{"id":5903,"date":"2025-06-09T08:52:34","date_gmt":"2025-06-09T08:52:34","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5903"},"modified":"2025-06-09T08:52:34","modified_gmt":"2025-06-09T08:52:34","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-surds-indices-and","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-surds-indices-and\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: SURDS AND INDICES"},"content":{"rendered":"

SURDS AND INDICES<\/span><\/h1>\n

I IMPORTANT FACTS AND FORMULAE <\/u>I<\/u><\/p>\n

    \n
  1. LAWS OF INDICES:<\/strong><\/li>\n<\/ol>\n

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

      \n
    • am<\/sup> <\/em>x an<\/sup> <\/em>= am <\/sup><\/em>+ n<\/em><\/sup><\/li>\n
    • am<\/sup>\u00ad<\/sub> \/ an<\/sup> = am-n<\/sup><\/em><\/li>\n
    • (am<\/sup>)n<\/sup> = amn<\/sup><\/em><\/li>\n
    • (ab)n<\/sup> = an<\/sup>bn<\/sup><\/em><\/li>\n
    • ( a\/ b )n<\/sup> = ( an<\/sup> \/ bn<\/sup> )<\/em><\/li>\n
    • a0<\/sup> = 1<\/em><\/li>\n<\/ul>\n

      \u00a0<\/em><\/p>\n

       <\/p>\n

        \n
      1. SURDS<\/strong>: Let a be a rational number and <\/em>n be a positive integer such that a1\/n<\/sup> = n<\/sup>sqrt(a)<\/em><\/li>\n<\/ol>\n

        \u00a0\u00a0 is irrational. Then n<\/sup><\/em>sqrt(a)\u00a0 is called a surd of order <\/em>n.<\/p>\n

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

          \n
        1. LAWS OF SURDS:<\/strong><\/li>\n<\/ol>\n

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

          (i)\u00a0 n<\/sup>\u221aa = a1\/2<\/sup><\/p>\n

          (ii) n <\/sup>\u221aab = n <\/sup>\u221aa * n <\/sup>\u221ab<\/p>\n

          (iii) n <\/sup>\u221aa\/b = n <\/sup>\u221aa\u00a0 \/\u00a0 n <\/sup>\u221ab<\/p>\n

          (iv) (n<\/sup> \u221aa)n<\/sup> = a<\/p>\n

          (v) m<\/sup>\u221a(n<\/sup>\u221a(a)) = mn<\/sup>\u221a(a)<\/p>\n

          (vi) (n<\/sup>\u221aa)m<\/sup> = n<\/sup>\u221aam<\/sup><\/p>\n

           <\/p>\n

          I SOLVED EXAMPLES <\/u><\/strong><\/p>\n

           <\/p>\n

           <\/p>\n

          Ex. 1. Simplify : (i) (27)2\/3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) (1024)-4\/5<\/sup>\u00a0\u00a0\u00a0 (iii)( 8 \/ 125 )-4\/3<\/sup><\/p>\n

           <\/p>\n

          Sol .\u00a0 \u00a0\u00a0 (i) (27)2\/3<\/sup> = (33<\/sup>)2\/3<\/sup> = 3( 3 * ( 2\/ 3))<\/sup> = 32<\/sup> = 9<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) (1024)-4\/5<\/sup> = (45<\/sup>)-4\/5 <\/sup>= 4 { 5 * ( (-4) \/ 5 )}<\/sup> = 4-4<\/sup> = 1 \/ 44<\/sup> = 1 \/ 256<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (iii) ( 8 \/ 125 )-4\/3<\/sup> = {(2\/5)3<\/sup>}-4\/3<\/sup> = (2\/5){ 3 * ( -4\/3)}<\/sup> = ( 2 \/ 5 )-4<\/sup> = ( 5 \/ 2 )4 <\/sup>\u00a0= 54 <\/sup>\/ 24<\/sup> = 625 \/ 16<\/p>\n

           <\/p>\n

           <\/p>\n

          Ex. 2. Evaluate: (i) (.00032)3\/5<\/sup> (ii)l <\/em>(256)0.16 <\/sup>x (16)0.18<\/sup>.<\/p>\n

           <\/p>\n

          Sol. \u00a0\u00a0\u00a0\u00a0 (i) (0.00032)3\/5<\/sup> = ( 32 \/ 100000 )3\/5<\/sup>. = (25<\/sup> \/ 105<\/sup>)3\/5 <\/sup>\u00a0=\u00a0 {( 2 \/ 10 )5<\/sup>}3\/5<\/sup> = ( 1 \/ 5 )(5 * 3 \/ 5)<\/sup> =\u00a0 (1\/5)3<\/sup>\u00a0 =\u00a0\u00a0 1 \/ 125<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) (256)0. 16<\/sup> * (16)0. 18<\/sup> = {(16)2<\/sup>}0. 16<\/sup> * (16)0. 18<\/sup> = (16)(2 * 0. 16)<\/sup> * (16)0. 18<\/sup><\/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 =(16)0.32<\/sup> * (16)0.18<\/sup>\u00a0 = (16)(0.32+0.18)<\/sup>\u00a0 = (16)0.5<\/sup> = (16)1\/2<\/sup> = 4.<\/p>\n

          \u00a0 196<\/strong><\/p>\n

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

           <\/p>\n

           <\/p>\n

          Ex. 3. What is the quotient when <\/em>(x-1<\/sup> – 1) is divided by <\/em>(x – 1) ?<\/p>\n

           <\/p>\n

          Sol.\u00a0\u00a0\u00a0\u00a0 x<\/u><\/em>-1<\/sup> -1 <\/u>= (1\/x)-1<\/u> = _1 -x<\/u> *\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0 <\/u>\u00a0= -1<\/u><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x <\/em>– 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x <\/em>– 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 x\u00a0\u00a0\u00a0\u00a0 (x <\/em>– 1)\u00a0\u00a0\u00a0\u00a0\u00a0 x<\/em><\/p>\n

          Hence, the required quotient is\u00a0\u00a0\u00a0 -1\/x<\/em><\/p>\n

           <\/p>\n

          Ex. 4. If 2x – 1<\/sup> + 2x + 1<\/sup> = 1280, then find the value <\/em>of\u00a0 x.<\/p>\n

          Sol.\u00a0\u00a0\u00a0\u00a0\u00a0 2x – 1<\/sup> + 2X+ <\/sup><\/em>1<\/sup> = 1280\u00a0 \u00f3 2x-1<\/sup> (1 +22<\/sup>) = 1280<\/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 \u00f3 2x-1<\/sup>\u00a0 =\u00a0\u00a0 1280 \/ 5 = 256 =\u00a0 28<\/sup>\u00a0\u00a0 \u00f3 x -1 = 8 \u00f3 x\u00a0 =\u00a0 9.<\/p>\n

           <\/p>\n

           <\/p>\n

          Hence, x = 9.<\/p>\n

          Ex. 5. Find <\/em>the value of [ 5 ( 81\/3<\/sup> + 271\/3<\/sup>)3<\/sup>]1\/ 4<\/sup><\/p>\n

           <\/p>\n

          Sol.\u00a0\u00a0\u00a0\u00a0\u00a0 [ 5 ( 81\/3<\/sup> + 271\/3<\/sup>)3<\/sup>]1\/ 4 <\/sup>\u00a0=\u00a0 [ 5 { (23<\/sup>)1\/3<\/sup> + (33<\/sup>)1\/3<\/sup>}3<\/sup>]1\/ 4<\/sup> =\u00a0\u00a0 [ 5 { (23 * 1\/3<\/sup>)1\/3<\/sup> + (33 *1\/3 <\/sup>)1\/3<\/sup>}3<\/sup>]1\/ 4<\/sup><\/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 = {5(2+3)3<\/sup>}1\/4<\/sup> = (5 * 53<\/sup>)1\/ 4<\/sup> =5(4 * 1\/ 4) <\/sup>\u00a0= 51<\/sup> = 5.<\/p>\n

           <\/p>\n

           <\/p>\n

          Ex. 6. Find the Value of {(16)3\/2<\/sup> + (16)-3\/2<\/sup>}<\/p>\n

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

          Sol.\u00a0\u00a0\u00a0\u00a0 [(16)3\/2<\/sup> +(16)-3\/2<\/sup> = (42<\/sup>)3\/2<\/sup> +(42<\/sup>)-3\/2<\/sup> = 4(2 * 3\/2)<\/sup> + 4{ 2* (-3\/2)}<\/sup><\/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 = 43<\/sup> + 4-3<\/sup> = 43<\/sup> + (1\/43<\/sup>) = ( 64 + ( 1\/64)) = 4097\/64.<\/p>\n

           <\/p>\n

          Ex. 7. If (1\/5)3y<\/sup> <\/em>= 0.008, then find the value of(0.25)y<\/sup>.<\/em><\/p>\n

          \u00a0<\/em><\/p>\n

          Sol. (1\/5)3y<\/sup> = 0.008 =\u00a0 8\/1000 =\u00a0 1\/125 = (1\/5)3<\/sup> \u00f3 3y <\/em>= 3 \u00f3 Y <\/em>= 1.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\ (0.25)y<\/sup> <\/em>= (0.25)1<\/sup> = 0.25.<\/p>\n

           <\/p>\n

          \u00a0<\/em>Ex. 8. Find the value <\/em>of\u00a0\u00a0\u00a0\u00a0 (243)n\/5 <\/sup><\/u>\u00b4 32n <\/sup><\/em>+ 1<\/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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9n <\/sup><\/em>\u00b4 3n <\/sup><\/em>-1<\/sup> .<\/p>\n

          \u00a0<\/em><\/p>\n

          Sol. <\/em>(243)n\/5<\/sup> x32n+l <\/sup><\/em><\/u>\u00a0\u00a0=\u00a0\u00a0 3 (5 * n\/5)<\/sup> <\/u>\u00b4 32n+l<\/sup> <\/u><\/em>_ = 3n\u00a0 <\/sup><\/u><\/em>\u00b432n+1<\/sup><\/u><\/em><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (32<\/sup>)n<\/sup> <\/em>\u00b4 3n <\/sup><\/em>– 1<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 32n<\/sup> <\/em>\u00b4 3n <\/sup><\/em>– 1<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 32n <\/em>\u00b4<\/em> 3n-l<\/sup><\/em><\/p>\n

          \u00a0<\/em><\/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= 3n <\/sup><\/u><\/em>+ (2n <\/em>+ 1)<\/sup><\/u> \u00a0\u00a0=\u00a0\u00a0 3(3n+1)<\/sup>\u00a0 <\/u>\u00a0=\u00a0 3(3n+l)-(3n-l)<\/sup> <\/em>= 32<\/sup> = 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 32n+n-1<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3(3n-1)<\/sup><\/p>\n

           <\/p>\n

          Ex. 9. Find the value Of (21\/4<\/sup>-1)(23\/4<\/sup>+21\/2<\/sup>+21\/4<\/sup>+1)<\/em><\/p>\n

           <\/p>\n

          Sol.<\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Putting 21\/4<\/sup> = x, we get :<\/p>\n

           <\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (21\/4<\/sup>-1) (23\/4<\/sup>+21\/2<\/sup>+21\/4<\/sup>+1)=(x-1)(x3<\/sup>+x2<\/sup>+x+1) , where x = 21\/4<\/sup><\/em><\/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 =(x-1)[x2<\/sup>(x+1)+(x+1)]<\/em><\/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 =(x-1)(x+1)(x2<\/sup>+1) = (x2<\/sup>-1)(x2<\/sup>+1)<\/em><\/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 =(x4<\/sup>-1) = [(21\/4<\/sup>)4<\/sup>-1] = [2(1\/4<\/sup>*4)<\/sup> \u20131] = (2-1) = 1.<\/em><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em><\/p>\n

          Ex. 10. Find the value of\u00a0\u00a0 62\/3<\/sup> <\/u>\u00b4\u00a0 3<\/sup>\u221a67<\/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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3<\/sup>\u221a66<\/sup><\/em><\/p>\n

          \u00a0<\/em><\/p>\n

          Sol.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>62\/3<\/sup> <\/u>\u00b4\u00a0 3<\/sup>\u221a67<\/sup>\u00a0 <\/u>\u00a0=\u00a0 62\/3 <\/sup>\u00a0<\/u>\u00b4 (67<\/sup>)1\/3<\/sup><\/u>\u00a0\u00a0 =\u00a0 62\/3<\/sup>\u00a0 <\/u>\u00b4\u00a0 6(7 * 1\/3)<\/sup><\/u>\u00a0 =\u00a0\u00a0 62\/3 <\/sup>\u00a0<\/u>\u00b4\u00a0 6(7\/3)<\/sup><\/u><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 3<\/sup>\u221a66<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (66<\/sup>)1\/3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6(6 * 1\/3)<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 62<\/sup><\/em><\/p>\n

          \u00a0<\/em><\/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 =62\/3<\/sup> <\/em>\u00b4 6((7\/3)-2)<\/sup> = 62\/3<\/sup> <\/em>\u00b4 61\/3<\/sup>\u00a0 = 61 <\/sup>= 6.<\/em><\/p>\n

          Ex. 11. If x= ya<\/sup>, y=zb<\/sup> and z=xc<\/sup>,then find the value of abc.<\/em><\/p>\n

          \u00a0<\/em><\/p>\n

          Sol.\u00a0\u00a0\u00a0 z1<\/sup>= xc<\/sup> =(ya<\/sup>)c<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [since x= ya<\/sup>]<\/em><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0=y(ac)<\/sup> = (zb<\/sup>)ac<\/sup>\u00a0\u00a0 [since y=zb<\/sup>]<\/em><\/p>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0=zb(ac)<\/sup>= zabc<\/sup><\/em><\/p>\n

          \\\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 abc = 1.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/sup><\/em><\/p>\n

          \u00a0<\/em>= 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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Ex. 12. Simplify [(xa<\/sup> \/ xb<\/sup>)^(a2<\/sup>+b2<\/sup>+ab)] * [(xb<\/sup> \/ xc<\/sup> )^ b2<\/sup>+c2<\/sup>+bc)] * [(xc<\/sup>\/xa<\/sup>)^(c2<\/sup>+a2<\/sup>+ca)]<\/p>\n

          Sol.<\/em><\/p>\n

          \u00a0<\/em>Given Expression<\/p>\n

          = [{x(o <\/sup><\/em>– b)<\/em><\/sup>}^(a2<\/sup> <\/em>+ b2<\/sup> <\/em>+ ob)].[‘(x(b <\/sup><\/em>– c)<\/em><\/sup>}^ (b2<\/sup> <\/em>+ c2<\/sup> + bc)].[‘(x(c <\/sup><\/em>– a)<\/em><\/sup>}^(c2<\/sup> <\/em>+ a2<\/sup> + ca])<\/em><\/p>\n

          = [x(a <\/sup><\/em>– b)(a2 <\/em>+ b2 <\/em>+ ab)<\/em><\/sup> . x(b <\/sup><\/em>– c) (b2 <\/em>+c2+ bc)<\/em><\/sup>.x(c<\/sup><\/em>– a) (c2 + a2 + ca)<\/em><\/sup>]<\/em><\/p>\n

          = [x^(a3<\/sup>-b3<\/sup>)].[x^(b3<\/sup>-e3<\/sup>)].[x^(c3<\/sup>-a3<\/sup>)] <\/em>= x^(a3<\/sup>-b3<\/sup>+b3<\/sup>-c3<\/sup>+c3<\/sup>-a3<\/sup>) <\/em>= x0<\/sup> <\/em>= 1.<\/p>\n

          Ex. 13. Which is larger \u221a2 <\/em>or 3<\/sup>\u221a3 ?<\/p>\n

          Sol. Given surds are of order 2 and 3. Their L.C.M. is 6. Changing each to a surd of order 6, we get:<\/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 \u221a2 = 21\/2<\/sup> = 2((1\/2)*(3\/2))<\/sup> =23\/6 <\/sup>=\u00a0 81\/6<\/sup> = 6<\/sup>\u221a8<\/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 3<\/sup>\u221a3= 31\/3<\/sup> = 3((1\/3)*(2\/2))<\/sup> = 32\/6<\/sup> = (32<\/sup>)1\/6<\/sup> = (9)1\/6<\/sup> = 6<\/sup>\u221a9.<\/p>\n

          Clearly, 6<\/sup>\u221a9 > 6<\/sup>\u221a8 and hence 3<\/sup>\u221a3\u00a0 > \u221a2.<\/p>\n

          Ex. 14. Find the largest from among <\/em>4\u221a6, \u221a2 and 3<\/sup>\u221a4.<\/em><\/p>\n

          Sol. Given surds are of order 4, 2 and 3 respectively. Their L.C,M, is 12, Changing each to a surd of order 12, we get:<\/p>\n

          4<\/sup>\u221a6 = 61\/4<\/sup> = 6((1\/4)*(3\/3))<\/sup> = 63\/12<\/sup> = (63<\/sup>)1\/12 <\/sup>\u00a0= (216)1\/12.<\/sup><\/p>\n

          \u221a2 = 21\/2<\/sup> = 2((1\/2)*(6\/6))<\/sup> = 26\/12<\/sup> = (26<\/sup>)1\/12 <\/sup>\u00a0= (64)1\/12<\/sup>.<\/p>\n

          3<\/sup>\u221a4 = 41\/3<\/sup> = 4((1\/3)*(4\/4))<\/sup>\u00a0 =\u00a0 44\/12 <\/sup>\u00a0= (44<\/sup>)1\/12<\/sup> = (256)1\/12<\/sup>.<\/p>\n

          Clearly, (256)1\/12<\/sup>\u00a0 >\u00a0 (216)1\/12 <\/sup>\u00a0>\u00a0 (64)1\/12<\/sup><\/p>\n

          Largest one is (256)1\/12<\/sup>.\u00a0 i.e. 3<\/sup>\u221a4 .<\/p>\n

          rs aggarwal quantitative aptitude surds and indices quantitative aptitude for competitive examinations rs aggarwal competition book rs aggarwal quantitative aptitude for competitive examinations surds and indices questions rs agarwal aptitude rs aggarwal aptitude book laws of surds rs aggarwal aptitude s chand quantitative aptitude rs aggarwal competition<\/span><\/p>\n

          “Quantitative Aptitude” by R.S. Aggarwal is a widely respected resource for mastering topics like Surds and Indices. To access this material legally and ensure the quality of the content, consider the following options:<\/p>\n

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            RS Aggarwal Quantitative Aptitude PDF Free Download: SURDS AND INDICES<\/h3>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

            SURDS AND INDICES I IMPORTANT FACTS AND FORMULAE I LAWS OF INDICES: \u00a0 am x an = am + n am\u00ad \/ an = am-n (am)n = amn (ab)n = anbn ( a\/ b )n = ( an \/ bn ) a0 = 1 \u00a0   SURDS: Let a be a rational number and n […]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[127],"tags":[2841,2241,2412,2414,2415,2842,2417,1923,2337,2419,2843,2844],"class_list":["post-5903","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude-pdf","tag-laws-of-surds","tag-quantitative-aptitude-for-competitive-examinations","tag-rs-agarwal-aptitude","tag-rs-aggarwal-aptitude","tag-rs-aggarwal-aptitude-book","tag-rs-aggarwal-competition","tag-rs-aggarwal-competition-book","tag-rs-aggarwal-quantitative-aptitude","tag-rs-aggarwal-quantitative-aptitude-for-competitive-examinations","tag-s-chand-quantitative-aptitude","tag-surds-and-indices","tag-surds-and-indices-questions"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5903","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/users\/41"}],"replies":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/comments?post=5903"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5903\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=5903"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=5903"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=5903"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}