{"id":5898,"date":"2025-06-09T06:56:51","date_gmt":"2025-06-09T06:56:51","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5898"},"modified":"2025-06-09T06:56:51","modified_gmt":"2025-06-09T06:56:51","slug":"rs-aggarwal-quantitative-aptitude-pdf-simplification-download","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-simplification-download\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free download: SIMPLIFICATION"},"content":{"rendered":"
    \n
  1. SIMPLIFICATION<\/u><\/strong><\/li>\n<\/ol>\n

    \u00a0<\/strong>IMPORTANT CONCEPTS<\/u><\/strong><\/p>\n

      \n
    1. \u2019BODMAS\u2019Rule: <\/strong>This rule depicts the correct sequence in which the operations are to be executed,so as to find out the value of a given expression.<\/li>\n<\/ol>\n

      Here, \u2018B\u2019 stands for \u2019bracket\u2019 ,\u2019O\u2019for \u2018of\u2019 , \u2018D\u2019 for\u2019\u00a0 division\u2019 \u00a0and \u2018M\u2019 for \u2018multiplication\u2019, \u2018A\u2019 for \u2018addition\u2019 and \u2018S\u2019 for \u2018subtraction\u2019.<\/p>\n

      Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order(), {} and [].<\/p>\n

      After removing the brackets, we must use the following operations strictly in the order:<\/p>\n

      (1)of (2)division (3) multiplication (4)addition (5)subtraction.<\/p>\n

      II. Modulus of a real number :<\/strong> Modulus of a real number a is defined as<\/h1>\n

      \u00a0\u00a0\u00a0\u00a0 |a| ={a, if a>0<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 -a, if a<0<\/p>\n

      \u00a0Thus, |5|=5 and |-5|=-(-5)=5.<\/p>\n

      III. Virnaculum (or bar): When an expression contains Virnaculum, before applying the \u2018BODMAS\u2019 rule, we simplify the expression under the Virnaculum.<\/p>\n

      SOLVED EXAMPLES<\/h2>\n

      Ex. 1. Simplify:<\/strong> (i)5005-5000+10\u00a0 (ii) 18800+470+20<\/p>\n

      Sol.<\/strong> (i)5005-5000+10=5005-(5000\/10)=5005-500=4505.<\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0(ii)18800+470+20=(18800\/470)+20=40\/20=2.<\/p>\n

      Ex. 2. Simplify: b-[b-(a+b)-{b-(b-a-b)}+2a]<\/h3>\n

      \u00a0Sol. Given expression=b-[b-(a+b)-{b-(b-a+b)}+2a]\u00a0<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =b-[b-a-b-{b-2b+a}+2a]<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =b-[-a-{b-2b+a+2a}]<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =b-[-a-{-b+3a}]=b-[-a+b-3a]<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =b-[-4a+b]=b+4a-b=4a.<\/p>\n

      Ex. 3. What value will replace the question mark in the following equation?<\/strong><\/h2>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4 1<\/u>+3 1<\/u>+?+2 1<\/u>=13 2<\/u>.<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0\u00a0\u00a0 5<\/p>\n

      Sol. Let 9\/2+19\/6+x+7\/3=67\/5<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then x=(67\/5)-(9\/2+19\/6+7\/3)\u00f3x=(67\/5)-((27+19+14)\/6)=((67\/5)-(60\/6)<\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00f3x=((67\/5)-10)=17\/5=3 2<\/u><\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/u>5<\/p>\n

      Hence, missing fractions =3 2<\/u><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5<\/p>\n

      \u00a0<\/strong>Ex.4. 4\/15 of 5\/7 of a number is greater than 4\/9 of 2\/5 of the same number by 8.<\/strong><\/p>\n

      What is half of that number?<\/strong><\/p>\n

       <\/p>\n

      Sol. Let the number be x. then 4\/15 of 5\/7 of x-4\/9 of 2\/5 of x=8\u00f34\/21x-8\/45x=8<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3(4\/21-8\/45)x=8\u00f3(60-56)\/315x=8\u00f34\/315x=8<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3x=(8*315)\/4=630\u00f31\/2x=315<\/p>\n

      Hence required number = 315.<\/p>\n

      Ex. 5. Simplify:\u00a0\u00a0\u00a0 3 1<\/u><\/strong>\u00b8{1 1<\/u>-1\/2(2 1<\/u>-1\/4-1\/6)}]<\/strong><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a02<\/strong><\/p>\n

      Sol. Given exp. =[13\/4\u00b8{5\/4-1\/2(5\/2-(3-2)\/12)}]=[13\/4\u00b8{5\/4-1\/2(5\/2-1\/12)}]<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =[13\/4\u00b8{5\/4-1\/2((30-1)\/12)}]=[13\/4\u00b8{5\/4-29\/24}]<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =[13\/4\u00b8{(30-39)\/24}]=[13\/4\u00b81\/24]=[(13\/4)*24]=78<\/p>\n

      Ex. 6. Simplify: 108<\/strong>\u00b836of 1<\/u>+2<\/u>*3 1<\/u><\/strong><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0 4<\/strong><\/p>\n

      Sol. <\/strong>Given exp.= 108\u00b89+2<\/u>*13<\/u>\u00a0 =108 <\/u>+13<\/u> =12+13\u00a0\u00a0 <\/u>=133 <\/u>= 13 3<\/u><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9\u00a0\u00a0\u00a0\u00a0\u00a0 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10\u00a0\u00a0\u00a0\u00a0\u00a0 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10<\/p>\n

      Ex.7 Simplify: (7\/2)<\/u><\/strong>\u00b8(5\/2)*(3\/2)\u00a0\u00a0\u00a0\u00a0 <\/u>\u00a0\u00a0<\/strong>\u00b8 5.25<\/strong><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (7\/2)<\/strong>\u00b8(5\/2)of (3\/2)<\/strong><\/p>\n

      sol.<\/strong><\/p>\n

      Given exp. (7\/2)<\/u>\u00b4(2\/5)<\/u>\u00b4(3\/2)\u00a0 <\/u>\u00b8\u00a0 5.25=(21\/10)<\/u>\u00b8(525\/100)=(21\/10)\u00b4(15\/14)<\/u><\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (7\/2)\u00b8(15\/4)<\/p>\n

      Ex. 8. Simplify: (i) 12.05*5.4+0.6\u00a0 (ii) 0.6*0.6+0.6*0.6 ( Bank P.O 2003)<\/strong><\/p>\n

      Sol. (i) Given exp. = 12.05*(5.4\/0.6) = (12.05*9) = 108.45<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) Given exp. = 0.6*0.6+(0.6*6) = 0.36+0.1 = 0.46<\/p>\n

      Ex. 9. Find the value of x in each of the following equation:<\/strong><\/p>\n

        \n
      • [(17.28\/x) \/ (3.6*0.2)] = 2<\/li>\n
      • 24 + 364.824 + x \u2013 36.4824 = 3794.1696<\/li>\n
      • 5 \u2013 { 5 \u00bd \u2013 [7 \u00bd\u00a0 + 2.8]\/x}*4.25\/(0.2)2<\/sup>\u00a0= 306 (Hotel Management,1997)<\/li>\n<\/ul>\n

        Sol. (i) (17.28\/x) = 2*3.6*0.2 \u00f3 x = (17.28\/1.44) = (1728\/14) = 12.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) (364.824\/x) = (3794.1696 + 36.4824) \u2013 3648.24 = 3830.652 \u2013 3648.24 = 182.412.<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3 x = (364.824\/182.412) =2.<\/p>\n

          \n
        • 5-{5.5-(7.5+(2.8\/x))}*(4.25\/0.04) = 306<\/li>\n<\/ul>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3 8.5-{5.5-{(7.5x+2.8)\/x)}*(425\/4) = 306<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3 8.5-{(5.5x-7.5x-2.8)\/x}*(425\/4) = 306<\/p>\n

          \u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3 8.5-{(-2x-2.8)\/x}*106.25\u00a0 = 306<\/p>\n

          \u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3 8.5-{(-212.5x-297.5)\/x} = 306<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3 (306-221)x = 297.5 \u00f3 x =(297.5\/85) = 3.5.<\/p>\n

          Ex. 10. If (x\/y)=(6\/5), find the value (x2<\/sup>+y2<\/sup>)\/(x2<\/sup>-y2<\/sup>) <\/strong><\/p>\n

          Sol.\u00a0 (x2<\/sup>+y2<\/sup>)\/(x2<\/sup>-y2<\/sup>) = ( x2<\/sup>\u00a0\/y2<\/sup>+ 1)\/ ( x2<\/sup>\u00a0\/y2<\/sup>-1) = [(6\/5)2<\/sup>+1] \/ [(6\/5)2<\/sup>-1]<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =\u00a0 [(36\/25)+1] \/ [(36\/25)-1] = (61*25)\/(25*11) = 61\/11<\/p>\n

          \u00a0Ex. 11. Find the value of 4 – \u00a0_____5_________\u00a0 <\/u><\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 1 + ___1___\u00a0 __<\/u><\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a03 + __1___<\/u>\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 + _1_<\/u>\u00a0\u00a0\u00a0 <\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 4<\/strong><\/p>\n

          Sol.\u00a0\u00a0\u00a0 Given exp. =\u00a0\u00a0\u00a0 4 – \u00a0_____5_______<\/u>\u00a0\u00a0 =\u00a0 4 – \u00a0____5________ <\/u>\u00a0= 4 – ____5_____<\/u>\u00a0<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 1 + ___1__\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01 + _____1____<\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 + ___1__\u00a0 <\/u><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a03 + __1___<\/u>\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3 + __4__<\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (31\/9)<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 + _1_<\/u>\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 4<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =\u00a0 4 – __5<\/u>____\u00a0\u00a0 =\u00a0 4 – __5___<\/u>\u00a0 =\u00a0 4 \u2013 (5*31)\/ 40 = 4 \u2013 (31\/8) = 1\/8<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 1 + _9\u00ad_<\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (40\/31)<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a031\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n

          Ex. 12. If\u00a0 _____2x______ <\/u>\u00a0\u00a0= 1 ., then find the value of x .<\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 1 + ___1___\u00a0 __<\/u><\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a01+ __x__<\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 –\u00a0 x<\/strong><\/p>\n

          \u00a0<\/strong>Sol. We have : _____2x______ _<\/u>\u00a0 = 1\u00a0 \u00f3\u00a0 _____2x_____<\/u> \u00a0\u00a0= 1 \u00f3\u00a0 __2x<\/u>____\u00a0\u00a0\u00a0 = 1<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 1 + ___1_____<\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 + ___1<\/u>____\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1+ (1 \u2013 x)<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 _(1 \u2013 x) \u2013 x<\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [1\/(1- x)]<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 – x\u00a0<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00f3 2x = 2-x \u00f3 3x = 2 \u00f3 x = (2\/3).\u00a0\u00a0<\/p>\n

          Ex.13.(i)If a\/b=3\/4 and 8a+5b=22,then find the value of a.<\/strong><\/p>\n

          (ii)if x\/4-x-3\/6=1,then find the value of x.<\/strong><\/p>\n

          \u00a0<\/u>Sol.<\/strong> (i) (a\/b)=3\/4 \u00deb=(4\/3) a.<\/p>\n

          \\ 8a+5b=22 \u00de 8a+5*(4\/3) a=22 \u00de 8a+(20\/3) a=22<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00de44a = 66 \u00de a=(66\/44)=3\/2<\/p>\n

          (ii) (x \/4)-((x-3)\/6)=1\u00db (3x-2(x-3) )\/12 = 1 \u00db 3x-2x+6=12 \u00db x=6.<\/p>\n

          Ex.14.If 2x+3y=34 and ((x + y)\/y)=13\/8,then find the value of 5y+7x.<\/strong><\/p>\n

          \u00a0<\/strong>Sol. <\/strong>The given equations are:<\/p>\n

          2x+3y=34 \u2026(i) and, ((x + y) \/y)=13\/8 \u00de 8x+8y=13y \u00de 8x-5y=0 \u2026(ii)<\/p>\n

          Multiplying (i) by 5,(ii) by 3 and adding, we get : 34x=170 or x=5.<\/p>\n

          Putting x=5 in (i), we get: y=8.<\/p>\n

          \\ 5y+7x=((5*8)+(7*5))=40+35=75<\/p>\n

          Ex.15.If 2x+3y+z=55,x-y=4 and y – x + z=12,then what are the values of x , y and z?<\/strong><\/p>\n

          \u00a0 <\/strong>Sol. <\/strong>The given equations are:<\/p>\n

          2x+3y+z=55 \u2026(i); x + z – y=4 \u2026(ii); y -x + z =12 \u2026(iii)<\/p>\n

          Subtracting (ii) from (i), we get: x+4y=51\u00a0 \u2026(iv)<\/p>\n

          Subtracting (iii) from (i), we get: 3x+2y=43\u00a0 \u2026(v)<\/p>\n

          \u00a0Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x=35 or x=7.<\/p>\n

          Putting x=7 in (iv), we get: 4y=44 or y=11.<\/p>\n

          Putting x=7,y=11 in (i), we get: z=8.<\/p>\n

          Ex.16.Find the value of (1-(1\/3))(1-(1\/4))(1-(1\/5))\u2026.(1-(1\/100)).\u00a0 <\/strong><\/p>\n

          Sol.<\/strong> Given expression = (2\/3)*(3\/4)*(4\/5) *\u2026\u2026.* (99\/100) = 2\/100 = 1\/50.<\/p>\n

          Ex.17. Find the value of (1\/(2*3))+(1\/(3*4))+(1\/(4*5))+(1\/(5*6))+\u2026..+ ((1\/(9*10)).<\/strong><\/p>\n

          \u00a0<\/strong>Sol. <\/strong>Given expression=((1\/2)-(1\/3))+((1\/3)-(1\/4))+((1\/4)-(1\/5))+<\/p>\n

          ((1\/5)-(1\/6))+\u2026.+ ((1\/9)-(1\/10))<\/p>\n

          =((1\/2)-(1\/10))=4\/10 = 2\/5.<\/p>\n

          Ex.18.Simplify: 9948<\/sup>\/49 <\/sub>\u00a0* 245.<\/strong><\/p>\n

          \u00a0<\/strong>Sol.<\/strong> Given expression = (100-1\/49) * <\/strong>245=(4899\/49) * 245 = 4899 * 5=24495.<\/p>\n

          Ex.19.A board 7ft. 9 inches long is divided into 3 equal parts . What is the length of each part?<\/strong><\/p>\n

          Sol<\/strong>. Length of board=7ft. 9 inches=(7*12+9)inches=93 inches.<\/p>\n

          \\Length of each part = (93\/3) inches = 31 inches = 2ft. 7 inches<\/p>\n

          20.A man divides Rs. Among 5 sons,4daughters and 2 nephews .If each daughter receives four times as much as each nephews and each son receives five times as much as each nephews ,how much does each\u00a0 daughter receive?<\/p>\n

          Let the share of each nephews be Rs.x.<\/p>\n

          Then,share of each daughter=rs4x;share of each son=Rs.5x;<\/p>\n

          So,5*5x+4*4x+2*x=8600<\/p>\n

          25x+16x+2x=8600<\/p>\n

          =43x=8600<\/p>\n

          x=200;<\/p>\n

          21.A man spends 2\/5 of his salary on house rent,3\/10 of his salary on food and 1\/8 of his salary on conveyence.if he has Rs.1400 left with him,find his expenditure on food and conveyence.<\/strong><\/p>\n

          Part of salary left=1-(2\/5+3\/10+1\/8)<\/p>\n

          Let the monthly salary be Rs.x<\/p>\n

          Then, 7\/40 of x=1400<\/p>\n

          X=(1400*40\/7)<\/p>\n

          =8600<\/p>\n

          Expenditure on food=Rs.(3\/10*800)=Rs.2400<\/p>\n

          Expenditure on conveyence=Rs.(1\/8*8000)=Rs.1000<\/p>\n

          \u00a0<\/strong>22.A third of Arun\u2019s marks in mathematics exeeds a half of his marks in english by 80.if he got 240 marks In two subjects together how many marks did he got inh english?<\/strong><\/p>\n

          Let Arun\u2019s marks in mathematics and english be x and y<\/p>\n

          Then 1\/3x-1\/2y=30<\/p>\n

          2x-3y=180\u2026\u2026>(1)<\/p>\n

          x+y=240\u2026\u2026.>(2)<\/p>\n

          solving (1) and (2)<\/p>\n

          x=180<\/p>\n

          and y=60<\/p>\n

          23.A tin of oil was 4\/5full.when 6 bottles of oil were taken out and four bottles of oil were poured into it, it was \u00be full. how many bottles of oil can the tin contain?<\/p>\n

          Suppose x bottles can fill the tin completely<\/p>\n

          Then4\/5x-3\/4x=6-4<\/p>\n

          X\/20=2<\/p>\n

          X=40<\/p>\n

          Therefore required no of bottles =40<\/p>\n

          \u00a0<\/strong>24.if 1\/8 of a pencil is black \u00bd of the remaining is white and the remaining 3 \u00bd\u00a0 is blue find the total length of the pencil?<\/strong><\/p>\n

          \u00a0Let the total length be xm<\/p>\n

          Then black part =x\/8cm<\/p>\n

          The remaining part=(x-x\/8)cm=7x\/8cm<\/p>\n

          White part=(1\/2 *7x\/8)=7x\/16 cm<\/p>\n

          Remaining part=(7x\/8-7x\/16)=7x\/16cm<\/p>\n

          7x\/16=7\/2<\/p>\n

          x=8cm<\/p>\n

          25.in a certain office 1\/3 of the workers are women \u00bd of the women are married and 1\/3 of the married women have children if \u00be of the men are married and 2\/3 of the married men have children what part of workers are without children?<\/p>\n

          \u00a0Let the total no of workers be x<\/p>\n

          No of women =x\/3<\/p>\n

          No of men =x-(x\/3)=2x\/3<\/p>\n

          No of women having\u00a0 children =1\/3 of \u00bd ofx\/3=x\/18<\/p>\n

          No of men having children=2\/3 of \u00be of2x\/3=x\/3<\/p>\n

          No of workers having children = x\/8 +x\/3=7x\/18<\/p>\n

          Workers having no children=x-7x\/18=11x\/18=11\/18 of all wprkers<\/p>\n

          26.a crate of mangoes contains one bruised mango for every thirty mango in the crate. If three out of every four bruised mango are considerably unsaleble and there are 12 unsaleable mangoes in the crate then how msny mango are there in the crate?<\/strong><\/p>\n

          \u00a0<\/strong>Let the total no of mangoes in the crate be x<\/p>\n

          Then the no of bruised mango = 1\/30 x<\/p>\n

          Let the no of unsalable mangoes =3\/4 (1\/30 x)<\/p>\n

          1\/40 x =12<\/p>\n

          x=480<\/p>\n

          \u00a0<\/strong>a train starts full of passengers at the first station it drops 1\/3 of the passengers and takes 280more at the second station it drops one half the new total and takes twelve more .on arriving at the third station it is found to have 248 passengers. Find the no of passengers in the beginning?<\/strong><\/p>\n

          Let no of passengers in the beginning be x<\/p>\n

          After first station no passengers=(x-x\/3)+280=2x\/3 +280<\/p>\n

          After second station no passengers =1\/2(2x\/3+280)+12<\/p>\n

          \u00bd(2x\/3+280)+12=248<\/p>\n

          2x\/3+280=2*236<\/p>\n

          2x\/3=192<\/p>\n

          x=288<\/p>\n

          28.if a2<\/sup>+b2<\/sup>=177and ab=54 then find the value of a+b\/a-b?<\/strong><\/p>\n

          (a+b)2<\/sup>=a2<\/sup>+b2<\/sup>+2ab=117+2*24=225<\/p>\n

          a+b=15<\/p>\n

          (a-b)2<\/sup>=a2<\/sup>+b2<\/sup>-2ab=117-2*54<\/p>\n

          a-b=3<\/p>\n

          a+b\/a-b=15\/3=5<\/p>\n

          29.find the value of (75983*75983- 45983*45983\/30000)<\/p>\n

          Given expression=(75983)2<\/sup>-(45983)2<\/sup>\/(75983-45983)<\/p>\n

          =(a-b)2<\/sup>\/(a-b)<\/p>\n

          =(a+b)(a-b)\/(a-b)<\/p>\n

          =(a+b)<\/p>\n

          =75983+45983<\/p>\n

          =121966<\/p>\n

          30.find the value of 343*343*343-113*113*113<\/u><\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0343*343+343*113+113*113<\/strong><\/p>\n

          Given expression= (a3<\/sup>-b3<\/sup>)<\/u><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 a2<\/sup>+ab+b2<\/sup><\/p>\n

          \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0=(a-b)<\/p>\n

          =(343-113)<\/p>\n

          .=230<\/p>\n

          \u00a0<\/strong>31.Village X has a population of 68000,which is decreasing at the rate of 1200 per year.VillagyY has a population\u00a0 of 42000,which is increasing <\/strong><\/p>\n

          at the rate of 800 per year .in how many years will the population of the two villages be equal?<\/strong><\/p>\n

          Let the population of two villages be equal after\u00a0 p years<\/p>\n

          Then,68000-1200p=42000+800p<\/p>\n

          2000p=26000<\/p>\n

          p=13<\/p>\n

          \u00a0<\/strong>32.From a group of boys and girls,15 girls leave.There are then left 2 boys for each girl.After this,45 boys leave.There are then 5 girls for each boy.Find <\/strong><\/p>\n

          the number of girls in the beginning?<\/strong><\/p>\n

          Let at present there be x boys.<\/p>\n

          Then,no of girls at present=5x<\/p>\n

          Before the boys had left:no of boys=x+45<\/p>\n

          And no of girls=5x<\/p>\n

          X+45=2*5x<\/p>\n

          9x=45<\/p>\n

          x=5<\/p>\n

          no of girls in the beginning=25+15=40<\/p>\n

          33.An employer pays Rs.20 for each day a worker works and for feits Rs.3 for each day is ideal at the end of sixty days a worker gets Rs.280 . for how many days did the worker remain ideal?<\/p>\n

          Suppose a worker remained ideal for x days then he worked for 60-x days<\/p>\n

          20*(60-x)-3x=280<\/p>\n

          1200-23x=280<\/p>\n

          23x=920<\/p>\n

          x=40<\/p>\n

          Ex 34.kiran had 85 currency notes in all , some of which were of Rs.100 denaomination and the remaining of Rs.50 denomination the total amount of all these currency note was Rs.5000.how much amount did she have in the denomination of Rs.50?<\/p>\n

          Let\u00a0 the no of fifty rupee notes be x<\/p>\n

          Then,no of 100 rupee notes =(85-x)<\/p>\n

          50x+100(85-x)=5000<\/p>\n

          x+2(85-x)=100<\/p>\n

          x=70<\/p>\n

          so,,required amount=Rs.(50*70)= Rs.3500<\/p>\n

          Ex. 35. When an amount was distributed among 14 boys, each of them got rs 80 more than the amount received by each boy when the same amount is distributed equally among 18 boys. What was the amount?<\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong> Let the total amount be Rs. X the,<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0x\u00a0\u00a0 –\u00a0\u00a0 x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 80 \u00f3 2 x\u00a0 = 80 \u00f3 x =63 x 80 = 5040.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 14\u00a0\u00a0\u00a0\u00a0 18\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 126\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 63<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence the total amount is 5040.<\/p>\n

          Ex. 36. Mr. Bhaskar is on tour and he has Rs. 360 for his expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by Rs. 3. for how many days is Mr. Bhaskar on tour?<\/strong><\/p>\n

          \u00a0<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong> Suppose Mr. Bhaskar is on tour for x days. Then,<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 360\u00a0 – 360\u00a0 = 3 \u00f3\u00a0 1\u00a0\u00a0 –\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0 =\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0 \u00f3\u00a0 x(x+4)\u00a0 =4 x 120 =480<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x+4 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sup>x+4\u00a0\u00a0\u00a0\u00a0\u00a0 120<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00f3 x2<\/sup> +4x \u2013480 = 0 \u00f3 (x+24) (x-20) = 0\u00a0 \u00f3 x =20.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence Mr. Bhaskar is on tour for 20 days.<\/p>\n

          Ex. 37. Two pens and three pencils cost Rs 86. four Pens and a pencil cost Rs. 112. find the cost of a pen and that of a pencil.<\/strong><\/p>\n

          \u00a0<\/strong>Sol.<\/strong> Let the cost of a pen ands a pencil be Rs. X and Rs. Y respectively.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, 2x\u00a0 + 3y = 86 \u2026.(i) and 4x + y =112.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Solving (i) and (ii), we get: x = 25 and y = 12.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Cost of a pen =Rs. 25 and the cost of a pencil =Rs. 12.<\/p>\n

          Ex. 38. Arjun and Sajal are friends . each has some money. If Arun gives Rs. 30 to Sajal, the Sajal will have twice the money left with Arjun. But, if Sajal gives Rs. 10 to Arjun, the Arjun will have thrice as much as is left with Sajal. How much money does each have?<\/strong><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Sol.<\/strong> Suppose Arun has Rs. X and Sjal has Rs. Y. then,<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2(x-30)= y+30\u00a0 => 2x-y =90\u00a0 \u2026(i)<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 and\u00a0 x +10 =3(y-10) => x-3y = – 40\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2026(ii)<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Solving (i) and (ii), we get x =62 and y =34.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Arun has Rs. 62 and Sajal has Rs. 34.<\/p>\n

          Ex. 39. In a caravan, in addition to 50 hens there are 45 goats and 8 camels with some keepers. If the total number of feet be 224 more than the number of heads, find the number of keepers.<\/strong><\/p>\n

          Sol.<\/strong> Let the number of keepers be x then,<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Total number of heads =(50 + 45 + 8 + x)= (103 + x).<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Total number of feet = (45 + 8) x 4 + (50 + x) x 2 =(312 +2x).<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (312 + 2x)-(103 + x) =224\u00f3 x =15.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, number of keepers =15.<\/p>\n

          “Quantitative Aptitude for Competitive Examinations” by R.S. Aggarwal is a widely acclaimed resource for aspirants preparing for various competitive exams. The book encompasses a comprehensive range of topics, including a dedicated section on “Simplification,” which is crucial for enhancing problem-solving efficiency.<\/p>\n

          While the complete book is available for purchase through authorized retailers, accessing specific sections like “Simplification” in PDF format can be challenging due to copyright restrictions. However, there are resources that provide focused practice on simplification problems:<\/p>\n

            \n
          1. \n

            Simplification Practice Questions<\/strong>:<\/a><\/p>\n