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<html><table height="500" width="1000" border="2"><TR height="5" width="1000"><strong><center><h2>APTITUDE</h2></center></strong></TR><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="numbers.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="chainrule1.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 contract is to be completed in 46 days and 117 men were set to work,each working 8 hours a day. After 33 days, 4/7 of the work is completed.How many additional men may be employed so that the work may be completedin time, each man now working 9 hours a day? Sol: 4/7 of work is completed . Remaining work=1- 4/7 =3/7 Remaining period= 46-33 =13 days Less work, less men(direct proportion) less days, more men(Indirect proportion) More hours/day, less men(Indirect proportion) work 4/7:3/7 Days 13:33 :: 117:x hrs/day 9:8 (4/7)*13*9*x=(3/7)*33*8*117 x=(3*33*8*117) / (4*13*9) x=198 men So, additional men to be employed=198 -117=812)A garrison had provisions for a certain number of days. After 10 days, 1/5 of the men desert and it is found that the provisions will now last just as long as before. How long was that? Sol: Let initially there be x men having food for y days. After, 10 days x men had food for ( y-10)days Also, (x -x/5) men had food for y days. x(y-10)=(4x/5)*y => (x*y) -50x=(4(x*y)/5) 5(x*y)-4(x*y)=50x x*y=50x y=503)A contractor undertook to do a certain piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completesthe work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?Sol: 40 days- 35 days=5 days =>(100*35)+(200*5) men can finish the work in 1 day. 4500 men can finish it in 4500/100= 45 days This s 5 days behind the schedule.4)12 men and 18 boys,working 7 陆 hors a day, can do a piece if work in 60 days. If a man works equal to 2 boys, then how many boys will be required to help 21 men to do twice the work in 50 days, working 9 hours a day? Sol: 1man =2 boys 12men+18boys=>(12*2+18)boys=42 boys let the required number of boys=x 21 men+x boys =>((21*2)+x) boys =>(42+x) boys less days, more boys(Indirect proportion) more hours per day, less boys(Indirect proportion) days 50:60 hrs/day 9:15/2 :: 42:(42+x) work 1:2 (50*9*1*(42+x))=60*(15/2)*2*42 (42+x)= (60*15*42)/(50*9)= 84 x=84-42= 42 =42 42 days behind the schedule it will be finished.<font size="3"><a href="chainrule1.html" target="right"><strong>BACK</strong></a><br><br></font></PRE></font></TD></TR></table></HTML>⌨️ 快捷键说明
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