution :
Tutorial9Pie /hart 7 A Eie )hart is a pictorial representation of a numerical data by non intersecting ad4ecent sectors of the circle such that area of each sector is proportional to the magnitude of the data represented by the sector. %a( he whole circle represents the total and the sectors$ indi,idual uantities. %b( he sectors$ are made considering the fact that the central angle is 08<_ %c( he central angle$ 08< _ can be di,ided in the ratio of uantities gi,en. %d( )entral angle or Angle of the sector is 7
he number of students studying in different faculties in the years ;<<1 and ;<<; from state A is as follows
5-+ In which faculty there was decrease in the number of students from ;<<1 to ;<<; ? %a( Arts %b(Agriculture %c( Eharmacy %d(6one %e( 6one of these Solution : or Arts$ 6umber of students in ;<<1 and ;<<;
or Agriculture 6umber of students in ;<<1 and ;<<;
or pharmacy Y is already more and total number of students are already more in ;<<;$ so correct option will be for Agriculture$ o(tion 8b is true 5-+ What is the ratio between the number of students studying pharmacy in the years ;<<1 and ;<<; respecti,ely ? %a( ' 7 0 %b( 0 7 ; %c( ; 7 0 %d( = 7 1; %e( 6one of these olution 7 atio between the number of students studying pharmacy in the years ;<<1 and ;<<; !
Dption %d( is true 5-+ What was the appro2imate percentage increase in the number of students of ngineering from the year ;<<1 to ;<<; ? %a( 1= %b( 15 %c( ;5 %d( ;0 %e( ;< Solution : 6umber of engineering students in ;<<1 !
6umber of engineering students in ;<<; !
otal increase ! =8<< - 80<< ! 10<<
5-+ In the year ;<<1$ the number of students studying Arts and )ommerce together is what percentage of the number of students studying these sub4ects together in ;<<; ? %a( =8 %b( >5 %c( >; %d( =: %e( 6one of these Solution : 6umber of students studying Arts and )ommerce together in ;<<1 !
6umber of students studying Arts and )ommerce together in ;<<; !
o$ option %b( is true 5-+ In which of the following faculties the percent increase in the number of students was minimum from ;<<1 to ;<<; ? %a( Arts %b( cience %c( )ommerce %d( *edicine %e( ngineering Solution : his uestion is uit lengthy$ irst find respecti,e number of students in the gi,en years according to sub4ects Arts
cience
)ommerce
*edicine
ngineering
;<<1
';<<
>'<<
==<<
0>5<
80<<
;<<;
''<<
>><<
:8<<
'<<<
=8<<
5-+ tudy the gi,en graph and table and answer the following uestions gi,en below. % otal population of all states gi,en in Eie chart is ;5 lakha ( In the year 1::> the data of different states regarding population of states
Se- and literac' wise (o(ulation ratio States
Se-
Fiterac'
M
>
Fiterate
,lliterate
"E
5
0
;
=
Bihar
0
1
1
'
AE
;
0
;
1
Oar
0
5
0
;
*@
0
'
5
1
6
0
0
=
;
Oerala
0
'
:
'
681 Appro2imately what is the total number of literate people in *@ and Oar together ? %a( '.5 lakhs %b( 8.5 Lakhs %c( 0 lakh %d( 0.5 lakhs %e( 8 lakhs Solution : Eopulation of Oar is 15 Y
o$ otal number of literate people in Oar is 7
Eopulation of *@ is 11 Y
o$ number of literate people in *@ and Oar together ! ; .;: & ;.;5 ! '.5' lakhs o$ option %a( is correct
682+ Appro2imately what will be the percentage of total male in "E $ *@ and kerala of the total population of the gi,en states ? %a( ;
%b( 1>Y
%c( ;>Y
%d( 0
%e( ;5Y
Solution : Eopulation in "E is ;5 Y
Eopulation in *@ ! ;.=5 lakhs from X%1(
Eopulation in Oerala ! > Y
o$ um of male population of "E$ *@ and kerala ! 0.:0 & 1.1= & <.>5 ! 5.:5 lakhs
option %e( correct
Alliagations And Mi-tures91 7 Alligation 9 In this we find the proportion in which ingredients at gi,en prices must be mi2ed to produce a mi2ture at a gi,en price$ this is termed as alligation. Alligation method can not be applied for absolute ,alues$ but it can be applied for Xestion related to atio$ ate$ Eercentage$ peed etc % *eans it can be applied for comparable ,alues like per hour$ per km $ per kg etc. ( %ule o* Alligation : If gradients are mi2ed in a ratio then we can write
5-+ In what proportion must weat at Erice '.1< per kg must be mi2ed with weat at Erice '.8< per kg$ so that the mi2ture be worth s '.0< a Og ? Solution : @ere$ )E of unit uantity of Kearer ! '8< per kg $ )E of unit uantity of )heaper ! '1< per kg o we can use abo,e formuls 7
o$ the reuired ratio be 0 7 ;
5-+ @ow many kg of rice at s. 8< per kg$ must be be mi2ed with 0< kg of rice at s ;5 per kg$ so that he may on selling the mi2ture at s 5< per kg gain ;5 Y on the outlay ? Solution : irst we ha,e to find cost price of mi2ture$ as seller is gaining ;5 Y profit on mi2ture so its cost price will be
6ow use the formula of alligation to find to find out uantity of dearer rice$ %6ote 7 @ere mean price will be )E of mi2ture$ do not get confused by elling price s 5< per kg (
5-+ A mi2ture of certain uantity of milk with ;< liters of water of worth s 1< per liter. If pure milk is of worth s 15 per liter$ how much milk is there in the mi2ture ? Solution : By rule of alligation
o$ Xuantity of milk will be '< liters
5-+ In what proportion must water be mi2ed with milk to gain ;< Y by selling it at cost price ? Solution : Let cost price of milk be s 1 per liter$ then .E of mi2ture is also s 1 per liter 6ow )E of mi2ture be ! 1 - % ;< Y of s 1 ( ! 1 - % ;< T 1 # 1<< ( ! >< # 1<< ! s ' # 5
euired ratio ! 1 7 ' 6uestion based on Mi-ture *rom two &essels 8 Note : >re=uentl' asked in MA entrance e-ams 5-+ *ilk and water are mi2ed in a ,essel A in the ratio 5 7 0 and in ,essel B in ratio : 7 = . In what ratio should uantities be taken from the two ,essels so as to form a mi2ture in which milk and water will be in the proportion of = 7 5 ? Solution : In ,essel A$ milk ! 5 # 5 & 0 ! 5 # > of the weight of mi2ture In ,essel B milk ! : # : & = ! : # 18 of the weight of mi2ture 6ow$ we ha,e to form a mi2ture in which milk be = # 1; of the weight of the mi2ture 6ow according to rule of alligation 7
5-+ A goldsmith has two qualities of gold, one of %9 carats and another of %* carates purity. n
what proportion should he mi! both to make an ornament of %' carats purity # Solution : 2y applying rule of alligations
1o both qualities of gold should be mi!ed in the ratio of ) 5 ' 5-+ (99 gm spirit solution has (9 @ spirit in it , "ow may grams of spirit should be added to make it 69 @ in the solution #
Solution : 2y applying rule of alligations and mi!tures
1o, the two mi!tures should be added in ratio ' 5 %
)oding # Kecoding is a method of transmitting a message between the sender and the recei,er which third person can not understand. Any word or message can be coded in many ways. In this candidate has to find the coding rule so that he can either code or decode the message in similar way here are se,eral ways of )DKI6+ letters of nglish alphabets$ some conseuently used are as follows81 5nglish al(habets Position *rom le*t to right :9 A
B
)
K
+
@
I
H
O
L
*
1
;
0
'
5
8
=
>
:
1<
11
1;
10
6
D
E
X
"
C
W
3
F
1'
15
18
1=
1>
1:
;<
;1
;;
;0
;'
;5
;8
82 5nglish al(habets (osition *rom %ight to le*t :9 F
3
W
C
"
X
E
D
6
1
;
0
'
5
8
=
>
:
1<
11
1;
10
*
L
O
H
I
@
+
K
)
A
B
1'
15
18
1=
1>
1:
;<
;1
;;
;0
;'
;5
;8
83 Series o* o((osite 5nglish Al(habets :9 A
B
)
K
+
@
I
H
O
L
*
F
3
W
C
"
X
E
D
6
5-+ In a certain code KIABL is written as HX*)B$ how is KI*6ID6 written in that code ? %a( 6H*HED %b( 6HDDEH %c( 6H*DEH %d( H6*DEH %e( 6one of these Solution : o sol,e these type of problems$ first put the letters one abo,e other and analyse carefully 7 K
I
A
B
L
H
X
*
)
B
6ow on identifying the pattern apply to word KI*6ID6 K
I
*
6
I
D
6
6
H
*
D
E
H
o correct option is %c( 6H*DEH 5-+ In a certain code AELD*B is written as )66*XB. @ow KIEL is written in that code ? %a( *DH %b( *DH %c( H*D %d( *DH %e( 6one of these Solution : Look out the words AELD*B and )66*XB $ analyse carefully and try to find out relation between the words position7
6ow code the word KIEL in similar way
o$ the correct option is %b( 5-+ In a certain code B"+D6 is written as XA+DE. @ow DB"K is written in that code ? %a( A*" %b( "6A %c( A*" %d( A6" %e( 6one of these Solution :9 Look out the words B"+D6 and XA+DE $ analyse carefully and try to find out relation between the words position7
6ow code the word DB"K in similar way
o the correct option is %d(
5-+ In a certain code A)DLiswritten as EKBO"F. @ow B"6I@ is written in that code ? %a( CA*+H
%b( )*IH
%c( C)*IH
%d( XCA*+@
%e( 6one of these
Solution :9 Look out the words B"+D6 and XA+DE $ analyse carefully and try to find out relation between the words position7
6ow code the word B"+D6 in similar way
o the correct option is %)( /oding @rou( o* words @ere a sentence or a group of words is coded and candidates ha,e to find the gi,en word . 5-+ In 5-+ In a certain code language Pom atP means Pleep )atP and Pat BallP means Poads leepP. What is the code for PleepP in that code language ? Solution :9 om :9 om at - leep )at at Ball - oads leep In both the sentences common word is leep and at$ so$ the code for leep is at. 5-+ In 5-+ In a certain code language P tik sa tu P means P he is what P G P sa pa na 4u P means P what can go there P G P rn na le P means P go and by P G What does Ptik means in that code language ? Solution : tik sa tu - he is what sa pa na 4u - what can go there rn na le - go and by o$ tik means he or is 5-+ In 5-+ In a certain code $ Ppick and chooseP is written as Pko ho poP and Ppick up and comeP is written as Pto no ko poP. @ow is pick written in that code ? %a( ko %b( po %c( ither ko or po %d( )annot be determined %e( 6one of these % Bank Eo 2am;<<8 ( ch oose - ko ho po Solution :9 pick :9 pick and choose
pick up and come - to no ko po o correct option is 8c 5ither ko or (o
5-+ In 5-+ In a certain code langauge Ppic na haP means Pwho is thereP $ P na ta ka P means P what is that P and P ha 4a pa P means P here and there P . Which of the following means P here P in the that code langauge ? %a( ha
%b( pa
%c( 4a
%d( pa or 4a
%e( 6one of these
Solution :9 pic :9 pic na ha ha - who is there
na ta ka - what is that
ha 4a ha 4a pa - here and there
o$ correct option is %d( pa or 4a erbal %easoning9 /oding o* word and s'mbols 5-+ In 5-+ In each uestion below a group of letters is gi,en followed by four combinations of digits or symbols numbered %a($ %b($ %c( and %d(. ou ha,e to find out which of the combinations correctly represents the group of letters based on the following coding system and the conditions those follow and gi,e the number of that combination as the answer. If none of the combination correctly represents gi,e option %e( none of these as the answer. Fetter : E
*
/ode 7
8
5
A 1
H `
:
;
O >
I
0
B R
" =
@ '
/ondition : 81 If 81 If the first letter is a ,owel and the last letter is a c onsonant$ both are to be coded as G
681 IOB"A 681 IOB"A %a( >R=;1
%b( T>R=;T
%c( Y>R=;Y
Solution : I
O
B
"
A
>
R
=
;
1
%d( T>R=;Y
%e( 6one of these
@ere first and last both letters are ,owels$ o option %a( is correct
682+ *EH@ 682+ *EH@ %a( :850`' Solution : T
%b( T:850` * 8
E 5
0
%c( Y850`Y
H
@
`
T
%d( T850`T
%e( 6one of these
@ere first letter is ,owel and last is consonant$ so condition %1( is applied
683 IEA"@* 683 IEA"@* %a( T51='T
%b( Y51='Y
%c( 51='8
Solution : I
E
A
"
@
*
T
5
1
=
'
T
%d( 51'=8
%e( 6one of these
@ere condition %1( is applied
684 @OHA 684 @OHA %a( 0'> %b( '>0`1 Solution : Y
@
O
'
%c( T'>`T H
>
%d( Y'>`Y
%e( 6one of these
A `
Y
@ere first letter is consonant and the last is ,owel$ so condition %;( is applied
68" *BH 68" *BH %a( T80RT Solution : ;
%b( Y80RY *
8
0
%c( ;08R`
B
H
R
`
%d( ;80R`
%e( 6one of these
Both %1( and %;( not get full filled so simple coding is to be done.
68# OH"A 68# OH"A %a( >;`=1
%b( T;`=T
Solution : O
H
Y
;
`
" =
%c( Y;`=Y
%d( ;`=1
%e( 6one of these
A Y
@ere condition %;( is satisfied.
68$ "*BOE 68$ "*BOE %a( T8R>5T
Solution : "
=
%b( =8R>5
*
8
B
O
E
R
>
5
%c( T8R>5
:
%d( =8R>5Y
%e( 6one of these
@ere first and lst both words are ,owels so none of the conditions get satisfied $ so normal coding is to be done. erbal reasoning tutorial9 irection and istance
%a( 6orth 6$ ast $ outh and West W are four directions. %b( While outh-ast %($ outh-West %W($ 6orth West %6W( and 6orth-ast %6( are four cardinal directions
5-+ Alok started walking towards outh. @e took right turn after walking 15 meters. @e again took a left turn after walking ;5 meters. Which direction he is facing now ? %a( outh %b( 6orth %c( West %d( ast %e( )an not be determined. Solution :9 let us understand by diagram
)learly he is facing outh direction$ so option %a( is correct 5-+ akesh walked 05 meters towards east and took a r ight turn and walked '5 meters. @e again took right turn and walked 55 meters. owards which direction is he from his starting point ? %a( outh %b( West %c( outh West %d( outh ast %e( 6one of these Solution :9 let us look at the diagram
o$ the correct option is towards outh-West %c(
5-+ A man facing towards ast. @e turns '5_ in anticlock wise direction and then another 1><_ in the same directionand then ;=<_ in clockwise direction . Which direction is he facing now ? %a( outh-ast %b( outh %c( outh-West %d( 6orth-West %e( 6one of these Solution :Let us look the diagram
o $ the correct option is owards option %a(
5-+ A boy in search of his pet dog$ he went :< meters in ast before turning to his right. @e went ;< meters before turning to his right again to look for his dog at his unclePs place 0< meters from this point. @is dog was not there$ from there he went 1<< meters to his north there found his dog. @ow far did the boy meet his dog from starting point ? %a( >< meters %b( 1<< meters %c( 1'< meters %d( ;8< meters %e( 6one of these Solution :9 Look at the diagram
o$ reuired distance A) can be calculated as 7
o$ correct option is %b(
5-+ A man walks 1 km towards ast and then he turns to outh and walks 5 km. Again he turns to ast and walks ; k$ after this he turns to 6orth and walks : km. 6ow$ how far is he from his starting point ? %a( 0 km %b( ' km %c( 5 km %d( = km %e( 6one of these Solution :9 LetPs look at the image
o$ the correct option is %c(
5-+ he door of ameshPs house faces the east. rom the back side of his house$ he walks straight 5< meters$ then turns to the right and walks 5< meters again. inally $ he turns towards left and stops after walking ;5 meters. 6ow$ amesh is in which direc ton from the starting point ? %a( outh-ast %b( 6orth-ast %c( outh-West %d( 6orth-West %e( 6one of these Solution :9 let us look at the diagram
o the reuired direction is 6orth-West$ option %d( is correct
5-+ ead the following information carefully and answer the uestions gi,en below 7%a( i2 flats on floor in two rows facing 6orth and outh are alloted to E$ X $ $ $ and ". %b( X gets a 6orth facing flat and is not ne2t to . %c( and " get diagonally opposite flats %d( $ ne2t to "$ gets a outh facing flat and gets a 6orth facing flat. 81 Which of the following combinations get south facing flats ? %a( X %b( " E %c( " E %d( Kata not sufficient %e( 6one of these Solution :9 In statement %b( X gets a 6orth facing flat and is not ne2t to means
In statement %c( and " get diagonally opposite flats means
In statement %d( ne2t to "$ gets a outh facing flat and gets a 6orth facing flat means
o$ outh acing flats are %c( " E 82 Whose flat is between X and ? %a( %b( " %c( %d( E %e( 6one of these Solution : $ option %a( correct
83 If the flats of and E are interchanged$ whose flat will be ne2t to that of " ? %a( E %b( X %c( %d( %e( none of these Solution : By this interchanged flat ne2t to " remain unchanged$ o option %c( is correct $ flat
5-+ he flats of which of the other pairs than "$ is diagonally opposite to each o ther ? %a( XE
%b( X
%c( E
%d(
%e( 6one of these
Solution :9 X E option %a( is correct erbal %easoning9PuHHle Test o sol,e the problems e2amining and re e2amining the problem situation from e,ery point of ,iew is to be done. uestions are of following types %a( )omprasion ype %b( eating Arrangements
%c( Eroblems based on blood relationship %d( )lassification ype %e( Humbled type prolems
ead the following information carefuly and answer the uestions gi,en below 7E$ X$ $ $ $ C and W are sitting around a circle facing at centre. is third to the right of C who is second to the right of E. is second to the left of X who is second to the left of W. C is sitting between and W. 681 Who is to the immediate left of ? %a( X %b( %c( E %d( Kata inediuate
%e( none of these
LetPs sol,e step by step7- is third to the right of who is second to the right o* P
is third to the right of C
C is sitting between and W.
is second to the left of X who is second to the left of W$ o in this case only this structure is possible.
o immediate left of is option %b( correct
682 Who is third to the left of ? %a( E %b( %c( C %d( W
%e( none of these
Solution:9 Dption is %d( W
683 Who is sitting between and C ? %a( %b( %c( 6one %d( data inadeuate
%e( none of these
Solution : )orrect option is %b(
684 Who is sitting between and W ? %a( X only %b( E only %c( E and
%d( data inadeuate
%e( none of these
Solution :9 Between and W EX and C are sitting so$ correct option is %e( none of these
5-+ In a school$ there were fi,e teachers. A and B were teaching @indi and nglish. ) and B were teaching nglish and +eography. K and A were teaching *athematics and @indi. and B were teaching @istory and rench. 681 Who among the teachers was teaching ma2imium number of sub4ects ? %a( A %b( B %c( ) %d( K %e( Solution :9 LetPs make a table for analysis nglish
@indi
*athematics
A
es
es
es
B
es
es
)
es
K
+eography
@istory
rench
es
es
es
es
es
es es
es
o$ ma2imum number of sub4ects was taught by B$ option %b( is correct
682 Which of the folowing pairs was teaching both +eography and @indi ? %a( A and B %b( B and ) %c( ) and A %d( K and B %e( 6one of these Solution :9 By seeing on table only B teaches +eography and @indi
683 *ore than two teachers were teahing which sub4ects ? %a( @istory %b( @indi %c( rench %d( +eography %e( *athematics Solution :9 @indi is being taught by three teachers A$ B and K o option %b( is correct
684+ K$ B and A were teaching which of the following sub4ects ? %a( nglish only %b( @indi and nglish %c( @indi only %d( nglish and +eography
%e( *athematics and @indi olution 7- @indi only $ option %c( is correct
68" Which among the teachers was teaching less than two sub4ects ? %a( A %b( B %c( K %d( Kata inadeuate %e( here is no such teachers olution 7- Dption %e( is correct$ all are teaching two or more than two sub4ects
tudy the following information carefully and and answer the uestions gi,en below 7E$ X$ $ $ $ C$ W and F are eight friends studying in three different engineering collegesA$ B and ) in three disciplines- *echanical$ lectrical and lectronics with not less than two and not more than three in any college. 6ot more than three of them study in any of the three disciplines. W studies lectrical in college B with only who s tudies *echnical. E and F do not study in college ) and study in same discipline but not lectrical. studies *echnical in college ) with C who studies lectrical. studies *echnical and does not study in the same college where studies. X does not study lectronics. 681 Which of the following combinations of college-students-specialiQation is correct ? %a( )--lectonics %b( A-F-lectrical %c( B-W-lectronics %d( B-W-lectrical %e( B-Flectronics Solution :9 riends kindly analyQe carefully *echnical
lectronics
lectrical
A B ) studies *echnical in college ) with C who studies lectrical. W studies lectrical in college B with only who studies *ec hnical. *echnical
lectronics
lectrical
A B
W
)
C
E and F do not study in college ) and s tudy in same discipline but not lectrical. studies *echnical and does not study in the same college where studies. *echnical
lectronics
A
E and F
B
E and F
)
lectrical
W C
X does not study lectronics. `` 6ot less than two and not more than three in any college.
``6ot more than three of them study in any of the three disciplines. *echnical
lectronics
lectrical
A
E$F
B
W$ X
)
C
o option$ B-W-lectrical is correct %d(
682 In which of the following colleges two students study in electrical discipline ? %a( A only %b( B only %c( ) only %d( )an not be determined %e( 6one of these Solution : Dption %b( W and X in college B
683 In which discipline does X study ? %a( lectrical %b( *echnical %c( lectrical or *echnical of these
%d( Kata inadeuate
%e( 6one
olution 7- Dption %a( lectrical is correct
684 In which of the clooeges at least one student in *echnical discipline ? %a( A only %b(B only %c( ) only %d( Both A and B %e( All A$ B and ) Solution :9 Dption %e( all A$ B and )
68" studies in which college ? %a( A
%b(B
%c( A or B
%d( Kata inadeuate
%e( 6one of these
Solution :9 Dption %a( A erbal %easoning9PuHHle Test92 . 5-+ ight persons $ $ +$ @$ I$ H and L are seated around a suare table -wo on each side. here are three lady menbers and they donot seated ne2t to each other. H is between L and + is between I and @$ a lady member is second to the left of H. a male member$ a lady member seated third to the left of here is a lady member between and I. 681 Who among the following is seated between and @ ? %a( %b( I %c( H %d( )an not be determined %e( 6one of these. Solution:9 Let L be at this place then H will be at this place see in fig
6ow$ + is between I and
@$ a lady member is second to the left of H
a male member$ a lady member seated third to the left of
here is a lady member between and I.
681 Who among the following is seated between and @ ? %a( %b( I %c( H %d( )an not be determined %e( 6one of these
Solution :9 O is between @ and $ so correct option is %e(
682 @ow many persons are seated between O and ? %a( Dne %b( wo %c( hree %d( )an not be determined
%e( 6one of these
Solution :9 $ I and + total three persons. option %c( is correct
683 Who among the following are the three lady members ? %a( $ + and H %b( $ @ and + %c( +$ @ and H %d( )an not be determined these
%e( 6one of
Solution :9$ +$ @ so option %b( is correct
684 Who among the following is to the immediate left of ? %a( + %b( I %c( H %d( )an not be determined %e( 6one of these Solution :9 H$ correct option is %c(
68" Which of the following is true about H ? %a( H is a male member %b( H is a female member %c( e2 of H can not be determined %d( Eosition of H can not be determined %e( 6one of these Solution :9here are only three female members % $ +$ @ ($ so rest of them are male members$ option %a( is correct
tudy the following information carefully and answer the uestions gi,en below 7 B$ ) $ K$ $ +$ @ and H are se,en students studying in three colleges E$ X and . here are three boys and four girls. here are at least one boy and one girl in each college. hree of them are in commerce discipline and two each in Arts and cience. B and her sister + are in science discipline but in different colleges. studies Arts in in college X and he does not study either with H or ). K is not in )ommerce and he studies in college only with B. All the three from )ommerce discipline do not study in the same college. @ studies in the same college with her friend +. Solution:9Let us analyse the uestion in seperate statements %1( hree of them are in commerce discipline and two each in Arts and cience. %;( B and her sister + are in science discipline but in different colleges. %0( studies Arts in in college X and he does not study either with H or ). %'( K is not in )ommerce and he studies in college only with B. %5( All the three from )ommerce discipline do not study in the same college. %8( @ studies in the same college with her friend +. irst we select simple condition %0( studies Arts in in college X and he does not study either with H or ). )ollege E
)ollege X
)ollege
K
Arts
Boy Boy
%'( K is not in )ommerce and he studies in college only with B. )ollege E )ollege X
)ollege
K B
Arts
Boy Boy
%;( B and her sister + are in science discipline but in different colleges. )ollege E )ollege X
)ollege
K B
Arts
Boy
cience
Boy +irl
%8( @ studies in the same college with her friend + means both are girls )ollege E )ollege X
)ollege
@ + K B
Arts cience
Boy +irl +irl
cience
Boy +irl
Again from statement %'( K is not in )ommerce and he studies in college only with B. o$ K will be from Arts$ as B and + are from sc ience )ollege E )ollege X
)ollege
@ +
Arts cience
Boy +irl +irl
K B
Arts cience
Boy +irl
emaning we can get from %0( studies Arts in in college X and he does not study either with H or ). H$ ) will be remaning )ommerce stream and study in college E but se2 is not determined. @ will also from commerce stream
)ollege E
H )
)ommerce )ommerce
Boy or +irl Boy or +irl
)ollege X
@ +
Arts )ommerce cience
Boy +irl +irl
)ollege
K B
Arts cience
Boy +irl
681 In which college do only )ommerece students study ? %a( 6one %b( X %c( %d( E %e( E or X Solution :9 )ollege E Dption %d(
682 If ) and B interchange their colleges satisfying all other conditions$ which of the following will definitely represent the girls ? %a( B)@K %b( B)@+ %c( B@+ %d( K)@ %e( 6one of these Solution:9 If ) and B interchange their position then ) must be a girl$ so B)@+ will be girls $ so option %b( correct
683 Which of the following represents the three in commerce discipline ? %a( @H) %b( @K) %c( @+ %d( Kata inadeuate %e( 6one of these Solution:9 option %a( @)H is correct
684 In which college do three of them study ? %a( E %b( %c( E or X %d( Kata inadeuate
%e( 6one of these
Solution :9 )orrect option is %e( none of these because all three students read in E and X
68" Which of the following represents three Boys ? %a( KH
%b( K)
%c( KH or K)
%d( Kata inadeuate
%e( 6one of these
Solution:9 )orrect option is %c( >inding Missing /haracters:9 In this section$ there are some figures gi,en which which reflects some patterns$ on basis of which you ha,e to find the missing characters$ always try to find the hidden EA6 to sol,e these problems. 681 ind out the missing number7'
8
1>
;
0
;
0
'
?
8
>
;=
Solution:9 In first column and second column we find
o missing character will be 0
6 82 ind the missing character in the figure
Solution:9 in first figure 7-
In second figure 7-
683 ind the missing character from the gi,en figure ?
Solution :9 Dn looking the figure we find the gi,en pattern
684 ind out the missing number in second triangle ?
Solution :9 Hust look the patterns of ouside numbers in triangle 1 and 0 we get the conclusion $
68" ind the missing number in figure 0 rd ?
Solution:9irst analyse the figure 1 and figure ; and try to ge t relationship
68# ind out the missing character ? :B
8A
0B
5)
;B
?
'A
>)
;A
%a( 1; A
%b( 5)
%c( =B
%d( ')
Solution:9 @ere each column contains A$B$)$ so one of missing character in 0rd column will be ) ach column has sum of two numbers eual to third$ ' & 5 ! : and 8 & ; ! >$ therefore 0&;! 5 $ so Ans is 5) 68$ ind the the missing characters in the gi,en table. ?
Solution:9 in any row multiple of first two column numbers is eual to third column number
68. ind the missing characters ? F'
30
C:
8
);
?
5
'
E15
%a( 1<
%b(1;
%c( A1<
%d( A1;
Solution:9 @ere each row letters are alternati,e alphabets$ F$ 3$ C D $ $ E imilarly $ ) then A
also numbers in each column are in arithmetic series$ '$ 5$ 8 and 0$ ;$ ' o :$ 15 then 1;
herefore reuired answer is A1;
Arithmetical %easoning Part91
681 A bus dri,er knows four different routes from Kelhi to Bareli$ he knows three different routes from Bareli to Lucknow and two different routes from Lucknow to +orakhpur. @ow many different routes he knows from Kelhi to +orakhpur ? %a( ;' routes
%b( 1; routes
%c( ;< routes
%d( > routes
%e( 6one of these
Solution:9 otal number of routes from Kelhi to +orakhpur! '0;!;'
682 Orishna was to earn s '<<< and a free holiday for eight weeks work. @e worked for only 0 weeks an earned and s. 1;<< and a free holiday. what was the ,alue of holiday ? %a( ;8<
%b( 05>
%c( '><
%d( ''<
%e( 6one of these
Solution:9Let the cost of holiday was s 2 then$ Eay for > weeks work ! '<<< & 2
2! '><$ option c is correct 683 he diagram gi,en below$ In this '<< candidates appeared in an e2amination. he diagram gi,es the number f students who failed in sub4ects *aths$ nglish and cience. What is the percentage of students who failed in at least two sub4ects ?
Solution :9 6umber of students who failed in at least two sub4ects ! number of students who failed in two or more su4ects ! 1; & > & > & 8 ! 0'
684 am has some mangoes to distributes among his students. If he keep '$ 5$ 8 in a pack he left with one mango$ But if he keeps 11 mangoes in a pack left with no mangoes. What is the minimum number of mangoes he has to distribute ? %a( 1'1
%b( 1<<
%c( 1;1
%d( 151
%e( 6one of these
Solution :9he reuired number will be such that it will produce a remainder 1 when di,ided by '$ 5 and 8 but gi,es no remainder when di,ided by 11$ o$ the reuired number of mangoes will be 1;1$ option %c( is correct. 68" In a school$ 85 Y students plays cricket$ ' < Y plays football and ;5 Y plays both foolball and cricket. What percentage of students neighter plays cricket nor plays football ? %a( 1<
%b( 5
%c( ;<
%d( 15
Solution:9Let the total number of students are 1<<
Let )ricket represents )$ ootball as hen ) & ;5 ! 85$ ∴ )!'<
& ;5 ! '<$ ∴ ! 15 ∴ 6umber
of students neighter plays cricket nor plays football ! 1<< - % ) & & ; 5 ( ! ;< Y $ Dption %c( is correct 68# hree uantities A$ B and ) are such that AB ! O)$ where O is a constant. When A is kept constant$ B ,aries directly as )$ When B is kept constant $ A ,aries directly ) and when ) is kept constant$ A ,aries in,ersly as B. Initially $ A was at 5 and A 7 B 7 ) was 1 7 0 7 5. ind the ,alue of A when B euals : at constant ). %a( >.00
%b( >
%c( :.;5
%d( :.55
Solution :9 Initially when A 7 B 7 ) ! 1 7 0 7 5 and A ! 5$ then B ! 15 and ) ! ;5 AB ! O) %gi,en($ o$ 5 15 ! O ;5 o O ! 0$ 6ow euation becomes AB ! 0) When B ! : at constant )$ A : ! 0 ;5 A ! >.00 option A is correct. 68$ A car tra,els from A to B at C1 km#hr$ tra,els back from B to A at C; km#hr and again goes back from A to B at C; km#hr. he a,erage s peed of the car is 7
Solution :9 Let the distance between two cities be A Om$ hen a,erage speed7
Dption %c( is correct 68. A$ B$ )$ K and play a game of cards. A says to B S If you gi,e me 0 cards$ you will ha,e as I ha,e at this moment while if K takes 5 cards from you$ he will ha,e as many as has S. A and ) together ha,e twice as many cards as has. B and K together also ha,e the same number of cards as A and ) taken together . If together they ha,e 15< cards. @ow many cards ) has ? %a( ;;
%b( ;'
%c( ;8
%d( ;>
Solution :9 @ere $ A & B & ) & K & ! 15< A!B-0 A & ) ! ; K&5!
%e( 0<
% asked in )A (
B&K!A&)!; ; & ; & ! 15<$ K & 5 ! 0< $
! 0<
K ! ;5
B & ;5 ! ; 0<
B ! 05
A ! 05 - 0 ! 0; 0; & ) ! ; 0<
) ! ;> option % d ( is correct
ne?ual!t0 bsolute alue
T%e absolute &alue of a number measures its distance to t%e origin on t%e real number line. Since < is at < units distance from t%e origin 8, t%e absolute &alue of < is <, <3<. Since '< is also at a distance of < units from t%e origin, t%e absolute &alue of '< is <, '<3< 5neuality e are ready for our first inequality. ?ind t%e set of solutions for x <. Translate into
Inglis%@ $e are looking for t%ose real numbers x $%ose distance from t%e origin is less t%an < units. b&iously $e are talking about t%e inter&al !'<,<"@ so all t%e &alues of U are '7 , '; , '- , '+ , 8 , +, - , ; , 7 . ?or t%e equation x VWX <, t%e solution $ill include < H '< $it% t%e set of ot%er numbers. Sol&ing linear inequalities is &ery similar to sol&ing linear equations, except for one detail@ you flip t%e inequality sign $%ene&er you multiply or di&ide t%e inequality by a negati&e. T%e easiest $ay to s%o$ t%is is $it% some examples U 0 ; - add '; to bot% t%e site ' ; '; ''''''''''''''''''' U '+ T%e only difference %ere is t%at you %a&e a Yless t%anY sign, instead of an YequalsY sign *asic Rules
a b implies t%at b 4 a implies t%at b'a 4 8 (nd a VWX b implies t%at b VWZ a implies t%at b'a 4 8 or b 3 a.
5ere are some rules for ordering real numbers. Try pro&ing t%ese yourself Let a, b, c and d be real numbers@ +"a 8 and b 8 implies t%at ab 4 8. -"a b and b c implies t%at a c. ;"a b implies t%at a 0 c b 0 c. 7"a b and c 4 8 implies t%at ac bc <"a b and c 8 implies t%at ac 4 bc. Kultiplying by a negati&e number re&erses t%e inequality. 9"a 4 + implies t%at a - 4 a. J"8 a + implies t%at a- a ="8 a b implies t%at a - b- F" 8 a, 8 b, and a- b- implies t%at a b. >o$ t%at $e %a&e all of t%ese rules, $e can start sol&ing inequalities. T%is is done by manipulating t%e inequality into a form t%at %as t%e &ariable on one side and %as a real expression on t%e ot%er side of t%e inequality. ?or instance, if t%e &ariable is x and t%e real expression is represented by a t%en t%e final form of t%e inequality is one of t%e follo$ing@ x4a xa xa xa 7olving 5,8925T587
#n t%is section, you $ill learn %o$ so sol&e inequalities. YSol&ing)) an inequality means finding all of its solutions. ( Ysolution)) of an inequality is a number $%ic% $%en substituted for t%e &ariable makes t%e inequality a true statement. 5ere is an example@ Consider t%e inequality U V[\ - 4 <. %en $e substitute = for x, t%e inequality becomes ='- 4 <. T%us, x 3= is a solution of t%e inequality. n t%e ot%er %and, substituting '- for x yields t%e false statement !'-"'- 4 <. T%us x 3 '- is >T a solution of t%e inequality. #nequalities usually %a&e many solutions. (s in t%e case of sol&ing equations, t%ere are certain manipulations of t%e inequality $%ic% do not c%ange t%e solutions. 5ere is a list of Ypermissible)) manipulations@ ule +. (dding:subtracting t%e same number on bot% sides.
8ample: T%e inequality x '-4< %as t%e same solutions as t%e inequality x 4 J. !T%e second inequality
$as obtained from t%e first one by adding - on bot% sides." ule -. S$itc%ing sides and c%anging t%e orientation of t%e inequality sign . 8ample: T%e inequality <' x 4 7 %as t%e same solutions as t%e inequality 7 < ' x . !e %a&e s$itc%ed
sides and turned t%e ]]4)) into a ]]))". Last, but not least, t%e operation $%ic% is at t%e source of all t%e trouble $it% inequalities@ ule ;a. Kultiplying:di&iding by t%e same PS#T#^I number on bot% sides. ule ;b. Kultiplying:di&iding by t%e same >IG(T#^I number on bot% sides (>D c%anging t%e orientation of t%e inequality sign. 8amples: T%is sounds %armless enoug%. T%e inequality -x9 %as t%e same solutions as t%e inequality
x;. !e di&ided by 0- on bot% sides". T%e inequality '- x 4 7 %as t%e same solutions as t%e inequality x '-. !e di&ided by !'-" on bot% sides and s$itc%ed Y4)) to Y))." But ule ; pro%ibits fancier mo&es @ T%e inequality x_` 4 x DIS >T %a&e t%e same solutions as t%e inequality x 4 +. !e $ere planning on di&iding bot% sides by x , but $e can)t, because $e do not kno$ at t%is point $%et%er x $ill be positi&e or negati&eN" #n fact, it is easy to c%eck t%at x 3 '- sol&es t%e first inequality, but does not sol&e t%e second inequality. nly ]]easy)) inequalities are sol&ed using t%ese t%ree rules most inequalities are sol&ed by using different tec%niques. Let)s sol&e some inequalities@ 8ample : /onsider the ineuality $ ; & < =
T%e basic strategy for inequalities and equations is t%e same@ isolate x on one side, and put t%e Yot%er stuffY on t%e ot%er side. ?ollo$ing t%is strategy, let)s mo&e 0< to t%e rig%t side. e accomplis% t%is by subtracting < on bot% sides !ule +" to obtain !-x 0 < " V[\ < J V[\ <, after simplification $e obtain -x nce $e di&ide by 0- on bot% sides !ule ;a", $e %a&e succeeded in isolating x on t%e left@ -x:- -:- or simplified, x + (ll real numbers less t%an + sol&e t%e inequality. e say t%at t%e Yset of solutions)) of t%e inequality consists of all real numbers less t%an +. #n interval notation , t%e set of solutions is t%e inter&al !'V , +".
%en t%e inequalities in&ol&e absolte &alues, you must be &ery careful $it% t%e use of t%e $ordsand . and or . T%ey end up gi&ing quite different results. Take a look %ere to see $%at $e mean. %en $e %a&e t%e absolute &alues, t%e and . condition applies $%en $e %a&e or signs and t%e or . condition applies $%en $e %a&e 4 or signs. So $e %a&e@ x a means 'a x a !$%ic% is t%e same as 'a x and x a". x a means V[\a VWX x VWZ a !$%ic% is t%e same as 'a x and x a" x 4 a means x 'a or x 4 a x a means x VWX 'a or x VWZ a ur met%od fails for more contri&ed examples. Let us consider t%e inequality x ';- x '7 #t)s back to basic algebra $it% a t$ist. T%e standard definition for t%e absolute &alue function is gi&en by@ x 3 x , if x VWZ 8 3 'x , if xVWX8 T%us $e can get rid of t%e sign in our inequality if $e kno$ $%et%er t%e expression inside, x ';, is positi&e or negati&e. e $ill do exactly t%atN Let)s first consider only t%ose &alues of x for $%ic% x V[\ ; VWZ 8. /ase 1 @ xVWZ; #n t%is case $e kno$ t%at x ';3 x ';, so our inequality becomes '() - x '7
Sol&ing t%e inequality, $e obtain x 4+
e %a&e found some solutions to our inequality@ x is a solution if x VWZ ;and x 4+ at t%e same timeN e)re talking about numbers U VWZ;. %at if x ';8 /ase $ @ x ; T%is time x ';8, so x ';3'! x ';"3;' x , so our inequality reads as
;' x - x '7.
(pplying t%e standard tec%niques, t%is can be simplified to x 4 J:; ur inequality %as some more solutions@ Onder our case assumption x ;, solutions are t%ose real numbers $%ic% satisfy x 4 J:; . e)re talking about numbers in t%e inter&al !J:;, ;" .Combining t%e solutions $e found for bot% cases, $e conclude t%at t%e set of solutions for t%e inequality x ';- x '7 are t%e numbers in t%e inter&al !J:; , 0V". 7olve x $ >?@ % x ; $ A 0
?actoring, $e get y 3 x - V[\ ; x 0 - 3 ! x V[\ -"! x V[\ +". Look at eac% of t%ese factors separately. T%e factor x V[\ + is positi&e for x 4 + similarly, x V[\ - is positi&e for x 4 - . T%inking back to $%en you first learned about negati&es, you kno$ t%at !plus"!plus" 3 !plus", !minus"!minus" 3 !plus", and !minus"!plus" 3 !minus". So, to compute t%e sign on y 3 x - V[\ ; x 0 - , $e only really need to kno$ t%e signs on t%e factors. So t%e solution of x - V[\ ; x 0 - 4 8 are t%e t$o inter&als -,egative in!inity( 1+ and -$( positive in!inity+.
ational #nequalities 7olve x B- x >?@ %+ < $ .
?irst off, you %a&e to remember t%at you can)t begin sol&ing until you %a&e t%e inequality in Y 3 8 Y format. 5ere)s %o$ t%e problem $orks@ x:!x';" VWX VW x : !x';" ' - VWX 8 >o$ con&ert to a common denominator@ x : !x V[\ ;" V[\ -!x';":!x';" VWX 8 ...and simplify@ x' - ! x V[\ ; " R : !x V[\ ; " VWX 8 same as !x' -x 0 9 ": !x V[\ ;" VWX 8 same as !'x 0 9 ": !x V[\ ;" VWX 8 T%e t$o factors are V[\ x 0 9 and x V[\ ; . >ote t%at x cannot equal ; , or else you $ould be di&iding by ero, $%ic% is not allo$ed. T%e first factor, V[\ x 0 9 , equals ero $%en x 3 9 . T%e ot%er factor, x V[\ ; , equals ero $%en x 3 ; . >o$, x cannot actually equal ; , so t%is endpoint $ill not be included in any solution inter&al !e&en t%oug% t%is is an Yor equal toY inequality", but $e need t%e &alue in order to figure out $%at our inter&als are. #n t%is case, our inter&als are !negati&e infinity, ; ", !;, 9R , and 9 , positi&e infinity". >ote t%e use of brackets to indicate t%at 9 can be included in t%e solution, but t%at ; cannot. To find the interest (I) and amount (A) of a given sum (P) in a given time (n) at simple interest The interest of * for one "ear is *r = and therefore for n "ears is *nr D
ie
5 ; Xnr
0+1
Also= A * - ( ie A ' - nr)
081
!rom 0+1 and 081= -e see that if of the quantities\ *, r, ( or *, n, r, and A/an" three #e gi$en the fourth ma" #e found To find the interest (I) and amount (A) of a given sum (P) in a given time (n) at compound interest The amount of X at the end of the first "ear is *0 D and since= this is the principal for the second "ear= the amount at the end of the second "ear is *0 1 0 or *0 2 imilarl"= the amount at the end of the third "ear is *0 3= and so onD hence the amount in n "ears is *0 n ie= A *0 n
therefore ( *'0 n$)
Note: 5f r denotes the interest on Re + for one "ear= -e ha$e3 0 - r
n business transactions, when the time contains a fraction of a year, it is usual to allow sim4le interest for the fraction of the year. Thus, the amount of e. % in = years is reckoned Y and the
amount of is years at compound interest is
1imilarly, the amount of in years is
5f the interest is pa"a#le more than once a "ear there is a distinction #et-een the nominal annual rate of interest and that actuall" recei$ed= -hich ma" #e called the true annual rateD thus if the interest is pa"a#le t-ice a "ear= and if r is the nominal annual rate of interest= the amount of Re + in half a "ear is = and therefore in the -hole "ear the amount of Re + is D so that the true annual rate of interest is 5f the interest is pa"a#le q times a "ear= ie compounding is done e$er" q months= and if r is the nominal annual rate= the interest on Re + for each inter$al is and therefore= the amount of X in n "ears is 5n this case= the interest is said to #e con$erted into principal] q times a "ear 5f the interest is con$erti#le into principal e$er" moment= then q #ecomes infinitel" great To find the $alue of the amount= put so that 5 $ rx D •
Thus the amount ; ; *enr 0since= % is infinite -hen q is infinite1
Tips and 7trategies to crack /ritical Reasoning in /ommon dmission Test -/T+ W#at Does a Cr!t!-al )eason!n" uest!on 'ook '!keJ After two years in reerse gear, when the com4etitors dented the market share of #78, the com4any has moed into to4 gear. 9rom a low of :2 4ercent in ;une 2<<<, the #78 market share swelled to 6< 4ercent in ;une 2<<.
Which among the following can be inferred from the aboe argument? 'a) #78 will again be the leading share holder in the market 'b) #78 will be the leading car share holder in the market 'c) #78 will be at the centre stage once again 'd) #78 may be at the centre stage once again
This is ho- a t"pical critical reasoning question looks like 5t has an argument= follo-ed #" the stem question= -hich is then follo-ed #" four options *ou are asked to choose the correct option
The first step to-ards tackling critical reasoning question is that of identif"ing the premise= conclusion and assumption of an argument ometimes a conclusion or a premise can #e identified #" the presence of special -ords that attach them to the different parts of an argument
The" are kno-n as conclusionIindicators ome of them are listed #elo- therefore hence thus so accordingl" 5n consequence consequentl" pro$es that as a result for this reason imilarl" there are certain phrases -hich points to the premise The" are kno-n as premiseIindicators ome of them are listed #elo-3 since #ecause for as as sho-n #" Note: Not e$er" argument -ould contain these indicators The meaning and conte%t of t he propositions ma" help us recogni/e the premise and the assumption
T0,!-al uest!ons There are some common question t"pes -hich can #e categori/ed as follo-s3 + Assumption 5 The assumption is the link #et-een the premise and the conclusion= hence should properl" fit #et-een them
8 5nference9Conclusion 5 An inference or a conclusion -ill not #e stated in the passage #ut has to #e #ased on the information gi$en in the passage ?trengthening 5n such questions "ou ha$e to look for the statement that -ould add more authenticit" to the argument : Weakening 5 5n -eakening questions= "ou need to look for the statement -hich questions the authenticit" of the argument \additional information that -ould nullif" the claim made in the argument Arithmetical %easoning Part91
681 A bus dri,er knows four different routes from Kelhi to Bareli$ he knows three different routes from Bareli to Lucknow and two different routes from Lucknow to +orakhpur. @ow many different routes he knows from Kelhi to +orakhpur ? %a( ;' routes
%b( 1; routes
%c( ;< routes
%d( > routes
%e( 6one of these
Solution:9 otal number of routes from Kelhi to +orakhpur! '0;!;'
682 Orishna was to earn s '<<< and a free holiday for eight weeks work. @e worked for only 0 weeks an earned and s. 1;<< and a free holiday. what was the ,alue of holiday ? %a( ;8<
%b( 05>
%c( '><
%d( ''<
%e( 6one of these
Solution:9Let the cost of holiday was s 2 then$ Eay for > weeks work ! '<<< & 2
2! '><$ option c is correct 683 he diagram gi,en below$ In this '<< candidates appeared in an e2amination. he diagram gi,es the number f students who failed in sub4ects *aths$ nglish and cience. What is the percentage of students who failed in at least two sub4ects ?
Solution :9 6umber of students who failed in at least two sub4ects ! number of students who failed in two or more su4ects ! 1; & > & > & 8 ! 0'
684 am has some mangoes to distributes among his students. If he keep '$ 5$ 8 in a pack he left with one mango$ But if he keeps 11 mangoes in a pack left with no mangoes. What is the minimum number of mangoes he has to distribute ? %a( 1'1
%b( 1<<
%c( 1;1
%d( 151
%e( 6one of these
Solution :9he reuired number will be such that it will produce a remainder 1 when di,ided by '$ 5 and 8 but gi,es no remainder when di,ided by 11$ o$ the reuired number of mangoes will be 1;1$ option %c( is correct. 68" In a school$ 85 Y students plays cricket$ ' < Y plays football and ;5 Y plays both foolball and cricket. What percentage of students neighter plays cricket nor plays football ? %a( 1<
%b( 5
%c( ;<
%d( 15
Solution:9Let the total number of students are 1<<
Let )ricket represents )$ ootball as hen ) & ;5 ! 85$ ∴ )!'< & ;5 ! '<$ ∴ ! 15 ∴ 6umber
of students neighter plays cricket nor plays football ! 1<< - % ) & & ; 5 ( ! ;< Y $ Dption %c( is correct 68# hree uantities A$ B and ) are such that AB ! O)$ where O is a constant. When A is kept constant$ B ,aries directly as )$ When B is kept constant $ A ,aries directly ) and when ) is kept constant$ A ,aries in,ersly as B. Initially $ A was at 5 and A 7 B 7 ) was 1 7 0 7 5. ind the ,alue of A when B euals : at constant ). %a( >.00
%b( >
%c( :.;5
%d( :.55
Solution :9 Initially when A 7 B 7 ) ! 1 7 0 7 5 and A ! 5$ then B ! 15 and ) ! ;5 AB ! O) %gi,en($ o$ 5 15 ! O ;5 o O ! 0$ 6ow euation becomes AB ! 0) When B ! : at constant )$ A : ! 0 ;5 A ! >.00 option A is correct. 68$ A car tra,els from A to B at C1 km#hr$ tra,els back from B to A at C; km#hr and again goes back from A to B at C; km#hr. he a,erage s peed of the car is 7
