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<html><table height="500" width="1000" border="2"><TR height="5" width="1000"><strong><center><marquee><font color="Green"><h1>APTITUDE</h1></center></strong></font></TR></marquee><TR><TD align="left" width="200" valign="top"><table><TR><a href="numbers.html"><strong>Numbers</strong></a></TR><br><TR><a href="hcf.html"><strong>H.C.F and L.C.M</strong></a></TR><br><TR><a href="dec.html" ><strong>Decimal Fractions</strong></a></TR><br><TR><a href="simplification.html"><strong>Simplification</strong></a></TR><br><TR><a href="squareandcuberoot.html" ><strong>Square and Cube roots</strong></a></TR><br><TR><a href="average.html" ><strong>Average</strong></a></TR><br><TR><a href="pnumbers.html" ><strong>Problems on Numbers</strong></a></TR><br><TR><a href="problemsonages.html"><strong>Problems on Ages</strong></a></TR><br><TR><a href="surdsandindices.html"><strong>Surds and Indices</strong></a></TR><br><TR><a href="percent.html" ><strong>Percentage</strong></a></TR><br><TR><a href="profitandloss.html" ><strong>Profit and Loss</strong></a></TR><br><TR><a href="ratioandproportion.html" ><strong>Ratio And Proportions</strong></a></TR><br><TR><a href="partnership.html"><strong>Partnership</strong></a></TR><br><TR><a href="chainrule1.html"><strong>Chain Rule</strong></a></TR><br><TR><a href="timeandwork.html" ><strong>Time and Work</strong></a></TR><br><TR><a href="pipesandcisterns.html" ><strong>Pipes and Cisterns</strong></a></TR><br><TR><a href="timeanddistance.html"><strong>Time and Distance</strong></a></TR><br><TR><a href="trains.html" ><strong>Trains</strong></a></TR><br><TR><a href="boats.html"><strong>Boats and Streams</strong></a></TR><br><TR><a href="alligation.html"><strong>Alligation or Mixture </strong></a></TR><br><TR><a href="simple.html" ><strong>Simple Interest</strong></a></TR><br><TR><a href="CI.html"><strong>Compound Interest</strong></a></TR><br><TR><a href=""><strong>Logorithms</strong></a></TR><br><TR><a href="areas.html" ><strong>Areas</strong></a></TR><br><TR><a href="volume.html"><strong>Volume and Surface area</strong></a></TR><br><TR><a href="races.html" ><strong>Races and Games of Skill</strong></a></TR><br><TR><a href="calendar.html" ><strong>Calendar</strong></a></TR><br><TR><a href="clocks.html" ><strong>Clocks</strong></a></TR><br><TR><a href="" ><strong>Stocks ans Shares</strong></a></TR><br><TR><a href="true.html" ><strong>True Discount</strong></a></TR><br><TR><a href="banker1.html" ><strong>Bankers Discount</strong></a></TR><br><TR><a href="oddseries.html"><strong>Oddmanout and Series</strong></a></TR><br><TR><a href=""><strong>Data Interpretation</strong></a></TR><br><TR><a href="probability.html"><strong>probability</strong></a></TR><br><TR><a href="percom1.html" ><strong>Permutations and Combinations</strong></a></TR><br><TR><a href="pinkivijji_puzzles.html" ><strong>Puzzles</strong></a></TR></table></TD><TD valign="top"><a href="timeandwork.html" target="right"><strong>BACK</strong></a><br><br><font size="5"><center><b><u>COMPLEX PROBLEMS</u></b></center></font><font size="4"><PRE>1)A and B undertake to do a piece of workfor Rs 600.A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they can finish it in 3 days, Find the share of each?Sol: C's one day's work=(1/3)-(1/6+1/8)=1/24 Therefore, A:B:C= Ratio of their one day鈥檚 work=1/6:1/8:1/24=4:3:1 A's share=Rs (600*4/8)=300 B's share= Rs (600*3/8)=225 C's share=Rs[600-(300+225)]=Rs 752)A can do a piece of work in 80 days. He works at it for 10 days & then B alonefinishes the remaining work in 42 days. In how much time will A and B, workingtogether, finish the work?Sol: Work done by A in 10 days=10/80=1/8 Remaining work=(1-(1/8))=7/8 Now, work will be done by B in 42 days. Whole work will be done by B in (42*8/7)=48 days Therefore, A's one day's work=1/80 B鈥檚 one day's work=1/48 (A+B)'s one day's work=1/80+1/48=8/240=1/30 Hence, both will finish the work in 30 days.3)P,Q and R are three typists who working simultaneously can type 216 pages in 4 hours In one hour , R can type as many pages more than Q as Q can type more than P. During a period of five hours, R can type as many pages as P can during seven hours. How many pages does each of them type per hour?Sol: Let the number of pages typed in one hour by P, Q and R be x,y and z respectively Then x+y+z=216/4=54 ---------------1 z-y=y-x => 2y=x+z -----------2 5z=7x => x=5x/7 ---------------3 Solving 1,2 and 3 we get x=15,y=18, and z=214)Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?Sol: Number of pages typed by Ronald in one hour=32/6=16/3 Number of pages typed by Elan in one hour=40/5=8 Number of pages typed by both in one hour=((16/3)+8)=40/3 Time taken by both to type 110 pages=110*3/40=8 hours.5)Two workers A and B are engaged to do a work. A working alone takes 8 hoursmore to complete the job than if both working together. If B worked alone, he would need 4 1/2 hours more to compete the job than they both working together. What time would they take to do the work together.Sol: (1/(x+8))+(1/(x+(9/2)))=1/x =>(1/(x+8))+(2/(2x+9))=1/x => x(4x+25)=(x+8)(2x+9) => 2x2 =72 => x2 = 36 => x=6 Therefore, A and B together can do the work in 6 days.6)A and B can do a work in12 days, B and C in 15 days, C and A in 20 days.If A,B and C work together, they will complete the work in how many days?Sol: (A+B)'s one day's work=1/12; (B+C)'s one day's work=1/15; (A+C)'s one day's work=1/20; Adding we get 2(A+B+C)'s one day's work=1/12+1/15+1/20=12/60=1/5 (A+B+C)'s one day work=1/10 So, A,B,and C together can complete the work in 10 days.7)A and B can do a work in 8 days, B and C can do the same wor in 12 days. A,B and C together can finish it in 6 days. A and C together will do it in how many days?Sol: (A+B+C)'s one day's work=1/6; (A+B)'s one day's work=1/8; (B+C)'s one day's work=1/12; (A+C)'s one day's work=2(A+B+C)'s one day's work-((A+B)'s one day work+(B+C)'s one day work) = (2/6)-(1/8+1/12) =(1/3)- (5/24) =3/24 =1/8 So, A and C together will do the work in 8 days.8)A can do a certain work in the same time in which B and C together can do it.If A and B together could do it in 10 days and C alone in 50 days, then B alonecould do it in how many days?Sol: (A+B)'s one day's work=1/10; C's one day's work=1/50 (A+B+C)'s one day's work=(1/10+1/50)=6/50=3/25 Also, A's one day's work=(B+C)鈥檚 one day's work From i and ii ,we get :2*(A's one day's work)=3/25 => A's one day's work=3/50 B's one day鈥檚 work=(1/10-3/50) =2/50 =1/25 B alone could complete the work in 25 days.9) A is thrice as good a workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:Sol: Ratio of times taken by A and B=1:3. If difference of time is 2 days , B takes 3 days If difference of time is 60 days, B takes (3*60/2)=90 days So, A takes 30 days to do the work=1/90 A's one day's work=1/30; B's one day's work=1/90; (A+B)'s one day's work=1/30+1/90=4/90=2/45 Therefore, A&B together can do the work in 45/2days10) A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes the remaining work in 42 days. In how much time will A&B,working together, finish the work?Sol: Work Done by A n 10 days =10/80=1/8 Remaining work =1-1/8=7/8 Now 7/8 work is done by B in 42 days Whole work will be done by B in 42*8/7= 48 days => A's one day's work =1/80 and B's one day's work =1/48 (A+B)'s one day's work = 1/80+1/48 = 8/240 = 1/30 Hence both will finish the work in 30 days.11) 45 men can complete a work in 16 days. Six days after they started working,so more men joined them. How many days will they now take to complete the remaining work?Sol: M1*D1/W1=M2*D2/W2 =>45*6/(6/16)=75*x/(1-(6/16)) => x=6 days12)A is 50% as efficient as B. C does half the work done by A&B together. If C alone does the work n 40 days, then A,B and C together can do the work in:Sol: A's one day's work:B's one days work=150:100 =3:2 Let A's &B's one day's work be 3x and 2x days respectively. Then C's one day's work=5x/2 => 5x/2=1/40 => x=((1/40)*(2/5))=1/100 A's one day's work=3/100 B's one day's work=1/50 C's one day's work=1/40 So, A,B and C can do the work in 13 1/3 days.13)A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days A alone can finish the remaining work?Sol: B's 10 day's work=10/15=2/3 Remaining work=(1-(2/3))=1/3 Now, 1/18 work is done by A in 1 day. Therefore 1/3 work is done by A in 18*(1/3)=6 days.14)A can finish a work in 24 days, B n 9 days and C in 12 days. B&C start the work but are forced to leave after 3 days. The remaining work done by A in:Sol: (B+C)'s one day's work=1/9+1/12=7/36 Work done by B & C in 3 days=3*7/36=7/12 Remaining work=1-(7/12)=5/12 Now , 1/24 work is done by A in 1 day. So, 5/12 work is done by A in 24*5/12=10 days15)X and Y can do a piece of work n 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of work. How long did the work last?Sol: work done by X in 4 days =4/20 =1/5 Remaining work= 1-1/5 =4/5 (X+Y)'s one day's work =1/20+1/12 =8/60=2/15 Now, 2/15 work is done by X and Y in one day. So, 4/5 work will be done by X and Y in 15/2*4/5=6 days Hence Total time taken =(6+4) days = 10 days16)A does 4/5 of work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work? Sol: Whole work is done by A in 20*5/4=25 days Now, (1-(4/5)) i.e 1/5 work is done by A& B in days. Whole work will be done by A& B in 3*5=15 days =>B's one day's work= 1/15-1/25=4/150=2/75 So, B alone would do the work in 75/2= 37 陸 days.17) A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work n 23 days. The number of days after which A left the work wasSol: (A+B)'s one day's work=1/45+1/40=17/360 Work done by B in 23 days=23/40 Remaining work=1-(23/40)=17/40 Now, 17/360 work was done by (A+B) in 1 day. 17/40 work was done by (A+B) in (1*(360/17)*(17/40))= 9 days So, A left after 9 days.18)A can do a piece of work in 10 days, B in 15 days. They work for 5 days. The rest of work finished by C in 2 days. If they get Rs 1500 for the whole work, the daily wages of B and C areSol: Part of work done by A= 5/10=1/2 Part of work done by B=1/3 Part of work done by C=(1-(1/2+1/3))=1/6 A's share: B's share: C's share=1/2:1/3:1/6= 3:2:1 A's share=(3/6)*1500=750 B's share=(2/6)*1500=500 C's share=(1/6)*1500=250 A's daily wages=750/5=150/- B's daily wages=500/5=100/- C's daily wages=250/2=125/- Daily wages of B&C = 100+125=225/-19)A alone can complete a work in 16 days and B alone can complete the same in 12 days. Starting with A, they work on alternate days. The total work willbe completed in how many days?(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 戮 daysSol: (A+B)'s 2 days work = 1/16 + 1/12 =7/48 work done in 6 pairs of days =(7/48) * 6 = 7/8 remaining work = 1- 7/8 = 1/8 work done by A on 13th day = 1/16 remaining work = 1/8 鈥?? 快捷鍵說明
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