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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" target="right"><strong>Decimal Fractions</strong></a></TR><br><TR><a href="simplification.html"><strong>Simplification</strong></a></TR><br><TR><a href="squareandcuberoot.html" target="right"><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" target="right"><strong>Percentage</strong></a></TR><br><TR><a href="profitandloss.html" target="right"><strong>Profit and Loss</strong></a></TR><br><TR><a href="ratioandproportion.html" target="right"><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" target="right"><strong>Time and Work</strong></a></TR><br><TR><a href="pipesandcisterns.html" target="right"><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" target="right"><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" target="right"><strong>Simple Interest</strong></a></TR><br><TR><a href="CI.html" target="right"><strong>Compound Interest</strong></a></TR><br><TR><a href=""><strong>Logorithms</strong></a></TR><br><TR><a href="areas.html" target="right"><strong>Areas</strong></a></TR><br><TR><a href="volume.html" target="right"><strong>Volume and Surface area</strong></a></TR><br><TR><a href="races.html" target="right"><strong>Races and Games of Skill</strong></a></TR><br><TR><a href="calendar.html" target="right"><strong>Calendar</strong></a></TR><br><TR><a href="clocks.html" target="right"><strong>Clocks</strong></a></TR><br><TR><a href="" target="right"><strong>Stocks ans Shares</strong></a></TR><br><TR><a href="true.html" target="right"><strong>True Discount</strong></a></TR><br><TR><a href="banker1.html" target="right"><strong>Bankers Discount</strong></a></TR><br><TR><a href="oddseries.html" target="right"><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" target="right" ><strong>Permutations and Combinations</strong></a></TR><br><TR><a href="pinkivijji_puzzles.html" target="right"><strong>Puzzles</strong></a></TR></table></TD><TD ><div><h2>TIME AND DISTANCE</h2></strong></div><font size="4"><strong><h3><u>Formula</u></h3></strong>I)Speed = Distance/Time<br><br>II)Time = Distance/speed<br><br>III) Distance = speed*time<br><br>IV) 1km/hr = 5/18 m/s<br>V)1 m/s = 18/5 Km/hr<br><br>VI)If the ratio of the speed of A and B is a:b,then the ratio of the <br>time taken by them to cover the same distance is 1/a : 1/b or b:a<br><br>VII) suppose a man covers a distance at x kmph and an equal distance at<br> y kmph.then the average speed during the whole journey is (2xy/x+y)kmph<br><br><strong><u>Solved Problems</u></strong><br><br><strong>1)A person covers a certain distance at 7kmph .How many meters <br>does he cover in 2 minutes.<br><br> Solution::</strong><br> speed=72kmph=72*5/18 = 20m/s<br> distance covered in 2min =20*2*60 = 2400m<br><br><strong>2)If a man runs at 3m/s. How many km does he run in 1hr 40min<br><br> Solution::</strong><br> speed of the man = 3*18/5 kmph<br> = 54/5kmph<br><br> Distance covered in 5/3 hrs=54/5*5/3 = 18km<br> <strong>3)Walking at the rate of 4knph a man covers certain distance <br>in 2hr 45 min. Running at a speed of 16.5 kmph the man will cover the <br>same distance in.<br><br>Solution::</strong><br> Distance=Speed* time<br> 4*11/4=11km<br><br> New speed =16.5kmph<br> therefore Time=D/S=11/16.5 = 40min<br><br><strong>1)A train covers a distance in 50 min ,if it runs at a speed<br> of 48kmph on an average.The speed at which the train must run to reduce<br> the time of journey to 40min will be.<br><br>Solution::</strong><br> Time=50/60 hr=5/6hr<br> Speed=48mph<br><br> distance=S*T=48*5/6=40km<br> time=40/60hr=2/3hr<br><br> New speed = 40* 3/2 kmph= 60kmph<br><br><strong>2)Vikas can cover a distance in 1hr 24min by covering 2/3 <br>of the distance at 4 kmph and the rest at 5kmph.the total distance is?<br><br>Solution::</strong><br><br> Let total distance be S<br> total time=1hr24min<br><br>A to T :: speed=4kmph<br> diistance=2/3S<br><br>T to S :: speed=5km<br> distance=1-2/3S=1/3S<br><br> 21/15 hr=2/3 S/4 + 1/3s /5<br>84=14/3S*3<br>S=84*3/14*3<br>= 6km<br><br><br><strong>3)walking at 戮 of his usual speed ,a man is late by 2 陸 hr.<br>the usual time is.<br><br>Solution::</strong><br> Usual speed = S<br> Usual time = T<br> Distance = D<br>New Speed is 戮 S<br>New time is 4/3 T<br>4/3 T 鈥?? 快捷鍵說明
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