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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="pnumbers.html" name="right"><b>BACK</b></a><font size="5"><center><b>PROBLEMS ON NUMBERS</b></center></font><br><br> <font size="5"><b>SOLVED PROBLEMS</b></font><br><br><br><br> <font size="4"> <b>Complex Problems</b>:<br><br>24.Six bells commence tolling together and toll at intervals <br>of 2,4,6,8,10,12 seconds respectively. In 30 minutes how many <br>times do they toll together? <br> Solution:<br>To find the time that the bells will toll together we have<br> to take L.C.M of 2,4,6,8,10,12 is 120.<br><br>So,the bells will toll together after every 120 seconds<br> i e, 2 minutes<br><br>In 30 minutes they will toll together [30/2 +1]=16 times<br><br>25.The sum of two numbers is 15 and their geometric mean is <br>20% lower than their arithmetic mean. Find the numbers?<br><br>a.11,4 b.12,3 c.13,2 d.10,5 <br><br>Solution:<br><pre>Sum of the two numbers is a+b=15.<br>their A.M = a+b / 2 and G.M = (ab)<sup>1/2</sup><br>Given G.M = 20% lower than A.M<br> =80/100 A.M<br> (ab)<sup>1/2</sup>=4/5 a+b/2 = 2*15/5= 6<br> (ab)<sup>1/2</sup>=6<br> ab=36 =>b=36/a<br> a+b=15<br> a+36/a=15<br> a2+36=15a<br> a2-15a+36=0<br> a2-3a-12a+36=0<br> a(a-3)-12(a-3)=0<br> a=12 or 3.<br></pre>If a=3 and a+b=15 then b=12.<br>If a=12 and a+b=15 then b=3. <br><br>Ans (b).<br><br><br>26.When we multiply a certain two digit number by the<br> sum of its digits 405 is achieved. If we multiply the<br> number written in reverse order of the same digits <br>by the sum of the digits,we get 486.Find the number?<br><br>a.81 b.45 c.36 d. none<br><br>Solution:<br>Let the number be x y.<br><br>When we multiply the number by the sum of its digit <br>405 is achieved.<br><br> (10x+y)(x+y)=405....................1<br><br>If we multiply the number written in reverse order by its<br> sum of digits we get 486.<br> (10y+x)(x+y)=486......................2<br><br>dividing 1 and 2<br> (10x+y)(x+y)/(10y+x)(x+y) = 405/486.<br> 10x+y / 10y+x = 5/6.<br> 60x+6y = 50y+5x<br> 55x=44y<br> 5x = 4y.<br> From the above condition we conclude that the above <br>condition is satisfied by the second option i e b. 45.<br><br>Ans (b).<br><br><br>27.Find the HCF and LCM of the polynomials x2-5x+6 and x2-7x+10?<br>a.(x-2),(x-2)(x-3)(x-5)<br>b.(x-2),(x-2)(x-3)<br>c.(x-3),(x-2)(x-3)(x-5)<br>d. none<br>Solution:<br>The given polynomials are <br> x2-5x+6=0................1 <br> x2-7x+10=0...............2<br>we have to find the factors of the polynomials<br><pre> x2-5x+6 and x2-7x+10 x2-2x-3x+6 x2-5x-2x+10 x(x-2)-3(x-2) x(x-5)-2(x-5) (x-3)(x-2) (x-2)(x-5)</pre><br>From the above factors of the polynomials we can easily<br> find the HCF as (x-3)and LCM as (x-2)(x-3)(x-5).<br><br>Ans (c)<br><br>28.The sum of all possible two digit numbers formed from <br>three different one digit natural numbers when divided by <br>the sum of the original three numbers is equal to?<br><br>a.18 b.22 c.36 d. none<br><br>Solution:<br>Let the one digit numbers x,y,z<br>Sum of all possible two digit numbers<br>=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y) = 22(x+y+z)<br>Therefore sum of all possible two digit numbers when <br>divided by sum of one digit numbers gives 22.<br><br><br>29.A number being successively divided by 3,5,8 leaves<br> remainders 1,4,7 respectively. Find the respective<br> remainders if the order of divisors are reversed?<br><br>Solution:<br>Let the number be x.<br><pre> 3 - x 5 y - 1 8 z - 4 1 - 7 </pre><br>z=8*1+7=15<br>y=5z+4 = 5*15+4 = 79<br>x=3y+1 = 3*79+1=238<br><pre>Now 8 238 5 29 - 6 3 5 - 4 1 - 2</pre><br>Respective remainders are 6,4,2.<br><br><br>30.The arithmetic mean of two numbers is smaller by 24<br> than the larger of the two numbers and the GM of <br>the same numbers exceeds by 12 the smaller of the numbers.<br> Find the numbers?<br><br>a.6,54 b.8,56 c.12,60 d.7,55<br>Solution:<br>Let the numbers be a,b where a is smaller and b <br>is larger number.<br>The AM of two numbers is smaller by 24 than the<br> larger of the two numbers.<br> AM=b-24<br><br>AM of two numbers is a+b/2.<br> a+b/2 = b-24<br> a+b = 2b-48<br> a = b-48...................1<br>The GM of the two numbers exceeds by 12 the smaller<br> of the numbers<br><br> GM = a+12<br><pre>GM of two numbers is (ab)<sup>1/2</sup><br> (ab) <sup>1/2</sup>= a+12<br> ab = a2+144+24a<br>from 1 b=a+48<br> a(a+48)= a2+144+24a<br> a2+48a = a2+144+24a<br> 24a=144=>a=6<br> Therefore b=a+48=54.<br><br></pre>Ans (a).<br><br>31.The sum of squares of the digits constituting a positive<br> two digit number is 13,If we subtract 9 from that number <br>we shall get a number written by the same digits in the <br>reverse order. Find the number?<br><br>a.12 b.32 c.42 d.52.<br><br>Solution:<br>Let the number be x y.<br>the sum of the squares of the digits of the number is 13<br> x2+y2=13<br>If we subtract 9 from the number we get the number<br> in reverse order<br> x y-9=y x.<br> 10x+y-9=10y+x.<br> 9x-9y=9<br> x-y=1<br> (x-y)2 =x2+y2-2x y<br> 1 =13-2x y<br> 2x y = 12<br> x y = 6 =>y=6/x<br> x-y=1<br> x-6/x=1<br> x2-6=x<br> x2-x-6=0 <br> x+2x-3x-6=0<br> x(x+2)-3(x+2)=0<br> x=3,-2.<br>If x=3 and x-y=1 then y=2.<br>If x=-2 and x-y=1 then y=-3.<br>Therefore the number is 32.<br><br>Ans (b).<br><br>32.If we add the square of the digit in the tens place <br>of the positive two digit number to the product of the <br>digits of that number we get 52,and if we add the square<br> of the digit in the unit's place to the same product <br>of the digits we get 117.Find the two digit number?<br><br>a.18 b.39 c.49 d.28<br>Solution:<br>Let the digit number be x y<br>Given that if we add square of the digit in the tens place <br>of a number to the product of the digits we get 52.<br><br> x2+x y=52.<br> x(x+y)=52....................1<br>Given that if we add the square of the digit in the unit's plac<br>e to the product is 117.<br><br> y2+x y= 117 <br> y(x+y)=117.........................2<br>dividing 1 and 2 x(x+y)/y(x+y) = 52/117=4/9<br> x/y=4/9<br>from the options we conclude that the two digit number is 49 <br>because the condition is satisfied by the third option.<br><br>Ans (c)<br><br>33.The denominators of an irreducible fraction is greater <br>than the numerator by 2.If we reduce the numerator of the<br> reciprocal fraction by 3 and subtract the given fraction <br>from the resulting one,we get 1/15.Find the given fraction?<br><br>Solution:<br>Let the given fraction be x / (x+2) because given that <br>denominator of the fraction is greater than the numerator by 2<br> 1 鈥⌨️ 快捷键说明
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