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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 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align="center"><b>CLOCKS</b></h2><br><br><font size="4">Medium Problems<br><br><br><u>Type3 </u>: At what time between 4 and 5 o'clock will the hands of a clock be at <br> rightangle?<br><br>Solution : In this type of problems the formulae is<br> (5*x + or -15)*(12/11)<br> Here x is replaced by the first interval of given time here i.e 4<br><br> Case 1 : (5*x + 15)*(12/11)<br> (5*4 +15)*(12/11)<br> (20+15)*(12/11)<br> 35*12/11=420/11=38 2/11 min.<br> Therefore they are right angles at 38 2/11 min .past4 <br><br> Case 2 : (5*x-15)*(12/11)<br> (5*4-15)*(12/11)<br> (20-15)*(12/11)<br> 5*12/11=60/11 min=5 5/11min<br> Therefore they are right angles at 5 5/11 min.past4.<br><br> <u>Another shortcut for type 3 is</u> :Here the given angle is right angle i.e 900.<br><br> Case 1 : The formulae is 6*x-(hrs*60+x)/2=Given angle<br> 6*x-(4*60+x)/2=90<br> 6*x-(240+x)/2=90<br> 12x-240-x=180<br> 11x=180+240<br> 11x=420<br> x=420/11= 38 2/11 min<br><br> Therefore they are at right angles at 38 2/11 min. past4.<br><br> Case 2 : The formulae is (hrs*60+x)/2-(6*x)=Given angle<br> (4*60+x)/2-(6*x)=90<br> (240+x)/2-(6*x)=90<br> 240+x-12x=180<br> -11x+240=180<br> 240-180=11x <br> x=60/11= 5 5/11 min<br><br> Therefore they art right angles at 5 5/11 min past4.<br><br><u>Type 4 </u> : Find at what time between 8 and 9 o'clock will the hands of a clock be<br><br> in the same straight line but not together ?<br>Solution : In this type of problems the formulae is<br><br> (5*x-30)*12/11<br> x is replaced by the first interval of given time Here i.e 8 <br> (5*8-30)*12/11<br> (40-30)*12/11<br> 10*12/11=120/11 min=10 10/11 min.<br> Therefore the hands will be in the same straight line but not<br> together at 10 10/11 min.past 8.<br><br><u>Another shortcut for type 4 is</u> : Here the hands of a clock be in the same<br> straight line but not together the angle is 180 degrees.<br> The formulae is (hrs*60+x)/2-(6*x)=Given angle<br> (8*60+x)/2-6*x=180<br> (480+x)/2-(6*x)=180<br> 480+x-12*x=360<br> 11x=480-360<br> x=120/11=10 10/11 min.<br> therefore the hands will be in the same straight line but not<br> together at 10 10/11 min. past8.<br><u>Type 5 </u> : At what time between 5 and 6 o鈥?? 快捷鍵說明
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